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https://www.2cto.com/kf/201701/593268.html	[ "2017-01-25 10:24:02 来源：萌萌程序员的博客\n\n1：表单的定义\n\n<form>表单标签\n\n```<form action=\"url\" method=get/post name=value target=\"目标窗口\" enctype=\"编码方式\">\n</form>```\n\n1：action\n\n2:method\n\npost---表单数据与URL是分开发的\n\n3:enctype\n\ntext/plain ------- 以纯文本的形式传送\n\napplication --- 默认的编码形式\n\nmultipart/form-data --- MIME编码，上传文件的表单必须选择该项\n\n====================\n\n2：表单元素\n\n1=添加表单控件\n\n```<form>\n<input name=\"控件名称\" type=\"控件类型\">\n</form>```\n\ninput表单类别\n\ntype=\"submit\" 将表单内容提交给服务器的按钮\n\ntype=\"reset\" 将表单内容全部清除，重新填写的按钮\n\ntype=\"image\" 图像提交按钮\n\ntype=\"hidden\" 隐藏域，隐藏域不显示在页面上，但会将内容传递到服务器中\n\ntype=\"file\" 文件域\n\n<input 属性1 属性2....>\n\ninput标签的常用属性\n\ntype 控件的类型\n\nname 控件的名称\n\nalign 对齐方式\n\nsize 控件的宽度\n\nmaxlength 允许输入的最大字符数\n\nvalue 用于指定输入默认值\n\n```<form>\n<input type=\"text\" size=\"20\">\n</form>```\n\n3：文件域\n\n(input type=\"file\")\n\n```<form name=\"form\" method=\"post\" action=\"https://www.baidu.com\">\n请上传你的照片到我的邮箱:\n<input type=\"file\" name=\"userpic\">\n<input type=\"submit\" value=\"发送\">\n</form>```\n\n4：====================\n\n```<select name=\"name\" size=\"value\" multipe>\n<option value=\"value\" selected>选项1</option>\n<option value=\"value\" >选项2</option>\n<option value=\"value\" >选项3</option>\n</select>```\n\nsize --- 显示选项的数目，当size为1时为下拉列表框的控件\n\nmultiple ----列表中的项目多选，用户用Ctrl键来实现多选\n\nvalue ----选项值\n\nselected === 默认选项\n\n```<form>\n请选择您的工作性质:<br>\n<select name=\"selected\" id=\"selected\">\n<option value=\"value\" selected>选项1</option>\n<option value=\"value\" >选项2</option>\n<option value=\"value\" >选项3</option>\n</select>\n<br>\n<input type=\"submit\" name=\"button\" id=\"button\" value=\"提交\">\n</form>```\n\n5=======\n\n<textarea name=\"field_name\" value=\"filed_value\" rows=value cols=value></textarea>\n\n```<form>\n<h2>用户调查</h2>\n请留言<br>\n<textarea name=\"textfiled\" cols=\"50\" rows=\"20\" id=\"textfild\"></textarea><br>\n<input type=\"submit\" name=\"button\" value=\"提交\">\n</form>```" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.52186215,"math_prob":0.41631132,"size":1830,"snap":"2019-26-2019-30","text_gpt3_token_len":840,"char_repetition_ratio":0.18017524,"word_repetition_ratio":0.07692308,"special_character_ratio":0.30054644,"punctuation_ratio":0.04347826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95163864,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-16T20:27:16Z\",\"WARC-Record-ID\":\"<urn:uuid:11158eec-0bdf-4aad-a21c-d224c4f863b4>\",\"Content-Length\":\"24228\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bbe2f754-107c-42f4-a563-e7b902f253b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:7bc8b8f8-3c34-4342-8bc4-eaaa2bd9f181>\",\"WARC-IP-Address\":\"218.90.201.27\",\"WARC-Target-URI\":\"https://www.2cto.com/kf/201701/593268.html\",\"WARC-Payload-Digest\":\"sha1:NIXWZOTJKJNUQXHTXO6OSXJDZGLMLBGY\",\"WARC-Block-Digest\":\"sha1:IGUHOOAESOZ3D2ODI7P2AU2YZJIRWW2C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195524879.8_warc_CC-MAIN-20190716201412-20190716223412-00293.warc.gz\"}"}
http://hrstc.org/node/26	[ "", null, "HRSTC [HeuRiSTiC]\n\n# Visualizing MovieLens Rating Matrix\n\nIt is interesting to look at the visualization of the MovieLens Rating Matrix", null, "Here is a random sub sample", null, "Here is the Matlab's code:\n\n% Multi-character delimiters are not supported;\n% so we get columns with delimeters too\nD = importdata('ratings.dat',':');\n\n% load data into sparse matrix\nX = sparse(D(:,3),D(:,1),D(:,5));\n\n% plot the whole matrix\nimagesc(X'); figure(gcf)\n\n% plot random sampling of matrix\nr1_idx = randint(1,100,[1,6040]);\nr2_idx = randint(1,100,[1,3592]);\nR = X(r2_idx,r1_idx);\nimagesc(R'); figure(gcf)" ]	[ null, "http://hrstc.org/themes/pushbutton/logo.png", null, "http://hrstc.org/sites/default/files/movielens_rating_matrix.png", null, "http://hrstc.org/sites/default/files/movielens_rating_matrix_sample.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71479553,"math_prob":0.99947506,"size":529,"snap":"2021-43-2021-49","text_gpt3_token_len":151,"char_repetition_ratio":0.0952381,"word_repetition_ratio":0.0,"special_character_ratio":0.3383743,"punctuation_ratio":0.22881356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988258,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T09:17:35Z\",\"WARC-Record-ID\":\"<urn:uuid:1752b94a-0df1-4283-8f37-6197c6060c7a>\",\"Content-Length\":\"6633\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f65f408-2fb7-4f6c-96b2-b30313179db9>\",\"WARC-Concurrent-To\":\"<urn:uuid:26a85523-6a0e-4852-8963-33394fe0b14a>\",\"WARC-IP-Address\":\"74.208.236.38\",\"WARC-Target-URI\":\"http://hrstc.org/node/26\",\"WARC-Payload-Digest\":\"sha1:TDPSKNLG2USHIJD3MR3LIMFCDRURECOI\",\"WARC-Block-Digest\":\"sha1:NYFMJPQ5YMIYLGRR6622BALRL4HDORUN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585504.90_warc_CC-MAIN-20211022084005-20211022114005-00607.warc.gz\"}"}
