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https://wiki.iaoa.org/index.php/Edu:Logic	[ "# Edu:Logic\n\n## Logic\n\n1. The combination of a formal language with a formal theory of truth or a proof theory (or both).\n2. The study of arguments, with the intention of describing how to distinguish good arguments from bad arguments. According to the generally accepted usage among philosophers, an argument is valid if (and only if) There are no cases in which the premises of the argument are true, but the conclusion of the argument false. A distinction is often made between valid arguments and cogent arguments. A cogent argument is a valid argument the premises of which are true. These terms only apply to what is called deductive logic, as opposed to inductive forms of reasoning.\n3. Some logics are classified as monotonic. Monotonic logics fit most closely what most people understand deductive logic to be. Roughly speaking, a logic is monotonic if (and only if) all deductively valid arguments formulated in that logic remain deductively valid, even if new premises are added to such arguments. Non-monotonic logics reflect what most people think of as inductive logic.\n4. Deductive reasoning is sometimes described as being most essentially inference from the general to the particular; inductive reasoning is sometimes described as being most essentially inference from the particular to the general. These descriptions are useful, but the two kinds of logic are best understood in terms of the degree of certainty conferred on the conclusion by the premises. Deductive arguments are those in which, in good arguments, the premises confer certainty on their conclusions. Deductive validity as described above embodies this requirement. Inductive arguments are those in which, in good arguments, the premises confer a degree of certainty less than total on their conclusions. In such arguments, the probability of the conclusion follows from the premises." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9357158,"math_prob":0.7472661,"size":1886,"snap":"2021-43-2021-49","text_gpt3_token_len":372,"char_repetition_ratio":0.16153029,"word_repetition_ratio":0.093959734,"special_character_ratio":0.18557794,"punctuation_ratio":0.1017964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95247364,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T03:17:19Z\",\"WARC-Record-ID\":\"<urn:uuid:163c119d-d5f5-4911-8a61-5f920927fb71>\",\"Content-Length\":\"20855\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62e270e5-ee18-4ecc-bd75-228124e9ca5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:117f8d1a-877a-4df7-a9f7-b81fa595633c>\",\"WARC-IP-Address\":\"188.166.12.201\",\"WARC-Target-URI\":\"https://wiki.iaoa.org/index.php/Edu:Logic\",\"WARC-Payload-Digest\":\"sha1:NKLML4DSAISR2MSRACUJMWUYPZ76JB7Q\",\"WARC-Block-Digest\":\"sha1:AKZT5DPKKWWMWM6TEM6NQCNSPOBFFCIS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585120.89_warc_CC-MAIN-20211017021554-20211017051554-00166.warc.gz\"}"}
https://origin.geeksforgeeks.org/energy-in-simple-harmonic-motion/?ref=lbp	[ "", null, "GFG App\nOpen App", null, "Browser\nContinue\n\n# Energy in Simple Harmonic Motion\n\nEach and every object possesses energy. In a simple harmonic motion, the object goes to the extreme and acquires potential energy. When the object comes back to the mean position, its velocity is at its maximum. Thus, in this case, the potential is converted to kinetic energy and vice versa. In an ideal simple harmonic motion, the energy is conserved. Even though it might change forms, the total energy remains constant. It is essential to study these changes in energy and the total energy to analyze the SHM and its properties. Let’s look at that in detail.\n\n### Simple Harmonic Motion\n\nSimple Harmonic Motion is a kind of periodic motion where the object moves to and fro around its mean position. The time period, in this case, remains constant. The time period is denoted by “T” and the distance of the mean position from the extreme position is called amplitude, it is denoted by A. The general equation for the position(x) of the object at any particular time is given by,", null, "Here,", null, "and", null, "denotes the phase shift.\n\nSimilarly, the equation for the velocity of the object in SHM can be found by differentiating this equation.", null, "Then, the equation for acceleration becomes,", null, "### Energy in Simple Harmonic Motion\n\nKinetic and potential energies in the SHM vary from zero to their maximum values. The equations mentioned above show that the velocity of the object follows the sinusoidal trajectory, which means that the velocity of the object increases and decreases. The velocity is zero at the extreme positions and maximum at the mean positions. The position where the velocity is maximum is the position where the kinetic energy of the object is also maximum.\n\nFor an object of mass “m”, with a velocity “v”. Kinetic energy is given by,\n\nKE =", null, "Since the object is in SHM, the value of the velocity can be substituted into the equation,\n\nKE =", null, "⇒ KE =", null, "⇒ KE =", null, "Notice that the KE is also in the form of a periodic equation.\n\nMaximum Value of K.E =", null, "(At the mean position)\n\nMinimum Value of K.E = 0 (At the extreme position)\n\nIn the case of conservative forces, which are directly proportional to the displacement. The potential energy is given by,\n\nU =", null, "Substituting the value of x(t) in SHM,\n\nU =", null, "⇒ U(t) =", null, "⇒ U(t) =", null, "Let’s find the total energy,\n\nE = U + K\n\n⇒ E =", null, "+", null, "⇒ E =", null, "⇒ E =", null, "", null, "The figure above shows the graph of KE and PE and the total energy E of the SHM. In this case, notice that the total energy of the system remains constant and independent of time.\n\n### Sample Problems\n\nQuestion 1: A particle of mass 10Kg is performing SHM where its position is given by the equation given below,\n\nx(t) = 3sin(5t)\n\nFind its kinetic energy at the mean position.\n\nThe KE at the mean position is given by,\n\nK.E =", null, "In this case, A = 3, m = 10Kg and", null, "K.E =", null, "⇒ K.E =", null, "⇒ K.E =", null, "⇒ K.E = 1225 J\n\nQuestion 2: A particle of mass 2Kg is performing SHM where its position is given by the equation given below,\n\nx(t) = cos(2t)\n\nFind its kinetic energy at the mean position.\n\nThe KE at the mean position is given by,\n\nK.E =", null, "In this case, A = 1, m = 2Kg and", null, "K.E =", null, "⇒ K.E =", null, "⇒ K.E = 4 J.\n\nQuestion 3: A particle of mass 2Kg is performing SHM connected to a spring (k = 100 N/m) where its position is given by the equation given below,\n\nx(t) = 20cos(t)\n\nFind its potential energy at the extreme position.\n\nThe KE at the mean position is given by,\n\nU(t) =", null, "In this case, A = 20, k =100 and", null, "U(t) =", null, "⇒ U(t) =", null, "⇒ U(t) = 20000 J\n\nQuestion 4: A particle of mass 1Kg is performing SHM connected to a spring (k = 10 N/m) where its position is given by the equation given below,\n\nx(t) = 2cos(t)\n\nFind its potential energy at the extreme position.\n\nThe KE at the mean position is given by,\n\nU(t) =", null, "In this case, A = 2, k =10 and", null, "U(t) =", null, "⇒ U(t) =", null, "⇒ U(t) = 20 J\n\nQuestion 5: A particle is performing SHM where its position is given by the equation given below,\n\nx(t) = 2sin(10t)\n\nFind its velocity at the mean position.\n\nThe KE at the mean position is given by,\n\nK.E =", null, "In this case, A = 2 and", null, "K.E =", null, "", null, "⇒", null, "", null, "⇒ v = 10 m/s\n\nQuestion 6: A particle is performing SHM where its position is given by the equation given below,\n\nx(t) = 10cos(10t + 5)\n\nFind its velocity at the mean position.\n\nThe KE at the mean position is given by,\n\nK.E =", null, "In this case, A = 10 and", null, "K.E =", null, "", null, "⇒", null, "⇒v2 = 102(10)2\n\n⇒ v = 100 m/s\n\nMy Personal Notes arrow_drop_up" ]	[ null, "https://media.geeksforgeeks.org/gfg-gg-logo.svg", null, "https://media.geeksforgeeks.org/auth-dashboard-uploads/web-2.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-2e7750f5fc029ba074fc47cd24ebb2e4_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8287021be16a1e6035f053917f628080_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-2dea1b55e27977d1a71f1e39cdeb3704_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ce546b3c951ef6f606ff261ecce8286d_l3.png", null, "https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-86a32bc6196d4e38eafe9ccd4e9cec69_l3.png", null, 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http://forums.wolfram.com/mathgroup/archive/2000/Jul/msg00261.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "3d-2d points\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg24463] 3d-2d points\n• From: Jeff Moran <prism at ulster.net>\n• Date: Tue, 18 Jul 2000 00:59:08 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Here's the problem:\n\nOn a 1:2 aspect ratio flat map of the Earth, using virtual colored gels\nover the geographic satellite imagery, we divided the globe\ninto a dodecahedron12 identical pentagons with 3 converging at either\npole and six around the equator. Using the 3D program\nLightwave, we rendered an image of each pentagon face on. We know the\ncoordinates, in latitude and longitude, of each point of\neach pentagon, so given the coordinates of a location somewhere on the\nglobe (e.g. New York City @ 40N42 and 74W00)--we can\ndetermine the pentagon to which the location belongs.\n\nOur problem is plotting an accurate representation of the location on\nthe rendered image. Our programmer (Aaron) was given very\nhelpful math on how to convert 3D coordinates to 2D which he translated\n(roughly) into this code:\n-------------------------------------------------\nloc = location coord\ncenter = pentagon center coord\n\nsloc = loc - center --subtract center from loc coord to make it relative\n\nto the center of the pentagon\n\ntheta = (PI * sloc.horz) / 180 convert coord to radians..\nphi = (PI * sloc.vert) / 180\n\nr = 3.5530 -- radius of the sphere in Lightwave\n\nx = r * sin(theta) * cos(phi)\ny = r * sin(phi)\nz = r * cos(theta) * cos(phi)\n\nf + .4696 -- focal length in Lightwave\nd = 6.3682 -- the distance between the center of the sphere and the\nviewer in Lightwave\n\nsx = (f * x) / (d - z) -- 2D coords of the loc..\nsy = (f * y) / (d - z)\n----------------------------------\n\nThe problem we are encountering now is that this doesn't account for the\n\nway latitude curves around the globe. This is especially\napparent around the poles (in the uppermost three and lowermost three\npentagons) where the curve is most pronounced. The\nformula treats all locations as if the pentagon were centered (none are)\n\non the equator and prime meridian.\n\n```\n\n• Prev by Date: Re: A strange bug in Solve\n• Next by Date: RE: fade to mauve\n• Previous by thread: Re: Need to reduce 2 lists so that only {x,y} pairs with same x remain" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8271604,"math_prob":0.9649059,"size":2027,"snap":"2020-24-2020-29","text_gpt3_token_len":527,"char_repetition_ratio":0.15274344,"word_repetition_ratio":0.0,"special_character_ratio":0.3103108,"punctuation_ratio":0.10427807,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951987,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T18:00:55Z\",\"WARC-Record-ID\":\"<urn:uuid:f2fdd917-6f07-44b4-b514-69446c959667>\",\"Content-Length\":\"45417\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe5a673a-d16f-4834-b63f-dd19e9484186>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1b661ea-2197-4d53-b9fe-ec47630daee5>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2000/Jul/msg00261.html\",\"WARC-Payload-Digest\":\"sha1:HTYWCUODOM5RXCL47Q6ZJN6SLSVI4ONY\",\"WARC-Block-Digest\":\"sha1:BVS6DETAHSV7XZNTF22OFT3DXZY4RIYJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881763.20_warc_CC-MAIN-20200706160424-20200706190424-00022.warc.gz\"}"}
https://www.teachoo.com/8799/2488/Ex-12.1--4/category/Ex-12.1/	[ "Ex 12.1\n\nChapter 12 Class 6 Ratio And Proportion\nSerial order wise", null, "Get live Maths 1-on-1 Classs - Class 6 to 12\n\n### Transcript\n\nEx 12.1, 4 Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar. Distance travelled by Hamid in 1 hour = 9 km ∴ Hamid’s speed = 9 km/hour Distance travelled by Akhtar in 1 hour = 12 km ∴ Akhtar’s speed = 12 km/hour Ratio = (𝐻𝑎𝑚𝑖𝑑^′ 𝑠 𝑠𝑝𝑒𝑒𝑑)/(𝐴𝑘ℎ𝑡𝑎𝑟^′ 𝑠 𝑠𝑝𝑒𝑒𝑑) = 9/12 ∴ Required ratio is 3 : 4", null, "" ]	[ null, "https://d1avenlh0i1xmr.cloudfront.net/1c07d37f-23c7-45cc-99d0-8c62758b5b01/slide9.jpg", null, "https://www.teachoo.com/static/misc/Davneet_Singh.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7616958,"math_prob":0.9799817,"size":924,"snap":"2023-14-2023-23","text_gpt3_token_len":466,"char_repetition_ratio":0.31413043,"word_repetition_ratio":0.05263158,"special_character_ratio":0.4491342,"punctuation_ratio":0.19622642,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9825308,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,9,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T10:24:10Z\",\"WARC-Record-ID\":\"<urn:uuid:5993992e-d0b7-40ad-98f9-1ca82c10d398>\",\"Content-Length\":\"172431\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50d460ca-b53b-4b4a-b3cd-9480fef2ad2e>\",\"WARC-Concurrent-To\":\"<urn:uuid:c18ac59d-1c2d-45be-96fb-a1ae3df886a8>\",\"WARC-IP-Address\":\"104.21.90.10\",\"WARC-Target-URI\":\"https://www.teachoo.com/8799/2488/Ex-12.1--4/category/Ex-12.1/\",\"WARC-Payload-Digest\":\"sha1:D2YRXJWTR7AXBBX24XROARKPY773RSZE\",\"WARC-Block-Digest\":\"sha1:JMK3QHNKSDDNX4RMMDF6NL7S2G3XV4CS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945323.37_warc_CC-MAIN-20230325095252-20230325125252-00308.warc.gz\"}"}
https://numbersdb.org/numbers/131180	[ "# The Number 131180 : Square Root, Cube Root, Factors, Prime Checker\n\nYou can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 131180. 131180 is written as One Hundred And Thirty One Thousand, One Hundred And Eighty. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.\n\n• Number 131180 is an even Number.\n• Number 131180 is not a Prime Number\n• Sum of all digits of 131180 is 14.\n• Previous number of 131180 is 131179\n• Next number of 131180 is 131181\n\n## Square, Square Root, Cube, Cube Root of 131180\n\n• Square Root of 131180 is 362.18779659177\n• Cube Root of 131180 is 50.810781599827\n• Square of 131180 is 17208192400\n• Cube of 131180 is 2257370679032000\n\n## Numeral System of Number 131180\n\n• Binary Representation of 131180 is 100000000001101100\n• Octal Representation of 131180 is 400154\n• Hexadecimal Representation of 131180 is 2006c\n\n## Sin, Cos, Tan of Number 131180\n\n• Sin of 131180 is 0.64278760968663\n• Cos of 131180 is -0.76604444311891\n• Tan of 131180 is -0.83909963117747\n\n## Multiplication Table for 131180\n\n• 131180 multiplied by 1 equals to 131,180\n• 131180 multiplied by 2 equals to 262,360\n• 131180 multiplied by 3 equals to 393,540\n• 131180 multiplied by 4 equals to 524,720\n• 131180 multiplied by 5 equals to 655,900\n• 131180 multiplied by 6 equals to 787,080\n• 131180 multiplied by 7 equals to 918,260\n• 131180 multiplied by 8 equals to 1,049,440\n• 131180 multiplied by 9 equals to 1,180,620\n• 131180 multiplied by 10 equals to 1,311,800\n• 131180 multiplied by 11 equals to 1,442,980\n• 131180 multiplied by 12 equals to 1,574,160\n\n## Division Table for 131180\n\n• 131180 divided by 1 equals to 131180\n• 131180 divided by 2 equals to 65590\n• 131180 divided by 3 equals to 43726.666666667\n• 131180 divided by 4 equals to 32795\n• 131180 divided by 5 equals to 26236\n• 131180 divided by 6 equals to 21863.333333333\n• 131180 divided by 7 equals to 18740\n• 131180 divided by 8 equals to 16397.5\n• 131180 divided by 9 equals to 14575.555555556\n• 131180 divided by 10 equals to 13118\n• 131180 divided by 11 equals to 11925.454545455\n• 131180 divided by 12 equals to 10931.666666667" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79422,"math_prob":0.9981366,"size":311,"snap":"2020-45-2020-50","text_gpt3_token_len":73,"char_repetition_ratio":0.104234524,"word_repetition_ratio":0.0,"special_character_ratio":0.23794213,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T07:19:10Z\",\"WARC-Record-ID\":\"<urn:uuid:eff21275-98f0-4825-b63a-05b3322893ee>\",\"Content-Length\":\"11036\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1458fbc0-428f-45f8-a33c-65be21d56277>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ec8f7cd-71bb-4e31-bfe1-850fdfd3f850>\",\"WARC-IP-Address\":\"68.66.216.43\",\"WARC-Target-URI\":\"https://numbersdb.org/numbers/131180\",\"WARC-Payload-Digest\":\"sha1:RLG3YULZSE4KNVPZNLKFPAHZLOJQAMHE\",\"WARC-Block-Digest\":\"sha1:XJAHCDFHQAQPCMPPHDHN67K7ZFV5VA2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141186761.30_warc_CC-MAIN-20201126055652-20201126085652-00592.warc.gz\"}"}
https://socratic.org/questions/how-do-you-solve-7-2-x-35-using-the-distributive-property	[ "# How do you solve 7(2+x)=35 using the distributive property?\n\nFeb 20, 2017\n\n$x = 3$\n\n#### Explanation:\n\nDistribute the 7 into the parentheses. Note that the coefficient in front of the x is 1\n\n$7 \\times 2 = 14$\n\n$7 \\times 1 = 7$\n\nYour problem can be rewritten as:\n\n$14 + 7 x = 35$\n\nNow solve for x. Subtract 14 on both sides and then divide by 7:\n\n$7 x = 21$\n\n$x = 3$\n\nYou can also do this dividing both sides of the equation in the beginning and simplifying:\n\n$\\left(\\frac{7}{7}\\right) \\left(2 + x\\right) = \\frac{35}{7}$\n\n$2 + x = 5$\n\n$x = 3$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8221307,"math_prob":0.9999894,"size":501,"snap":"2020-34-2020-40","text_gpt3_token_len":121,"char_repetition_ratio":0.104627766,"word_repetition_ratio":0.0,"special_character_ratio":0.23952095,"punctuation_ratio":0.08080808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998856,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T05:57:44Z\",\"WARC-Record-ID\":\"<urn:uuid:56c37bdb-6b62-45b6-aa32-95ad6d117cea>\",\"Content-Length\":\"33253\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d232ada3-b105-466c-9aa3-726b7347f13e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2bd42162-b2c4-42e6-a0b8-bb672f534941>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-7-2-x-35-using-the-distributive-property\",\"WARC-Payload-Digest\":\"sha1:TMS3AG762PMRITUOMLAJIVMH4QDYPOYA\",\"WARC-Block-Digest\":\"sha1:QPOYKY55KE4CFCDUALCYBMOYU37ZXZ5K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738609.73_warc_CC-MAIN-20200810042140-20200810072140-00490.warc.gz\"}"}
https://de.mathworks.com/matlabcentral/cody/problems/406-back-to-basics-16-byte-order/solutions/1266270	[ "Cody\n\n# Problem 406. Back to basics 16 - byte order\n\nSolution 1266270\n\nSubmitted on 11 Sep 2017 by Andrew Dobrovolc\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\nx = uint16(12345); y_correct = 14640; assert(isequal(byte_order(x),y_correct))\n\n2 Pass\nx = uint32(12345); y_correct = 959447040; assert(isequal(byte_order(x),y_correct))\n\n3 Pass\nx = int32(-12345); y_correct = -942669825; assert(isequal(byte_order(x),y_correct))\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":0.5206078,"math_prob":0.9469969,"size":547,"snap":"2020-45-2020-50","text_gpt3_token_len":163,"char_repetition_ratio":0.1510129,"word_repetition_ratio":0.0,"special_character_ratio":0.3857404,"punctuation_ratio":0.11956522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9727711,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T03:46:10Z\",\"WARC-Record-ID\":\"<urn:uuid:3f463e08-f119-42d7-a060-859390277766>\",\"Content-Length\":\"80079\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df83db59-faea-46a2-814f-ea8425de4289>\",\"WARC-Concurrent-To\":\"<urn:uuid:da231cd6-82bd-4a6f-a6d9-295c16424a23>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/406-back-to-basics-16-byte-order/solutions/1266270\",\"WARC-Payload-Digest\":\"sha1:6KLK6T2QJ7GXNF2543KR37MTWZ6COFLZ\",\"WARC-Block-Digest\":\"sha1:XY4XHA4JOPYDCKWOG65I7SO5ZO66C6IJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141189038.24_warc_CC-MAIN-20201127015426-20201127045426-00577.warc.gz\"}"}
https://coolconversion.com/weight/kg-lbs-oz/_3.09_kilograms_in_lbs_and_oz_	[ "# 3.09 Kilograms in lbs and oz\n\n3.09 kg = 6 lb 12 15/16 oz()\n\n() This result may be rounded to the nearest 1/16 of an ounce.\n\nor\n+\n\n## How many pounds and ounces in 3.09 kilograms?\n\nHow many lbs and oz in 3.09 kilograms? There are 6 lb 12 15/16 oz (ounces) in 3.09 kg. Use our calculator below to transform any kg or grams value in lbs and ounces.\n\nUsing this converter you can get answers to questions like:\n\n• How many lb and oz are in 3.09 kilogramss?\n• 3.09 kilogramss is equal to how many pounds and ounces?\n• How to convert kilograms or grams to pounds and ounces?\n• How do I convert kilograms to pounds in baby weight?\n\n## How to convert 3.09 kilograms to pounds and ounces step-by-step\n\nOne kilogram is a unit of masss (not weight) and equals approximately 2.2 pounds. One pound equals 16 ounces exactly. If you need to be super precise, you can use one kilogram = 2.2046226218488 pounds. Once this is very close to 2.2 pounds, you will almost always want to use the simpler number to make the math easier.\n\n### Step 1: Convert from kilograms to pounds\n\n1 kilogram = 2.2 x pounds, so,\n\n3.09 x 1 kilogram = 3.09 x 2.2 pounds (rounded), or\n\n3.09 kilograms = 6.798 pounds.\n\n### Step 2: Convert the decimal part of pounds to ounces\n\nAn answer like \"6.798 pounds\" might not mean much to you because you may want to express the decimal part, which is in pounds, in ounces which is a smaller unit.\n\nSo, take everything after the decimal point (0.8), then multiply that by 16 to turn it into ounces. This works because one pound = 16 ounces. Then,\n\n6.8 pounds = 6 + 0.8 pounds = 6 pounds + 0.8 x 16 ounces = 6 pounds + 12.8 ounces. So, 6.8 pounds = 6 pounds and 12 ounces (when rounded). Obviously, this is equivalent to 3.09 kilograms.\n\n### Step 3: Convert from decimal ounces to an usable fraction of once\n\nThe previous step gave you the answer in decimal ounces (12.8), but how to express it as an usable (practical) fraction? See below a procedure, which can also be made using a calculator, to convert the decimal ounces to the nearest usable fraction:\n\na) Subtract 12, the number of whole ounces, from 12.8:\n\n12.8 - 12 = 0.8. This is the fractional part of the value in ounces.\n\nb) Multiply 0.8 times 8 (it could be 8, 16, 32, 64, ... depending on the exactness you want) to get the number of 16th's ounces:\n\n0.8 x 16 = 12.8.\n\nc) Take the integer part:\n\nint(12.8) = 13. This is the number of 16th's of a pound and also the numerator of the fraction.\n\nFinally, 3.09 kilograms = 6 pounds 12 3/4 ounces.\n\nAs 12/16 is not in the simplest form, it should be reduced to 3/4 in order to get a simpler fraction.\n\nIn short:\n\n3.09 kg = 6 pounds 12 3/4 ounces\n\nImportant! This result may differ from the calculator above because we've assumed here that 1 kilogram equals 2.2 pounds instead of 2.2046226218488 pounds.\n\n## Other units also called ounce\n\nOne avoirdupois ounce is equal to approximately 28.3 g (grams). The avoirdupois ounce is used in US and British systems. Our converter uses this unit. There other units also called ounce:\n\n• The troy ounce of about 31.1 g (grams) which is is used only for measuring the mass of precious metals like gold, silver, platinum and palladium.\n• The fluid ounce (fl oz, fl. oz. or oz. fl). It is not a unit of mass but volume. It is equivalent to about 30 ml.\n\n## Examples of kilograms to pounds and ounces conversions\n\n### Disclaimer\n\nWhile every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8750306,"math_prob":0.99136245,"size":3207,"snap":"2021-21-2021-25","text_gpt3_token_len":897,"char_repetition_ratio":0.16453326,"word_repetition_ratio":0.003327787,"special_character_ratio":0.30246335,"punctuation_ratio":0.1565452,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9925111,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-18T21:03:03Z\",\"WARC-Record-ID\":\"<urn:uuid:91a989c1-a237-468c-885d-f469a208c7cd>\",\"Content-Length\":\"115164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:89993317-f215-4e9d-88ec-a2dad6a92e15>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b32b094-d284-42a0-b317-733d9d93b1e0>\",\"WARC-IP-Address\":\"172.67.145.41\",\"WARC-Target-URI\":\"https://coolconversion.com/weight/kg-lbs-oz/_3.09_kilograms_in_lbs_and_oz_\",\"WARC-Payload-Digest\":\"sha1:CAQEAVXIQWSTQIPSDZILRFNP27T5W7KF\",\"WARC-Block-Digest\":\"sha1:X6YJ4VKDK5RPZU76GMFPY2AZH7W5YNF6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487641593.43_warc_CC-MAIN-20210618200114-20210618230114-00541.warc.gz\"}"}
https://proofwiki.org/wiki/Euler%27s_Number_is_Irrational	[ "# Euler's Number is Irrational\n\n## Theorem\n\nEuler's number $e$ is irrational.\n\nAiming for a contradiction, suppose that $e$ is rational.\n\nThen there exist coprime integers $m$ and $n$ (and we can choose $n$ to be positive) such that:\n\n$\\dfrac m n = e = \\displaystyle \\sum_{i \\mathop = 0}^\\infty \\frac 1 {i!}$ from the definition of Euler's number.\n\nMultiplying both sides by $n!$, observe that:\n\n$\\dfrac m n n! = n! \\displaystyle \\sum_{i \\mathop = 0}^\\infty \\frac 1 {i!} = \\paren {\\dfrac {n!} {0!} + \\dfrac {n!} {1!} + \\dfrac {n!} {2!} + \\cdots + \\dfrac {n!} {n!} } + \\paren {\\frac {n!} {\\paren {n + 1}!} + \\dfrac {n!} {\\paren {n + 2}!} + \\dfrac {n!} {\\paren {n + 3}!} + \\cdots}$\n\nHence:\n\n $\\ds m \\paren {n - 1}! - \\paren {\\frac {n!} {0!} + \\frac {n!} {1!} + \\frac {n!} {2!} + \\cdots + \\frac {n!} {n!} }$ $=$ $\\ds \\frac 1 {\\paren {n + 1} } + \\frac 1 {\\paren {n + 1} \\paren {n + 2} } + \\frac 1 {\\paren {n + 1} \\paren {n + 2} \\paren {n + 3} } + \\cdots$ $\\ds$ $<$ $\\ds \\frac 1 {\\paren {n + 1} } + \\frac 1 {\\paren {n + 1} \\paren {n + 1} } + \\frac 1 {\\paren {n + 1} \\paren {n + 1} \\paren {n + 1} } + \\cdots$ $\\ds$ $=$ $\\ds \\sum_{i \\mathop = 0}^\\infty \\paren {\\frac 1 {n + 1} }^{\\paren {i + 1} }$ $\\ds$ $=$ $\\ds \\frac {\\frac 1 {n + 1} } {1 - \\frac 1 {n + 1} } = \\frac 1 n \\le 1$ from Sum of Infinite Geometric Sequence \n\nObserve that the quantity on the left hand side must be an integer, as it is composed entirely of sums and differences of integer terms.\n\nIt must be strictly positive, as it is equal to:\n\n$\\dfrac 1 {\\paren {n + 1} } + \\dfrac 1 {\\paren {n + 1} \\paren {n + 2} } + \\dfrac 1 {\\paren {n + 1} \\paren {n + 2} \\paren {n + 3} } + \\cdots$\n\nwhich is strictly positive.\n\nThus:\n\n$m \\paren {n - 1}! - \\paren {\\dfrac {n!} {0!} + \\dfrac {n!} {1!} + \\dfrac {n!} {2!} + \\cdots + \\dfrac {n!} {n!} }$\n\nmust be a strictly positive integer less than $1$.\n\nFrom this contradiction it follows that $e$ must be irrational.\n\n$\\blacksquare$\n\n## Historical Note\n\nThe proof that Euler's number $e$ is irrational was demonstrated by Leonhard Paul Euler in $1737$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55300754,"math_prob":1.0000072,"size":3035,"snap":"2021-31-2021-39","text_gpt3_token_len":1187,"char_repetition_ratio":0.21840976,"word_repetition_ratio":0.25383306,"special_character_ratio":0.4682043,"punctuation_ratio":0.22824302,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999981,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T07:49:53Z\",\"WARC-Record-ID\":\"<urn:uuid:d023add0-3bd2-4b3a-831b-52768f9440a5>\",\"Content-Length\":\"45035\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68dc9576-1d41-4a99-adfc-07a7064f1aeb>\",\"WARC-Concurrent-To\":\"<urn:uuid:e39b2936-c4ba-42d2-aa6a-bd749f354248>\",\"WARC-IP-Address\":\"104.21.84.229\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Euler%27s_Number_is_Irrational\",\"WARC-Payload-Digest\":\"sha1:B5FIEO77Q6EHRPT47VENXX2S5BMXSKEC\",\"WARC-Block-Digest\":\"sha1:RA7CKWXVL6CF6NNHWO2OQFTFCO6FD7SJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057033.33_warc_CC-MAIN-20210920070754-20210920100754-00701.warc.gz\"}"}