https://accord-framework.net/docs/html/M_Accord_Math_Matrix_DivideByDiagonal_33.htm	[ " Matrix.DivideByDiagonal Method (Byte[][], Single[], Single[][])", null, "Computes the product Ainv(B) of matrix A and diagonal matrix B.\n\nNamespace: Accord.Math\nAssembly: Accord.Math (in Accord.Math.dll) Version: 3.8.0", null, "Syntax\npublic static float[][] DivideByDiagonal(\nthis byte[][] a,\nfloat[] diagonal,\nfloat[][] result\n)\n\nParameters\n\na\nType: SystemByte\nThe left matrix A.\ndiagonal\nType: SystemSingle\nThe diagonal vector of inverse right matrix B.\nresult\nType: SystemSingle\nThe matrix R to store the product R = AB of the given matrices A and B.\n\nType: Single\n\nUsage Note\n\nIn Visual Basic and C#, you can call this method as an instance method on any object of type . When you use instance method syntax to call this method, omit the first parameter. For more information, see Extension Methods (Visual Basic) or Extension Methods (C# Programming Guide).", null, "See Also" ]	[ null, "https://accord-framework.net/docs/icons/logo.png", null, "https://accord-framework.net/docs/icons/SectionExpanded.png", null, "https://accord-framework.net/docs/icons/SectionExpanded.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.53745323,"math_prob":0.8927517,"size":1031,"snap":"2022-05-2022-21","text_gpt3_token_len":254,"char_repetition_ratio":0.12950341,"word_repetition_ratio":0.0,"special_character_ratio":0.24054316,"punctuation_ratio":0.15675676,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99235374,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T19:43:29Z\",\"WARC-Record-ID\":\"<urn:uuid:5449d767-5af4-49a5-84f2-4aad42ecaf49>\",\"Content-Length\":\"85365\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6bd435bf-ce72-4297-8367-8846f7421993>\",\"WARC-Concurrent-To\":\"<urn:uuid:171afc2e-7833-4f80-859b-97a8fa29d27c>\",\"WARC-IP-Address\":\"192.30.252.153\",\"WARC-Target-URI\":\"https://accord-framework.net/docs/html/M_Accord_Math_Matrix_DivideByDiagonal_33.htm\",\"WARC-Payload-Digest\":\"sha1:B5ENUMD4AUDR27YCO2I4FHYOYXLDUWLD\",\"WARC-Block-Digest\":\"sha1:BKLUXTI4IFLJJCYXWJULFDVTAANR6EFZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306335.77_warc_CC-MAIN-20220128182552-20220128212552-00376.warc.gz\"}"}
https://www.physicsoverflow.org/44477/how-to-find-solution-of-friedman-equation-numerically	[ "#", null, "How to find solution of Friedman equation numerically ?\n\n+ 0 like - 0 dislike\n241 views\n\nI am trying to solve friedman acceleration equation and continuity equation numerically using Runge Kutta Dormand Prince Method . My code was doing excellent for solving coupled differential equations. But, when I am trying to solve this equation , I am getting wrong results , I am getting same variation of scale factor for both matter dominated and radiation dominated case while one should vary as a2/3 and other should vary as a1/2. Can someone please tell me the possible reasons why this is happen ?\n\n$\\Ddota= -1/6(\\rho + 3p)$\n\nClosed by author request" ]	[ null, "https://www.physicsoverflow.org/qa-plugin/po-printer-friendly/print_on.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94121367,"math_prob":0.95112795,"size":534,"snap":"2023-14-2023-23","text_gpt3_token_len":118,"char_repetition_ratio":0.12075472,"word_repetition_ratio":0.023255814,"special_character_ratio":0.21910113,"punctuation_ratio":0.072916664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9810235,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T14:02:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f796c451-1f82-4b96-9ad3-d22cfead7642>\",\"Content-Length\":\"90331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce91913b-0b68-43bf-9aa2-be42ddee3c5d>\",\"WARC-Concurrent-To\":\"<urn:uuid:0c20d648-c69e-4ca6-8aad-5b5de157d6d9>\",\"WARC-IP-Address\":\"129.70.43.86\",\"WARC-Target-URI\":\"https://www.physicsoverflow.org/44477/how-to-find-solution-of-friedman-equation-numerically\",\"WARC-Payload-Digest\":\"sha1:XFDKOEFDNCBVOAMRHABVAJUIN2ZPYF2F\",\"WARC-Block-Digest\":\"sha1:CNT4VUUYLJ3MH2GKJ6FW64C2SUMZWRSW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948632.20_warc_CC-MAIN-20230327123514-20230327153514-00352.warc.gz\"}"}
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/each-count-to.html	[ "# eachCountTo\n\nCommon\nJVM\nJS\nNative\n1.1\n`fun <T, K, M : MutableMap<in K, Int>> Grouping<T, K>.eachCountTo( destination: M): M`\n(source)\n\nGroups elements from the Grouping source by key and counts elements in each group to the given destination map.\n\nIf the destination map already has a value corresponding to the key of some group, that value is used as an initial value of the counter for that group.\n\n``````\n\nfun main(args: Array<String>) {\n//sampleStart\nval words = \"one two three four five six seven eight nine ten\".split(' ')\nval frequenciesByFirstChar = words.groupingBy { it.first() }.eachCount()\nprintln(\"Counting first letters:\")\nprintln(frequenciesByFirstChar) // {o=1, t=3, f=2, s=2, e=1, n=1}\n\nval moreWords = \"eleven twelve\".split(' ')\nval moreFrequencies = moreWords.groupingBy { it.first() }.eachCountTo(frequenciesByFirstChar.toMutableMap())\nprintln(moreFrequencies) // {o=1, t=4, f=2, s=2, e=2, n=1}\n//sampleEnd\n}``````\n\nReturn the destination map associating the key of each group with the count of elements in the group." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65128946,"math_prob":0.9696859,"size":992,"snap":"2022-27-2022-33","text_gpt3_token_len":275,"char_repetition_ratio":0.12044534,"word_repetition_ratio":0.0,"special_character_ratio":0.26612905,"punctuation_ratio":0.17098446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96058583,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T19:39:32Z\",\"WARC-Record-ID\":\"<urn:uuid:97fe0039-1425-4059-94bc-9a856447d061>\",\"Content-Length\":\"25804\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99955d64-1ba2-4cf5-87cd-bea0f84e4ec5>\",\"WARC-Concurrent-To\":\"<urn:uuid:023f3340-d566-4bbe-b65b-c827c6b6d6a8>\",\"WARC-IP-Address\":\"18.67.65.38\",\"WARC-Target-URI\":\"https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/each-count-to.html\",\"WARC-Payload-Digest\":\"sha1:FOZEQVLP6Y23H7DJQW4LND4DCMCHJMGK\",\"WARC-Block-Digest\":\"sha1:B7DRNX6YDDNV4CA3GO7MR2DV4FTTIGIZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104204514.62_warc_CC-MAIN-20220702192528-20220702222528-00117.warc.gz\"}"}
https://wiki.phagocytes.ca/index.php/Bradford_Assay	[ "Introduction\n\nThe Bradford assay should be used to quantify total protein within a sample relative to a standard curve. This is important when running proteins on SDS-PAGE/Western blot, where each lane must contain an identical amount of total protein. This way, the intensity of the bands can be compared between different lanes, as standardizing the amount of protein loaded per lane will account for any variation in cell count from which the proteins were derived.\n\nProtocol\n\nFor a 96 well plate\n\n1. Using a 0.2 or 0.45 μm filter and 15 mL syringe, filter Bio-Rad Protein Assay Dye Reagent Concentrate (Bradford Dye) into a 15 mL falcon tube. Dilute 1:5 with ddH2O (1 part concentrate, 4 parts ddH2O). Invert to mix. Let warm to room temperature.\n2. Collect lysate or supernatant containing protein. Keep on ice.\n3. Label eight 1.5 mL tubes as: 1, 2, 3, 4, 5, 6, 7, 8. Set up the BSA standard by following Table 1. (Use the Bio-Rad 2 mg/mL stock. Do not make BSA yourself)\n• Make sure the diluent is the same as buffer that the protein of interest is in. For example, if in PBS, use PBS to dilute the standard. If in RIPA buffer, use RIPA, etc.\n• Mix by pipetting up and down. Avoid creating bubbles in the tube\n4. Set up 500 μL dilutions of your sample of interest. Make dilutions of 1:5, 1:10, and 1:25.\n• If you think your sample is very concentrated, instead make 1:10, 1:25, and 1:25 dilutions\n• If you think your sample is very dilute, use undiluted, 1:2, and 1:5 dilutions. (This uses a lot of sample however!)\n5. Add 150 μL of sample or standard to each well following Figure 1. Avoid bubbles by only pipetting to the first stop.\n6. Pour the dye into a multichannel pipette reservoir. Using a multichannel pipette, pipet 150 μL of dye into each well. Mix by pipetting carefully up and down.\n7. Incubate for at least 5 minutes at room temperature. Do not exceed 1 hr.\n8. Take off the lid of the plate, and place it into a plate reader/spectrometer.\n9. Measure absorbance at 595 nm.\nTable 1.\nTube # Standard volume (μL) Source of standard Diluent volume (μL) Final concentration (μg/mL)\n1 10 2 mg/mL stock 790 25\n2 10 2 mg/mL stock 990 20\n3 6 2 mg/mL stock 794 15\n4 500 Tube 2 500 10\n5 500 Tube 4 500 5\n6 500 Tube 5 500 2.5\n7 500 Tube 6 500 1.25\n8 - - 500 0\n\n-\n\nAnalysis using the standard curve\n\n1. Export results from spectrometer to Excel.\n2. Subtract the average absorbance of each standard/sample by the average absorbance of the blank. Essentially, you are setting the blank absorption to 0 and making all readings relative to that.