https://forums.toadworld.com/t/check-for-null-value-in-tdp-transformation-calculated-column/47021	[ "# Check For {null} Value in TDP Transformation Calculated Column\n\nHello,\n\nI want to create a calculated column in the transformation tool of TDP where I check if a value is {null} (or not {null}) and return something accordingly. I cannot figure out how TDP represents the null value when creating a calculated column. Appreciate any help.\n\nThanks,\nTroy\n\ninstead of calculated field try using Find and Replace. That rule is set up to recognize Nulls and allow you to set value.\n\nWhat I have is one column with dates/nulls and another column with a status of that project. I would like to create a column which checks if I have a date in the one column and a status of \"On Air\" in the other column. I don't want to replace the null values in this case. Is there a way to check a value is null in an if statement?\n\nLet me play around with it and let you know.\n\nPlease try Group Column if you aim to check a column value is null or not null and return something, not create calculated column.\n\nI have the same question. Besides using an Nvl command to do a null check is there a specific format to see if a column is equal/not equal to null? Example if I wanted to say\n\nIf([Test_Column] = null, 'Null Value', 'Not a Null Value')\n\nHow would be represent the null in the comparison statement to have it check? I could always write\n\nIf(Nvl([Test_Column],0) = 0, 'Null Value', 'Not a Null Value')\n\nbut if I do that then I have to make sure that the replacement value for the null matches the column data type or it will generate an error.\n\nAny info is appreciated.\n\nMy understanding is you want to check null value for one column, if it's null keep null value, if not equal null replace with other value, please try Group Column, is this you are looking for?\n\nThe Grouping makes sense when just looking at transforming the values of a single column, but I think the question here is how can we in building a formula evaluate a column to see if a value is null. For example the formula below is checking a column to see if a value is null, if it is not a calculation takes place, if it is null text is entered in the new calculated field.\n\nIf([TestColumn] != {null}, Sum([Sales]+[SalesTax]), 'No Sale')\n\nAs of right now the only way I can write this formula to work as intended is to do it as shown below\n\nIf(Nvl([TestColumn],0) != 0, Sum([Sales]+[SalesTax]), 'No Sale')\n\nI know the Nvl function can do the check for the null in this calculation, but is there a specific syntax to be used to make the first formula work? If not, working forward with the Nvl is acceptable. It would just help to streamline the formulas in a Nested If scenario.\n\nThanks\n\nThe only way I know how to make a calculated value based on another field is to use a case statement in the SQL. See Query Builder example below for adding Case statement. Here I am checking for Null in Amount_Billed and created a new column and setting value to 0.0 if null or original value if not null.\n\nI think this would work. Can I created a calculated field in the query builder between two different tables? I have one field that shows a project status and another field that shows the date that project was completed. These columns are from two different tables. I want to create a third column which says \"TRUE\" if there is a date and the status shows Complete but \"FALSE\" if there is a date but the status does not show Complete.", null, "" ]	[ null, "https://aws1.discourse-cdn.com/quest/original/2X/a/a33dca33d879063c5abbe8e85d151baadfc1ab90.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8876574,"math_prob":0.87379223,"size":885,"snap":"2019-26-2019-30","text_gpt3_token_len":211,"char_repetition_ratio":0.11577752,"word_repetition_ratio":0.050314464,"special_character_ratio":0.24067797,"punctuation_ratio":0.09139785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96352047,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T21:10:10Z\",\"WARC-Record-ID\":\"<urn:uuid:87b28839-72e5-4c07-bb17-3354ad47fc93>\",\"Content-Length\":\"31134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f780a89-8018-449c-a3db-53fbe9686828>\",\"WARC-Concurrent-To\":\"<urn:uuid:1306207c-c49b-461e-ab55-8dd678609023>\",\"WARC-IP-Address\":\"216.218.159.21\",\"WARC-Target-URI\":\"https://forums.toadworld.com/t/check-for-null-value-in-tdp-transformation-calculated-column/47021\",\"WARC-Payload-Digest\":\"sha1:B4NC7VUW4TSLYT34V4Q4K5PEJPAQTUGZ\",\"WARC-Block-Digest\":\"sha1:O2URVCQ5CBWYVQN2H3TPSUICUVYZVVTL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527204.71_warc_CC-MAIN-20190721205413-20190721231413-00489.warc.gz\"}"}
https://www.scala-lang.org/files/archive/api/3.2.2/scala/math/Ordering$$Unit$.html	[ "# Unit\n\nscala.math.Ordering\\$.Unit\\$\nobject Unit extends UnitOrdering\n\n## Attributes\n\nSource\nOrdering.scala\nGraph\nSupertypes\ntrait UnitOrdering\ntrait Ordering[Unit]\ntrait Equiv[Unit]\ntrait Serializable\ntrait Comparator[Unit]\nclass Object\ntrait Matchable\nclass Any\nSelf type\nUnit.type\n\n## Members list\n\n### Type members\n\n#### Inherited classlikes\n\nclass OrderingOps(lhs: T)\n\nThis inner class defines comparison operators available for `T`.\n\nThis inner class defines comparison operators available for `T`.\n\nIt can't extend `AnyVal` because it is not a top-level class or a member of a statically accessible object.\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\nSupertypes\nclass Object\ntrait Matchable\nclass Any\n\n### Value members\n\n#### Inherited methods\n\ndef compare(x: Unit, y: Unit): Int\n\nReturns an integer whose sign communicates how x compares to y.\n\nReturns an integer whose sign communicates how x compares to y.\n\nThe result sign has the following meaning:\n\n- negative if x < y - positive if x > y - zero otherwise (if x == y)\n\n## Attributes\n\nInherited from:\nUnitOrdering\nSource\nOrdering.scala\noverride def equiv(x: Unit, y: Unit): Boolean\n\nReturn true if `x` == `y` in the ordering.\n\nReturn true if `x` == `y` in the ordering.\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\noverride def gt(x: Unit, y: Unit): Boolean\n\nReturn true if `x` > `y` in the ordering.\n\nReturn true if `x` > `y` in the ordering.\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\noverride def gteq(x: Unit, y: Unit): Boolean\n\nReturn true if `x` >= `y` in the ordering.\n\nReturn true if `x` >= `y` in the ordering.\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef isReverseOf(other: Ordering[_]): Boolean\n\nReturns whether or not the other ordering is the opposite ordering of this one.\n\nReturns whether or not the other ordering is the opposite ordering of this one.\n\nEquivalent to `other == this.reverse`.\n\nImplementations should only override this method if they are overriding reverse as well.\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\noverride def lt(x: Unit, y: Unit): Boolean\n\nReturn true if `x` < `y` in the ordering.\n\nReturn true if `x` < `y` in the ordering.\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\noverride def lteq(x: Unit, y: Unit): Boolean\n\nReturn true if `x` <= `y` in the ordering.\n\nReturn true if `x` <= `y` in the ordering.\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef max[U <: Unit](x: U, y: U): U\n\nReturn `x` if `x` >= `y`, otherwise `y`.\n\nReturn `x` if `x` >= `y`, otherwise `y`.\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef min[U <: Unit](x: U, y: U): U\n\nReturn `x` if `x` <= `y`, otherwise `y`.\n\nReturn `x` if `x` <= `y`, otherwise `y`.\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef on[U](f: U => Unit): Ordering[U]\n\nGiven f, a function from U into T, creates an Ordering[U] whose compare function is equivalent to:\n\nGiven f, a function from U into T, creates an Ordering[U] whose compare function is equivalent to:\n\n``````def compare(x:U, y:U) = Ordering[T].compare(f(x), f(y))\n``````\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef orElse(other: Ordering[Unit]): Ordering[T]\n\nCreates an Ordering[T] whose compare function returns the result of this Ordering's compare function, if it is non-zero, or else the result of `other`s compare function.\n\nCreates an Ordering[T] whose compare function returns the result of this Ordering's compare function, if it is non-zero, or else the result of `other`s compare function.\n\n## Value parameters\n\nother\n\nan Ordering to use if this Ordering returns zero\n\n## Attributes\n\nExample\n\n``````case class Pair(a: Int, b: Int)\nval pairOrdering = Ordering.by[Pair, Int](_.a)\n.orElse(Ordering.by[Pair, Int](_.b))\n``````\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef orElseBy[S](f: Unit => S)(implicit ord: Ordering[S]): Ordering[T]\n\nGiven f, a function from T into S, creates an Ordering[T] whose compare function returns the result of this Ordering's compare function, if it is non-zero, or else a result equivalent to:\n\nGiven f, a function from T into S, creates an Ordering[T] whose compare function returns the result of this Ordering's compare function, if it is non-zero, or else a result equivalent to:\n\n``````Ordering[S].compare(f(x), f(y))\n``````\n\nThis function is equivalent to passing the result of `Ordering.by(f)` to `orElse`.\n\n## Attributes\n\nExample\n\n``````case class Pair(a: Int, b: Int)\nval pairOrdering = Ordering.by[Pair, Int](_.a)\n.orElseBy[Int](_.b)\n``````\nInherited from:\nOrdering\nSource\nOrdering.scala\noverride def reverse: Ordering[T]\n\nReturn the opposite ordering of this one.\n\nReturn the opposite ordering of this one.\n\nImplementations overriding this method MUST override isReverseOf as well if they change the behavior at all (for example, caching does not require overriding it).\n\n## Attributes\n\nDefinition Classes\nInherited from:\nOrdering\nSource\nOrdering.scala\ndef reversed(): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparing[U <: Comparable[_ >: U <: <FromJavaObject>]](x\\$0: Function[_ >: Unit <: <FromJavaObject>, _ <: U]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparing[U <: <FromJavaObject>](x\\$0: Function[_ >: Unit <: <FromJavaObject>, _ <: U], x\\$1: Comparator[_ >: U <: <FromJavaObject>]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparing(x\\$0: Comparator[_ >: Unit <: <FromJavaObject>]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparingDouble(x\\$0: ToDoubleFunction[_ >: Unit <: <FromJavaObject>]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparingInt(x\\$0: ToIntFunction[_ >: Unit <: <FromJavaObject>]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef thenComparingLong(x\\$0: ToLongFunction[_ >: Unit <: <FromJavaObject>]): Comparator[T]\n\n## Attributes\n\nInherited from:\nComparator\ndef tryCompare(x: Unit, y: Unit): Option[Int]\n\nReturns whether a comparison between `x` and `y` is defined, and if so the result of `compare(x, y)`.\n\nReturns whether a comparison between `x` and `y` is defined, and if so the result of `compare(x, y)`.\n\n## Attributes\n\nInherited from:\nOrdering\nSource\nOrdering.scala\n\n### Implicits\n\n#### Inherited implicits\n\nimplicit def mkOrderingOps(lhs: Unit): OrderingOps\n\nThis implicit method augments `T` with the comparison operators defined in `scala.math.Ordering.Ops`.\n\nThis implicit method augments `T` with the comparison operators defined in `scala.math.Ordering.Ops`.\n\nInherited from:\nOrdering\nSource\nOrdering.scala" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62621814,"math_prob":0.44290262,"size":4874,"snap":"2023-14-2023-23","text_gpt3_token_len":1202,"char_repetition_ratio":0.2073922,"word_repetition_ratio":0.6515151,"special_character_ratio":0.23327862,"punctuation_ratio":0.14722537,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98682576,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-28T12:23:01Z\",\"WARC-Record-ID\":\"<urn:uuid:4720150a-63b0-4b78-b0c6-f6ad7799e9ca>\",\"Content-Length\":\"374061\",\"Content-Type\":\"application/http; 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https://calcpercentage.com/what-is-142-percent-of-345	[ "# PercentageCalculator, What is 142% of 345?\n\n## What is 142 percent of 345? 142% of 345 is equal to 489.9\n\n%\n\n### How to Calculate 142 Percent of 345?\n\n• F\n\nFormula\n\n(142 ÷ 100) x 345 = 489.9\n\n• 1\n\nConvert percent to decimal\n\n142% to decimal is 142 ÷ 100 = 1.42\n\n• 2\n\nMultiply the decimal number with the second number\n\n1.42 x 345 = 489.9\n\n#### Example\n\nFor example, John needs 142 percent of the shares to be in power. The number of shares is 345. How many shares does John need to buy? What is 142 percent of 345? 142 percent to decimal is equal 142 / 100 = 1.42 1.42 x 345 = 489.9 so John need to buy 489.9 shares" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8848921,"math_prob":0.9994729,"size":429,"snap":"2022-27-2022-33","text_gpt3_token_len":142,"char_repetition_ratio":0.15058823,"word_repetition_ratio":0.023255814,"special_character_ratio":0.43356642,"punctuation_ratio":0.1262136,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99951386,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T21:15:45Z\",\"WARC-Record-ID\":\"<urn:uuid:82b8c29c-4dce-489c-abbc-1062b96290d1>\",\"Content-Length\":\"11937\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:981e5e34-4ead-486f-ab9e-10e6ef38b4b8>\",\"WARC-Concurrent-To\":\"<urn:uuid:62a397d0-2cae-498d-a71e-3f36a470fc31>\",\"WARC-IP-Address\":\"76.76.21.9\",\"WARC-Target-URI\":\"https://calcpercentage.com/what-is-142-percent-of-345\",\"WARC-Payload-Digest\":\"sha1:GYXRN5BTGD655M6ZKYK5MKHLF32OYB4D\",\"WARC-Block-Digest\":\"sha1:KCRBBBVG7AA2SB3IWVM5TMCE5OUZJO3X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572212.96_warc_CC-MAIN-20220815205848-20220815235848-00123.warc.gz\"}"}
https://questioncove.com/updates/4e5819370b8b1f45b475cbc6	[ "Mathematics", null, "OpenStudy (anonymous):\n\nI need help with gcf and lcm", null, "OpenStudy (aroub):\n\npost ur question! :)", null, "OpenStudy (anonymous):\n\nhow do I find the gcf of 76", null, "OpenStudy (aroub):\n\nand?", null, "OpenStudy (aroub):\n\ni mean 76 and what?", null, "OpenStudy (anonymous):\n\n56", null, "OpenStudy (anonymous):\n\nTo find the LCM of two numbers that are easy to factor I like to make a chart of the factors (and their multiplicity). \\begin{array}{\|c\|ccc}&\\\\ Number &2 & 7 & 19\\\\ \\hline 56 & 3 & 1 & 0\\\\ \\hline 76 & 2 & 0 &1\\end{array} So we have $2^3 \\cdot 7 = 8 \\cdot 7 = 56$$2^2 \\cdot 19 = 4 \\cdot 19 = 76$ To find the LCM we just take the product of each of the factors raised to the max multiplicity of that factor (the biggest number in the column). $LCM = 2^3\\cdot 7^1 \\cdot 19^1 = 1064$", null, "OpenStudy (aroub):\n\nif u want to find the gcf of the numbers u have 56 and 76 u write the factors of the number for 56 there is 1 and 56, 28 and 2 , 14 and 4 , 7 and 8 , okay i think this for the 56, 76= 38 and 2, 19 and 4 , 76 and 1 , okay i guess thats it now u see whats common u have 1,2,4 so 4 is the greatest so its the gcf", null, "OpenStudy (anonymous):\n\nYou can also use the same chart I made for the LCM. The GCF is the product of the minimum values for all the columns. $GCF = 2^2 \\cdot 7^0 \\cdot 19^0 = 4$", null, "OpenStudy (aroub):\n\nyes..", null, "OpenStudy (anonymous):\n\nIf it's two numbers that aren't easy to factor I have to use Euclid's method for finding the GCD, then use the equation: $LCM(a,b) = \\frac{\|a\\cdot b\|}{GCD(a,b)}$", null, "OpenStudy (anonymous):\n\nBut it doesn't generalize for more than 2 numbers.\n\nLatest Questions", null, "Victorya: why do you use math in this world?\n7 minutes ago 7 Replies 3 Medals", null, "PapaLazy: How you work dis ._.\n21 minutes ago 68 Replies 6 Medals", null, "Gorillaz1002: i need your opinion on this i updated it a little\n1 hour ago 23 Replies 3 Medals", null, "xXQuintonXx: @aevans182 ur dm's r off -_-\n1 hour ago 0 Replies 0 Medals", null, "MrsHero: Help ss down below ;/..\n1 hour ago 7 Replies 1 Medal", null, "MrsHero: Help ss down below ;/\n1 hour ago 0 Replies 0 Medals", null, "nettym: math help please.\n1 hour ago 10 Replies 2 Medals", null, "MrsHero: Can yall rate my pic i took in Meepcity/roblox SS down bellow\n1 hour ago 8 Replies 0 Medals", null, "MrsHero: Helpp ss down bellow i suck at math thts wy im here u270c\n1 hour ago 5 Replies 0 Medals", null, "DONTPISSWITHME: What is 45^2 please help me\n1 hour ago 21 Replies 1 Medal" ]	[ null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/images/default-avatar.svg", null, "https://questioncove.com/assets/users/victorya/avatar/small", null, "https://questioncove.com/assets/users/papalazy/avatar/small", null, "https://questioncove.com/assets/users/gorillaz1002/avatar/small", null, "https://questioncove.com/assets/users/xxquintonxx/avatar/small", null, "https://questioncove.com/assets/users/mrshero/avatar/small", null, "https://questioncove.com/assets/users/mrshero/avatar/small", null, "https://questioncove.com/assets/users/nettym/avatar/small", null, "https://questioncove.com/assets/users/mrshero/avatar/small", null, "https://questioncove.com/assets/users/mrshero/avatar/small", null, "https://questioncove.com/images/default-avatar.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89293194,"math_prob":0.8789483,"size":1362,"snap":"2021-21-2021-25","text_gpt3_token_len":472,"char_repetition_ratio":0.13770251,"word_repetition_ratio":0.0,"special_character_ratio":0.37077826,"punctuation_ratio":0.08074534,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9809582,"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,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,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T03:26:51Z\",\"WARC-Record-ID\":\"<urn:uuid:0a416ce5-f5e8-4795-b6cc-825795e57654>\",\"Content-Length\":\"24532\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb058d6a-912a-4930-adab-88a10192c3c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3eb533b-e9e8-4ba1-9a0d-e89a4f067121>\",\"WARC-IP-Address\":\"192.249.125.12\",\"WARC-Target-URI\":\"https://questioncove.com/updates/4e5819370b8b1f45b475cbc6\",\"WARC-Payload-Digest\":\"sha1:QCEM7TR64MENJBD2CB5CSYJII3WRGCWW\",\"WARC-Block-Digest\":\"sha1:ZRUJFGQKESNM3VFHICINN6VADQP4GPRC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991641.5_warc_CC-MAIN-20210511025739-20210511055739-00068.warc.gz\"}"}
https://mathworld.wolfram.com/EdgeColoring.html	[ "", null, "TOPICS", null, "", null, "# Edge Coloring", null, "An edge coloring of a graph", null, "is a coloring of the edges of", null, "such that adjacent edges (or the edges bounding different regions) receive different colors. An edge coloring containing the smallest possible number of colors for a given graph is known as a minimum edge coloring.\n\nA (not necessarily minimum) edge coloring of a graph can be computed using EdgeColoring[g] in the Wolfram Language package Combinatorica` .\n\nThe edge chromatic number gives the minimum number of colors with which graph's edges can be colored.\n\nChromatic Number, Edge Chromatic Number, Graph Coloring, k-Coloring, Labeled Graph, Minimum Edge Coloring, Minimum Vertex Coloring, Vertex Coloring\n\n## Explore with Wolfram\|Alpha", null, "More things to try:\n\n## References\n\nFiorini, S. and Wilson, R. Edge-Colourings of Graphs. Pittman, 1977.Nemhauser, G. L. and Park, S. \"A Polyhedral Approach to Edge Coloring.\" Operations Res. Lett. 10, 315-322, 1991.Saaty, T. L. and Kainen, P. C. The Four-Color Problem: Assaults and Conquest. New York: Dover, p. 13, 1986.Skiena, S. \"Edge Colorings.\" §5.5.4 in Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. Reading, MA: Addison-Wesley, p. 216, 1990.\n\nEdge Coloring\n\n## Cite this as:\n\nWeisstein, Eric W. \"Edge Coloring.\" From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/EdgeColoring.html" ]	[ null, "https://mathworld.wolfram.com/images/header/menu-icon.png", null, "https://mathworld.wolfram.com/images/sidebar/menu-close.png", null, "https://mathworld.wolfram.com/images/header/search-icon.png", null, "https://mathworld.wolfram.com/images/eps-svg/EdgeColoring_850.svg", null, "https://mathworld.wolfram.com/images/equations/EdgeColoring/Inline1.svg", null, "https://mathworld.wolfram.com/images/equations/EdgeColoring/Inline2.svg", null, "https://mathworld.wolfram.com/images/wolframalpha/WA-logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70903707,"math_prob":0.65346414,"size":1379,"snap":"2023-40-2023-50","text_gpt3_token_len":373,"char_repetition_ratio":0.15345454,"word_repetition_ratio":0.009803922,"special_character_ratio":0.23930384,"punctuation_ratio":0.23104693,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97288287,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T21:48:23Z\",\"WARC-Record-ID\":\"<urn:uuid:d181d3db-0639-462c-894a-6b93c4047a36>\",\"Content-Length\":\"68620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7426a2c-339a-46a6-8377-fddab8b94785>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a14ffd1-fb69-46f0-b0dc-bb428bbdba7c>\",\"WARC-IP-Address\":\"140.177.8.25\",\"WARC-Target-URI\":\"https://mathworld.wolfram.com/EdgeColoring.html\",\"WARC-Payload-Digest\":\"sha1:NBBBRN4XTMTGFTCLUTY3KMP24O7SRT3O\",\"WARC-Block-Digest\":\"sha1:PN3DDILEEONVIYGTOX43PSXGCJFIZZRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100452.79_warc_CC-MAIN-20231202203800-20231202233800-00585.warc.gz\"}"}
https://irmar.univ-rennes1.fr/seminaire/cryptographie/andre-schrottenloher	[ "# Improved Classical and Quantum Algorithms for Subset-Sum\n\nWe present new classical and quantum algorithms for solving random hard instances\nof the subset-sum problem, in which we are given n integers on n bits and try\nto find a subset of them that sums to a given target. This classical NP-complete\nproblem has several applications in cryptography and underlies the security\nof some proposed post-quantum cryptosystems.\n\nAt EUROCRYPT 2010, Howgrave-Graham and Joux (HGJ) introduced the representation\ntechnique and presented an algorithm running in time $\\bigOt{2^{0.337 n}}$.\nThis asymptotic time was improved by Becker, Coron, Joux (BCJ) at EUROCRYPT 2011.\nWe show how to improve this further.\n\nWe then move to the context of quantum algorithms. The two previous\nquantum speedups in the literature are given by Bernstein, Jeffery, Lange\nand Meurer (PQCRYPTO 2013) and Helm and May (TQC 2018), which are respectively\nquantum versions of HGJ and BCJ. They both rely on the framework of quantum\nwalks, use exponential quantum memory with quantum random-access and require\nan unproven conjecture on quantum walk updates.\n\nWe devise a new algorithm, using quantum search only,\nthat achieves the first quantum speedup in the model of \\emph{classical}\nmemory with quantum random access. Next, we study improvements for the\nquantum walks. We show how to avoid the quantum walk conjecture and give\na better quantum walk time complexity for subset-sum." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83768857,"math_prob":0.93407905,"size":1440,"snap":"2021-31-2021-39","text_gpt3_token_len":331,"char_repetition_ratio":0.13579386,"word_repetition_ratio":0.0,"special_character_ratio":0.20347223,"punctuation_ratio":0.089147285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9787686,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T16:25:34Z\",\"WARC-Record-ID\":\"<urn:uuid:2381b128-41a5-452a-84e6-05bdd9f89131>\",\"Content-Length\":\"58171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48a87639-63ca-4398-ae95-33a1a813cb22>\",\"WARC-Concurrent-To\":\"<urn:uuid:975fa53f-5059-4a09-b745-28fd326cda4a>\",\"WARC-IP-Address\":\"129.20.126.134\",\"WARC-Target-URI\":\"https://irmar.univ-rennes1.fr/seminaire/cryptographie/andre-schrottenloher\",\"WARC-Payload-Digest\":\"sha1:J7CMQIUWCUCCLXOQSX4UUYIETGEBTFJH\",\"WARC-Block-Digest\":\"sha1:PB6JUIWOWO3C624MUTFXTAGEQ24NKHUT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056892.13_warc_CC-MAIN-20210919160038-20210919190038-00344.warc.gz\"}"}
https://www.physicsforums.com/threads/simple-dimensional-analysis.831246/	[ "# Simple Dimensional Analysis\n\n## Homework Statement\n\nIs the following equation dimensionally correct?\n\nE = (1/2) mv\nwhere:\nE = energy\nm = mass\nv = speed\n\n## The Attempt at a Solution\n\n1. I understand that the 1/2 is irrelevant.\n2. I broke everything down into length, time, and mass.\n3. I got ML^2/T^2 = ML/T\n4. My confusion is that if you square L and T on the left side, you still have length and time so I mean essentially, you have the same types of quantities on both sides of the equation in the same order, so I want to say it is correct, however the fact that L and T are squared on the left side but not on the right side bugs me and I'm doubtful.\n\n## Homework Statement\n\nIs the following equation dimensionally correct?\n\nBy \"the following equation\" do you mean E=(1/2)mv^2 ?\n\nBy \"the following equation\" do you mean E=(1/2)mv^2 ?\nI was referring to the equation under Relevant Equations (my apologies, should've clarified that), which is E=(1/2)mv\n\nI was referring to the equation under Relevant Equations (my apologies, should've clarified that), which is E=(1/2)mv\n\nYou are missing ^2 at the end of the equation: it should be E = (1/2)mv^2.\n\nYou are missing ^2 at the end of the equation: it should be E = (1/2)mv^2.\nOkay so it isn't dimensionally correct? The book just gives me an equation (it's not necessarily supposed to be right or wrong) and I'm just supposed to say whether the given equation is dimensionally correct. But since the equation itself is wrong then I'm going to assume that the correct version is dimensionally correct and this one is not.\n\nOkay so it isn't dimensionally correct? The book just gives me an equation (it's not necessarily supposed to be right or wrong) and I'm just supposed to say whether the given equation is dimensionally correct. But since the equation itself is wrong then I'm going to assume that the correct version is dimensionally correct and this one is not.\n\nAbsolutely right. As you have pointed out from the beginning LHS has extra L/T that RHS does not have, therefore it cannot be dimensionally correct.\n\n•", null, "Michele Nunes" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9572578,"math_prob":0.99040514,"size":651,"snap":"2021-21-2021-25","text_gpt3_token_len":179,"char_repetition_ratio":0.10664606,"word_repetition_ratio":0.0,"special_character_ratio":0.2780338,"punctuation_ratio":0.10273973,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996629,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T14:21:36Z\",\"WARC-Record-ID\":\"<urn:uuid:e6c77bf0-943c-415b-8964-9a9c1c283fb8>\",\"Content-Length\":\"78004\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d50e874a-c751-4ec9-9e3f-a6388d61ab8a>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f8282d0-33a1-43be-a82b-a5682ac29e4a>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/simple-dimensional-analysis.831246/\",\"WARC-Payload-Digest\":\"sha1:4ZYRWO4IVAGJZE7V67IB3UR3L7CBLXUQ\",\"WARC-Block-Digest\":\"sha1:BY4S34TVVP4WWC2OWKUWOWJE5WRDGF5R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488273983.63_warc_CC-MAIN-20210621120456-20210621150456-00294.warc.gz\"}"}
https://www.igi-global.com/chapter/complex-valued-symmetric-radial-basis/6768	[ "", null, "# Complex-Valued Symmetric Radial Basis Function Network for Beamforming\n\nSheng Chen\nDOI: 10.4018/978-1-60566-214-5.ch007\nOnDemand:\n(Individual Chapters)\nAvailable\n\\$37.50\nNo Current Special Offers\n\n## Abstract\n\nThe complex-valued radial basis function (RBF) network proposed by Chen et al. (1994) has found many applications for processing complex-valued signals, in particular, for communication channel equalization and signal detection. This complex-valued RBF network, like many other existing RBF modeling methods, constitutes a black-box approach that seeks typically a sparse model representation extracted from the training data. Adopting black-box modeling is appropriate, if no a priori information exists regarding the underlying data generating mechanism. However, a fundamental principle in practical data modelling is that if there exists a priori information concerning the system to be modeled it should be incorporated in the modeling process. Many complex-valued signal processing problems, particularly those encountered in communication signal detection, have some inherent symmetric properties. This contribution adopts a grey-box approach to complex-valued RBF modeling and develops a complex-valued symmetric RBF (SRBF) network model. The application of this SRBF network is demonstrated using nonlinear beamforming assisted detection for multiple-antenna aided wireless systems that employ complex-valued modulation schemes. Two training algorithms for this complex-valued SRBF network are proposed. The first method is based on a modified version of the cluster-variation enhanced clustering algorithm, while the second method is derived by modifying the orthogonal-forward-selection procedure based on Fisher ratio of class separability measure. The effectiveness of the proposed complex-valued SRBF network and the efficiency of the two training algorithms are demonstrated in nonlinear beamforming application.\nChapter Preview\nTop\n\n## Introducion\n\nThe radial basis function (RBF) network is a popular artificial neural network (ANN) architecture that has found wide-ranging applications in many diverse fields of engineering, see for example, (Chen et al., 1990; Leonard & Kramer, 1991; Chen et al., 1993; Caiti & Parisini, 1994; Gorinevsky et al., 1996; Cha & Kassam, 1996; Rosenblum & Davis, 1996; Refaee et al., 1999; Muraki et al., 2001; Mukai, et al., 2002; Su et al., 2002; Li et al., 2004; Lee & Choi, 2004; Ng et al., 2004; Oyang et al., 2005; Acir et al., 2005; Tan et al., 2005). The RBF method is a classical numerical technique for nonlinear functional interpolation with real-valued data (Powell, 1987). A renewed interest in the RBF method coincided with a recent resurgence in the field of ANNs. Connections between the RBF method and the ANN was made and the RBF model was re-interpreted as a one-hidden-layer feedforward network (Broomhead & Lowe, 1988; Poggio & Girosi, 1990). Specifically, by adopting the ANN interpretation, a RBF model can be considered as a processing structure consisting of a hidden layer and an output layer. Each node in the hidden layer has a radially symmetric response around a node parameter vector called a centre, with the hidden node’s response shape determined by the chosen basis function as well as a node width parameter, while the output layer is a set of linear combiners with linear connection weights.\n\nThe parameters of the RBF network include its centre vectors and variances or covariance matrices of the basis functions as well as the weights that connect the RBF nodes to the network output. All the parameters of a RBF network can be learned together via nonlinear optimisation using the gradient based algorithms (Chen et al., 1990a; An et al., 1993; McLoone et al., 1998; Karayiannis et al., 2003; Peng et al., 2003), the evolutionary algorithms (Whitehead & Choate, 1994; Whitehead, 1996; Gonzalez et al., 2003) or the expectation-maximisation algorithm (Yang & Chen, 1998; Mak & Kung, 2000). Generally, learning based on such a nonlinear approach is computationally expensive and may encounter the problem of local minima. Additionally, the network structure or the number of RBF nodes has to be determined via other means, typically based on cross validation. Alternatively, clustering algorithms can be applied to find the RBF centre vectors as well as the associated basis function variances (Moody & Darken, 1989; Chen et al., 1992; Chen, 1995; Uykan, 2003). This leaves the RBF weights to be determined by the usual linear least squares solution. Again, the number of the clusters has to be determined via other means, such as cross validation. One of the most popular approaches for constructing RBF networks however is to formulate the problem as a linear learning one by considering the training input data points as candidate RBF centres and employing a common variance for every RBF node. A parsimonious RBF network is then identified using the orthogonal least squares (OLS) algorithm (Chen et al., 1989; Chen et al., 1991; Chen et al., 1999; Chen et al., 2003; Chen et al., 2004a).\n\n## Complete Chapter List\n\nSearch this Book:\nReset" ]	[ null, "https://coverimages.igi-global.com/cover-images/covers/9781605662145.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.881207,"math_prob":0.92099667,"size":4856,"snap":"2023-40-2023-50","text_gpt3_token_len":1091,"char_repetition_ratio":0.1420033,"word_repetition_ratio":0.010810811,"special_character_ratio":0.22817133,"punctuation_ratio":0.15906563,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96223897,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T17:51:33Z\",\"WARC-Record-ID\":\"<urn:uuid:4e616343-9c5d-4e4e-884d-5b2c8beb3428>\",\"Content-Length\":\"96972\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f61a6f7a-0292-4104-bd6a-cf25c1e0eef5>\",\"WARC-Concurrent-To\":\"<urn:uuid:93006451-e6f8-4fe6-b153-6e5e69690961>\",\"WARC-IP-Address\":\"40.121.212.165\",\"WARC-Target-URI\":\"https://www.igi-global.com/chapter/complex-valued-symmetric-radial-basis/6768\",\"WARC-Payload-Digest\":\"sha1:YR4GXADPR4VR5RGBSNBHTDQPTAU2MYY7\",\"WARC-Block-Digest\":\"sha1:TAYCQMJD445QCOUGIY5UH5TYLLQY4GGC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506528.19_warc_CC-MAIN-20230923162848-20230923192848-00633.warc.gz\"}"}
https://converter.ninja/volume/metric-teaspoons-to-us-tablespoons/507-brteaspoon-to-ustablespoon/	[ "# 507 metric teaspoons in US tablespoons\n\n## Conversion\n\n507 metric teaspoons is equivalent to 171.437095098344 US tablespoons.\n\n## Conversion formula How to convert 507 metric teaspoons to US tablespoons?\n\nWe know (by definition) that: $1\\mathrm{brteaspoon}\\approx 0.33814022701843\\mathrm{ustablespoon}$\n\nWe can set up a proportion to solve for the number of US tablespoons.\n\n$1 ⁢ brteaspoon 507 ⁢ brteaspoon ≈ 0.33814022701843 ⁢ ustablespoon x ⁢ ustablespoon$\n\nNow, we cross multiply to solve for our unknown $x$:\n\n$x\\mathrm{ustablespoon}\\approx \\frac{507\\mathrm{brteaspoon}}{1\\mathrm{brteaspoon}}*0.33814022701843\\mathrm{ustablespoon}\\to x\\mathrm{ustablespoon}\\approx 171.437095098344\\mathrm{ustablespoon}$\n\nConclusion: $507 ⁢ brteaspoon ≈ 171.437095098344 ⁢ ustablespoon$", null, "## Conversion in the opposite direction\n\nThe inverse of the conversion factor is that 1 US tablespoon is equal to 0.00583304330621302 times 507 metric teaspoons.\n\nIt can also be expressed as: 507 metric teaspoons is equal to $\\frac{1}{\\mathrm{0.00583304330621302}}$ US tablespoons.\n\n## Approximation\n\nAn approximate numerical result would be: five hundred and seven metric teaspoons is about one hundred and seventy-one point four three US tablespoons, or alternatively, a US tablespoon is about zero point zero one times five hundred and seven metric teaspoons.\n\n## Footnotes\n\n The precision is 15 significant digits (fourteen digits to the right of the decimal point).\n\nResults may contain small errors due to the use of floating point arithmetic.\n\nWas it helpful? Share it!" ]	[ null, "https://converter.ninja/images/507_brteaspoon_in_ustablespoon.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81511545,"math_prob":0.98577154,"size":970,"snap":"2021-21-2021-25","text_gpt3_token_len":198,"char_repetition_ratio":0.17287785,"word_repetition_ratio":0.013071896,"special_character_ratio":0.21958762,"punctuation_ratio":0.09195402,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9547343,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T22:44:17Z\",\"WARC-Record-ID\":\"<urn:uuid:5856db03-d070-4f59-a852-dde2c019d500>\",\"Content-Length\":\"18801\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d085c1b9-51ba-42a1-aa0a-d542734dd1d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba8ae67b-51d6-447a-a2a8-a247fe6f6bd7>\",\"WARC-IP-Address\":\"172.67.167.118\",\"WARC-Target-URI\":\"https://converter.ninja/volume/metric-teaspoons-to-us-tablespoons/507-brteaspoon-to-ustablespoon/\",\"WARC-Payload-Digest\":\"sha1:OPMKOMOMNHRZYVM53A3TYK7LHZFD4MLJ\",\"WARC-Block-Digest\":\"sha1:YJMKJFRL4IRU7CMHJPLF6K7BCNPRH5NB\",\"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-20210617222646-20210618012646-00325.warc.gz\"}"}
https://emresahin.net/pgm-course-notes/	[ "# Preliminaries\n\n## Distributions\n\nSuppose A has 2, B has 2 and C has 3 possible values. Their Joint Probability Distribution will contain 2x2x3=12 values.\n\nWe can condition the values by setting a variable to a certain value.\n\nWe can also marginalize the values to a certain variable and check the distribution of this single variable.\n\n# Factors\n\nA factor $\\phi$ is a function that takes values for A, B and C and returns a real value.\n\nThe arguments that the factor takes are called the scope of the factor.\n\nBoth normalized and unnormalized measures are factors. But the scope of each can change.\n\nCPD (Conditional Probability Distribution) is another example of a factor." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82435495,"math_prob":0.9466026,"size":660,"snap":"2023-14-2023-23","text_gpt3_token_len":147,"char_repetition_ratio":0.14786585,"word_repetition_ratio":0.0,"special_character_ratio":0.20454545,"punctuation_ratio":0.08943089,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-10T20:30:45Z\",\"WARC-Record-ID\":\"<urn:uuid:e7b78b7c-34b7-47f3-b634-5c137ad754d7>\",\"Content-Length\":\"3717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bf06a06-8440-41d3-883a-aac0ddc44e56>\",\"WARC-Concurrent-To\":\"<urn:uuid:434e9a4c-d498-42d3-9131-abb9afcdc6d1>\",\"WARC-IP-Address\":\"75.2.60.5\",\"WARC-Target-URI\":\"https://emresahin.net/pgm-course-notes/\",\"WARC-Payload-Digest\":\"sha1:EGTOJFO5J2AISZWIQYNQK472CSZ6XUZ4\",\"WARC-Block-Digest\":\"sha1:26CRMXEER34ER72B6IMRLRIQK2FVQVRN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224646350.59_warc_CC-MAIN-20230610200654-20230610230654-00317.warc.gz\"}"}
https://convertoctopus.com/444-cubic-centimeters-to-teaspoons	[ "## Conversion formula\n\nThe conversion factor from cubic centimeters to teaspoons is 0.20288413535365, which means that 1 cubic centimeter is equal to 0.20288413535365 teaspoons:\n\n1 cm3 = 0.20288413535365 tsp\n\nTo convert 444 cubic centimeters into teaspoons we have to multiply 444 by the conversion factor in order to get the volume amount from cubic centimeters to teaspoons. We can also form a simple proportion to calculate the result:\n\n1 cm3 → 0.20288413535365 tsp\n\n444 cm3 → V(tsp)\n\nSolve the above proportion to obtain the volume V in teaspoons:\n\nV(tsp) = 444 cm3 × 0.20288413535365 tsp\n\nV(tsp) = 90.080556097022 tsp\n\nThe final result is:\n\n444 cm3 → 90.080556097022 tsp\n\nWe conclude that 444 cubic centimeters is equivalent to 90.080556097022 teaspoons:\n\n444 cubic centimeters = 90.080556097022 teaspoons\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 teaspoon is equal to 0.011101174807613 × 444 cubic centimeters.\n\nAnother way is saying that 444 cubic centimeters is equal to 1 ÷ 0.011101174807613 teaspoons.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that four hundred forty-four cubic centimeters is approximately ninety point zero eight one teaspoons:\n\n444 cm3 ≅ 90.081 tsp\n\nAn alternative is also that one teaspoon is approximately zero point zero one one times four hundred forty-four cubic centimeters.\n\n## Conversion table\n\n### cubic centimeters to teaspoons chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to teaspoons\n\ncubic centimeters (cm3) teaspoons (tsp)\n445 cubic centimeters 90.283 teaspoons\n446 cubic centimeters 90.486 teaspoons\n447 cubic centimeters 90.689 teaspoons\n448 cubic centimeters 90.892 teaspoons\n449 cubic centimeters 91.095 teaspoons\n450 cubic centimeters 91.298 teaspoons\n451 cubic centimeters 91.501 teaspoons\n452 cubic centimeters 91.704 teaspoons\n453 cubic centimeters 91.907 teaspoons\n454 cubic centimeters 92.109 teaspoons" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6722153,"math_prob":0.9919124,"size":2081,"snap":"2021-21-2021-25","text_gpt3_token_len":517,"char_repetition_ratio":0.2715455,"word_repetition_ratio":0.013157895,"special_character_ratio":0.32532436,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99025756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-22T17:25:18Z\",\"WARC-Record-ID\":\"<urn:uuid:1d33f068-9bee-4d7d-a380-446beb85115c>\",\"Content-Length\":\"30980\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e3a0031-8861-4c63-ae78-1abcbbfc9672>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4616d66-fc3d-4bf9-a32b-56bae24954ea>\",\"WARC-IP-Address\":\"104.21.61.99\",\"WARC-Target-URI\":\"https://convertoctopus.com/444-cubic-centimeters-to-teaspoons\",\"WARC-Payload-Digest\":\"sha1:B6NIEPJWWRNZ6EYLM4VTNPEIVCIZ4MTY\",\"WARC-Block-Digest\":\"sha1:BPREY2Q2XY3PL3LW23B3U6PCX55EUAAX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488519183.85_warc_CC-MAIN-20210622155328-20210622185328-00254.warc.gz\"}"}
https://algebra-help.com/algebra-help-factor/angle-complements/how-to-figure-an-interest.html	[ "", null, "# Our users:\n\nDon Woodward, ND\n\nThis new version is a vast improvement over the old one.\nGus Taylor, AZ\n\nAlgebrator is simply amazing. Who knew that such an inexpensive program would make my sons grades improve so much.\nDavid Brown, CA\n\nI think this is a great piece of software. I applaud your efforts to bring a product to struggling math students such as myself.\nGeorge Tereckski, MA\n\nIts been a long time since I needed to understand algebra and when it came time to helping my son, I couldnt do it. Now, with your algebra software, we are both learning together.\nJ.S., Alabama\n\n# Students 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\nSearch phrases used on 2011-06-29:\n\n• negative fractions\n• online calculators to find greatest common factor of monomials in algebra equations\n• aptitude questions on cubes\n• clep algebra study\n• Bitsize KS2 algebra\n• If you are looking at a graph of a qaudratic equation , how do you determine where the solutions are?\n• online inequality calculator\n• second order differential equation system of first order\n• dependent probability ti-83\n• free percentage worksheet\n• cube root 25\n• ti 89 pre calculus software\n• Nelson Math Worksheets For grade sevens\n• matlab non-linear equations\n• blank scale worksheet + prealgebra\n• multiply and simplify rational expressions calculator\n• cost accouting + problem solutions\n• kumon algebra\n• parabola standard form\n• combining like terms lesson plans\n• algebra printouts\n• find equation of line parallel solver\n• ti-89+solve complex number matrices\n• example math trivia question algebra\n• maths sheets for 7 year olds\n• Hash calc SH1\n• multiplication of cuberoot by a squareroot\n• algebra calculator free\n• formula of substitution method\n• square root in excel\n• adding subtracting decimal story problem worksheet\n• \"aleks cheat\"\n• worksheet about subtraction of integers\n• intermediate algebra by washington\n• collegealgebrahelp\n• algebra 1 solving linear equations worksheet\n• estimate evaluate difference\n• online calculate gcd\n• ti-89+pdf\n• beginning algebra worksheets for 6th grade\n• activity +powers and square roots\n• solve \"quadratic equation\" vertex two points\n• calculator complex number equations sistems\n• one proportion z test problems\n• trivia on vector measurement\n• sample test paper for apptitude\n• invention of mathematical value 'pie'\n• finding the slope in a grid\n• FACTORING BY THE SAME CUBES\n• math trivia high school\n• multiplying standard form\n• math test pdf\n• exercise on permutations\n• forth grade stuff to do online\n• Simultaneous equations for dummies\n• extracting square root\n• graphing calculator answers into factions\n• cost accounting basics\n• PLANE WORKSHEETS\n• What is the basic principle that can be used to simplify a polynomial\n• permutation and combination +books\n• Simplify the expression with square roots\n• range of a quadratic equation\n• examples of math trivias\n• base 8 conversion ti-89\n• factor expressions calculator\n• real solution of the equation by factoring\n• group theory problems and answers\n• different poems in algebra\n• math trivia questions on sets,fractions and integer\n• math trivias\n• teaching of permutation and combination made easy\n• formula for fractions to decimal" ]	[ null, "https://algebra-help.com/images/template/phone.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8259951,"math_prob":0.98153013,"size":3959,"snap":"2021-31-2021-39","text_gpt3_token_len":911,"char_repetition_ratio":0.12515803,"word_repetition_ratio":0.0,"special_character_ratio":0.20813337,"punctuation_ratio":0.044701986,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996045,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T01:16:35Z\",\"WARC-Record-ID\":\"<urn:uuid:3d615c89-cdff-4f21-96a7-b2005d06de54>\",\"Content-Length\":\"12717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc106195-4805-419e-9968-707ffbd8b748>\",\"WARC-Concurrent-To\":\"<urn:uuid:974e1243-f7ef-4955-a77c-15d12f47157c>\",\"WARC-IP-Address\":\"54.197.228.212\",\"WARC-Target-URI\":\"https://algebra-help.com/algebra-help-factor/angle-complements/how-to-figure-an-interest.html\",\"WARC-Payload-Digest\":\"sha1:I2T2VD5F2YHF6GZSZHZRRFATOA4ELD7K\",\"WARC-Block-Digest\":\"sha1:4B36O7V5HP4HHQ2BQ6K6C7AEYG2GDLRW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153899.14_warc_CC-MAIN-20210729234313-20210730024313-00254.warc.gz\"}"}
https://www.nagwa.com/en/lessons/303165635278/	[ "# Lesson: Expected Values of Discrete Random Variables Mathematics • 10th Grade\n\nIn this lesson, we will learn how to calculate the expected value of a discrete random variable from a table, a graph, and a word problem." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8815653,"math_prob":0.941712,"size":679,"snap":"2022-27-2022-33","text_gpt3_token_len":137,"char_repetition_ratio":0.13185185,"word_repetition_ratio":0.0,"special_character_ratio":0.19145803,"punctuation_ratio":0.11382114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98525935,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T10:51:13Z\",\"WARC-Record-ID\":\"<urn:uuid:d9f09481-ba7f-4414-8d70-b1e58bb65505>\",\"Content-Length\":\"249463\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa0fa048-c2aa-4f9b-a153-82a326356609>\",\"WARC-Concurrent-To\":\"<urn:uuid:66e8303b-e549-433e-8f6c-1044bf80f0cf>\",\"WARC-IP-Address\":\"76.223.114.27\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/lessons/303165635278/\",\"WARC-Payload-Digest\":\"sha1:WX5VRHFC75PFCXSARR5A5PAGGDXHZW3P\",\"WARC-Block-Digest\":\"sha1:R7AXAWZB3KR4WFEOZDWUYMSDETXF4BIG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570793.14_warc_CC-MAIN-20220808092125-20220808122125-00485.warc.gz\"}"}
https://library.wolfram.com/infocenter/Courseware/7591/	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Title", null, "", null, "", null, "", null, "Rational Functions: Slant Asymptotes", null, "", null, "", null, "Author", null, "", null, "", null, "", null, "Andy Dorsett\n Organization: Wolfram Research, Inc.", null, "", null, "", null, "Education level", null, "", null, "", null, "", null, "Precollege", null, "", null, "", null, "Description", null, "", null, "", null, "", null, "Students will predict when a rational function will have a slant (or oblique) asymptote and demonstrate how that affects the behavior of the graph.\n\nThis notebook covers topics that fulfill the following NCTM standards:\n• Algebra 9 - 12: Understand patterns, relations, and functions\n• Algebra 9 - 12: Represent and analyze mathematical situations and structures using algebraic symbols", null, "", null, "", null, "Subjects", null, "", null, "", null, "", null, "", null, "Education > Precollege", null, "Mathematics > Algebra", null, "Mathematics > Calculus and Analysis", null, "Mathematics > Calculus and Analysis > Calculus", null, "Mathematics > Geometry", null, "Mathematics > Geometry > Analytic Geometry", null, "Teaching", null, "", null, "", null, "Keywords", null, "", null, "", null, "", null, "Algebra 2, College Algebra, Pre-Calculus, Analytic Geometry, Rational Functions, Slant Asymptotes, Oblique, Graphing, High School, Precollege, NCTM", null, "", null, "", null, "URL", null, "", null, "", null, "", null, "http://www.wolfram.com/solutions/precollege", null, "", null, "", null, "Downloads", null, "", null, "", null, "", null, "", null, "RationalFunctionsSlantAsymptotes.nb (389.5 KB) - Mathematica Notebook", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]	[ null, 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https://mcqclass10.com/if-the-power-of-a-lens-is-4-0d-then-it-means-that-the-lens-is-a/	[ "# If the power of a lens is – 4.0D then it means that the lens is a\n\nQ.) If the power of a lens is – 4.0D then it\nmeans that the lens is a\n\n(a) concave lens of focal length -50 m\n\n(b) convex lens of focal length +50 cm\n\n(c) concave lens of focal length -25 cm\n\n(d) convex lens of focal length -25 m\n\nAns. (c) concave lens of focal length -25 cm" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70449096,"math_prob":0.99946445,"size":270,"snap":"2022-27-2022-33","text_gpt3_token_len":85,"char_repetition_ratio":0.24060151,"word_repetition_ratio":0.21818182,"special_character_ratio":0.33333334,"punctuation_ratio":0.048387095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.989586,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T20:37:06Z\",\"WARC-Record-ID\":\"<urn:uuid:408716af-6d80-4e4e-966d-a122efa0e7f4>\",\"Content-Length\":\"83870\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:deebd710-f338-477c-bfc6-ff2880f6f174>\",\"WARC-Concurrent-To\":\"<urn:uuid:9dedc986-1bfe-4aa4-a197-34fef6c2f6a7>\",\"WARC-IP-Address\":\"104.21.69.107\",\"WARC-Target-URI\":\"https://mcqclass10.com/if-the-power-of-a-lens-is-4-0d-then-it-means-that-the-lens-is-a/\",\"WARC-Payload-Digest\":\"sha1:KKQILJDLGMPE3236QCVR2XABT6F3E7HR\",\"WARC-Block-Digest\":\"sha1:QVXJEPIO5FLP3BI6HJN2IJR37LTXLXKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571758.42_warc_CC-MAIN-20220812200804-20220812230804-00677.warc.gz\"}"}
https://se.mathworks.com/matlabcentral/answers/400044-how-do-i-write-the-code-for-navierstokes-equation-in-matlab-comsol	[ "# How do i write the code for navierstokes equation in matlab/comsol?\n\n9 views (last 30 days)\nBartholomew Osegbe on 10 May 2018\nAnswered: Bartholomew Osegbe on 13 May 2018\nThe idea is to write the code for the navierstokes equation with the laplacian velocity and pressure term\nKSSV on 10 May 2018\nYou want it in comsol? So this is not the forum. You want in MATLAB? What have you attempted?\n\nPrecise Simulation on 10 May 2018\nThe incompressible Navier-Stokes equations is also available as a built-in pre-defined Navier-Stokes physics mode in the FEATool FEM Matlab toolbox. In this case the equations are in 2D defined as\nrho_nsu' - miu_ns(2ux_x + uy_y + vx_y) + rho_ns(uux_t + vuy_t) + p_x = Fx_ns\nrho_nsv' - miu_ns(vx_x + uy_x + 2vy_y) + rho_ns(uvx_t + vvy_t) + p_y = Fy_ns\nux_t + vy_t = 0\nthe Navier-Stokes equations in axisymmetry (2D cylindrical coordinates) and 3D are defined similarly.\n\nBartholomew Osegbe on 10 May 2018\nI have attempted the weak method fem.equ.weak = { ... { ... {'(-etaux)test(ux)+(-etauy)test(uy)-test(u)px'} ... {'(-etavx)test(vx)+(-etavy)test(vy)-test(v)py'} ... {'-test(px)u-test(py)v'} ... } ... }; but I have errors and when I sourced online I found a code that uses the poisson equation but just wanted a simple code with a laplacian velocity and pressure term.\n\nAmeer Hamza on 10 May 2018\nYou may find these FEX submission useful:\n\nBartholomew Osegbe on 13 May 2018\n\n### Categories\n\nFind more on Structural Mechanics in Help Center and File Exchange\n\n### Tags\n\nNo tags entered yet.\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":0.81019676,"math_prob":0.85418695,"size":1187,"snap":"2022-40-2023-06","text_gpt3_token_len":349,"char_repetition_ratio":0.10819949,"word_repetition_ratio":0.010152284,"special_character_ratio":0.26790228,"punctuation_ratio":0.05116279,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9829859,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-07T17:59:55Z\",\"WARC-Record-ID\":\"<urn:uuid:6e6faf1a-52b0-49e9-a999-bbd9bbc5c241>\",\"Content-Length\":\"157971\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6580d7a5-0f59-4437-855c-512d42ab41e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6ec20e4-77bb-4a97-9fb7-68c10dc7f620>\",\"WARC-IP-Address\":\"104.86.80.92\",\"WARC-Target-URI\":\"https://se.mathworks.com/matlabcentral/answers/400044-how-do-i-write-the-code-for-navierstokes-equation-in-matlab-comsol\",\"WARC-Payload-Digest\":\"sha1:JIKBA27WFF3Q3WXZRDFGQ5GYXUVBPE6B\",\"WARC-Block-Digest\":\"sha1:XCE5T77ITKNHRLBN67FK4GVI6R4OWNPJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500628.77_warc_CC-MAIN-20230207170138-20230207200138-00428.warc.gz\"}"}