\n3. Plot the known standard concentrations (X-axis: 0, 1.25, 2.5, 5, 10, 15, 20, and 25 μg/mL) with the average absorbance at each concentration (Y-axis). Add a line of best fit and display the equation with R2 value. Ideally, the R2 value should be ≥ 0.99. R2 < 0.94 curves should not be used.\n4. Calculate the average absorbance for the samples of interest by using the equation of the line graph y = mx+b (Absorbance = Slope*Concentration + b) and solve for x.\n5. If the calculated concentration falls within the linear range of the standard curve (1.25 - 25 μg/mL), then it is accurate. Determine the concentration of the undiluted sample by multiplying by the dilution factor." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8359916,"math_prob":0.93784827,"size":3293,"snap":"2022-05-2022-21","text_gpt3_token_len":970,"char_repetition_ratio":0.121009424,"word_repetition_ratio":0.013490725,"special_character_ratio":0.29456422,"punctuation_ratio":0.13694721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9526173,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T00:36:51Z\",\"WARC-Record-ID\":\"<urn:uuid:f8ef2edc-5e56-4c2b-884f-72a2302d094e>\",\"Content-Length\":\"19115\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78d3d835-a91f-4eee-b69c-de11ce01638c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ba65644-74ec-470c-b20b-4324b9484963>\",\"WARC-IP-Address\":\"149.56.155.8\",\"WARC-Target-URI\":\"https://wiki.phagocytes.ca/index.php/Bradford_Assay\",\"WARC-Payload-Digest\":\"sha1:NAW5QVSUA5RRYWLOMAQTDWGFG7H66QEK\",\"WARC-Block-Digest\":\"sha1:UYRH4XM4JHTXM4PWRPXH6X27Y5VKHKAF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662534693.28_warc_CC-MAIN-20220520223029-20220521013029-00263.warc.gz\"}"}
http://robmorton.20m.com/science/velocity.html	[ "The Nature of Light: Measuring The Velocity of Light\n\nThe apparent distance travelled and time taken by light from its source to an observer differ according to whether we look at it from the point of view of the source or that of the observer. [Português] [PDF].\n\nBasis For The Velocity of Light\n\nThe velocity of light in free space is usually represented by the letter c. It is understood to be an invariant quantity. In other words it is a natural constant, which is built into the fundamental fabric of the universe. Why should the velocity of light have this fixed finite value? Why shouldn't the velocity of light be infinite? It is because light is an electromagnetic disturbance, and space has a natural reluctance to being electromagnetically disturbed.\n\nConsequently, when an electric field is applied at a particular place, the immediate space around it takes time to polarize. It has a kind of electrical inertia that impedes the electric field's effort to polarize it. The surrounding space lags more and more behind in adopting a polarized state the further it is from the place where the electric field is applied. The adoption of this polarized state therefore \"travels\" outwards spherically from the place where the electric field is applied. Due to the natural quantitative value of space's reluctance to polarize, the velocity of this \"travel\" (or propagation) is c, the velocity of light.\n\nSuppose the applied electric field starts off with zero strength. In other words, it doesn't exist yet. Then it starts to grow. It grows slowly at first. Then its rate of growth gradually increases to a maximum. Its rate of growth then reduces, becoming zero again as the strength of the field reaches its maximum. Then the field starts to decay (reduce). Its rate of decay is slow at first, speeds up and then decays back to zero as the strength of the electric field reaches zero again. This process forms an electric field pulse.\n\nA complementary property of space is that the rate of change of an electric field becomes manifested as the magnitude of what is generally known as a magnetic field. Space does not polarize magnetically as it does electrically. A magnetic field doesn't have \"ends\" or poles like an electric field. Instead, it acts in a circular sense, adopting something that is perhaps best conceptualized as a kind of rotational inertia. This results in a situation where, as the electric field is growing or collapsing most rapidly, the associated magnetic field is at its strongest. Conversely, as the magnetic field is changing (growing or collapsing) most rapidly, the electric field is at its strongest.\n\nThe result is that the electric and magnetic fields exchange energy repeatedly rather like the bob of a pendulum continually exchanges its potential energy (due to its height) with its kinetic energy (due to its speed). The result is that these two types of \"force fields\" fall away from their source as an ever-expanding sphere, playing a game of \"throw and catch\" with their energy. This energy becomes ever more thinly spread over the area of an ever-expanding spherical surface. The frequency with which the electric and magnetic fields exchange energy is the frequency of the electromagnetic disturbance.