http://web.gauhati.ac.in/research/statistics	[ "### Statistics", null, "Bayesian inference for queuing model Bayesian inference & prediction for queuing model Amit Choudhury and his collaborator estimate traffic intensity for single server queuing model using Bayesian inference. Bayesian inference is an important technique in mathematical statistics, which uses Bayes' theorem to update the probability for a hypothesis as more evidence or information becomes available. This research works is published in the journal Communications in Statistics - Simulation and Computation. Authors Arpita Basak and Amit Choudhury Department of Statistics, GU Abstract This paper is concerned with the problem of estimating traffic intensity, ρ for single server queuing model in which inter-arrival and service times are exponentially distributed (Markovian) using data on queue size (number of customers present in the queue) observed at any random point of time. Here, it is assumed that ρ is unknown but random quantity. Bayes estimator of ρ are derived under squared error loss function assuming two forms of prior information on ρ. The performance of the proposed Bayes estimators is compared with that of the corresponding classical version estimator based on max- imum likelihood principle. The model comparison criterion based on Bayes factor is used to select a suitable prior for Bayesian analysis. Journal Reference Commun Stat - Simul Comput (2019).", null, "" ]	[ null, "http://web.gauhati.ac.in/research/_/rsrc/1554797349101/people/arrow.png", null, "http://web.gauhati.ac.in/research/_/rsrc/1554797349393/people/pulse-line.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9246059,"math_prob":0.65178525,"size":1180,"snap":"2020-24-2020-29","text_gpt3_token_len":219,"char_repetition_ratio":0.09948979,"word_repetition_ratio":0.011494253,"special_character_ratio":0.16694915,"punctuation_ratio":0.05851064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97683346,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T08:46:42Z\",\"WARC-Record-ID\":\"<urn:uuid:be82bf55-89d2-4878-bbc3-7268929468a4>\",\"Content-Length\":\"32254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ce11037-59b0-427c-8424-d6856916af8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:29512f9e-7f78-4436-8bc6-43904306c1c0>\",\"WARC-IP-Address\":\"172.217.7.211\",\"WARC-Target-URI\":\"http://web.gauhati.ac.in/research/statistics\",\"WARC-Payload-Digest\":\"sha1:LKDNNAMAWDRCGZQSBCX2UQ3DQRDH2QKV\",\"WARC-Block-Digest\":\"sha1:5NVJFVD3PPTAHSR3RTXWKRSHPB72H2OD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657143354.77_warc_CC-MAIN-20200713064946-20200713094946-00097.warc.gz\"}"}
https://wiki.secondlife.com/wiki/LlGetDate	[ "# llGetDate\n\n LSL Portal\nFunction: string llGetDate( );\n 204 Function ID 0 Forced Delay 10 Energy\n\nReturns a string that is the current date in the UTC time zone in the format \"YYYY-MM-DD\".\n\nIf you wish to know the time as well use: llGetTimestamp which uses the format \"YYYY-MM-DDThh:mm:ss.ff..fZ\"\n\n## Examples\n\n```// Birthday surprise\ndefault\n{\nstate_entry()\n{\nllSetTimerEvent(0.1);\n}\n\ntimer()\n{\nif(llGetDate() == \"2009-02-15\")\nllSetText(\"HAPPY BIRTHDAY!\", <0,1,0>, 1.0);\nelse\nllSetText(\"A surprise is coming...\", <0,1,0>, 1.0);\n\nllSetTimerEvent(3600.0); // check every hour.\n}\n}\n```\n```// Function to calculate the numeric day of year\ninteger dayOfYear(integer year, integer month, integer day)\n{\nreturn day + (month - 1) * 30 + (((month > 8) + month) / 2)\n- ((1 + (((!(year % 4)) ^ (!(year % 100)) ^ (!(year % 400))) \| (year <= 1582))) && (month > 2));\n}\n\ndefault\n{\ntouch_end(integer count)\n{\nlist dateComponents = llParseString2List(llGetDate(), [\"-\"], []);\ninteger year = (integer) llList2String(dateComponents, 0);\ninteger month = (integer) llList2String(dateComponents, 1);\ninteger day = (integer) llList2String(dateComponents, 2);\nllSay(0, \"The current day of the year is \" + (string) dayOfYear(year, month, day));\n}\n}\n```\n```// Function to calculate whether a current year is a leap year\n\ninteger is_leap_year( integer year )\n{\nif( year % 4 ) return FALSE; // Not a leap year under any circumstances\nif( year <= 1582 ) return TRUE; // In the Julian calender before 24 February 1582, every fourth year was a leap year\nif( !( year % 400 )) return TRUE; // A leap century is a leap year if divisible by 400\nif( !( year % 100 )) return FALSE; // Any other century is not a leap year\nreturn TRUE; // It is divisible by 4 and not a century and not Julian, therefore it is a leap year\n}\n```\n\nThe previous script is way unnecessary for Avatar age, SL sales history etc. Here is all that is needed for 99% of SL applications.\n\nThis code is in fact valid for all years from 1901 to 2099, as 2000 was a leap year.\n\n``` if (year % 4) // TRUE if NOT a leap year\n```\n\n## Useful Snippets\n\n### Functions\n\n • llGetTimestamp – Same format but with the time.\n\n### Articles\n\n • ISO 8601\n\n## Deep Notes\n\n#### Signature\n\n```function string llGetDate();\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59953976,"math_prob":0.9426568,"size":2659,"snap":"2023-40-2023-50","text_gpt3_token_len":794,"char_repetition_ratio":0.12316384,"word_repetition_ratio":0.0088495575,"special_character_ratio":0.3407296,"punctuation_ratio":0.16837782,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96176726,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T19:25:36Z\",\"WARC-Record-ID\":\"<urn:uuid:d262acb5-f6e9-4439-91b2-5f848356480e>\",\"Content-Length\":\"54748\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc49da98-8ae9-425c-83e3-052980ee8fc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:d990d147-bf47-4a4d-9ab0-1e24db2f9019>\",\"WARC-IP-Address\":\"23.207.202.144\",\"WARC-Target-URI\":\"https://wiki.secondlife.com/wiki/LlGetDate\",\"WARC-Payload-Digest\":\"sha1:AH7NSNQXRR4POF5T7YEL2KT2KTJGT54E\",\"WARC-Block-Digest\":\"sha1:VFXO2UVMVRW2RHGRBZQQNFLWUUOOUJDD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510219.5_warc_CC-MAIN-20230926175325-20230926205325-00311.warc.gz\"}"}
https://zxi.mytechroad.com/blog/math/leetcode-1588-sum-of-all-odd-length-subarrays/	[ "Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.\n\nA subarray is a contiguous subsequence of the array.\n\nReturn the sum of all odd-length subarrays of arr.\n\nExample 1:\n\nInput: arr = [1,4,2,5,3]\nOutput: 58\nExplanation: The odd-length subarrays of arr and their sums are:\n = 1\n = 4\n = 2\n = 5\n = 3\n[1,4,2] = 7\n[4,2,5] = 11\n[2,5,3] = 10\n[1,4,2,5,3] = 15\nIf we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58\n\nExample 2:\n\nInput: arr = [1,2]\nOutput: 3\nExplanation: There are only 2 subarrays of odd length, and . Their sum is 3.\n\nExample 3:\n\nInput: arr = [10,11,12]\nOutput: 66\n\n\nConstraints:\n\n• 1 <= arr.length <= 100\n• 1 <= arr[i] <= 1000\n\n## Solution 0: Brute Force\n\nEnumerate all odd length subarrys: O(n^2), each take O(n) to compute the sum.\n\nTotal time complexity: O(n^3)\nSpace complexity: O(1)\n\n## Solution 1: Running Prefix Sum\n\nReduce the time complexity to O(n^2)\n\n## Solution 2: Math\n\nCount how many times arr[i] can be in of an odd length subarray\nwe chose the start, which can be 0, 1, 2, … i, i + 1 choices\nwe chose the end, which can be i, i + 1, … n – 1, n – i choices\nAmong those 1/2 are odd length.\nSo there will be upper((i + 1) * (n – i) / 2) odd length subarrays contain arr[i]\n\nans = sum(((i + 1) * (n – i) + 1) / 2 * arr[i] for in range(n))\n\nTime complexity: O(n)\nSpace complexity: O(1)\n\n## C++\n\nIf you like my articles / videos, donations are welcome.\n\nBuy anything from Amazon to support our website", null, "" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56101614,"math_prob":0.9995389,"size":2035,"snap":"2021-04-2021-17","text_gpt3_token_len":829,"char_repetition_ratio":0.11127523,"word_repetition_ratio":0.14351852,"special_character_ratio":0.42800984,"punctuation_ratio":0.16973415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997656,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T13:00:00Z\",\"WARC-Record-ID\":\"<urn:uuid:d0218f96-5842-4f4e-9266-1f4c9ee39083>\",\"Content-Length\":\"65391\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e669b18d-2946-46b9-89ba-c307dbe342a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:04a2f805-09c2-426a-bf0a-fe840f18b727>\",\"WARC-IP-Address\":\"107.180.12.40\",\"WARC-Target-URI\":\"https://zxi.mytechroad.com/blog/math/leetcode-1588-sum-of-all-odd-length-subarrays/\",\"WARC-Payload-Digest\":\"sha1:OOGSPXLIAJCQWM3HNLA6OPYEER2MCD26\",\"WARC-Block-Digest\":\"sha1:X2BMJAXYV6L5VZYGKZH4KXH73K5MLILL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703514796.13_warc_CC-MAIN-20210118123320-20210118153320-00688.warc.gz\"}"}
https://debraborkovitz.com/0020-algebra-as-generalized-arithmetic/	[ "Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time.\n\nI used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view random questions on a variety of topics or to download paper practice tests.\n\n## MTEL General Curriculum Mathematics Practice\n\n Question 1\n\n#### Some children explored the diagonals in 2 x 2 squares on pages of a calendar (where all four squares have numbers in them). They conjectured that the sum of the diagonals is always equal; in the example below, 8+16=9+15.", null, "#### Which of the equations below could best be used to explain why the children's conjecture is correct?\n\n A $$\\large 8x+16x=9x+15x$$Hint: What would x represent in this case? Make sure you can describe in words what x represents. B $$\\large x+(x+2)=(x+1)+(x+1)$$Hint: What would x represent in this case? Make sure you can describe in words what x represents. C $$\\large x+(x+8)=(x+1)+(x+7)$$Hint: x is the number in the top left square, x+8 is one below and to the right, x+1 is to the right of x, and x+7 is below x. D $$\\large x+8+16=x+9+15$$Hint: What would x represent in this case? Make sure you can describe in words what x represents.\nQuestion 1 Explanation:\nTopic: Recognize and apply the concepts of variable, equality, and equation to express relationships algebraically (Objective 0020).\n Question 2\n\n#### The result is always the number that you started with! Suppose you start by picking N. Which of the equations below best demonstrates that the result after Step 6 is also N?\n\n A $$\\large N2+205-100\\div 10=N$$Hint: Use parentheses or else order of operations is off. B $$\\large \\left( \\left( 2N+20 \\right)5-100 \\right)\\div 10=N$$ C $$\\large \\left( N+N+20 \\right)5-100\\div 10=N$$Hint: With this answer you would subtract 10, instead of subtracting 100 and then dividing by 10. D $$\\large \\left( \\left( \\left( N\\div 10 \\right)-100 \\right)5+20 \\right)2=N$$Hint: This answer is quite backwards.\nQuestion 2 Explanation:\nTopic: Recognize and apply the concepts of variable, function, equality, and equation to express relationships algebraically (Objective 0020).\n Question 3\n\n A $$\\large 0.8p=\\8.73$$Hint: 80% of the regular price = $8.73. B $$\\large \\8.73+0.2\\8.73=p$$Hint: The 20% off was off of the ORIGINAL price, not off the$8.73 (a lot of people make this mistake). Plus this is the same equation as in choice c. C $$\\large 1.2\\8.73=p$$Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice b. D $$\\large p-0.2\\8.73=p$$Hint: Subtract p from both sides of this equation, and you have -.2 x 8.73 =0. Question 3 Explanation: Topics: Use algebra to solve word problems involving percents and identify variables, and derive algebraic expressions that represent real-world situations (Objective 0020). Question 4 #### Taxicab fares in Boston (Spring 2012) are$2.60 for the first $$\\dfrac{1}{7}$$ of a mile or less and $0.40 for each $$\\dfrac{1}{7}$$ of a mile after that. #### Let d represent the distance a passenger travels in miles (with $$d>\\dfrac{1}{7}$$). Which of the following expressions represents the total fare? A $$\\large \\2.60+\\0.40d$$Hint: It's 40 cents for 1/7 of a mile, not per mile. B $$\\large \\2.60+\\0.40\\dfrac{d}{7}$$Hint: According to this equation, going 7 miles would cost$3; does that make sense? C $$\\large \\2.20+\\2.80d$$Hint: You can think of the fare as $2.20 to enter the cab, and then$0.40 for each 1/7 of a mile, including the first 1/7 of a mile (or $2.80 per mile). Alternatively, you pay$2.60 for the first 1/7 of a mile, and then $2.80 per mile for d-1/7 miles. The total is 2.60+2.80(d-1/7) = 2.60+ 2.80d -.40 = 2.20+2.80d. D $$\\large \\2.60+\\2.80d$$Hint: Don't count the first 1/7 of a mile twice. Question 4 Explanation: Topic: Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020), and select the linear equation that best models a real-world situation (Objective 0022). Question 5 #### Cell phone plan A charges$3 per month plus $0.10 per minute. Cell phone plan B charges$29.99 per month, with no fee for the first 400 minutes and then $0.20 for each additional minute. #### Which equation can be used to solve for the number of minutes, m (with m>400) that a person would have to spend on the phone each month in order for the bills for plan A and plan B to be equal? A $$\\large 3.10m=400+0.2m$$Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should. B $$\\large 3+0.1m=29.99+.20m$$Hint: Doesn't account for the 400 free minutes. C $$\\large 3+0.1m=400+29.99+.20(m-400)$$Hint: Why would you add 400 minutes and$29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should. D $$\\large 3+0.1m=29.99+.20(m-400)$$Hint: The left side is $3 plus$0.10 times the number of minutes. The right is $29.99 plus$0.20 times the number of minutes over 400.\nQuestion 5 Explanation:\nIdentify variables and derive algebraic expressions that represent real-world situations (Objective 0020).\n Question 6\n\n#### A sales companies pays its representatives $2 for each item sold, plus 40% of the price of the item. The rest of the money that the representatives collect goes to the company. All transactions are in cash, and all items cost$4 or more. If the price of an item in dollars is p, which expression represents the amount of money the company collects when the item is sold?\n\n A $$\\large \\dfrac{3}{5}p-2$$Hint: The company gets 3/5=60% of the price, minus the $2 per item. B $$\\large \\dfrac{3}{5}\\left( p-2 \\right)$$Hint: This is sensible, but not what the problem states. C $$\\large \\dfrac{2}{5}p+2$$Hint: The company pays the extra$2; it doesn't collect it. D $$\\large \\dfrac{2}{5}p-2$$Hint: This has the company getting 2/5 = 40% of the price of each item, but that's what the representative gets.\nQuestion 6 Explanation:\nTopic: Use algebra to solve word problems involving fractions, ratios, proportions, and percents (Objective 0020).\n Question 7\n\n#### The student‘s solution is correct.\n\nHint:\nTry plugging into the original solution.\n\n#### The student did not correctly use properties of equality.\n\nHint:\nAfter $$x=-2x+10$$, the student subtracted 2x on the left and added 2x on the right.\n\n#### The student did not correctly use the distributive property.\n\nHint:\nDistributive property is $$a(b+c)=ab+ac$$.\n\n#### The student did not correctly use the commutative property.\n\nHint:\nCommutative property is $$a+b=b+a$$ or $$ab=ba$$.\nQuestion 7 Explanation:\nTopic: Justify algebraic manipulations by application of the properties of equality, the order of operations, the number properties, and the order properties (Objective 0020).\n Question 8\n\n#### Solve for x: $$\\large 4-\\dfrac{2}{3}x=2x$$\n\n A $$\\large x=3$$Hint: Try plugging x=3 into the equation. B $$\\large x=-3$$Hint: Left side is positive, right side is negative when you plug this in for x. C $$\\large x=\\dfrac{3}{2}$$Hint: One way to solve: $$4=\\dfrac{2}{3}x+2x$$ $$=\\dfrac{8}{3}x$$.$$x=\\dfrac{3 \\times 4}{8}=\\dfrac{3}{2}$$. Another way is to just plug x=3/2 into the equation and see that each side equals 3 -- on a multiple choice test, you almost never have to actually solve for x. D $$\\large x=-\\dfrac{3}{2}$$Hint: Left side is positive, right side is negative when you plug this in for x.\nQuestion 8 Explanation:\nTopic: Solve linear equations (Objective 0020).\n Question 9\n\n#### $$\\large A-B+C\\div D\\times E$$?\n\n A $$\\large A-B-\\dfrac{C}{DE}$$Hint: In the order of operations, multiplication and division have the same priority, so do them left to right; same with addition and subtraction. B $$\\large A-B+\\dfrac{CE}{D}$$Hint: In practice, you're better off using parentheses than writing an expression like the one in the question. The PEMDAS acronym that many people memorize is misleading. Multiplication and division have equal priority and are done left to right. They have higher priority than addition and subtraction. Addition and subtraction also have equal priority and are done left to right. C $$\\large \\dfrac{AE-BE+CE}{D}$$Hint: Use order of operations, don't just compute left to right. D $$\\large A-B+\\dfrac{C}{DE}$$Hint: In the order of operations, multiplication and division have the same priority, so do them left to right\nQuestion 9 Explanation:\nTopic: Justify algebraic manipulations by application of the properties of order of operations (Objective 0020).\n Question 10\n\n#### The commutative property is used incorrectly.\n\nHint:\nThe commutative property is $$a+b=b+a$$ or $$ab=ba$$.\n\n#### The associative property is used incorrectly.\n\nHint:\nThe associative property is $$a+(b+c)=(a+b)+c$$ or $$a \\times (b \\times c)=(a \\times b) \\times c$$.\n\n#### The distributive property is used incorrectly.\n\nHint:\n$$(x+3)(x+3)=x(x+3)+3(x+3)$$=$$x^2+3x+3x+9.$$\nQuestion 10 Explanation:\nTopic: Justify algebraic manipulations by application of the properties of equality, the order of operations, the number properties, and the order properties (Objective 0020).\nThere are 10 questions to complete.\n\nIf you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here." ]	[ null, "http://debraborkovitz.com/wp-content/uploads/2012/03/Q55-calendar.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8730516,"math_prob":0.9979289,"size":9091,"snap":"2021-31-2021-39","text_gpt3_token_len":2722,"char_repetition_ratio":0.14988445,"word_repetition_ratio":0.160632,"special_character_ratio":0.32724673,"punctuation_ratio":0.119201995,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996363,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T00:16:28Z\",\"WARC-Record-ID\":\"<urn:uuid:53b368df-763f-4318-9b31-18a32b6d9943>\",\"Content-Length\":\"105914\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c05e0a5-f950-4365-96d1-84459e3adbec>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd4fe0bb-3b41-4464-8e76-ff248493d753>\",\"WARC-IP-Address\":\"173.236.186.133\",\"WARC-Target-URI\":\"https://debraborkovitz.com/0020-algebra-as-generalized-arithmetic/\",\"WARC-Payload-Digest\":\"sha1:KHODW23OBB2VUYEX6FPH5FUINO2LMRZ3\",\"WARC-Block-Digest\":\"sha1:JYYUKKLFR6IC47HEJTWAKTN74QY5EGCJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151531.67_warc_CC-MAIN-20210724223025-20210725013025-00351.warc.gz\"}"}
https://placement.freshersworld.com/upsc-question-papers/question-paper/33134720	[ "UPSC Question-Paper Contributed by sanalKumar updated on Oct 2020\n• Apply to Premium Jobs from Top MNCs\n• Priority Shortlisting & Call-letters for drives\nApply to Premium Jobs from Top MNCs\nPriority Shortlisting & Call-letters for drives\nAccess of Complete Question and Solution of the Test Series\nTake Test Now\n\nUPSC IES/ISS General ability, Aptitude questions with answers and explanation, UPSC previous years solved questions with answers\nPart - C UPSC questions with answers and explanations\nNumerical Aptitude\n\n101. If p = 124, 3√p (p2 + 3p + 3) + 1 = ?a. 5\nb. 7\nc. 123\nd. 125 (Ans)\nAns :\n3√p (p2 + 3p + 3) + 1\n= 3√p3 + 3p2 + 3p + 1\n= 3√(p + 1)3= (p + 1)\n= 125\n\n102. If √1 - x3/100 = 3/5, then x equals —\na. 2\nb. 4 (Ans)\nc. 16\nd. (136)1/3\n\nAns : √1 - x3/100 = 3/5⇒ 1 - x3/100 = 9/25∴ x3/100 = 16/25⇒ x3 = 16100/25 = (4)3∴ x = 4 103. I multiplied a natural number by 18 and another by 21 and added the products. Which one of the following could be the sum?a. 2007 (Ans)\nb. 2008\nc. 2006\nd. 2002\nAns : 18x + 21y = 3 (6x + 7y)Only 2007 is divisible by 3. Hence it is only possible.\n\n104. The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is —a. 135\nb. 240\nc. 73\nd. 106 (Ans)Ans : x y = 45 x - y = 4∴ x2 + y2 = (x - y)2 + 2xy = 16 + 90 = 106\n\n105. √8 + √57 + √38 + √108 + √169 = ?a. 4 (Ans)\nb. 6\nc. 8\nd. 10\nAns : ? = √8 + √57 + √ 38 +√108 + √ 169\n\n= √ 8 + √ 57 + √38 + √108 +13\n= √ 8 + √57 + 7 = √ 8 + 8 = 4\n\n106. The square root of 14 + 6√5 is —\na. 2 + √5\nb. 3 + √5 (Ans)\nc. 5 + √3\nd. 3 + 2√5\nAns : √ 14 + 6 √ 5 = √9 + 5 + 23* √5= 3 + √5\n\n107. When 231 is divided by 5 the remainder is —\na. 4\nb. 3 (Ans)\nc. 2\nd. 1\nAns : The unit digit in (2)31 will be same as the unit digit in (2)3, because the last digit is repeated after each 4 index∴ (2)31 → 228 + 3 = 23 = 8∴ After dividing by 5, the remainder will be = 8 - 5 = 3.\n108. The value of 1 + 1 is 1 + 1 1 + 1 1 + 1 1 + 2/3a. 21/13\nb. 17/3\nc. 34/21 (Ans)\nd. 8/5\nAns : 1 + 1 is 1 + 1 1 + 1 1 + 1 1 + 2 /3\n\n= 1 + 1 1 + 1 1 + 1 1 + 3/5\n= 1 + 1 1 + 1 1 + 5/8\n= 1 + 1 1 + 8/13 = 1 + 13/21 = 34/21109. The unit digit in the product (122)173 is —\na. 2 (Ans)\nb. 4\nc. 6\nd. 8\nAns : A\n\n110. The value of 2 + √3 / 2 - √3 + 2 - √3 / 2 + √3 + √3 + 1 / √3 - 1 is—a. 16 + √3 (Ans)b. 4 - √3c. 2 - √3d. 2 + √3\nAns : 2 + √ 3 / 2 - √3 + 2 - √3 / 2 + √3 + √3 + 1 / √3 - 1= (2 + √3)2 + (2 - √3)2 / ( 2 - √3) + (2 + √3) + (√3 + 1) (√3 + 1) / (√3 - 1) (√3 + 1)= 14/1 + 3 + 1 + 2√3/2\n= 14 + 2 + √3\n= 16 + √3\n111. If ab = 2a + 3b - ab, then the value of (35 + 53) is—a. 10 (Ans)\nb. 6\nc. 4\nd. 2\nAns : a b = 2a + 3b - ab∴ (3 * 5 + 5 * 3) = (2 * 3 + 3 * 5 - 3 * 5) + (2 * 5 + 3 * 3 - 5 * 3)= (6 + 15 - 15) + (10 + 9 - 15)\n= 6 + 4\n= 10\n\n112. Simplify : 0.0347 * 0.0347 * 0.0347 + (0.9653)3 / (0.0347)2 - (0.347) (0.09653) + (0.9653)2a. 0.9306\nb. 1.0009\nc. 1.0050\nd. 1 (Ans) (Ans)\nAns : 0.0347 * 0.0347 * 0.0347 + (0.9653)3 / (0.0347)2 - (0.347) (0.09653) + (0.9653)2= (0.0347)3 + (0.9653)3 / (0.0347)2 - (0.0347) (0.9653) + (0.9653)2 (0.0347 + 0. 9653)= [(0.0347)2 - (0.0347 * 0.9653) + (0.9653)2] / [(0.0347)2 - 0.0347 * 0.9653 + (0.9653)2]\n= 1\n\n113. A copper wire is bent in the form of an equilateral triangle and has area 121 √3 cm2, If the same wire is bent into the form of a circle, the area (in cm2) enclosed by the wire is ( Take = 22/7)—\na. 364.5\nb. 693.5\nc. 346.5 (Ans)\nd. 639.5\nAns : Area of equilateral ? = √3/ 4 (side)2∴ 121 √3 = √3/4 (side)2∴ (side) = √4 121 = 22 cm∴ Length of wire = 3 22 = 66cm∴ 2 * 22/7 * r = 66r = 66 * 7 / 2 * 22\n= 21 / 2 cm\n∴ Area of the circle = 22/7 * 21/2 * 21/2= 346.5 cm2\n\n114. A child reshapes a cone made up of clay of height 24 cm and radius 6 cm into a sphere. The radius (in cm) of the sphere is—a. 6 (Ans)\nb. 12\nc. 24\nd. 48\nAns : 3/4 r3 = 1/3 (6)2 * 24∴ r3 = 216 = (6)3∴ r = 6 cm\n\n115. Water flows into a tank which is 200 m long and 150 m wide, through a pipe of crosssection 0.3 m * 0.2 m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8 m is—\na. 50\nb. 120\nc. 150\nd. 200 (Ans)\nAns : Reqd. time = 200 * 150 * 8/ 0.3 * 0.2 * 20000 hrs.= 200 hrs.\n\n116. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is—\na. 1\nb. 2 (Ans)\nc. 3\nd. 4\nAns : Let the numbers be 13x and 13y.∴ 13x * 13y = 2028\n∴ xy = 12∴ x = 3 and y = 4or x = 1 and y = 12∴ The number of pairs are = 2\n\n117. Two equal vessels are filled with the mixtures of water and milk in the ratio of 3 : 4 and 5 : 3 respectively. If the mixtures are poured into a third vessel, the ratio of water and milk in the third vessel will be—a. 15 : 12b. 53 : 59c. 20 : 9d. 59 : 53 (Ans)Ans : Reqd. ratio = (3/7 + 5/8) / (4/7 + 3/8)= (24 + 35)/ (32 + 21)\n= 59 : 53\n118. I am three times as old as my son. 15 years hence, I will be twice as old as my son. The sum of our ages is—\na. 48 years\nb. 60 years (Ans)\nc. 64 years\nd. 72 years\nAns : Let the present age of the son be x years∴ My present age = 3x years∴ 3x + 15 / x + 15 = 2/1\n∴ Reqd. sum = 45 + 15 = 60 years\n\n119. Three bells ring simultaneously at 11 a.m. They ring at regular intervals of 20 minutes, 30 minutes, 40 minutes respectively. The time when all the three ring together next is—a. 2 p.m.\nb. 1 p.m. (Ans)\nc. 1.15 p.m.\nd. 1.30 p.m.\nAns : L.C.M. of 20, 30 and 40 minutes= 120 minutes= 2 hours.\n∴ Reqd. time = 1 p.m.\n\n120. A and B together can do a work in 12 days. B and C together do it in 15 days, If A's efficiency is twice that of C, then the days required for B alone of finish the work is—\na. 60\nb. 30\nc. 20 (Ans)\nd. 15\nAns : Let the time taken by A to finish the work be x days∴ C will take 2x days to finish the work\n∴ Work of B for 1 day = 1/12 - 1/x\nand work of B for 1 day = 1/15 - 1/2x∴ 1/12 -1/x = 1/15 - 1/2x\n∴ 1/2x = 1/12 - 1/15\n= 1/60\n∴ x = 30\n∴ Work of B for 1 day = 1/12 - 1/30= 1/20\n∴ B alone will finsh the work in 20 days. 121. A and B can do a work in 12 days, B and C can do the same work in 15 days, C and A can do the same work in 20 days. The time taken by A, B and C to do the same work is—a. 5 days\nb. 10 days (Ans)\nc. 15 days\nd. 20 days\nAns : Reqd. time= 12 * 15 * 20 * 2/ (12 * 15 + 15 * 20 + 12 * 20)\n= 7200/(180 + 300 + 240)\n= 10 days\n\n122. A is 50% as efficient as B . C does half of the work done by A and B together. If C alone does the work in 20 days, then A, B and C together can do the work in—a. 5 2/3 days\nb. 6 2/3 days (Ans)\nc. 6 days\nd. 7 days\nAns : Let the time taken by B to complete the work be x days∴ Time ,, ,, by A ,, ,,= 2x days\n∴ Work done by by C for 1 day\n= 1/2 (1/x + 1/2x)\n= 3/4x\n∴ 3/4x = 1/20⇒ x = 15∴ Work done by (A + B + C) for 1 day = 1/30 + 1/15 + 1/20 = 9/60= 3/20\n∴ Reqd. time = 6 2/3 days\n\n123. The ratio of the volumes of water and glycerine in 240 cc of mixture is 1 : 3. The quantity of water (in cc) that should be added to the mixture so that the new ratio of added to the mixture so that the new ratio of the volumes of water and glycerine becomes 2 :3 is—a. 55\nb. 60 (Ans)\nc. 62.5\nd. 64\nAns : The quantity of water to be added= 240 (3-2)/ (1 + 3)\n= 60 c.c.\n\n124. At present, the ratio of the ages of Maya and Chhaya is 6 : 5 and fifteen years from now, the ratio will get changed to 9 : 8 Maya's present age is—a. 21 years\nb. 24 years\nc. 30 years (Ans)\nd. 40 years\nAns : Let the present ages of Maya and Chhaya be 6x and 5x years respectively.6x + 15/ 5x + 15 = 9/8\n48x + 120 = 45x + 135\nx = 5\n∴ Present age of Maya is = 30 years.\n\n125. The ratio of the income to the expenditure of a family is 10 : 7. If the family's expenses are Rs.10,500, then savings of the family is—a. Rs.4,500 (Ans)\nb. Rs.10,000\nc. Rs.4,000\nd. Rs.5,000\nAns : Let the income and expenditures be Rs.10x and Rs.7x respectively∴ 7x = 10500⇒ x = Rs.1500∴ Savings = (10-7) * 1500= Rs.4500\n\n126. The average mathematics marks of two Sections A and B of Class IX in the annual examination is 74. The average marks of Section A is 77.5 and that of Section B is 70. The ratio of the number of students of Section A and B is—a. 7 : 8\nb. 7 : 5\nc. 8 : 7 (Ans)\nd. 8 : 5\nAns : Let the numbers of A and B be n and P respectively∴ n * 77.5 + P * 70 = (n + P) * 7477.5n - 74n = 74P - 70P\n3.5n = 4P\n∴ n/P = 4 : 3.5= 8 : 7\n\n127. The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg instead of 87 kg. The correct average weight is—a. 88.95 kg\nb. 89.25 kg\nc. 89.55 kg\nd. 89.85 kg (Ans)\nAns : The correct average wt.\n= 20 * 89.4 + 87 - 78/ 20\n= 1788 +9/20\n= 89.85 kg.\n\n128. The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is—\na. 500 (Ans)\nb. 600\nc. 700\nd. 800\nAns : No. of revolutions= 1540 * 100 * 7/98 * 22\n= 500\n\n129. In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is—\na. 3√7\nb. 7√3\nc. 10√7/3 (Ans)\nd. 7√10/3Ans : cos 60 = (10)2 + (10/3)2 - AD2/ 2 * 10 * 10/3\n1/2 = 100 + 100/9 - AD2 / 200/31/2 * 200/3 = 1000 - 9 AD2 /9\n\n130. The cost price of an article is Rs.800. After allowing a discount of 10% , a gain of 12.5% was made. Then the marked price of the article is—a. Rs.1,000 (Ans)\nb. Rs.1,100\nc. Rs.1,200\nd. Rs.1,300\nAns : S. P. of the article = 800 * 112.5/100= Rs.900\n∴ M.P. of the article = 100 * 900/ (100 - 10)\n= Rs.1000\n\n131. A man bought an article listed at Rs.1,500 with a discount of 20% offered on the list price. What additional discount must be offered to the man to bring the net price to Rs.1,104 ?a. 8% (Ans)\nb. 10%\nc. 12%d. 15%\nAns : S. P. the article after a discount of 20%\n= 1500 * 80/100\n= Rs.1200\n∴ Additional discount = (1200 - 1104) * 100/1200= 8%\n\n132. If a/b = c/d = e/f = 3, then 2a2 + 3c2 + 4e2/ 2b2 + 3d2 + 4f2 = ?a. 2\nb. 3\nc. 4\nd. 9 (Ans)\nAns : a/b = c/d = e/f = 3∴ a = 3bc = 3d\nand e = 3f∴ ? = 2a2 + 3c2 + 4e2/ 2b2 +3d2 + 4f2= 2 * 9b2 + 3 * 9d2 + 4 * 9f2/ 2b2 + 3d2 + 4f2\n= 9\n\n133. The floor of a room is of size 4 m * 3m and its height is 3 m. The walls and ceiling of the room require painting. The area to be painted is—\na. 66 m2\nb. 54 m2 (Ans)\nc. 43 m2\nd. 33 m2\nAns : Reqd. area = 2(4 + 3) * 3 + 4 * 3\n= 42 + 12\n= 54 m2\n\n134. When the price of an article was reduced by 20%, its sale increased by 80%. What was the net effect on the sale?