\n\nThe important thing to note from this rather over-simplistic description of electromagnetic propagation in free space is that the effective electromagnetic wave propagates spherically at this fundamental velocity c away from its source.\n\nNotwithstanding, an observer has no way of sensing the approach of a light-pulse. He has no way of sensing when it left its source. He has no way of sensing how far it has travelled. He has no way of sensing how much time it took to reach him. He therefore has no way of sensing how fast it travelled towards him. He can only sense it when it eventually \"hits\" him. Even then, he has no way to sense its velocity of impact.\n\nA Two-Way Trip is Necessary\n\nTo acquire some idea of how \"fast\" light travels, an observer must use the principle of radar. He must set up a source of light that he can control. He must have a stop-clock to measure time. He must have a distant object (ideally a mirror) that can reflect the light emitted by his source. He must have a means of detecting the light reflected from the distant object. He can use his eyes, of course. However, an electronic detector will help him to make a more accurate measurement. He must know accurately the distance x that the distant object is from him. His light-source must be wired so that it automatically starts his clock when it emits a pulse of light. His light-detector must be wired so that it stops the clock immediately it detects the arrival of the light-pulse reflected from the distant object.", null, "The observer triggers his light-source to emit a short pulse of light. At the same time, the light-source starts the clock. The light-pulse travels to the distant object (mirror). The mirror then reflects the light pulse back towards the observer. The observer's light detector detects the arrival of the returning light-pulse and immediately stops the clock. The clock reveals the amount of time t the light-pulse took to travel the distance 2x to the distant object and back again. The velocity of light is thus revealed as c = 2x/t.\n\nThis is simply an illustrative experiment. Nowadays, much more refined techniques and apparatus are used to provide more accurate measurements of the velocity of light.\n\nFrames of Reference\n\nI proposed in my previous article that the frame of reference relative to which light travels at its universal velocity c be exclusively that of its source. So what happens in this case when light is reflected back to the original source? In whose frame of reference is the light travelling at the universal velocity c? Isn't it that of the observer who was the original source? What would be the case if the distant object (the mirror) were moving away from the source/observer at a velocity v that was a significant fraction of the velocity of light?\n\nLet me suggest the following. When light is reflected from an object such as a mirror, the process is not the same as that of a ball bouncing off a wall. The original light is not reflected back. Instead, the original light is absorbed by the atoms of the material of which the mirror is made. These atoms then use that energy to generate new light. In other words, a mirror is itself a separate light-source that is \"powered\" by the energy from the incident light. This is consistent with the observation that an object rarely reflects the same colour of light that it receives. It radiates a colour that is charactistic of the material of which it is made.\n\nIf this be the case, then the outbound light pulse is travelling at velocity c with respect to the source/observer frame of reference and the return light-pulse is travelling at velocity c with respect to the distant object's (the mirror's) frame of reference. This situation is illustrated below.", null, "When the light-pulse is emitted by the source, the mirror is at a distance x. The light-pulse travels towards the mirror. It travels at velocity c relative to the frame of reference of the source. By the time the light-pulse reaches the mirror, the mirror has travelled a further distance ½vt away from the source. The total distance travelled by the light on the outbound journey is therefore x + ½vt. The mirror absorbs the energy of the light-pulse. It uses this energy to generate another light-pulse. This new light-pulse travels in the direction of the observer. It does so at velocity c relative to the frame of reference of the mirror. By the time the new light-pulse reaches the observer, the observer has travelled yet a further distance of ½vt away from the mirror. The returning light-pulse therefore travels a distance x + vt. The total distance travelled by the outbound and return light-pulses is therefore 2x + 1·5vt. The distance x is therefore given by x = ½(c − 1·5v)t.\n\nWhen the returning light-pulse reaches the observer, it will appear to him to have originated from where the mirror is at that instant. The mirror will appear to be at a distance d given by the formula:\n\nd = x + vt = ½(c - 1·5v)t + vt = (½c + ¼v)t;\n\nwhere t is the amount of time that has elapsed since the observer's light-source emitted the original outbound light-pulse and v is the relative velocity at which the observer and the mirror are receding from each other. This reasoning requires that the \"velocity\" with which wave-crests approach the receding observer is necessarily c − v, and for an approaching observer, c + v. However, this does not mean that anything is materially travelling faster than light. Nor does it mean that information is travelling faster than light.\n\nAgain, although this double source-centred view appears to work, it is not - as mentioned in the previous essay - consistent with any view of gravitational phenomena. Notwithstanding, if the observer-centred Ætherial View is applied to the thought experiment above, the mathematical reasoning is essentially the same. And it is consistent with a means of explaining gravity.\n\n© 17 October 2006 Robert John Morton \| PREV \| NEXT" ]	[ null, "http://robmorton.20m.com/science/images/velocity2.gif", null, "http://robmorton.20m.com/science/images/velocity1.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9561874,"math_prob":0.94016266,"size":8994,"snap":"2019-43-2019-47","text_gpt3_token_len":1855,"char_repetition_ratio":0.15494995,"word_repetition_ratio":0.044980444,"special_character_ratio":0.19979987,"punctuation_ratio":0.07955215,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95616317,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T21:59:25Z\",\"WARC-Record-ID\":\"<urn:uuid:08e7eb9b-1f17-42f9-93d0-17bbff9a79ea>\",\"Content-Length\":\"14776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d14a527a-d573-4d91-bd6b-18a1f0533307>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3d82db7-7775-42a3-9af8-2e28760935ba>\",\"WARC-IP-Address\":\"64.136.20.53\",\"WARC-Target-URI\":\"http://robmorton.20m.com/science/velocity.html\",\"WARC-Payload-Digest\":\"sha1:KOGUD3LYOZFTMHMBB5QW3BT6NGY4DRSQ\",\"WARC-Block-Digest\":\"sha1:5QE2GDWSZE64NPJOGK4YGRC6V7ESFGO3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660323.32_warc_CC-MAIN-20191015205352-20191015232852-00381.warc.gz\"}"}
https://www.doubtnut.com/question-answer/simplify-each-of-the-following-and-express-the-result-as-a-rational-number-in-standard-form-19-36xx--642588771	[ "Getting Image", null, "", null, "", null, "", null, "Register now for special offers", null, "+91\n\nHome\n\n>\n\nEnglish\n\n>\n\nClass 8\n\n>\n\nMaths\n\n>\n\nChapter\n\n>\n\nRational Numbers\n\n>\n\nSimplify each of the following...\n\nUpdated On: 27-06-2022", null, "UPLOAD PHOTO AND GET THE ANSWER NOW!", null, "Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.", null, "Transcript\n\n11 today question is simplify each of the following and Express as Express the result is a rational number in standard form in first part Vinay to simplify -19 upon 36 X 16in second part -13.9 X 27 - 26 so let's solve this question taking first part in first we need to simplify -19 upon 36 X 16 which can be done as 16 by one so making it more simple by cancellation method 16 and 36 will be cancelled 4 times a section will be cancelled for X and 3936 will be cancelled 9 time from 4 now it is not further we cannot make it more simple and more simplified so just multiplying numerator parts 19 x 4 19 multiplayer for will be\n\n76 and minus will be there and 9 X 1 is 9 only soon -76.9 is the answer for the first part moving on to the second part in second part we need to simplify -13.9 X 27 upon -26 so making it more simple by cancellation method so 2626 will be cancelled - 2 X from 13 - 13 so it will be -2 and -1 and 27 will be cancelled 3 x 9 so now it is not we cannot simply for that I cancellations to moving to the next tab 2 - 1 x 3 will be -3 and denominator 1 x minus phase -2 so answer will be 3\n\nto Fir the second part in the standard form I hope you understood the solution thank you\n\n## Related Videos\n\n642587141\n\n200\n\n7.2 K\n\n2:06\nSimplify each of the following and express the result as a rational number in standard from: <br>(i)(-19)/(36)xx16 <br>(ii) (-13)/9xx(27)/(-26)\n1531585\n\n200\n\n2.6 K\n\n1:04\nSimplify each of the following