\na. 44% increase (Ans)\nb. 44% decrease\nc. 66% increase\nd. 75% increase\nAns : Reqd. % effect = [80 - 20 - 8020/100]%= 44% increase\n\n135. The price of sugar goes up by 20%. If a housewife wants the expenses on sugar to remain the same, she should reduce the consumption by—\na. 15 1/5%b. 16 2/3% (Ans)c. 20%\nd. 25%\nAns : Reqd. % reduction =20 100/ (100 + 20)%\n= 16 2/3%\n\n136. In a factory 60% of the workers are above 30 years and of these 75% are males and the rest are females. If there are 1350 male workers above 30 years, the total number of workers in the factory is—a. 3000 (Ans)\nb. 2000\nc. 1800\nd. 1500\nAns : Let the total no.of workers be x.\n∴ No. of males above 30 years\n⇒ x * 60/100 * 75/100 = 1350\n∴ x = 1350 * 100 * 100/ 60 * 75= 3000\n\n137. Walking at 3/4 of his usual speed, a man is 1 1/2 hours late. His usual time to cover the same distance, in hours is—a. 4 1/2 (Ans)\nb. 4\nc. 5 1/2\nd. 5\nAns : Let the usual time be x hours x * usual speed= 3/4 * usual speed * (x + 3/2)\nx = 3/4 x + 9/8\n∴ x = 9/8 * 4= 4 1/2 hours\n\n138. The selling price of 10 oranges is the cost price of 13 oranges. Then the profit percentages is—a. 30% (Ans)\nb.10%\nc. 13%\nd. 3%\nAns : Profit % = (13-10)/10 * 100%\n= 30%\n\n139. The marked price of a radio is Rs.480. The shopkeeper allows a discount of 10% and gains 8%. If no discount is allowed, his gain percent would be—a. 18%\nb. 18.5%\nc. 20.5%\nd. 20% (Ans)\nAns : C.P. of the radio = 480 * 90/ 100 * 100/108= Rs.400\n∴ New profit % = 480 - 400/400 * 100%=20%\n\n140. A man sold 20 apples for Rs.100 and gained 20%. How many apples did he buy for Rs.100?a. 20\nb. 22\nc. 24 (Ans)\nd. 25\nAns : ? C.P. of 1 apple = Rs. 100/20 * 100/120∴ No. of reqd. apples = 100 * 20 * 120/ 100 * 100= 24\n\n141. A rectangular sheet of metal is 40 cm by 15 cm. Equal squares of side 4 cm are cut off at the corners and the remainder is folded up to form an open rectangular box. The volume of the box is—a. 896 cm3 (Ans)\nb. 986 cm3\nc. 600 cm3\nd. 916 cm3\nAns : Vol. of the open box\n= (40 - 8) * (15 - 8) * 4\n= 32 * 7 * 4\n= 896 cm3\n142. If 78 is divided into three parts which are in the ratio 1 : 1/3 : 1/6, the middle part is—a. 9 1/3\nb. 13\nc. 17 1/3 (Ans)\nd. 18 1/3\nAns : Ratio = 1 : 1/3 : 1/6= 6 : 2 : 1\n∴ The middle part = 2 * 78/(6 + 2 + 1)= 52/3\n= 17 1/3\n\n143. The simple interest on a sum of money is 1/9 of the principal and the number of years is equal to rate per cent per annum. The rate per annum is—a. 3%\nb. 1/3%\nc. 3 1/3% (Ans)\nd. 3/10%\nAns : ? 1/9 * P = P * R * R/100∴ R = 10/3%= 3 1/3%\n\n144. The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs.48. Then the sum is—a. Rs.1,000\nb. Rs.1,200 (Ans)\nc. Rs.1,500\nd. Rs.2,000\nAns : ? D = P (r/100)2⇒ 48 = P (20/100)2\n∴ P = 48 * 25= Rs.1200\n\n145. Shri X goes to his office by scooter at a speed of 30 km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance to his office is—a. 20 km\nb. 21 km\nc. 22 km (Ans)\nd. 24 km\nAns : Reqd. distance = 30 * 24/(30 - 24) * (6 + 5)/60= 720 * 11/ 6 * 60\n= 22 km\n\n146. A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?a. 6 years\nb. 4 years (Ans)\nc. 8 years\nd. 5 years\nAns : Reqd. time = a log q/ log p= 3 * log 16/ log 8\n= 3 * 4 log 2/ 3 log 2\n= 4 years\n\nDirections - The pie chart given below shows the spending of a family on various heads during a month. Study the graph and answer questions 147 to 150.(Image)\n\n147. If the total income of the family is Rs.25,000, then the amount spent on Rent and Food together is—a. Rs.17,250\nb. Rs.14,750 (Ans)\nc. Rs.11,250\nd. Rs.8,500\nAns : Reqd. expenditure = (45 + 14) * 25000/100\n= Rs.14750\n\n148. What is the ratio of the expenses on Education to the expenses on Food?\na. 1 : 3 (Ans)\nb. 3 : 1\nc. 3 : 5\nd. 5 : 3\nAns : Reqd. ratio = 15 : 45\n= 1 : 3\n\n149. Expenditure on Rent is what per cent of expenditure on Fuel ?\na. 135%\nb. 156% (Ans)\nc. 167%\nd. 172%\nAns : Reqd. ratio % = 14 * 100/9%= 156%\n150. Which three expenditures together have a central angle of 1080 ?a. Fuel, Clothing and Others\nb. Fuel, Education and Others (Ans)\nc. Clothing, Rent and Others\nd. 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https://www.cut-the-knot.org/do_you_know/fraction.shtml	[ "The word 'fraction' derives from the Latin fractio - to break. However, there are continuous fractions.", null, "Well, not quite. The accepted terminology is continued fraction. However, one of the Russian terms for what I want to discuss on this page is indeed continuous fraction. Continued fraction are extremely important in the theory of rational approximation. They also provide a simple way to construct a variety of transcendental numbers. Here's the definition.\n\nIn general, continued fraction is an expression in the form", null, "where ai and bi may be any kind of numbers, variables, or functions. With all bi = 1 we have simple continued fractions.", null, "We shall only be concerned with the latter variety. A more convenient notation is r = [a0,a1,a2,a3,...]. To end with notations, if a0 is an integer, most often it's separated from the rest of the coefficients with a semicolon: r = [a0;a1,a2,a3,...]. In what follows I shall assume that all ai are integers and, for i > 0, ai > 0.\n\nTo make the terminology and notations more intuitive let's apply Euclid's algorithm to finding the greatest common divisor of, say, 1387 and 3796. Begin by dividing the smaller number 1387 into the larger one (3796) and keeping track of the remainder.\n\n3796 = 1387·2 + 1022\n\nwhich also can be written as\n\n 3796/1387 = 2 + 1022/1387 = 2 + 1/(1387/1022).\n\nThe key step of the algorithm is exactly this: \"Keep dividing the smaller number 1387 into the larger one.\" Thus, let's continue\n\n 1387/1022 = 1 + 365/1022 = 1 + 1/(1022/365).\n\nIn other words,\n\n 3796/1387 = 2 + 1022/1387 = 2 + 1/(1 + 1/(1022/365)).\n\nFurther\n\n1022 = 365·2 + 292\n365 = 292·1 + 73\n292 = 73·4\n\nOmitting parentheses (as is customary), we may write\n\n 3796/1387 = 2 + 1022/1387 = 2+1/1+ 1/(1022/365) = 2+1/1+1/2+ 1/(365/292) = 2+1/1+1/2+1/1+1/4.\n\nFinally, 3796/1387 = [2;1,2,1,4]. The above explains why the terms ai are usually called quotients. Also, it shows that every rational number can be associated with a finite continued fraction. On the other hand, given a finite fraction [a0;a1,a2,a3,...,an], we may as well carry out the arithmetic operations in the reverse order. Which will ultimately lead to a rational number r. Since, obviously, [a0;a1,a2,a3,...,an,1] = [a0;a1,a2,a3,...,an+1], let's agree to exclude the finite fractions with the last quotient equal to 1. With this convention, the correspondence between rational numbers and finite continued fractions becomes 1-1.\n\nIrrational numbers can also be uniquely associated with (simple) continued fractions. Take an irrational r and denote a0 = [r], where [] is the floor function. Next write r = a0 + 1/r1, where r1 = 1/(r - a0). Continue in this fashion: a1 = [r1] and r2 = 1/(r1 - a1), and more generally\n\nak = [rk]\nrk = 1/(rk-1 - ak-1),\n\nwhere k > 1 (with the convention that r0 = r.) We can write\n\nr = [a0;r1] = [a0;a1,r2] = [a0;a1,a2,r3] ...\n\nQuite naturally the terms ri are called remainders. Note that, since, for an irrational r, 1 > r - [r] > 0, r1 = 1/(r - [r]) is also irrational and r1>1. The same is true for k > 1: all rk are irrational and rk > 1. Therefore the process never stops and leads to a uniquely defined infinite continued fraction for which ak > 0 for k > 0. Note that with thus defined correspondence r1 = [a1,r2] = [a1,a2,r3] = ..., r2 = [a2,r3] = ..., and so on.\n\nAs we see, the association between real numbers and continued fractions is defined recursively. It will come then as no surprise that many of the features of the continued fractions are expressed in recursive terms. For example, for a continued fraction (either finite or infinite) one defines a family of finite segments sk = [a0;a1,a2,...,ak], each sk being a rational number: sk= pk/qk with qk > 0. Then we have the following fundamental theorem that can be proved by induction.\n\n## Theorem 1\n\nFor all k ≥ 2,\n\n (1) pk = akpk-1 + pk-2 (2) qk = akqk-1 + qk-2\n\nSet p-1 = 1 and q-1 = 0. First subtracting (2) multiplied by pk-1 from (1) multiplied by qk-1 and then subtracting (2) multiplied by pk-2 from (1) multiplied by qk-2 we further get\n\n## Theorem 2\n\nFor all k ≥ 0,\n\n (3) qkpk-1 - pkqk-1 = (-1)k (4) qkpk-2 - pkqk-2 = (-1)k-1ak\n\n(3) and (4) can be rewritten as\n\n (3') sk - sk-1 = (-1)k-1/(qkqk-1) (4') sk - sk-2 = (-1)kak/(qkqk-2)", null, "It's about time I start drawing some conclusions.\n\n1. From (2), qi grows with i.\n2. All fractions sk with k >1 are irreducible. This follows from (3); for a common factor of pk and qk would divide 1.\n3. From (4'), if k is even, sk is an increasing sequence. For k odd, sk decreases.\n4. From (3'), if k is even, sk < sk-1. If k is odd, the reverse inequality holds.\n5. Combining 3. and 4. we get\n\ns0 < s2 < s4 < ... < s2n < ... < s2n-1 < ... < s5 < s3 < s1\n\n6. Therefore, sk converges both for even and odd indices. From 1. and (3') the limits coincide. Thus, sk is a convergent sequence. It's a gratifying fact that the sequence converges to r. For this reason the fractions sk are known as Convergents. Let's prove this. As we know,\n\nr = [a0;a1,...,ak-1,rk]\n\nTherefore,\n\n (5) r = (pk-1rk + pk-2)/(qk-1rk + qk-2), k > 1.\n\nOn the other hand, obviously\n\n (6) sk = (pk-1ak + pk-2)/(qk-1ak + qk-2), k>1\n\nFrom (5) and (6)", null, "which implies", null, "which, in view of 1., not only proves that sk converges to r but also gives an estimate for the rate of convergence. Note that this is the same inequality we once proved using the Pigeonhole principle. The new proof supplies a constructive way to approximate an irrational number.", null, "## Examples\n\n1. r = [1;1,1,1,...] = 1 + 1/r. Therefore, r is the positive root of r 2 = r+1. Thus the golden ratio (1 + 5)/2 has the simplest continued fraction representation possible.\n2. Similarly, you can find [2;2,2,2,...] and, in general, [n;n,n,...].\n3. r = [1;2,2,2,...] can be found once you know [2;2,2,...]. But there is another way. r-1 = [0;2,2,2,...] whereas r+1 = [2;2,2,2,...]. Therefore (r - 1)(r + 1) = 1. Or r = 2.\n4. r = [1;1,2,1,2,1,2,...]. Similar to the above, r+1 = [2;1,2,1,2,...] whereas r-1 = [0;1,2,1,2,...]. Let r1 = [1;2,1,2,...]. Then (r+1) = [2;r1] = 2·r1. On the other hand, (r-1) = [0; r1] = 1/r1. Thus we get (r+1)(r-1) = 2. Finally, r = 3.\n\nAll of the above fractions are periodic in the sense we apply to decimal fractions. Le comte J. L. Lagrange (1736-1813) proved that this is a characteristic property of roots of quadratic polynomials.\n\n5. The next two examples are due to S. Ramanujan, one of the greatest mathematical geniuses. Being a poor uneducated clerk, in 1913 Ramanujan sent a letter to G. H. Hardy who was by this time a world famous mathematician. He was accustomed to receiving letters from cranks. So he was bored more than anything else. Then it dawned on him that \"They must be true because, if they were not true, no one would have had the imagination to invent them.\" The examples below were among 120 theorems from the Ramanujan's letter and are specifically referred to in the above passage.", null, "6. The following couple belongs to L. Euler", null, "", null, "" ]	[ null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction1.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction2.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction3.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction4.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction5.gif", null, "https://www.cut-the-knot.org/do_you_know/cfraction6.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8956979,"math_prob":0.99859464,"size":5610,"snap":"2019-51-2020-05","text_gpt3_token_len":1816,"char_repetition_ratio":0.10631467,"word_repetition_ratio":0.012108981,"special_character_ratio":0.36292335,"punctuation_ratio":0.19429453,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994122,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,5,null,5,null,null,null,5,null,5,null,null,null,5,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-14T16:23:54Z\",\"WARC-Record-ID\":\"<urn:uuid:b6824077-afec-4ff1-8d4d-dd5c5aa5018b>\",\"Content-Length\":\"22336\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2cf24552-688c-4be2-83bd-fddd073fc00f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6ba9c2e-34c4-46ae-9ea8-fe9c92dc304c>\",\"WARC-IP-Address\":\"107.180.50.227\",\"WARC-Target-URI\":\"https://www.cut-the-knot.org/do_you_know/fraction.shtml\",\"WARC-Payload-Digest\":\"sha1:SSKYQEMDS63UZADFUPK4QFMSN3ORLOKH\",\"WARC-Block-Digest\":\"sha1:EZHUFWPVFWLWBUUNM5UTRISVESMWE3WT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541281438.51_warc_CC-MAIN-20191214150439-20191214174439-00437.warc.gz\"}"}
https://www.wikitechy.com/technology/java-programming-bellman-ford-algorithm/	[ "# Java Programming – Bellman–Ford Algorithm\n\nJava Programming - Bellman–Ford Algorithm - Dynamic Programming Given a graph and a source vertex src in graph, find shortest paths from src to all vertices\n\nGiven a graph and a source vertex src in graph, find shortest paths from src to all vertices in the given graph. The graph may contain negative weight edges.\nWe have discussed Dijkstra’s algorithm for this problem. Dijksra’s algorithm is a Greedy algorithm and time complexity is O(VLogV) (with the use of Fibonacci heap). Dijkstra doesn’t work for Graphs with negative weight edges, Bellman-Ford works for such graphs. Bellman-Ford is also simpler than Dijkstra and suites well for distributed systems. But time complexity of Bellman-Ford is O(VE), which is more than Dijkstra.\n\nAlgorithm\nFollowing are the detailed steps.\n\nInput: Graph and a source vertex src\nOutput: Shortest distance to all vertices from src. If there is a negative weight cycle, then shortest distances are not calculated, negative weight cycle is reported.\n\n• This step initializes distances from source to all vertices as infinite and distance to source itself as 0. Create an array dist[] of size \|V\| with all values as infinite except dist[src] where src is source vertex.\n• This step calculates shortest distances. Do following \|V\|-1 times where \|V\| is the number of vertices in given graph.\n• …..Do following for each edge u-v\n………………If dist[v] > dist[u] + weight of edge uv, then update dist[v]\n………………….dist[v] = dist[u] + weight of edge uv\n• This step reports if there is a negative weight cycle in graph. Do following for each edge u-v\n……If dist[v] > dist[u] + weight of edge uv, then “Graph contains negative weight cycle”\nThe idea of step 3 is, step 2 guarantees shortest distances if graph doesn’t contain negative weight cycle. If we iterate through all edges one more time and get a shorter path for any vertex, then there is a negative weight cycle\n\nHow does this work? Like other Dynamic Programming Problems, the algorithm calculate shortest paths in bottom-up manner. It first calculates the shortest distances for the shortest paths which have at-most one edge in the path. Then, it calculates shortest paths with at-nost 2 edges, and so on. After the ith iteration of outer loop, the shortest paths with at most i edges are calculated. There can be maximum \|V\| – 1 edges in any simple path, that is why the outer loop runs \|v\| – 1 times. The idea is, assuming that there is no negative weight cycle, if we have calculated shortest paths with at most i edges, then an iteration over all edges guarantees to give shortest path with at-most (i+1) edges.\n\nExample\nLet us understand the algorithm with following example graph. The images are taken from this source.\n\nLet the given source vertex be 0. Initialize all distances as infinite, except the distance to source itself. Total number of vertices in the graph is 5, so all edges must be processed 4 times.", null, "Let all edges are processed in following order: (B,E), (D,B), (B,D), (A,B), (A,C), (D,C), (B,C), (E,D). We get following distances when all edges are processed first time. The first row in shows initial distances. The second row shows distances when edges (B,E), (D,B), (B,D) and (A,B) are processed. The third row shows distances when (A,C) is processed. The fourth row shows when (D,C), (B,C) and (E,D) are processed.", null, "The first iteration guarantees to give all shortest paths which are at most 1 edge long. We get following distances when all edges are processed second time (The last row shows final values).", null, "The second iteration guarantees to give all shortest paths which are at most 2 edges long. The algorithm processes all edges 2 more times. The distances are minimized after the second iteration, so third and fourth iterations don’t update the distances.\n\nJava\n``````// A Java program for Bellman-Ford's single source shortest path\n// algorithm.\nimport java.util.;\nimport java.lang.;\nimport java.io.*;\n\n// A class to represent a connected, directed and weighted graph\nclass Graph\n{\n// A class to represent a weighted edge in graph\nclass Edge {\nint src, dest, weight;\nEdge() {\nsrc = dest = weight = 0;\n}\n};\n\nint V, E;\nEdge edge[];\n\n// Creates a graph with V vertices and E edges\nGraph(int v, int e)\n{\nV = v;\nE = e;\nedge = new Edge[e];\nfor (int i=0; i<e; ++i)\nedge[i] = new Edge();\n}\n\n// The main function that finds shortest distances from src\n// to all other vertices using Bellman-Ford algorithm. The\n// function also detects negative weight cycle\nvoid BellmanFord(Graph graph,int src)\n{\nint V = graph.V, E = graph.E;\nint dist[] = new int[V];\n\n// Step 1: Initialize distances from src to all other\n// vertices as INFINITE\nfor (int i=0; i<V; ++i)\ndist[i] = Integer.MAX_VALUE;\ndist[src] = 0;\n\n// Step 2: Relax all edges \|V\| - 1 times. A simple\n// shortest path from src to any other vertex can\n// have at-most \|V\| - 1 edges\nfor (int i=1; i<V; ++i)\n{\nfor (int j=0; j<E; ++j)\n{\nint u = graph.edge[j].src;\nint v = graph.edge[j].dest;\nint weight = graph.edge[j].weight;\nif (dist[u]!=Integer.MAX_VALUE &&\ndist[u]+weight<dist[v])\ndist[v]=dist[u]+weight;\n}\n}\n\n// Step 3: check for negative-weight cycles. The above\n// step guarantees shortest distances if graph doesn't\n// contain negative weight cycle. If we get a shorter\n// path, then there is a cycle.\nfor (int j=0; j<E; ++j)\n{\nint u = graph.edge[j].src;\nint v = graph.edge[j].dest;\nint weight = graph.edge[j].weight;\nif (dist[u]!=Integer.MAX_VALUE &&\ndist[u]+weight<dist[v])\nSystem.out.println(\"Graph contains negative weight cycle\");\n}\nprintArr(dist, V);\n}\n\n// A utility function used to print the solution\nvoid printArr(int dist[], int V)\n{\nSystem.out.println(\"Vertex Distance from Source\");\nfor (int i=0; i<V; ++i)\nSystem.out.println(i+\"\\t\\t\"+dist[i]);\n}\n\n// Driver method to test above function\npublic static void main(String[] args)\n{\nint V = 5; // Number of vertices in graph\nint E = 8; // Number of edges in graph\n\nGraph graph = new Graph(V, E);\n\n// add edge 0-1 (or A-B in above figure)\ngraph.edge.src = 0;\ngraph.edge.dest = 1;\ngraph.edge.weight = -1;\n\n// add edge 0-2 (or A-C in above figure)\ngraph.edge.src = 0;\ngraph.edge.dest = 2;\ngraph.edge.weight = 4;\n\n// add edge 1-2 (or B-C in above figure)\ngraph.edge.src = 1;\ngraph.edge.dest = 2;\ngraph.edge.weight = 3;\n\n// add edge 1-3 (or B-D in above figure)\ngraph.edge.src = 1;\ngraph.edge.dest = 3;\ngraph.edge.weight = 2;\n\n// add edge 1-4 (or A-E in above figure)\ngraph.edge.src = 1;\ngraph.edge.dest = 4;\ngraph.edge.weight = 2;\n\n// add edge 3-2 (or D-C in above figure)\ngraph.edge.src = 3;\ngraph.edge.dest = 2;\ngraph.edge.weight = 5;\n\n// add edge 3-1 (or D-B in above figure)\ngraph.edge.src = 3;\ngraph.edge.dest = 1;\ngraph.edge.weight = 1;\n\n// add edge 4-3 (or E-D in above figure)\ngraph.edge.src = 4;\ngraph.edge.dest = 3;\ngraph.edge.weight = -3;\n\ngraph.BellmanFord(graph, 0);\n}\n}``````\n\nOutput :\n\n```Vertex Distance from Source\n0 0\n1 -1\n2 2\n3 -2\n4 1```\n\nNotes\n\n• Negative weights are found in various applications of graphs. For example, instead of paying cost for a path, we may get some advantage if we follow the path.\n• Bellman-Ford works better (better than Dijksra’s) for distributed systems. Unlike Dijksra’s where we need to find minimum value of all vertices, in Bellman-Ford, edges are considered one by one.", null, "#### Venkatesan Prabu\n\nWikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.\n\nX" ]	[ null, "https://www.wikitechy.com/technology/wp-content/uploads/2017/06/Bellman-Algorithm.png", null, "https://www.wikitechy.com/technology/wp-content/uploads/2017/06/Bellman-ford.png", null, "https://www.wikitechy.com/technology/wp-content/uploads/2017/06/Bellman-ford-algorithm.png", null, "https://www.wikitechy.com/technology/wp-content/uploads/2018/10/IMG-20160229-WA0017-150x150.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8838418,"math_prob":0.9912552,"size":4065,"snap":"2020-34-2020-40","text_gpt3_token_len":940,"char_repetition_ratio":0.14159074,"word_repetition_ratio":0.07402032,"special_character_ratio":0.23296434,"punctuation_ratio":0.12348668,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995376,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,7,null,7,null,7,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T00:48:56Z\",\"WARC-Record-ID\":\"<urn:uuid:f2ed3f04-2b80-4307-92ea-6de670158c7c>\",\"Content-Length\":\"129318\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76b90eee-dbdb-4ff2-92f9-6829e287673e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a87e719-3f02-4a64-888b-c05c59c43781>\",\"WARC-IP-Address\":\"172.67.155.59\",\"WARC-Target-URI\":\"https://www.wikitechy.com/technology/java-programming-bellman-ford-algorithm/\",\"WARC-Payload-Digest\":\"sha1:WNSF3OXJIHMODUFGPCWLHBBC27HFYH4M\",\"WARC-Block-Digest\":\"sha1:NJR4ZWK747TLNMO4E3KWSA56MXZWNBXV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738723.55_warc_CC-MAIN-20200810235513-20200811025513-00032.warc.gz\"}"}
https://balbhartisolutions.com/maharashtra-board-class-12-economics-solutions-chapter-3b/	[ "Balbharti Maharashtra State Board Class 12 Economics Solutions Chapter 4 Supply Analysis Textbook Exercise Questions and Answers.\n\n## Maharashtra State Board Class 12 Economics Solutions Chapter 3B Elasticity of Demand\n\n1. Complete the following statements:\n\nQuestion 1.\nPrice elasticity of demand on a linear demand curve at the X-axis is ……………\na) zero\nb) one\nc) infinity\nd) less than one\na) zero", null, "Question 2.\nPrice elasticity of demand on a linear demand curve at the Y-axis is equal to\na) zero\nb) one\nc) infinity\nd) greater than one\nc) infinity\n\nQuestion 3.\nDemand curve is parallel to X axis, in case of …………..\na) perfectly elastic demand\nb) perfectly inelastic demand\nc) relatively elastic demand\nd) relatively inelastic demand\na) perfectly elastic demand\n\nQuestion 4.\nWhen percentage change in quantity demanded is more than the percentage change in price, the demand curve is ………………..\na) flatter\nb) steeper\nc) rectangular\nd) horizontal\na) flatter\n\nQuestion 5.\nEd = 0 in case of ………………\na) luxuries\nb) normal goods\nc) necessities\nd) comforts\nc) necessities\n\n2. Give et onomic terms:\n\nQuestion 1.\nDegree of responsiveness of quantity demanded o change in income only.\nIncome elasticity\n\nQuestion 2.\nDegree of responsiveness of a change in quantity demanded of one commodity due to change in the price of another commodity.\nCross elasticity", null, "Question 3.\nDegree of responsiveness of a change of quantity demanded of a good to a change in its price.\nElasticity of demand\n\nQuestion 4.\nElasticity resulting from infinite change in quantity demanded.\nPerfectly elastic demand\n\nQuestion 5.\nElasticity resulting from a proportionate change in quantity demanded due to a proportionate change in price.\nPrice elasticity\n\n3. Complete the correlation:\n\n1) Perfectly elastic demand: Ed = ∞ :: ……………. : Ed = 0\n2) Rectangular hyperbola : ………………. : Steeper demand curve : Relatively inelastic demand.\n3) Straight line demand curve : Linear demand curve:: …………….. non linear demand curve.\n4) Pen and ink : …………….. :: Tea or Coffee: Substitutes.\n5) Ratio method : Ed = $$\\frac{\\% \\Delta \\mathbf{Q}}{\\% \\Delta \\mathrm{P}}$$ :: …………… : Ed = $$\\frac{\\text { Lower segment }}{\\text { Upper segment }}$$\n\n1. Perfectly inelastic demand\n2. Unitary elastic demand\n3. Unitary elastic (convex to origin)\n4. Complementary goods\n5. Point or Geometric method\n\n4. Assertion and Reasoning type questions:\n\nQuestion 1.\nAssertion (A) : Elasticity of demand explains that one variable is influenced by another variable.\nReasoning (R) : The concept of elasticity of demand indicates the effect of price and changes in other factors on demand.\nOptions: 1) (A) is True, but (R) is False\n2) (A) is False, but (R) is True\n3) Both (A) and (R) are True and (R) is the correct explanation of (A)\n4) Both (A) and (R) are True and (R) is not the correct explanation of (A)\n3) Both (A) and (R) are True and (R) is the correct explanation of (A)", null, "Question 2.\nAssertion (A) : A change in quantity demanded of one commodity due to a change in the price of other commodity is cross elasticity.\nReasoning (R) : Changes in consumers income leads to a change in the quantity demanded.\nOptions:\n1) (A) is True, but (R) is False\n2) (A) is False, but (R) is True\n3) Both (A) and (R) are True and (R) is the correct explanation of (A)\n4) Both (A) and (R) are True and (R) is not the correct explanation of (A)\n4) Both (A) and (R) are True and (R) is not the correct explanation of (A)\n\nQuestion 3.\nAssertion (A) : Degree of price elasticity is less than one in case of relatively inelastic demand.\nReasoning (R): Change in demand is less then the change in price.\nOptions: 1) (A) is True, but (R) is False\n2) (A) is False, but (R) is True\n3) Both (A) and (R) are True and (R) is the correct explanation of (A)\n4) Both (A) and (R) are True and (R) is not the correct explanation of (A)\n3) Both (A) and (R) are True and (R) is the correct explanation of (A)\n\n5. Distinguish between:\n\nQuestion 1.\nRelatively elastic demand and Relatively inelastic demand.\nRelatively Elastic Demand\n\n1. When percentage change in quantity demanded is greater than the percentage change in price then demand is said to be Relatively Elastic demand.\n2. The numerical co-efficient is greater than one (e > 1).\n3. Demand curve slopes flatter.\n4.", null, "5. Example : luxury goods like LCD, TV, Car etc.\n\nRelatively inelastic demand.\n\n1. When percentage change in quantity demanded is less than percentage change in price then demand is said to be Relatively Inelastic demand.