and express the result as a rational number in standard from: (-19)/(36)xx16 (ii) (-13)/9xx(27)/(-26)\n1533529\n\n0\n\n1.8 K\n\n1:06\nSimplify each of the following and express the result as a rational number in standard form: (-19)/(36)x 16 (ii) (-13)/9x(27)/(-26)\n642588773\n\n0\n\n6.1 K\n\n1:21\nSimplify each of the following and express the result as a rational number in standard form: (-11)/9xx(-81)/(-88) (ii) (-5)/9xx(72)/(-25)\n642588772\n\n0\n\n900\n\n2:05\nSimplify each of the following and express the result as a rational number in standard form: (-9)/(16)xx(-64)/(-27) (ii) (-50)/7xx(14)/3\n1533530\n\n300\n\n9.2 K\n\n1:28\nSimplify each of the following and express the result as a rational number in standard form: (-9)/(16)x(-64)/(-27) (ii) (-50)/7x(14)/3", null, "", null, "" ]	[ null, "https://d10lpgp6xz60nq.cloudfront.net/images/doubtnut_header_logo_white_new.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/mweb-us-icon-small-camera.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/video-page-ic-search-white.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/mweb-home-white-hamburger-icon.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/mweb-india-country-flag-small.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/sticky_menu_camera_icon.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/mweb-icon-video-page-icon-small-text-book.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/mweb-icon-video-page-icon-small-text-book.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/user-comment.svg", null, "https://d10lpgp6xz60nq.cloudfront.net/images/loader.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9107969,"math_prob":0.9770329,"size":1302,"snap":"2022-40-2023-06","text_gpt3_token_len":343,"char_repetition_ratio":0.1540832,"word_repetition_ratio":0.04477612,"special_character_ratio":0.29185867,"punctuation_ratio":0.02846975,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9782894,"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,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\":\"2022-10-05T19:35:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c5a6159e-81b1-4bb2-b56b-e4ca4258e88b>\",\"Content-Length\":\"185446\",\"Content-Type\":\"application/http; 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https://www.talks.cam.ac.uk/talk/index/59061	[ "# Nesting statistics in the O(n) loop model on random lattices\n\nRandom Geometry\n\nCo-author: Jeremie Bouttier (CEA Saclay and ENS Paris)\n\nWe investigate how deeply nested are the loops in the O(n) model on random maps. In particular, we find that the number P of loops separating two points in a planar map in the dense phase with V >> 1 vertices is typically of order c(n) ln V for a universal constant c(n), and we compute the large deviations of P. The formula we obtain shows similarity to the CLE _{kappa} nesting properties for n = 2\u000fspi(1 - 4/kappa). The results can be extended to all topologies using the methods of topological recursion.\n\nThis talk is part of the Isaac Newton Institute Seminar Series series." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82674104,"math_prob":0.91948605,"size":1803,"snap":"2022-27-2022-33","text_gpt3_token_len":390,"char_repetition_ratio":0.086158976,"word_repetition_ratio":0.050179213,"special_character_ratio":0.18968385,"punctuation_ratio":0.06472492,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9783402,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T10:30:30Z\",\"WARC-Record-ID\":\"<urn:uuid:04b3da3e-5596-421d-b23d-2acebbfd12d0>\",\"Content-Length\":\"11434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da2668ad-42b7-40e7-997a-d637715756de>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bce335f-9a63-4359-9319-aab59a453373>\",\"WARC-IP-Address\":\"131.111.150.181\",\"WARC-Target-URI\":\"https://www.talks.cam.ac.uk/talk/index/59061\",\"WARC-Payload-Digest\":\"sha1:LHNCZQJF3O24H2VRSYTD3PSAEU2EO6PQ\",\"WARC-Block-Digest\":\"sha1:JAZC32E47Q2VQUWMANNUDAO4W3TDMFI2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572898.29_warc_CC-MAIN-20220817092402-20220817122402-00257.warc.gz\"}"}