\n2. The numerical co-efficient is less than one (e < 1).\n3. Demand curve slopes steeper.\n4.", null, "5. Example : foodgrains.", null, "Question 2.\nPerfectly elastic demand and Perfectly inelastic demand.\nPerfectly elastic demand :\n\n1. When a small change in price brings an infinite change in quantity demanded, then demand is said to be Perfectly Elastic demand.\n2. The numerical value of Perfectly Elastic demand is infinite i.e. e = ∞\n3. The demand curve is horizontal straight line parallel to X-axis.\n4. Such a demand is a myth or theoretical.\n5.", null, "Perfectly inelastic demand.\n\n1. When a change in price does not bring any change in quantity demanded, then demand\nis said to be Perfectly Inelastic demand.\n2. The numerical value of Perfectly Inelastic demand is zero i.e. e = 0.\n3. The demand curve is a vertical straight line parallel to Y—axis.\n4. Such demand is found in case of life saving drugs, salt, etc.\n5.", null, "Question 1.\nExplain the factors influencing elasticity of demand.\nThe concept of Price Elasticity was developed i by great neo-classical economist Dr. Alfred \\ Marshall in the year 1890.\nAccording to Dr. Alfred Marshall, “The elasticity or responsiveness of demand in a market is great or small, according to the amount demanded which increases much or little for a given fall in price, and diminishes much or little for a given rise in price. ”\nElasticity of demand in fact refers to the £ degree of responsiveness of the quantity demanded of a commodity to change in the variable on which demand depends.\n\nQuestion 2.\nExplain the total outlay method of measuring elasticity of demand?\nTotal Outlay Method : This method was introduced by Dr. Alfred Marshall. The limitation of this method is that in this method unlike ratio method, the exact numerical value of the elasticity of demand cannot be determined. According, to this method, the elasticity of demand is measured on the basis of expenditure incurred by consumer when the price of a commodity changes.\n\nTotal outlay or total expenditure can be calculated by multiplying the price with the quantity demanded (Price x Quantity demand = Total Expenditure). Depending upon the kind of change in total outlay, whether it increases, or decreases, or remain constant with the change in price we will be able to decide the type of elasticity. This can be explained with the following example:-", null, "1. If the total outlay remains the same with a rise or fall in price then the demand is said to be unitary (e = 1) elastic.\n2. If the total outlay decreases with a rise in price and increases with a fall in price, the elasticity of demand is greater than one or Relatively Elastic e > 1.\n3. If the total outlay increases with a rise in price and decreases with a fall in price, then elasticity is less than one or relatively inelastic, e < 1.", null, "", null, "Question 3.\nExplain importance of elasticity of demand.", null, "• Nature of Commodity : By nature, commodities are classified as necessaries, comforts and luxuries. Normally demand for j necessaries like food grains are relatively inelastic and for comforts and luxuries like diamond, perfumes, etc is relatively elastic.\n• Availability of Substitutes : The larger the number of substitutes available for a commodity, the greater would be the elasticity. Demand for products like soap, soft drinks, detergents, tooth paste, etc. have many substitute so demand is elastic, ‘j However, salt, garlic, onions have no substitute so demand is inelastic.\n• Durability of the Commodity : The demand for durable goods like T.V., car, fridge, etc is relatively inelastic in the short run and elastic in the long run. Whereas the demand for perishable goods is relatively inelastic.\n• Uses of Commodity : Single use commodities have less elastic demand and multi-use goods like coal, electricity, sugar, etc. have relatively elastic demand.\n• Range of Price : The demand for commodities which are highly priced and will have a inelastic demand like AC, car, etc. Even very low priced goods have inelastic demand.\n• Consumer’s Income : Generally if income is very high, the demand for over allcommodities tends to be relatively inelastic. The demand pattern of the rich people is rarely affected even when there is significant price change.\n• Influence of Habits and Customs : When a person is habituated to consume a certain commodity, the demand will be inelastic for that commodity. E.g. demand for cigarettes to a chain smoker is inelastic.\n• Time Period : The demand for goods is less elastic in the short period and more elastic in the long period. This is because (1) in the long period consumer are better informed about their price (2) habits of consumer’s change in the long run (3) durable goods get worn out in the long period.\n• Proportion of Income Spend : If consumer spends a very small proportion of his income on a commodity, the demand for it will be relatively inelastic & vice-versa. For e.g. demand for salt, newspaper, pins are inelastic.\n• Urgency and Postponement : If the demand for a commodity is urgent then demand for it will be inelastic. E.g. demand for medicine for a patient. Whereas, if the demand for a commodity can be postponed it will have elastic demand.\n• Complementary Goods : Complementary goods are those goods which are demanded jointly such as car and petrol, mobile and sim cards, etc. Demand for petrol will be inelastic as car cannot run without petrol.\n\n7. Observe the following figure and answer the questions:\n\nQuestion 1.\nIdentify and define the degrees of elasticity of demand from the following demand curves.", null, "", null, "Concept: Perfectly Inelastic demand (Ed = 0) Explanation : When change in price has no effect on the quantity demanded of that commodity, then it is called as perfectly inelastic demand. Demand curve ‘DD’ is a vertical straight line parallel to ‘Y’ – axis.", null, "", null, "Concept: Perfectly Elastic demand (Ed = ∞) (infinity)\nExplanation: When a change in price leads to infinite change in quantity demanded of a commodity then it is called as perfectly) (d) elastic demand.\nDemand curve is horizontal straight line ( parallel to ‘X’ – axis.", null, "Concept: Ed = 1 Unitary elastic demand Explanation : When proportionate or percentage change in quantity demanded is exactly equal to proportionate or percentage change in price, then it is called as Unitary Elastic demand. Demand curve is called as rectangular hyperbola.", null, "Concept: Relatively Elastic Demand (Ed > 1)\nExplanation : When proportionate or percentage change in quantity demanded is more than proportionate change it its price, then it is called as Relatively Elastic Demand. Demand curve is called as flatter curve.\n\nQuestion 2.\nIn the following diagram AE is the linear demand curve of a commodity. On the basis of the given diagram state whether the following statements are True or False. Give reasons to your answer.", null, "1) Demand at point ‘C’ is relatively elastic demand.\n2) Demand at point ‘B’ is unitaiy elastic demand.\n3) Demand at point ‘D’ is perfectly inelastic demand.\n4) Demand at point ‘A’ is perfectly elastic demand." ]	[ null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-3.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-4.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-1.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-2.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-11.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-12.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-10.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-5.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-6.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Solutions.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-7.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-8.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-9.png", null, "https://maharashtraboardsolutions.guru/wp-content/uploads/2021/07/Maharashtra-Board-Class-12-Economics-Solutions-Chapter-3B-Elasticity-of-Demand-13.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8605303,"math_prob":0.9199964,"size":11838,"snap":"2022-05-2022-21","text_gpt3_token_len":2760,"char_repetition_ratio":0.19891837,"word_repetition_ratio":0.18060367,"special_character_ratio":0.23230276,"punctuation_ratio":0.12866817,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99729145,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,3,null,3,null,null,null,3,null,3,null,3,null,3,null,null,null,3,null,3,null,3,null,null,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T21:09:52Z\",\"WARC-Record-ID\":\"<urn:uuid:eb3f7bbe-95f7-4087-83c0-ed3512e6b541>\",\"Content-Length\":\"50236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24835290-46d7-408e-8bd8-730885bcf102>\",\"WARC-Concurrent-To\":\"<urn:uuid:369cdab3-db13-44a9-b7d3-51f2cffbaf76>\",\"WARC-IP-Address\":\"172.105.50.94\",\"WARC-Target-URI\":\"https://balbhartisolutions.com/maharashtra-board-class-12-economics-solutions-chapter-3b/\",\"WARC-Payload-Digest\":\"sha1:VEGO3YOMXR42VE4KMX6Y2TJR4TTJFSG2\",\"WARC-Block-Digest\":\"sha1:H36MAIWIGTP53GMOZIZWYUGHR73265BH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662541747.38_warc_CC-MAIN-20220521205757-20220521235757-00596.warc.gz\"}"}
https://answers.everydaycalculation.com/divide-fractions/24-3-divided-by-10-90	[ "Solutions by everydaycalculation.com\n\n## Divide 24/3 with 10/90\n\n1st number: 8 0/3, 2nd number: 10/90\n\n24/3 ÷ 10/90 is 72/1.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 10/90: 90/10\n2. Now, multiply it with the dividend\nSo, 24/3 ÷ 10/90 = 24/3 × 90/10\n3. = 24 × 90/3 × 10 = 2160/30\n4. After reducing the fraction, the answer is 72/1\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6969796,"math_prob":0.9618313,"size":364,"snap":"2020-10-2020-16","text_gpt3_token_len":164,"char_repetition_ratio":0.19444445,"word_repetition_ratio":0.0,"special_character_ratio":0.50549453,"punctuation_ratio":0.08988764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96108484,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-28T12:51:03Z\",\"WARC-Record-ID\":\"<urn:uuid:dd8f9658-1a8a-40e6-b5da-d1a34e4038ee>\",\"Content-Length\":\"7748\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54ee6eeb-272c-40b0-86d5-2787ce56e6a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f989748-99f2-4db2-b35d-9d16deb1ea7a>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/divide-fractions/24-3-divided-by-10-90\",\"WARC-Payload-Digest\":\"sha1:UOOWNW7OH5NC2AOKUGUXDERK52TEGOEG\",\"WARC-Block-Digest\":\"sha1:YODNGMBFHYT7M6QOZXMMXZ37QITPDXF4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875147154.70_warc_CC-MAIN-20200228104413-20200228134413-00049.warc.gz\"}"}
https://solvedlib.com/n/8-points-sketch-the-graph-of-f-r-12-in-r-on-the-domain-1,1592175	[ "# [8 points] Sketch the graph of f(r) 12 . In(r) on the domain 1 < I < 2 and then\n\n###### Question:\n\n[8 points] Sketch the graph of f(r) 12 . In(r) on the domain 1 < I < 2 and then evaluate Jz? . In(r) dr. [6 points] Sketch the graph of f(r) sin(r? ) on the domain 0 < 1 < Vz,and then evaluate sin(2? ) dx:", null, "", null, "#### Similar Solved Questions\n\n##### C6 Use the data set in WAGEZ RAW for this problemAs usual; be sure all of the follow- ing regressions contain an intercept. Run a simple regression of IQ on educ t0 obtain the slope coefficient say, & . Run the simple regression of log(wage) on educ, and obtain the slope coefficient 8_ (iii) Run the multiple regression of log(wage) on educ and IQ, and obtain the slope coefficients 8 , and Bz. respectively_ (iv) Verify that B, 8 8,6 ;.\nC6 Use the data set in WAGEZ RAW for this problemAs usual; be sure all of the follow- ing regressions contain an intercept. Run a simple regression of IQ on educ t0 obtain the slope coefficient say, & . Run the simple regression of log(wage) on educ, and obtain the slope coefficient 8_ (iii) Run...\n##### I6 3 8 NW L 1 } J WI 1 1 J [ L 8 J U 1 HHY 8 L V Is U 1 1 1 8 1 1 1 1 8 [ 1 1 i 8 8 0 1 8 8 ;1 I { 3 1\ni6 3 8 NW L 1 } J WI 1 1 J [ L 8 J U 1 HHY 8 L V Is U 1 1 1 8 1 1 1 1 8 [ 1 1 i 8 8 0 1 8 8 ; 1 I { 3 1...\n##### A night before your exam, should you re-read information or test yourself? Where does re-reading information...\nA night before your exam, should you re-read information or test yourself? Where does re-reading information store, in your short-term or long-term memory?...\n##### Question 8In the dchyurohalogsnation of bromobutane which confonnation bclov: leads drectly to the (onation of cls-2-buteng?CH; H;c_ BrCH; HSC HH H;c_ BrH H HM Br HHS HCH;O7kWITenid \|Only \|Onty II\nQuestion 8 In the dchyurohalogsnation of bromobutane which confonnation bclov: leads drectly to the (onation of cls-2-buteng? CH; H;c_ Br CH; HSC H H H;c_ Br H H H M Br H HS H CH; O7kWI Tenid \| Only \| Onty II...\n##### The following is a special case of a model, called the Ehrenfestmodel, that has been used to explain diffusion of gases. We havetwo urns that, between them, contain four balls. At each step, oneof the four balls is chosen at random and moved from the urn thatit is in into the other urn. We choose as possible states of theprocess the number of balls in the first urn. (a) Define thetransition matrix P of a Markov chain modelling this process. (b)Show that this Markov chain is ergodic but not regul\nThe following is a special case of a model, called the Ehrenfest model, that has been used to explain diffusion of gases. We have two urns that, between them, contain four balls. At each step, one of the four balls is chosen at random and moved from the urn that it is in into the other urn. We choos...\n##### Q 4 (Bounce question) SolveX' =(1 3)x\nQ 4 (Bounce question) Solve X' = (1 3)x...\n##### Tn = {tk kzn} and Prove - Rbe that inf(Tn) < sup(Tm) for all n,m. Let Tn â‚¬ sequence m and n 2 m. In each case_ work out whether Tn â‚¬ Tm r Tm C Tn, then (Hint: Two cases n < compare their supremums and infimums: -\nTn = {tk kzn} and Prove - Rbe that inf(Tn) < sup(Tm) for all n,m. Let Tn â‚¬ sequence m and n 2 m. In each case_ work out whether Tn â‚¬ Tm r Tm C Tn, then (Hint: Two cases n < compare their supremums and infimums: -...\n##### 1. list the usefulness and likitations of a blance sheet - be specific 2. why ars...\n1. list the usefulness and likitations of a blance sheet - be specific 2. why ars disclosure notes important ? list some examples of items that should be disclosed ....\nAccounting Please Help Keith opens a business, Keith S.A. and opens a bank account in the name of the business. For the first week, the following transactions take place: Day 1: Keith puts £ 1, 000 of his own money in the bank account Day 2: The commercial buys 100 shirts, each costing &po...\n##### The intersection of X2= Y2 + Z2 and the plane Y= 2\nthe intersection of X2= Y2 + Z2 and the plane Y= 2...\n##### 66. U.S. Imports. The amount of imports to the United States has increased exponentially since 1980...\n66. U.S. Imports. The amount of imports to the United States has increased exponentially since 1980 - (Sources: U.S. Census Bureau; U.S. Bureau of Economic Analysis; U.S. Department of Commerce). The exponential function 1(x) = 297.539(1.075), where x is the number of years after 1980, can be used ...\n##### USc (0) Thc Ira-Pezaidal Pul (b) Ihe midpoint Rule , and (C) Simpsons Rule 4 approximatc tc Given_intcrval witn mc Jpec Fica velue af n (Rouna Your amswcr + Six dccma \| Places)3c0 (22) dx ns4Tra-pczoidal RaleMc_mid poin t Rulc(C) I Simpeans Rule\nUSc (0) Thc Ira-Pezaidal Pul (b) Ihe midpoint Rule , and (C) Simpsons Rule 4 approximatc tc Given_intcrval witn mc Jpec Fica velue af n (Rouna Your amswcr + Six dccma \| Places) 3c0 (22) dx ns4 Tra-pczoidal Rale Mc_mid poin t Rulc (C) I Simpeans Rule...\nQuestion 30 1.5 pts A firm with cost of capital of 10 percent is evaluating two mutually exclusive projects and constructed the following NPV profile: $100 NPV(A) .......NPV(B) 50 + Net Present Value “ .....$0 ....sete 0% 5% 10% 15% 20% 25% 30% 35% 40% -$50 Dis count Rate True or false: Proje... 1 answer ##### 1 G. Gram invested$43,000 cash in the company. 1 The company rented a furnished office...\n1 G. Gram invested $43,000 cash in the company. 1 The company rented a furnished office and paid$2,400 cash for May’s rent. 3 The company purchased $1,930 of office equipment on credit. 5 The company paid$740 cash for this month’s cleaning services. 8 The company p...\n##### Please write this code in c++ // a: an array of ints. size is how many...\nplease write this code in c++ // a: an array of ints. size is how many ints in array 7/ Return the largest value in the list. 7/ This value may be unique, or may occur more than once // You may assume size >= 1 int largestValue(int a, int size) { assert(size >= 1); return -42; // STUB !!!...\n##### Pz) lel S= { v, 3, 6, 9 '2 } 77~ Skow Hu+ 5 sbri 75 unfer Jachi 044 mu (Kplahio mau T5 .0 Lis+ aU (Je~pobL in 8:6v\nPz) lel S= { v, 3, 6, 9 '2 } 77~ Skow Hu+ 5 sbri 75 unfer Jachi 044 mu (Kplahio mau T5 . 0 Lis+ aU (Je~pobL in 8:6v...\n##### (II) A $15-\\mathrm{cm}$ -long tendon was found to stretch 3.7 $\\mathrm{mm}$ by a force of 13.4 $\\mathrm{N}$ . The tendon was approximately round with an average diameter of 8.5 $\\mathrm{mm}$ . Calculate the Young's modulus of this tendon.\n(II) A $15-\\mathrm{cm}$ -long tendon was found to stretch 3.7 $\\mathrm{mm}$ by a force of 13.4 $\\mathrm{N}$ . The tendon was approximately round with an average diameter of 8.5 $\\mathrm{mm}$ . Calculate the Young's modulus of this tendon....\n##### QuuestionNot yet answered1. 0) Indicate whether the following is true or false. Justify your answer.(2 marks) b}) Consider the universal set U = {1,2,3,4,5,6,7}and the sets A,B,CEU , where A={2,4,7} B = {1, 3, 4} and â‚¬= {4,6,7}Marked out of 16.00Flag questionFind J(A 0 C) where J is the powerset of the set An â‚¬(3 marks)(ii) Write the elements of the following set B X (A n C)(3 marks}Let the universal set be the set R of all real numbers and let A = {x eR 0<s5} B = {xâ‚¬R 2s<9}Find\nquuestion Not yet answered 1. 0) Indicate whether the following is true or false. Justify your answer. (2 marks) b}) Consider the universal set U = {1,2,3,4,5,6,7}and the sets A,B,CEU , where A={2,4,7} B = {1, 3, 4} and â‚¬= {4,6,7} Marked out of 16.00 Flag question Find J(A 0 C) where J is the...\n##### Solve these systems of equations a\n1. Solve these systems of equations a. 3x + 6y + 3z= 3 2x - y + 4z= 14 x + 10y + 8z= -8 b. x + y + 3z= 10 x + 2y + 3z= 4 x + 4y + 3z= -6 2. Solve these systems of equations word problems. a. A movie theater sold 450 tickets to a movie. Adult tickets cost $8 and children's tickets cost$6....\n##### How do you find the definite integrals for the lengths of the curves, but do not evaluate the integrals for y=x^3, 0<=x<=1?\nHow do you find the definite integrals for the lengths of the curves, but do not evaluate the integrals for y=x^3, 0<=x<=1?...\n##### DENTAL HYGIENE QUESTION Scenario 3 - The current protocol in Dr. Grains offices utilizes a medical...\nDENTAL HYGIENE QUESTION Scenario 3 - The current protocol in Dr. Grains offices utilizes a medical history form that is slightly larger than a postcard to gather all the patient data. As a fairly recent dental hygiene graduate, the inadequacy of the questions on the history form has bothered Hann...\n##### P.3 Let A =(1) Find the row space and column space of A. (2)Determine the and rank and nullity of A.\nP.3 Let A = (1) Find the row space and column space of A. (2) Determine the and rank and nullity of A....\n##### Assume that we want to find the probability that when five consumers are randomly selected, exactly two of them are comfortable with delivery by drones. Also assume that $42 \\%$ of consumers are comfortable with the drones (based on a Pitney Bowes survey). Identify the values of $n, x, p,$ and $q$.\nAssume that we want to find the probability that when five consumers are randomly selected, exactly two of them are comfortable with delivery by drones. Also assume that $42 \\%$ of consumers are comfortable with the drones (based on a Pitney Bowes survey). Identify the values of $n, x, p,$ and $q$...." ]	[ null, "https://i.imgur.com/jG4Kdh9.jpeg", null, "https://cdn.numerade.com/previews/5d6788ed-6a92-4eb1-ab48-a4dfafa8da8f_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8550499,"math_prob":0.97613055,"size":14826,"snap":"2022-27-2022-33","text_gpt3_token_len":4481,"char_repetition_ratio":0.0956686,"word_repetition_ratio":0.48349655,"special_character_ratio":0.3101983,"punctuation_ratio":0.16373755,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98364276,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T13:09:08Z\",\"WARC-Record-ID\":\"<urn:uuid:b12260ea-ed16-492d-bec9-e718a62949f8>\",\"Content-Length\":\"94399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0d28bff-8520-4032-a19c-c9cd11aa9f0f>\",\"WARC-Concurrent-To\":\"<urn:uuid:daf17bcc-86c6-4e07-9d72-cd82e4edbea3>\",\"WARC-IP-Address\":\"104.21.12.185\",\"WARC-Target-URI\":\"https://solvedlib.com/n/8-points-sketch-the-graph-of-f-r-12-in-r-on-the-domain-1,1592175\",\"WARC-Payload-Digest\":\"sha1:TF4SXLNRSLM2TWR7I3JMIJZZPJYXBTCV\",\"WARC-Block-Digest\":\"sha1:3JNG2GKY5U23IM4VQCPTE7YKDOHX4NHW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572174.8_warc_CC-MAIN-20220815115129-20220815145129-00541.warc.gz\"}"}
http://arfer.net/projects/builder/notebook	[ "# Builder notebook\n\n## Parameter sensitivity\n\nI notice that, at least under a simple exponential model, choice probabilities are much more sensitive to the discount rate for larger delay intervals between the two choices.\n\nd = data.frame(\nssr = c(10, 10),\nssd = c( 0, 0),\nllr = c(13, 45),\nlld = c( 5, 30))\n\ndo.call(rbind, lapply(1 : nrow(d), function (dr)\nwith(list(disc = c(.055, .050, .045)),\ntransform(d[dr,], disc = disc, p = round(digits = 2,\nilogit(llr * exp(-disc * lld) - ssr * exp(-disc * ssd)))))))\n\nssr ssd llr lld disc p\n1 10 0 13 5 0.055 0.47\n2 10 0 13 5 0.050 0.53\n3 10 0 13 5 0.045 0.59\n4 10 0 45 30 0.055 0.20\n5 10 0 45 30 0.050 0.51\n6 10 0 45 30 0.045 0.84\n\nThis makes sense when you consider that we're talking about discount rate: the longer the delay, the more time the reward has been discounted, so a given discount rate will have a greater total effect as the delay lengthens.\n\nFrom a Bayesian perspective, this means that there will be larger uncertainty in choice probabilities for futher-apart pairs, since longer delays will magnify uncertainity in the parameter values.\n\n## Adaptive estimation of discount rate and ρ\n\nBut the aforementioned phenomenon could be useful from the perspective of choosing quartets that optimize our ability to estimate parameters. Given a proposed discount rate k, we can construct a quartet that can precisely distinguish whether the true rate is above or below k, by just choosing a quartet on the indifference curve for k that has a big delay interval. It follows that in the context of the generalized hyperbolic model (GHM), a natural way to adaptively estimate parameters is to first estimate the discount rate with a binary search of candidates, keeping the smaller-sooner delay 0 so that curvature doesn't muck things up, then try to estimate curvature.\n\nSo can I similarly zero in on ρ and curvature values? What sorts of quartets are most diagnostic for values of these parameters?\n\nIt looks like zeroing in on ρ will be in general much harder than for the discount rate. No particular quartets are terribly diagnostic, because the general effect of lower ρs is to shrink choice probabilities towards 1/2, and the effect of ρ is strongest in the far-off regions of ilogit's domain where the argument to ilogit makes the least difference in choice probabilities. I guess the thing to do is to consider two possible values of ρ at a time and try to pick the difference in discounted values that will maximize the difference in choice probabilities for these two ρs. Let's try analytically optimizing this difference the good-ol'-fashioned Calc I way, by finding roots of the derivative. I'll use Maxima.\n\nilogit(x) :=\n1 / (1 + exp(-x));\nchoicep(ssr, ssd, llr, lld, disc, rho) :=\nilogit(rho * (\n(llr * exp(-disc * lld)) - (ssr * exp(-disc * ssd))));\nreward_with_v(val, disc, delay) :=\nexp(disc * delay) * val;\nchoicep_valdiff(ssr, ssd, lld, valdiff, disc, rho) :=\nchoicep(\nssr, ssd,\nreward_with_v(ssr + valdiff, disc, lld),\nlld, disc, rho);\n\nexpr: choicep_valdiff(ssr, ssd, lld, D, disc, rho) -\nchoicep_valdiff(ssr, ssd, lld, D, disc, rho);\n/* We want to choose the value of D that maximizes expr (labeling\nrho, rho so that this difference is positive). /\nexpr: diff(expr, D);\n/ Now let's try to clean that expression up a bit. /\nexpr: subst(v, D - exp(-disc ssd)ssr + ssr, expr);\n\n\nUnfortunately, I don't think we can solve\n\n$\\frac{ \\rho_1 e^{\\rho_1 v} }{ (e^{\\rho_1 v} + 1)^2 } - \\frac{ \\rho_2 e^{\\rho_2 v} }{ (e^{\\rho_2 v} + 1)^2 } = 0$\n\nfor v. So we'll need to find each root numerically, or maximize the original expression numerically.\n\nNow let's try implementing an adaptive parameter-estimating procedure with what I've thought through so far.\n\n## Back to real data\n\nBut let's put development of this adaptive procedure on hold for a moment to look at how different sorts of models do when applied to real data. In particular, I'd like to know if a uniform noise parameter (postulating a certain fixed-per-subject probability that the subject behaves randomly each trial) can explain the data as well as or better than ρ. Such a parameter would be much easier to estimate than ρ.\n\nSo I tried fitting a whole bunch of models to the entire audTemp dataset (70 trials for each of 93 subjects). I split the dataset into training and testing halves, not at random, but with a striped scheme that made sure the training set included 7 trials from each of the 14-trial bins and that these 7 trials included the 1st and 14th trials (in indifference-k order) so the models wouldn't need to extrapolate, as it were. Here was each model's proportion of test-set choices predicted correctly:\n\nmodel correct\nmodel.diff 0.850\nmodel.diffrho 0.850\nmodel.ghmrho 0.840\nmodel.exprho 0.838\nmodel.expdelta 0.828\nmodel.exp 0.825\nmodel.rewards 0.816\nmodel.hyprho 0.812\nmodel.ratio 0.810\nmodel.ll 0.766\nmodel.delays 0.694\nmodel.constant 0.686\n\nThe ranking is somewhat surprising, particularly in how model.diff did better than the GHM. Christian points out that model.diff is similar to the model suggested in Scholten and Read (2010).\n\nThis exercise with the audTemp data has arguably provided more questions than answers. I'm thinking that what I now want to do is create adaptive estimation procedures for each model so they can all be fit reasonably well, then put subjects through these procedures along with a model-neutral test set of trials. By such means, we can begin to separate the questions of how well we can estimate model parameters and how well a model describes behavior, that is, how well it can predict choice behavior given training data that's appropriate for the model.\n\nRather than invent a different adaptive procedure for every model, we can divide the models into nested sets, or at least families, and create adaption procedures for the most general member of the set. Such a procedure should also work for other members of the set.\n\nWe'll start by considering the following families:\n\n• Degenerate\n• constant\n• Simple\n• fullglm\n• ll\n• ss (This one isn't worth trying to train with our existing data, since SS delays vary too little.)\n• rewards\n• delays\n• diff\n• diffrho\n• diffdelta\n• rd\n• rdrho\n• rddelta\n• Discounting\n• ghmk.rho\n• ghmk.delta\n• hyprho\n• hypdelta\n\nIn the case of the constant model, the only sense in which we can be adaptive is to choose how many training trials we administer. And how many training trials we'd need is partly a function of where the true probability lies, but also largely a function of the desired precision. And I've yet to specify how much precision I want. So let's put that aside for now, and return to the GHM.\n\nAs of 6 October 2012, I've created an adaptive procedure for ghmk.rho, discount.adapt.est, that gets much tighter 95% credible intervals in 60 trials than naive.est, a procedure that picks quartets non-adaptively. Hooray! Handling delta instead of rho shouldn't be too hard—it's obvious what trials are most diagnostic for it—so now let's turn to the simple family. I had trouble fitting fullglm with JAGS, but I ought to be able to do it with Stan.\n\nBut, hmm, the technique I used to estimate rho in discount.adapt.est suggests a model-general technique to adaptively estimate all the parameters simultaneously. For each round, you pick two parameter vectors from the current posterior that are in the credible region but are far from each other, which is a multidimensional equvialent of picking .025 and .975 quantiles. Then, out of a few thousand possible quartets, you pick the one that maximizes the difference in choice probability for these two parameter vectors. I think I'll try implementing this to see if it can handle the GHM just as well as the fancier procedure. If so, this procedure may be suitable for any model.\n\nThe most mysterious part here is picking the two parameter vectors. How is one to do that? Well, you start with however many distinct parameter vectors you got out of MCMC, on the order of 500. Think of the vectors geometrically, as points. (You'd want to rescale each parameter to the interval [0, 1] so parameters are treated equally.) Suppose you find the two distantmost points A and B. Then you sort all the points into a list by their distance from A and B such that A is the first item, B is the last item, and points equidistant from both go in the middle. Then you take the points 2.5% and 97.5% of the way along the list. And that's it. This algorithm sounds slow, but with reasonably small posterior samples, it might be fast enough.\n\nIt looks like this procedure, even with a simplification to just pick the farthest two points in the posterior (instead of a pair of points with the .95-quantile distance, or something), works quite well. It fits ghmk.rho not significantly worse than discount.adapt.est. Now, how well can it handle fullglm?\n\nTurns out that it can't handle fullglm well, in that Stan fits fullglm slowly, and once the adaptive session is over, the 95% posterior interval for each parameter isn't as likely as it should be to contain the true value.\n\nmodel.rewards—actually, now model.rewards.logprior—when sent through the adaptive procedure—actually, a new parameter-by-parameter version, general.adapt.est.1patatime—yields wide 95% intervals. The model has severe collinearity issues; the parameters are usually correlated about -.99 in the posterior. However, the posterior medians tend to be close to the true values. Roughly the same can be said for model.diff.\n\nOne thing that's bugging me is that my criterion of precision, which is how tight the 95% posterior intervals for the parameters are, may not be a good one. In particular, I fear that low precision in this sense may be perfectly good precision in applications. I guess what I should try is examining the tightness of posterior intervals for some choice probabilities. Say, take 500 trials from adaptive.quartets and then compare true with predicted choice probabilities. For a measure of precision per se, take the mean of the widths of the 95% credible intervals. To check that these intervals are reasonably accurate, compute their coverage. Hence, check.model.precision in models.R.\n\nApplying check.model.precision has been interesting. With model.ghmk.rho, I get mostly tight intervals with perfect coverage, and the same for model.rewards.logpriors, even if the intervals are perhaps looser in that case. (And yes, general.adapt.est.1patatime is worth something for model.rewards.logpriors: its training quartets yield tighter intervals than a random sample of 60 adaptive.quartets.) model.diff also seems to have good coverage (at least, most of the time, I think), although the intervals are even looser.\n\nBy the way, I tried using check.model.precision with the whole of adaptive.quartets as a test set, although this is, of course, slow and takes like a gig of RAM. For model.diff, I got this (the first few rows are quantiles and the mean of the widths of 95% posterior intervals around all the choice probabilities; the last row is the proportion of these intervals that contained the true value):\n\n 2.5% 0.001 25.0% 0.015 meanwidth 0.117 75.0% 0.203 97.5% 0.391 coverage 1\n\nFor model.ghmk.rho, I got this:\n\n 2.5% 0 25.0% 0 meanwidth 0.011 75.0% 0 97.5% 0.182 coverage 0.947\n\nAnyway, the moral is that the adaptive procedures I have may be good enough for model.rewards.logprior and model.diff (although the latter probably needs better priors), even if the estimates of individual parameters appear imprecise.\n\n## Using a model-fitting server\n\nWhen using the adaptive procedure with human subjects, I'd like to have a computer other than the lab Windows boxes (when running Stony Brook students) or my VPS (when using MTurk) doing the computation. Getting Stan to work on a Windows box looks like a pain, and my VPS doesn't have the RAM for it and likely also won't have enough CPU cycles for it. So I'll run a sort of server program, written in R and interacting with Stan, that the client program (either a task program in lab, or a CGI program on my VPS) talks to in order to get each quartet during the adaptive procedure.\n\nSo now I need to design how the inter-process communication (IPC) is going to work. Here's the order of events:\n\n• The client gets a new decision from the subject. It saves the decision to the database, reads the decision from the database (why write and then read? the write would be INSERT OR IGNORE, so this scheme makes sure only one of the two possible choices for each trial ever gets sent to the server), does something to send this decision to the server, and waits until it gets a response. The message sent to the server identifies the subject or session, the model, the trial the decision was about, and the chosen reward (SS or LL).\n• The server has a master process that listens to a socket and hands off requests to child processes. (It arranges for child processes to be automatically reaped.) When it receives a request, it makes a new child and passes down the job description. Each child begins by creating a lockfile (a blank temporary file) and trying to insert a row for the job and the name of the lockfile in a database.\n\nIf it succeeds, it does the job. When it's done, it writes the job result to the database and removes the lockfile.\n\nks.test(scurve, punif)$statistic}))) print(mi <- which.min(vals)) print(vals[mi]) list(g, vals, mi)} The result was okay for v30 but not for scurve. I then tried analytic methods: Suppose we want the random variable SCURVE to be uniform. Then 1/(1 - 10 E) must be uniform. Then what distribution must E have? E = 1/-(10/(1 - scurve) - 10) = (1 - 1/scurve) * (1/10) For each y (i.e., value taken by E), F_{E}(y) = prob(E ≤ y) = prob((1 - 1/SCURVE) * (1/10) ≤ y) = prob(SCURVE ≤ 1 / (1 - 10y)) = F_{SCURVE}(1 / (1 - 10y)) = 1 / (1 - 10y) Now, how would we sample E in the context of Stan? Basically, by using the quantile function, which adds up to pretty much the same thing as parametrizing the model in terms of scurve, which is what I was doing to start with! Let's stick to the v30–scurve–rho parametrization of the GHM for now. ## Leave-one-out cross-validation for the majority model How well would the majority model do in leave-one-out cross-validation with the test set? For each subject, there would be 100 cross-validation rounds, one for each excluded quartet. Pick a subject and let N be their number of LL choices in the test set. Suppose we're doing a given round. In the part of the test set we're training from, there are n LL choices and (99 - n) SS choices. We predict LL for the left-out trial when and only when (modulo off-by-one errors, which I'm ignoring, as I usually do for the majority model) n >= (99 - n) 2n >= 99 n >= 49.5 n > 49 Now consider the two cases for the two possible values of the left-out trial. • If the left-out trial is LL, then n = N - 1, and we predict LL iff N - 1 > 49 or N > 50, meaning we predict correctly iff N > 50. • If the left-out trial is SS, then n = N, and we predict LL iff N > 49, meaning we predict correctly iff N < 50. (Yes, we always get it wrong when N = 50.) So now let's try characterizing overall cross-validation performance of the majority model as a function of N. • Suppose N > 50. Then we predict each left-out trial correctly when and only when it's LL. So overall, we get N of 100 rounds right. • Suppose N < 50. Then we predict each left-out trial correctly when and only when it's SS. So overall, we get (100 - N) of 100 rounds right. • When N = 50, we get it wrong every time. The upshot of this is that the majority model's performance when peeking at the data is equivalent to its performance under leave-one-out cross-validation, with the unimportant exception of the case when N = 50. ## Test-set consistency Of course, the point of randomly assigning people to models was to equate test-set performance as much as possible. Let's see how well this worked. Our sample sizes are: table(factor(sb[good.s, \"model\"])) count diff.rho 43 expk.rho 47 ghmk.rho 52 sr.rho 45 test.set.consistency(good.s)$plot\n\n\nIn this figure, you can see for each quartet and model group the proportion of subjects who picked LL. Those ranges look pretty wide, don't they?\n\nround(test.set.consistency(good.s)$quantiles, 3) value 2.5% 0.037 50% 0.104 97.5% 0.211 So the median of these ranges is 10%. That's not great. ## Parameter restriction s0175 may've suffered from the lower bound of diff.rho's prior for dr being too high. On the other hand, of the tight posteriors for dr, none go below -.4, suggesting that intervals that hit -.8 or so are just underconstrained. sr.rho seems to have a more substantial problem with the priors being too restricted. Look at show.param(l$sr.rho, \"dr\") and show.param(l$sr.rho, \"rho\"). But let's take a closer look at the subjects whose dr s seem cut off above. • s0040 and s0112 chose LL for every training trial, including$100 now vs. $105 in 4 months, so they were (almost?) more patient than we could've measured. • s0104 made some weird combinations of choices in the training trials, such as LL for$100 today vs. $110 in 4 months followed by SS for$90 today vs. $95 in 4 months. s0029 didn't seem to do anything weird, and judging from qplot(model.sr.rho$sample.posterior(ss(ts, s == \"s0029\" & type == \"train\"), 3e5)[,\"rho\"], geom = \"density\")\n\n\nhis rho parameter badly wants to be higher.\n\nI've tried widening the priors of sr.rho (I computed gamma as (1 / (1000 rho)) instead of (1 / (100 * rho)), and tau as (exp(100 * dr) * gamma) instead of (exp(100 * dr) * gamma)) and running fit.all again. Surprisingly, although this did indeed solve the problem of, e.g., the posterior of rho for s0029 butting up against the end, prediction in the test set was generally worse than before. So so much for that.\n\n## Comparing train-to-test fits with test-to-test fits\n\nThe goal of this study has been to fit several models with a training procedure that's as ideal as possible, then use the fits thus obtained to predict performance in a separate, model-neutral test set. Perhaps we can illustrate the strengths of this approach by comparing the conclusions we get this way to conclusions we'd get from just fitting all the models directly to the test set, which is much more like the usual practice in psychology and behavioral economics. For ease of implementation, at least to begin with, let's do this test-to-test procedure the Bayesian way (albeit, again, fitting subjects indpendently, without hierarchal models) rather than using a more common technique like MLE.\n\nfitall.train2test = cached(fit.all.train2test(good.s))\nfitall.test2test = cached(fit.all.test2test(good.s))\n\n\n(When I last ran the above, on 15 Jul 2013, there was only one bailout, for s0237 in train2test, who was assigned ghmk.rho.)\n\nFirst, here's the mean number of correct predictions in the test set for the usual (train-to-test) procedure. The way to interpret these tables is that each row gives you the bootstrap confidence that the mean of m1 is greater than the mean of m2.\n\nboot.correctness(fitall.train2test)\n\nm1 m1.mean m2 m2.mean confidence\n1 expk.rho 0.859 ghmk.rho 0.840 0.88\n2 ghmk.rho 0.840 sr.rho 0.809 0.96\n3 sr.rho 0.809 diff.rho 0.742 1.00\n\nSo sr.rho made more correct predictions than diff.rho, the discounting models made more correct predictions than sr.rho and diff.rho, and ghmk.rho didn't make more correct predictions than expk.rho.\n\nboot.correctness(fitall.test2test)\n\nm1 m1.mean m2 m2.mean confidence\n1 ghmk.rho 0.868 sr.rho 0.861 0.80\n2 sr.rho 0.861 expk.rho 0.858 0.68\n3 expk.rho 0.858 diff.rho 0.855 0.63\n\nCompared to the train-to-test procedure, the means are more homogeneous (confidences for the ranking are lower). Generally, they've increased, but some increases have been more dramatic.\n\ndo.call(rbind, lapply.names(models.by.name, function (m)\n{train2test = fitall.train2test[[m]]$total test2test = fitall.test2test[[m]]$total\ndata.frame(train2test, test2test,\ndiff = test2test - train2test)}))\n\ntrain2test test2test diff\nexpk.rho 0.859 0.858 -0.001\nghmk.rho 0.840 0.868 0.028\ndiff.rho 0.742 0.855 0.113\nsr.rho 0.809 0.861 0.052\n\nNotice how the better each model did in the train-to-test procedure, the smaller the improvement in its performance by letting it peek at the test set. This is nice. It shows how the train-to-test procedure, by protecting against overfitting, did a better job than the test-to-test procedure of distinguishing among the models. It also implies that expk.rho is less readily overfit than the other models.\n\nNow let's look at the RMSE of all the predictions. For each prediction, we calculate the error as p - t, where p is the predicted probability of choosing the larger-later option (LL), and t is 1 if the the actual choice was LL and 0 if it was SS.\n\nboot.rmse(fitall.train2test)\n\nm1 m1.mean m2 m2.mean confidence\n1 diff.rho 0.420 sr.rho 0.369 0.99\n2 sr.rho 0.369 ghmk.rho 0.331 0.99\n3 ghmk.rho 0.331 expk.rho 0.329 0.55\n\nThis is essentially the same ranking as for the correctness scores (except it's backwards because smaller means are better this time). Hooray, consistency.\n\nboot.rmse(fitall.test2test)\n\nm1 m1.mean m2 m2.mean confidence\n1 diff.rho 0.312 expk.rho 0.308 0.66\n2 expk.rho 0.308 sr.rho 0.299 0.86\n3 sr.rho 0.299 ghmk.rho 0.295 0.67\n\nLikewise, this ranking is consistent with the one for test-to-test correctness scores.\n\ntmm.rmse = function (tmm)\nmean(daply(ss(tmm, !is.na(trial)), .(factor(s)), function(slice)\nsqrt(mean((slice$mean - (slice$actual == \"ll\"))^2))))\n\ndo.call(rbind, lapply.names(models.by.name, function (m)\n{train2test = tmm.rmse(fitall.train2test[[m]]$tmm) test2test = tmm.rmse(fitall.test2test[[m]]$tmm)\ndata.frame(\ntrain2test = round(train2test, 3),\ntest2test = round(test2test, 3),\ndiff = round(test2test - train2test, 3))}))\n\ntrain2test test2test diff\nexpk.rho 0.329 0.308 -0.021\nghmk.rho 0.331 0.295 -0.036\ndiff.rho 0.420 0.312 -0.108\nsr.rho 0.369 0.299 -0.070\n\nWe have largely the same pattern of differences as for correctness rates.\n\nHere are plots of the v30 parameter for expk.rho with the test-to-test procedure and train-to-test procedure. It looks like the 95% credible intervals are generally narrower for train-to-test, perhaps because of the adaptive procedure.\n\ntmm.param.plot(fitall.train2test$expk.rho$tmm, \"v30\")\n\ntmm.param.plot(fitall.test2test$expk.rho$tmm, \"v30\")\n\n\nAnd now for some nigh-illegible plots of every prediction.\n\nshow.all.predictions(fitall.train2test)\n\nshow.all.predictions(fitall.test2test)\n\n\nThe models look more different from each other under train-to-test, as expected. By the way, it looks like Christian was right about people not having a good sense of what 8 weeks is like, judging from the yellow horizontal stripe in every panel.\n\n### Comparing parameter estimates for expk.rho and ghmk.rho\n\nFor the exponential model and the GHM, let's see how parameter estimates differed between test-to-test and train-to-test. In these figures, for legibility, only the 95% credible intervals are shown.\n\nttvtt = function(model, param)\ntmms.param.plot(param.name = param, rbind(\ntransform(fitall.train2test[[model]]$tmm, group = \"train2test\"), transform(fitall.test2test[[model]]$tmm,\ngroup = \"test2test\")))\n\nttvtt(\"expk.rho\", \"v30\") +\nscale_y_continuous(limits = c(0, 1), expand = c(0, 0))\n\nttvtt(\"ghmk.rho\", \"v30\") +\nscale_y_continuous(limits = c(0, 1), expand = c(0, 0))\n\n\nFor v30, it looks like train-to-test produced tighter intervals than test-to-test. But the intervals for ghmk are wider than the intervals for expk. Also, it looks like the distribution of v30s is more bimodal for expk than ghmk.\n\nttvtt(\"expk.rho\", \"rho\") + scale_rho\n\nttvtt(\"ghmk.rho\", \"rho\") + scale_rho\n\n\nFor rho, it looks like train-to-test predicted much larger individual differences than test-to-test, and this effect is greater for expk than ghmk. Perhaps as a result, there's greater agreement between the procedures for ghmk.\n\nSo one general impression I'm getting is that expk is more variable between subjects than ghmk. I guess that's a weak argument that expk is better, if we take for granted that individual differences in temporal discounting are large.\n\nLastly, for laughs, here's a plot for ghmk's curvature parameter.\n\nttvtt(\"ghmk.rho\", \"scurve\") + scale_rho\n\n\nHere I guess the chief thing to observe is that there was a lot of uncertainty about curvature, with perhaps a bit more for train-to-test. Notice that most intervals contain .1 (hyperbolic discounting) and a number less than .13 (0 is exponential discounting).\n\n### Parameter correlations\n\nReading Peters, Miedl, and Büchel (2012) gave me the idea of checking correlations between parameters. Let's look at the between-subject correlations of posterior means. We'll use Spearman correlations to avoid scale issues.\n\nround(digits = 3,\ntmm.param.means.cor(fitall.train2test$expk.rho$tmm))\n\nvalue\n0.646\n\nSo more patient people were noisier. That's an annoyingly strong correlation, actually.\n\nround(digits = 3,\ntmm.param.means.cor(fitall.train2test$ghmk.rho$tmm))\n\nscurve rho\nv30 0.752 0.509\nscurve 0.235\n\nThe correlation of v30 and rho is still here, if somewhat lesser. The correlation of v30 and scurve is substantial: more patient people had flatter discount functions.\n\n## Test-to-test AIC\n\naic.test2test = cached(aics.by.model(\nss(ts, s %in% good.s & type == \"test\"),\nfitall.test2test))\n\nround(digits = 2,\nboot.ranks(aic, model, data = aic.test2test, n = 5e4))\n\nsr.rho expk.rho diff.rho\nghmk.rho 0.61 0.77 0.95\nsr.rho 0.68 0.91\nexpk.rho 0.82\n\n## Paper thinking\n\n### Potential paper outline\n\n• Introduce the idea of intertemporal choice\n• Discuss past controversies on what models are appropriate\n• Titration or matching procedures in preference to shuffled choices\n• Test-set deficiencies\n• Small size\n• Ignoring front-end delays\n• Using a very limited assortment of durations (e.g., just 15 days and 30 days)\n• Using weird durations that people don't have a feel for (e.g., 17 days)\n• Training-set deficiencies: a given set of training data may be more helpful for one model than another\n• The danger of overfitting that arises from training and testing with the same data\n• No explicit modeling of noise\n• Using point estimates of parameters rather than marginalizing over all uncertainty in Bayesian fashion\n• Introduce the four specific models tested in this study\n• Present results of the test-to-test procedure (after explaining just enough of the study design so that this can be understood)\n• Present results of the train-to-test procedure (after explaining the rest of the study design), contrasting them with the test-to-test results\n• Toubia, Johnson, Evgeniou, and Delquié (2013)\n\n### Completeness vs. accuracy\n\nWhat we'd really like is a complete model of intertemporal choice. That's why observation of anomalies motivates switching to weirder models. But we can't get a complete model soon. Such a model would need to incorporate essentially everything that can influence a judgment or decision, and practically anything can:\n\n• Some particular amounts or delays might have special meaning to some people\n• If you're paid biweekly, you might compare a 2-week (but not 1-week or 3-week) reward to your paycheck\n• Perhaps you need $110 for a bill • Perhaps you think 40 is a lucky number • Choices in one trial can be influenced by choices in past trials • Someone might take an option because it's similar to options they took before • Someone might apply a deliberate rule (e.g., regarding pay rate). • Display format can influence choices •$1 versus 100 cents\n• 7 days versus 1 week\n• Left side versus right side\n• Non-fluent options may be rounded in different ways by different subjects\n• Someone might take an option because it's easy to conceive of (fluency)\n• Physiological state or other apparent influences on patience (e.g., intelligence)\n• We might hope that the effect of these things can be described with the same parameters useful for describing individual differences, but that's not guaranteed.\n\nSince completeness is out of reach, at least for the foreseeable future, it seems reasonable to ask, rather than which model is most complete, which model is most accurate overall. That is, which model can actually do the best job of predicting people's choices?\n\n(Besides, what's the point of a complete but inaccurate model?)\n\nA good answer to this question could be of great practical value. If there's any real connection between what intertemporal-choice tasks measure and the real-life DVs mentioned earlier, models that do a better job of predicting lab behavior should also do a better job of predicting real-life DVs.\n\nSo I set out to do a model-comparison study. There have been a lot of such studies, but mine differs because of the focus on predictive accuracy. (Also, the model set is unusually diverse.)\n\n## Within-subject consistency\n\nReviewer 2 for the first submission to JBDM suggested \"including the same binary choices twice, and checking for consistency\". The test set, by construction, had no duplicates, but we can look for duplicates in the training sets.\n\ndodge(factor(model), unique.train, data = sb[good.s,]) + boxp +\ncoord_cartesian(ylim = c(10, 50))\n\n\nIt appears that duplicates are actually very common. In fact, nobody saw 50 unique trials. Let's see how often people chose consistently across duplicates.\n\ndupes = ddply(ss(ts, s %in% good.s),\n.(s, q = paste(ssr, ssd, llr, lld)),\nfunction(slice)\nif (nrow(slice) == 1)\nNULL\nelse\nc(n = nrow(slice), chose.ll = sum(slice$choice == \"ll\"))) Note that dupes, unlike the plot above, includes the test set as well; although the test set doesn't contain any duplicates internally, the training and test sets may share quartets. table(dupes$n)\n\ncount\n2 1041\n3 303\n4 138\n5 44\n\nWe see that most duplicates appeared twice or thrice, although a significant minority appeared four or five times. (We're treating the same quartet as different when different subjects saw it.)\n\nround(digits = 3, c(\n\"all\" = mean(with(dupes, (chose.ll/n) %in% c(0, 1))),\n\"2 or 3 reps\" = mean(with(ss(dupes, n %in% c(2, 3)), (chose.ll/n) %in% c(0, 1))),\n\"2 reps only\" = mean(with(ss(dupes, n == 2), (chose.ll/n) %in% c(0, 1)))))\n\nvalue\nall 0.676\n2 or 3 reps 0.673\n2 reps only 0.695\n\nThese are the proportion of quartets for which people were perfectly consistent. I would characterize these figures as… unimpressive. To get a sense of how unimpressive, we can try something like a significance test.\n\nround(digits = 3,\nqbinom(.999999, size = nrow(ss(dupes, n == 2)), prob = .5)/nrow(ss(dupes, n == 2)))\n\nvalue\n0.573\n\nOkay, given that that's the .999999 quantile of the relevant distribution, a boneheaded model of null responding is not, in fact, tenable. That's comforting, at least.\n\nNow, reviewer 2 was (rightly) particularly concerned about differental random responding, or effects of random responding, between the model conditions: \"If the procedure introduces more random or heuristic responding into the dataset, how this would affect the relative predictive accuracy of the models under consideration?\" Let's first see how consistency compares between the model conditions.\n\nround(digits = 3, daply(dupes,\n.(model = factor(sb[char(s), \"model\"])),\nfunction(slice)\nmean((slice$chose.ll/slice$n) %in% c(0, 1))))\n\nvalue\ndiff.rho 0.727\nexpk.rho 0.688\nghmk.rho 0.664\nsr.rho 0.619\n\nThat's not great. I hoped all the figures would be about the same. I guess we have to acknowledge that it is possible the adaptive procedure induces different degrees of response noise for different models.\n\n## Test set-only validation\n\nReviewer 1 for JBDM wanted an analysis using half of the test set as training data and the other half as test data, and also for us to include a hyperbolic model. We can improve on this proposal by randomly splitting the data into training and test sets separately per subject.\n\nfitall.holdout = cached(fit.all.holdout(good.s,\nholdout.factor = 1/2,\nmodels = c(models.by.name, list(hypk.rho = model.hypk.rho))))\n\ndodge(Var2, value, data = melt(sapply(c(names(models.by.name), \"hypk.rho\"), function(m) fitall.holdout[[m]]$each.s$correct)), discrete = T) + boxp\n\nrx = simplify2array(lapply.names(fitall.holdout, function(m) daply(ss(fitall.holdout[[m]]$tmm, !is.na(trial)), .(factor(s)), function(slice) with(slice, sqrt(mean( (mean - (actual == \"ll\"))^2 )))))) dodge(Var2, value, data = melt(rx)) + boxp The correct-count and RMSE ratings shown here are for all models similar to the performance of Exp under train-to-test. So it seems that the effect of using the adaptive procedure for training, rather than more trials of the kind that appear the test set, is only to worsen the performance of Diff and SR, and with this obstacle removed, all the models perform about the same. Oops? ### Cross-validation Now let's try 5-fold cross-validation within the test set. fitall.xvalid = cached(fit.all.xvalid( good.s, folds = 5, models = c(models.by.name, list(hypk.rho = model.hypk.rho)), ts)) fitall.xvalid.correct = ss( melt(sapply(c(names(models.by.name), \"hypk.rho\"), function(m) with(fitall.xvalid[[m]], tapply((mean >= .5) == (actual == \"ll\"), s, sum)))), !is.na(value)) names(fitall.xvalid.correct) = qw(s, model, correct) dodge(model, correct, discrete = T, data = fitall.xvalid.correct) + scale_y_continuous(breaks = seq(50, 100, 5)) + boxp rx = simplify2array(lapply.names(fitall.xvalid, function(m) daply(fitall.xvalid[[m]], .(factor(s)), function(slice) with(slice, sqrt(mean( (mean - (actual == \"ll\"))^2 )))))) dodge(Var2, value, data = melt(rx)) + boxp These results for correct-count and RMSE look very similar to the results for one-round validation presented previously. How about within-subjects comparisons of these cross-validations? x = ss( data.frame(sapply(c(names(models.by.name), \"hypk.rho\"), function(m) with(fitall.xvalid[[m]], tapply((mean >= .5) == (actual == \"ll\"), s, sum)))), !is.na(expk.rho)) dodge(Var2, value, discrete = T, stack.width = 10, data = melt(ss(t(maprows(x, function(v) v - v[\"diff.rho\"])), select = -diff.rho))) + boxp + scale_y_continuous(name = \"Improvement over Diff\", breaks = seq(-10, 16, 2)) + xlab(\"Model\") x = simplify2array(lapply.names(fitall.xvalid, function(m) daply(fitall.xvalid[[m]], .(factor(s)), function(slice) with(slice, sqrt(mean( (mean - (actual == \"ll\"))^2 )))))) dodge(Var2, value, data = melt(ss(t(maprows(x, function(v) v[\"diff.rho\"] - v)), select = -diff.rho))) + boxp + scale_y_continuous(name = \"Improvement over Diff\") + xlab(\"Model\") Here is a table of for each model, the number of subjects whose lowest RMSE was for that model. sort(table(maprows(x, function(v) names(v)[which.min(v)])), dec = T) value sr.rho 70 diff.rho 42 expk.rho 28 hypk.rho 24 ghmk.rho 23 sr.rho (BT) is the clear winner by this metric. But here is a measure of within-subject variability of RMSE: the quantiles of the ranges of RMSE between the five models. round(d = 3, quantile( maprows(x, function(v) max(v) - min(v)), c(.025, .25, .5, .75, .975))) value 2.5% 0.008 25% 0.022 50% 0.038 75% 0.055 97.5% 0.105 This seems to imply that within-subject differences in model accuracy are generally small. ## Other learners ### Rationale We now seem to be in the surprising situation where all the models are performing about the same, with accuracy around 86% and RMSE around .325. There are all sorts of things we could look at in order to try to explain why all the models are performing similarly to each other and why cross-validated test error is hardly more than the training error. But we have little ability to distinguish plausible explanations from correct explanations without collecting new data for the purpose. Which is one of the problems with explanatory science, of course. Let's instead back up a bit and frame the question of this project (and the eventual paper) as \"What sorts of predictive procedures are good in this situation?\", where \"this situation\" is now prediction of the test set with (disjoint pieces of) itself. We'll investigate a few other predictive procedures, including very simple things like the majority model (analyzed predictively, not prophetically as in the first manuscript) and perhaps naive Bayes, and also heavier-duty machine-learning techniques like decision trees. Non-Bayesian applications of the existing models would also make sense. The goal is to identify what is the best predictive accuracy achievable (and with which procedures) and what is the simplest procedure one can use that still achieves respectable accuracy. We might also try changing the training-set size to see how it affects these matters. ### SVM tuning svm.explore = function(gamma, cost) do.call(rbind, lapply(good.s, function(the.s) {message(the.s); t = ss(ts, char(s) == the.s & type == \"test\"); if (any(table(t$choice) < 4)) return(NULL); iv = t[qw(ssr, ssd, llr, lld)]; for (col in seq_len(ncol(iv))) iv[,col] = zscore(iv[,col]); data.frame(s = the.s, tune.svm(x = iv, y = t$choice, scale = F, type = \"C-classification\", cost = cost, gamma = gamma)$performances[qw(gamma, cost, error)])}))\n\n# svm.optim = function(the.s)\n# {message(the.s);\n# t = ss(ts, char(s) == the.s & type == \"test\");\n# if (any(table(t$choice) < 4)) # return(NULL); # iv = t[qw(ssr, ssd, llr, lld)]; # for (col in seq_len(ncol(iv))) iv[,col] = zscore(iv[,col]) # dv = t$choice\n# o = optim(c(-13, 19),\n# fn = function(p)\n# tune.svm(x = iv, y = dv,\n# scale = F, type = \"C-classification\",\n# gamma = 2^p, cost = 2^p)$best.performance, # # method = \"BFGS\", # control = list(trace = 1, maxit = 25, reltol = 1e-2)) # o} svm.perfs = rbind( cached(do.call(rbind, lapply(good.s, function(the.s) {message(the.s); t = ss(ts, char(s) == the.s & type == \"test\"); if (any(table(t$choice) < 4)) return(NULL); iv = t[qw(ssr, ssd, llr, lld)]; for (col in seq_len(ncol(iv))) iv[,col] = zscore(iv[,col]); data.frame(s = the.s, tune.svm(x = iv, y = t$choice, scale = F, type = \"C-classification\", cost = 2^seq(-5, 15, 2), gamma = 2^seq(-15, 3, 2))$performances[qw(gamma, cost, error)])}))),\ncached(do.call(rbind, lapply(good.s, function(the.s) {message(the.s); t = ss(ts, char(s) == the.s & type == \"test\"); if (any(table(t$choice) < 4)) return(NULL); iv = t[qw(ssr, ssd, llr, lld)]; for (col in seq_len(ncol(iv))) iv[,col] = zscore(iv[,col]); data.frame(s = the.s, tune.svm(x = iv, y = t$choice, scale = F, type = \"C-classification\", cost = 2^seq(17, 21, 2), gamma = 2^seq(-15, 5, 2))$performances[qw(gamma, cost, error)])}))), cached(svm.explore(gamma = 2^c(-17, -19, -21), cost = 2^c(23, 25, 27))), cached(svm.explore(gamma = 2^c(-17, -19, -21), cost = 2^c(17, 19, 21))), cached(svm.explore(gamma = 2^c(-15, -13, -11), cost = 2^c(23, 25, 27)))) svm.perfs = transform(svm.perfs, gamma = log(gamma, 2), cost = log(cost, 2)) ggplot(aggregate(excess ~ gamma + cost, ddply(svm.perfs, .(s), function(slice) transform(slice, excess = error - min(error))), function (v) sum(v^2))) + geom_raster(aes(gamma, cost, fill = excess)) + scale_fill_gradient(low = \"gray80\", high = \"black\") + geom_text(aes(gamma, cost, label = round(100*excess))) + geom_point(aes(x, y), data = expand.grid(x = seq(-25, 5, len = 5), y = seq(-5, 25, len = 5)), color = \"blue\") I don't really want to \"overfit\" my SVM tuning to the results here, not least because the surface could look quite different when there's lots of noise or the training set is small. And I guess I want to stick to a grid search, because that's standard practice, and designing an effective yet fast method of adaptively tuning SVM hyperparameters is probably not worth my time for this project. So I'll do a grid search on the blue points. ### Comparison of robustness We will repeat the cross-validation analyses from before, except that in each fold, we'll randomly tamper with a fixed portion of the training set. For the choice we want to corrupt, we'll choose at random whether to replace it with SS or LL rather than always changing it to its reverse (which I suspect is too harsh to more complex models). correct.by.integrity = cached(metaparam.xvalid( corrupt.training.data.mxw, good.s, c(90, 70, 60, 50, 30, 15, 10))) correct.by.train.size.plot(T, aggregate( correct ~ learner + train.size, correct.by.integrity, median)) + coord_cartesian(xlim = c(10, 90), ylim = c(50, 90)) + xlab(\"Uncorrupted training trials\") + ylab(\"Median correct predictions\") Here diff.glm seems to be the clear winner. Although its performance deteriorates as the training set is corrupted, it tends to do much better than other models facing the same amount of noise. correct.by.train.size.plot(T, aggregate( correct ~ learner + train.size, correct.by.integrity, function (v) quantile(v, .1))) + coord_cartesian(xlim = c(10, 90), ylim = c(40, 80)) + xlab(\"Uncorrupted training trials\") + ylab(\"First decile of correct predictions\") diff.glm isn't as consistently good when we look at worse cases, the first decile of performance. Here there's a less clear winner, although sr.mle and hyp.mle are good across-the-board performers, and diff.glm only gets bad when train.size = 5. ### Comparison of training-set sizes correct.by.train.size = cached(metaparam.xvalid( shrink.training.data.mxw, good.s, c(90, 70, 60, 50, 30, 15, 10))) correct.by.train.size.plot(T, aggregate( correct ~ learner + train.size, correct.by.train.size, median)) + coord_cartesian(xlim = c(10, 90), ylim = c(50, 90)) + ylab(\"Median correct predictions\") sr.mle is slightly better than diff.glm and hyp.mle, which are about evenly matched. correct.by.train.size.plot(T, aggregate( correct ~ learner + train.size, correct.by.train.size, function (v) quantile(v, .1))) + coord_cartesian(xlim = c(10, 90), ylim = c(40, 80)) + ylab(\"First decile of correct predictions\") Now diff.glm isn't doing so hot. sr.mle and hyp.mle continue to do well. ## New analyses for BJDSM submission 2 ### Differences between MTurk and lab dodge(ifelse(is.na(worker), \"Lab\", \"MTurk\"), test.lls, data = sb[good.s,]) + boxp Looks like MTurk subjects are more patient. correct.lab = cached(metaparam.xvalid( shrink.training.data.mxw, row.names(ss(sb, good & is.na(worker))), 90)) correct.mturk = cached(metaparam.xvalid( shrink.training.data.mxw, row.names(ss(sb, good & !is.na(worker))), 90)) ggplot(prettify.lnames(ss(correct.lab, train.size == 90))) + stat_summary(aes(learner, correct), geom = \"crossbar\", fun.data = function(y) names<-(quantile(y, c(.25, .5, .75)), qw(ymin, y, ymax))) + stat_summary(aes(learner, correct), geom = \"point\", fun.y = function(y) quantile(y, .1), shape = \"X\", size = 5) + stat_summary(aes(learner, correct), geom = \"point\", fun.y = mean, color = \"red\", size = 3) + scale_y_continuous(breaks = seq(0, 100, 2), minor_breaks = NULL) + coord_cartesian(ylim = c(50, 92)) + xlab(\"Model\") + ylab(\"Accuracy\") + theme( panel.grid.major.x = element_blank(), panel.grid.major.y = element_line(colour = \"gray50\", size = 0.07)) ggplot(prettify.lnames(ss(correct.mturk, train.size == 90))) + stat_summary(aes(learner, correct), geom = \"crossbar\", fun.data = function(y) names<-(quantile(y, c(.25, .5, .75)), qw(ymin, y, ymax))) + stat_summary(aes(learner, correct), geom = \"point\", fun.y = function(y) quantile(y, .1), shape = \"X\", size = 5) + stat_summary(aes(learner, correct), geom = \"point\", fun.y = mean, color = \"red\", size = 3) + scale_y_continuous(breaks = seq(0, 100, 2), minor_breaks = NULL) + coord_cartesian(ylim = c(50, 92)) + xlab(\"Model\") + ylab(\"Accuracy\") + theme( panel.grid.major.x = element_blank(), panel.grid.major.y = element_line(colour = \"gray50\", size = 0.07)) These are both comparable to the original figure. ### Prediction overlap indpred = cached(ordf(individual.xvalid(good.s), learner, s, qn)) indpred$predll = indpred$pred == \"ll\" indpred.ary = acast(indpred, learner ~ qn ~ s, value.var = \"predll\") overlap.plot = function(focal.lname) {focus = indpred.ary[focal.lname,,] d = prettify.lnames(do.call(rbind, lapply(setdiff(names(learners), focal.lname), function(comparison.lname) data.frame( learner = comparison.lname, qn = 1 : nrow(focus), agree = rowMeans(focus == indpred.ary[comparison.lname,,]))))) ggplot(d) + ggtitle(paste(\"Agreement with predictions of\", names(pretty.learner.names)[pretty.learner.names == focal.lname])) + geom_tile(aes(learner, qn, fill = agree)) + scale_fill_gradient2(name = \"Agreement\", limits = c(0, 1), midpoint = .5, low = \"#ff7f00\", high = \"blue\") + scale_y_continuous(name = \"Quartet\", expand = c(0, 0), breaks = c(1, 25, 50, 75, 100)) + scale_x_discrete(name = \"Comparison model\", expand = c(0, 0))} overlap.plot(\"majority\") overlap.plot(\"knn\") overlap.plot(\"diff.glm\") overlap.plot(\"exp.mle\") overlap.plot(\"hyp.mle\") overlap.plot(\"sr.mle\") overlap.plot(\"ghm.mle\") overlap.plot(\"full_logit\") overlap.plot(\"rf\") overlap.plot(\"svm\") ### Saturated logit model xvalid.saturated = cached(individual.xvalid(good.s, lnames = \"saturated_logit\", extra.learners = list(saturated_logit = extra.learner.saturated_logit))) ggplot(data.frame(correct = daply(xvalid.saturated, .(factor(s)), function(x) sum(x$choice == x\\$pred)))) +\nstat_summary(aes(1, y = correct), geom = \"crossbar\",\nfun.data = function(y)\nnames<-(quantile(y, c(.25, .5, .75)), qw(ymin, y, ymax))) +\nstat_summary(aes(1, y = correct), geom = \"point\",\nfun.y = function(y) quantile(y, .1),\nshape = \"X\", size = 5) +\nstat_summary(aes(1, y = correct), geom = \"point\",\nfun.y = mean,\ncolor = \"red\", size = 3) +\nscale_y_continuous(breaks = seq(0, 100, 2), minor_breaks = NULL) +\nscale_x_continuous(name = \"\", breaks = c()) +\ncoord_cartesian(ylim = c(50, 92)) +\nylab(\"Accuracy\") +\ntheme(\npanel.grid.major.x = element_blank(),\npanel.grid.major.y = element_line(colour = \"gray50\", size = 0.07))\n\n\n## References\n\nBurks, S. V., Carpenter, J. P., Goette, L., & Rustichini, A. (2009). Cognitive skills affect economic preferences, strategic behavior, and job attachment. Proceedings of the National Academy of Sciences, 106(19), 7745–7750. doi:10.1073/pnas.0812360106\n\nChabris, C. F., Laibson, D., Morris, C. L., Schuldt, J. P., & Taubinsky, D. (2008). Individual laboratory-measured discount rates predict field behavior. Journal of Risk and Uncertainty, 37(2, 3), 237–269. doi:10.1007/s11166-008-9053-x\n\nDemurie, E., Roeyers, H., Baeyens, D., & Sonuga‐Barke, E. (2012). Temporal discounting of monetary rewards in children and adolescents with ADHD and autism spectrum disorders. Developmental Science, 15(6), 791–800. doi:10.1111/j.1467-7687.2012.01178.x\n\nKhurana, A., Romer, D., Betancourt, L. M., Brodsky, N. L., Giannetta, J. M., & Hurt, H. (2012). Early adolescent sexual debut: The mediating role of working memory ability, sensation seeking, and impulsivity. Developmental Psychology, 48(5), 1416–1428. doi:10.1037/a0027491\n\nLuhmann, C. C. (2013). Discounting of delayed rewards is not hyperbolic. Journal of Experimental Psychology: Learning, Memory, and Cognition, 39(4), 1274–1279. doi:10.1037/a0031170\n\nMadden, G. J., Petry, N. M., Badger, G. J., & Bickel, W. K. (1997). Impulsive and self-control choices in opioid-dependent patients and non-drug-using control participants: Drug and monetary rewards. Experimental and Clinical Psychopharmacology, 5(3), 256–262. doi:10.1037/1064-1297.5.3.256\n\nMcKerchar, T. L., & Renda, C. R. (2012). Delay and probability discounting in humans: An overview. Psychological Record, 62(4), 817–834.\n\nMeier, S., & Sprenger, C. D. (2010). Present-biased preferences and credit card borrowing. American Economic Journal: Applied Economics, 2(1), 193–210. doi:10.1257/app.2.1.193\n\nPeters, J., Miedl, S. F., & Büchel, C. (2012). Formal comparison of dual-parameter temporal discounting models in controls and pathological gamblers. PLOS ONE. doi:10.1371/journal.pone.0047225" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8515669,"math_prob":0.97945017,"size":56879,"snap":"2022-27-2022-33","text_gpt3_token_len":15753,"char_repetition_ratio":0.116043955,"word_repetition_ratio":0.074611396,"special_character_ratio":0.28477997,"punctuation_ratio":0.18592331,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9934604,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T09:30:53Z\",\"WARC-Record-ID\":\"<urn:uuid:1ca7bf1b-340c-4f7c-a283-d8617712e555>\",\"Content-Length\":\"118469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8c8dc7b-d21c-4275-960d-e00fd9209a80>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9370e2e-00a6-4924-b82c-14f44eac3e6d>\",\"WARC-IP-Address\":\"71.19.145.174\",\"WARC-Target-URI\":\"http://arfer.net/projects/builder/notebook\",\"WARC-Payload-Digest\":\"sha1:WFNUDJBAQG2YICD2EBEYXAROUAVVBPSP\",\"WARC-Block-Digest\":\"sha1:M2IBM5EH2PRG5C7F7KUOIOAZ4EHAJKRP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104364750.74_warc_CC-MAIN-20220704080332-20220704110332-00053.warc.gz\"}"}
http://atcoder.noip.space/contest/abc043/a	[ "# Home\n\nScore : $100$ points\n\n### Problem Statement\n\nThere are $N$ children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give $1$ candy to the first child in the line, $2$ candies to the second child, ..., $N$ candies to the $N$-th child. How many candies will be necessary in total?\n\n### Constraints\n\n• $1≦N≦100$\n\n### Input\n\nThe input is given from Standard Input in the following format:\n\n$N$\n\n\n### Output\n\nPrint the necessary number of candies in total.\n\n### Sample Input 1\n\n3\n\n\n### Sample Output 1\n\n6\n\n\nThe answer is $1+2+3=6$.\n\n### Sample Input 2\n\n10\n\n\n### Sample Output 2\n\n55\n\n\nThe sum of the integers from $1$ to $10$ is $55$.\n\n### Sample Input 3\n\n1\n\n\n### Sample Output 3\n\n1\n\n\nOnly one child. The answer is $1$ in this case." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8735187,"math_prob":0.9946242,"size":534,"snap":"2021-43-2021-49","text_gpt3_token_len":158,"char_repetition_ratio":0.14339623,"word_repetition_ratio":0.0,"special_character_ratio":0.32022473,"punctuation_ratio":0.14634146,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973427,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T12:28:58Z\",\"WARC-Record-ID\":\"<urn:uuid:4f3c7835-86d4-4300-95ea-cefdc567158f>\",\"Content-Length\":\"3860\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f46d369-b88a-49d6-97a2-01754dde2fec>\",\"WARC-Concurrent-To\":\"<urn:uuid:0df78334-04f8-4815-8635-0b4f5a74c5d5>\",\"WARC-IP-Address\":\"43.128.40.193\",\"WARC-Target-URI\":\"http://atcoder.noip.space/contest/abc043/a\",\"WARC-Payload-Digest\":\"sha1:HB3X3LQZNCZD53MVIAQYILSBMTOWCMFZ\",\"WARC-Block-Digest\":\"sha1:V4IMA4HBLV7XRT6I22KZGFYT5YRCI62X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358180.42_warc_CC-MAIN-20211127103444-20211127133444-00572.warc.gz\"}"}
http://lists.suckless.org/hackers/0109/13234.html	[ "# [PATCH 1/5] ed: remove infinite loops in join() and getindex()\n\nFrom: Thomas Mannay <audiobarrier_AT_openmailbox.org>\nDate: Sun, 9 Oct 2016 23:10:20 +0000\n\n```---\ned.c \| 8 ++++++--\n1 file changed, 6 insertions(+), 2 deletions(-)\ndiff --git a/ed.c b/ed.c\nindex 184ed30..f552234 100644\n--- a/ed.c\n+++ b/ed.c\n_AT_@ -192,7 +192,9 @@ getindex(int line)\nstruct hline lp;\nint n;\n\n-\tfor (n = 0, lp = zero; n != line; ++n)\n+\tif (line == -1)\n+\t\tline = 0;\n+\tfor (n = 0, lp = zero; n != line; n++)\nlp = zero + lp->next;\n\nreturn lp - zero;\n_AT_@ -806,9 +808,11 @@ join(void)\nstatic char s;\n\nfree(s);\n-\tfor (s = NULL, i = line1; i <= line2; i = nextln(i)) {\n+\tfor (s = NULL, i = line1;; i = nextln(i)) {\nfor (t = gettxt(i); (c = t) != '\\n'; ++t)\ns = addchar(t, s, &cap, &len);\n+\t\tif (i == line2)\n+\t\t\tbreak;\n}\n\ns = addchar('\\n', s, &cap, &len);\n--\n2.10.0\n--Multipart=_Sun__9_Oct_2016_23_20_20_+0000_Tw0GY2bnCixoZsD5\nContent-Type: text/x-patch;" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5242937,"math_prob":0.9954465,"size":1077,"snap":"2019-51-2020-05","text_gpt3_token_len":426,"char_repetition_ratio":0.09040075,"word_repetition_ratio":0.093023255,"special_character_ratio":0.48653668,"punctuation_ratio":0.24505928,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9788389,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T07:23:38Z\",\"WARC-Record-ID\":\"<urn:uuid:cbf8d5c8-7260-40f6-8b3a-ec77c85364dc>\",\"Content-Length\":\"6411\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:816091de-4468-437d-a443-569bf152295c>\",\"WARC-Concurrent-To\":\"<urn:uuid:be56f722-7ff9-4367-a5c8-a8277b5e0d35>\",\"WARC-IP-Address\":\"195.201.112.86\",\"WARC-Target-URI\":\"http://lists.suckless.org/hackers/0109/13234.html\",\"WARC-Payload-Digest\":\"sha1:Q4NS346EOHUFJTRPHPPXTGIG2HQUXASZ\",\"WARC-Block-Digest\":\"sha1:WUG4KVX35CPJ65PBR2HS3THVRCQLCF7D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540518337.65_warc_CC-MAIN-20191209065626-20191209093626-00359.warc.gz\"}"}
https://forum.allaboutcircuits.com/threads/15-led-parallel-circuit.64032/#post-437419	[ "# 15 LED parallel circuit\n\n#### 1ntegr0\n\nJoined Dec 29, 2011\n2\nProbably a stupid question, but I'm not sure so I have to ask\n\nI have 15 LEDs (TSAL6400 TR Diode @ 100 mA) in parallel, divided into 2 groups. First group is 4 LEDs, second group is 11 LEDs. The first group is always on, the second group can be switched on and off by a switch. Wiring is all done, no problems there.\n\nThe problem is my power source. Previously, I used a special device that allowed me to send exactly 100 mA through the entire system, so no resistors were needed. Now, I can no longer access that device so I need to use a 9V battery (the square one). This also means that I have to use resistors now.\n\nI read somewhere that it is necessary for parallel circuits to use a different resistor for EACH LED (is this correct?). I also read that for parallel circuits Req = 1/R1 + 1/R2 ... 1/Rn but I'm not sure how to interpret this.\n\nBy doing R = U / I I find that R = 90 Ω. Does this mean that I need to give every LED a 90 Ω resistor, or do I have to devide 90 by 15? If so, wouldn't that give problems with my switch? Because if the switch is off, only 4 LEDs are powered so I would have to divide 90 by 4 instead of 15..\n\nJoined Dec 26, 2010\n2,148\nIf you read up on basic series and parallel circuits this should all become obvious to you. It isn't rocket science, as they say.\n\nUntil then, here are a few pointers.\n\n1. A separate resistor for each LED is very desirable. It is possible to use a single resistor for a group, but the brightness of the LEDs may be less consistent, and the reliability will be poorer.\n2. The resistor for each LED or group of LEDs must be calculated for the current and voltage concerned. Why in the name of... would anyone imagine that a single LED would require the same series resistor as a group of several in parallel?\n3. If you have two parallel groups switched separately, the minimum requirement would be a separate series resistor for each group. This will avoid a current change when the second set turns on.\n4. The resistor value needed for a single LED or parallel group is (Vs-Vf)/I, where Vs is the supply voltage, Vf is the LED forward voltage, and I is the total current for the number of LEDs concerned. Therefore, if a single LED requires R ohms, a group of ten LEDs in parallel would require R/10 ohms.\n5. Running large parallel groups of a little 9V battery is a very bad plan. You won't get 100mA for long at all - maybe a few hours. Better results might be obtained with the LEDs rearranged in series pairs [with less series resistance: R=(Vs-2Vf)/I ].\n6. Better still, use a battery made out of AA cells - depending on the LED voltage you may be able to use three or four cells to run LEDS - not in series but singly.\n\nJoined Dec 26, 2010\n2,148\nGood grief, these things are rated at 100mA (max) each ! http://www.vishay.com/docs/81011/tsal6400.pdf\n\nThere is no way that you could run 15 of these contraptions off a small 9V battery. At full power, a parallel group of 15 would draw 1.5 amps. Even putting them into series strings of two or three would still need more juice than a battery like that could provide.\n\nEdit: You might well choose to run each LED at less than 100mA, especially if you use the riskier option of not having individual resistors, but this still looks way beyond what this sort of battery can handle. http://en.wikipedia.org/wiki/Nine-volt_battery\n\nWhat sort of battery did you have in mind, and how long do you need this light to run for?\n\nLast edited:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9706216,"math_prob":0.7641602,"size":2207,"snap":"2022-27-2022-33","text_gpt3_token_len":586,"char_repetition_ratio":0.10894235,"word_repetition_ratio":0.96230596,"special_character_ratio":0.28092432,"punctuation_ratio":0.11660079,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96746355,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T16:14:21Z\",\"WARC-Record-ID\":\"<urn:uuid:ef8a5afc-99c8-45d7-ab00-787b3bd48a2b>\",\"Content-Length\":\"101012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e1387d0-5357-474e-a7b0-9bb126d5d708>\",\"WARC-Concurrent-To\":\"<urn:uuid:82af523d-49ed-4b38-980d-293aaa736c78>\",\"WARC-IP-Address\":\"104.20.234.39\",\"WARC-Target-URI\":\"https://forum.allaboutcircuits.com/threads/15-led-parallel-circuit.64032/#post-437419\",\"WARC-Payload-Digest\":\"sha1:MZVRLJP2DO5ZISTT3MGENHY5ERSJWWPJ\",\"WARC-Block-Digest\":\"sha1:ERASNCW3DUCJUALNOBECZ5WBHZM27WI6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104585887.84_warc_CC-MAIN-20220705144321-20220705174321-00155.warc.gz\"}"}
https://research.nu.edu.kz/en/publications/on-the-singularity-analysis-of-intersecting-separatrices-in-near-	[ "# On the singularity analysis of intersecting separatrices in near-integrable dynamical systems\n\nTassos Bountis, V. Papageorgiou, M. Bier\n\nResearch output: Contribution to journalArticlepeer-review\n\n19 Citations (Scopus)\n\n## Abstract\n\nIt has been recently proved by S. L. Ziglin that transversal intersections of separatrices (invariant manifolds) in near-integrable Hamiltonian systems of two degrees of freedom imply the existence of multi-valued solutions with infinitely many Riemann sheets. Ziglin's theorem is illustrated here on a simple example and then extended and applied to non-Hamiltonian, analytic flows x ̇=f(x) + εg(x,t), with x≡(x,y) and g(x,t)=g(x,t+2π), which for ε = 0 possess the Painlevé property. On the other hand, the theoretically expected logarithmic singularities for ε ≠ 0 are obtained explicitly in solutions near the intersecting separatrices. Thus, we conjecture that dynamical systems with the Painlevé property, can have no separatrix intersections and hence no strange attractors, etc. These singularities are then numerically located and found to form certain very interesting \"chimney\" patterns in the complex t-plane, on which they accumulate densely. The upper part of the chimneys (away from the Re t axis) is asymptotically quite insensitive to changes in parameter values or initial conditions. The singularity pattern itself, however, becomes \"denser\" and each chimney is seen to gradually \"collapse\" towards the Re t axis, as the amplitude of the driving term increases and the motion becomes more chaotic.\n\nOriginal language English 292-304 13 Physica D: Nonlinear Phenomena 24 1-3 https://doi.org/10.1016/0167-2789(87)90081-9 Published - 1987\n\n## ASJC Scopus subject areas\n\n• Statistical and Nonlinear Physics\n• Mathematical Physics\n• Condensed Matter Physics\n• Applied Mathematics" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87687343,"math_prob":0.5910889,"size":3578,"snap":"2021-21-2021-25","text_gpt3_token_len":943,"char_repetition_ratio":0.09233352,"word_repetition_ratio":0.7306968,"special_character_ratio":0.24455003,"punctuation_ratio":0.13323353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95089644,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T09:57:43Z\",\"WARC-Record-ID\":\"<urn:uuid:b69fa141-3c4f-4a93-a1c6-eaabf8b1ec07>\",\"Content-Length\":\"54599\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5357d11-5b06-4418-83dc-783eb60fdf8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:111ed89f-ede4-4ab0-a01a-7894de2773ba>\",\"WARC-IP-Address\":\"13.228.199.194\",\"WARC-Target-URI\":\"https://research.nu.edu.kz/en/publications/on-the-singularity-analysis-of-intersecting-separatrices-in-near-\",\"WARC-Payload-Digest\":\"sha1:2GDCFLFTXWTMABBTPAXXQ3TOJNSOT642\",\"WARC-Block-Digest\":\"sha1:ZUMTBU2L2BRLSJKN3SUPAYRG772ITL6L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988753.91_warc_CC-MAIN-20210506083716-20210506113716-00071.warc.gz\"}"}