https://grinebiter.com/Calculators/Factors/Factors-of-776.html	[ "Factors of 776\n\nHere we have calculated and created a list of all the factors of seven hundred seventy-six.\nThe factors of 776 are:\n\n1, 2, 4, 8, 97, 194, 388, 776\n\nFactors of 776 are integers that you can multiply together to get 776. Here are all the combinations of factors of 776 that you can multiply together to get 776:\n\n1 x 776 = 776\n2 x 388 = 776\n4 x 194 = 776\n8 x 97 = 776\n97 x 8 = 776\n194 x 4 = 776\n388 x 2 = 776\n776 x 1 = 776\n\nIf you divide 776 by a factor of 776, you will get one of the other factors of 776. Here are all combinations of dividing 776 by its factors:\n\n776 / 1 = 776\n776 / 2 = 388\n776 / 4 = 194\n776 / 8 = 97\n776 / 97 = 8\n776 / 194 = 4\n776 / 388 = 2\n776 / 776 = 1\n\nIn summary, we have not only listed all the factors of 776, but also illustrated that our solution is correct by multiplying and dividing the factors of seven hundred seventy-six.\n\nFactoring Calculator\nDo you need the factors for another number? No problem - just enter it below and press \"Factor\".\n\nFactors of 777\nHere is the next number on our list for which we have calculated the factors." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8875111,"math_prob":0.98897564,"size":1110,"snap":"2021-31-2021-39","text_gpt3_token_len":337,"char_repetition_ratio":0.18625678,"word_repetition_ratio":0.041666668,"special_character_ratio":0.4063063,"punctuation_ratio":0.084033616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99989426,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T05:27:55Z\",\"WARC-Record-ID\":\"<urn:uuid:1e432fd4-baa5-4ab2-9d97-8cedb4102722>\",\"Content-Length\":\"6894\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1500492-0615-4df3-86ae-2f6cbc02a927>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebfb9cb1-b9a2-42ed-9f07-5ae34dc6d477>\",\"WARC-IP-Address\":\"99.86.230.101\",\"WARC-Target-URI\":\"https://grinebiter.com/Calculators/Factors/Factors-of-776.html\",\"WARC-Payload-Digest\":\"sha1:GYORKKPDKVG3JQNSQOACSRYBRMZAA7AC\",\"WARC-Block-Digest\":\"sha1:TDEV7SWRJKGTRAMBZWWB2VVGGT56OI34\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057329.74_warc_CC-MAIN-20210922041825-20210922071825-00608.warc.gz\"}"}
https://www.physicsforums.com/threads/sum-of-independent-uniform-distribution-conditional-on-uniform.392882/	[ "# Sum of independent uniform distribution conditional on uniform\n\ngrossgermany\n\n## Homework Statement\n\nLet X and Y be independent and normal, then we know that\nIt must be the case that X+Y and X are jointly normal\nTherefore we can apply the projection theorem:\nwhich states that if A and B are jointly normal then VAR(A\|B)=VAR(B)-$$\\rho$$^2VAR(B) , apply the theorem to A=X+Y, B=Y to find\nVAR(X+Y\|X)\n\nThere is a similar procedure of finding E(X+Y\|X)\n\nI know how to do the above. However, what I don't know is what if X and Y are independent but each are UNIFORMLY distributed on [-1,1]\nWhat is:\n1.VAR(X+Y\|X)\n2.E(X+Y\|X)\n\n## The Attempt at a Solution\n\nLast edited:\n\nHomework Helper\nthis may help (or may not), with bits borrowed form wiki\nhttp://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables\nthough it should be independent of the probabilty distribution used\n\nlet Z = X + Y, and consider the joint probability density $f_{X,Y,Z}(x,y,z)$ the probability density of z is given by\n$$f_{Z}(z) = \\int f_{X,Y,Z}(x,y,z)dxdy$$\n\nyou can re-write the joint distribution using conditional probabilities & the independence of X & Y, following that through leads to the convolution\n$$f_{Z}(z) = \\int f_{X,Y,Z}(x,y,z)dxdy = \\int f_{X}(x)f_{Y}(y\|x)f_{Z}(z\|x,y)dxdy = \\int f_{X}(x)f_{Y}(y)\\delta(z-x-y)dxdy = \\int f_{X}(x)f_{Y}(z-x)dx$$\n\nThe conditional probability given $X = x_0$ will be\n$$f_{Z}(z\|x_0) = \\int f_{X}(x\|x_0)f_{Y}(y\|x_0)f_{Z}(z\|x_0,y)dxdy = \\int \\delta(x-x_0)f_{Y}(y)\\delta(z-x_0-y)dxdy=\\int f_{Y}(y)\\delta(z-x_0-y)dy= f_{Y}(z-x_0)$$\n\nThen take the expectation\n$$E(Z\|x_0) = \\int z f(z\|x_0)dz = \\int z f_{Y}(z-x_0)dz$$\n\nand similar for the variance\n\ngrossgermany\nUnfortunately I am already familiar with this topic with regard to normally distibuted variables.\nWhat I need is the mean and variance of sum of two independent Uniformly distributed variables conditional on one of the uniformly distributed variable.\nI know that X+Y should be a triangular distributed, if X and Y are independent uniformly distributed.\nHowever, what is\nE(x+y\|x)\nVar(x+y\|x)\n\nLast edited:\nHomework Helper\nthe derivation does not assume normal variables & should work for any distribution\n\nthe convolution integral will give you the triangular distribution\n\nthe last integral should give you the expectation with conditional X=x_0 (try it, it should be pretty simple)\n\nthen you should be able to use the conditional distiribution for the finding the variance as well\n\nThats rigorous... but if you just think about what is happening when X=x_0 is known you should be able to guess the mean & variance based on the distribution of Y easily enough\n\nLast edited:\nHomework Helper\nPS, hint - as X is a set value, the only random variable is Y, so in effect Z = constant + Y...\n\nLast edited:\ngrossgermany\nThank you very much for your reply. I mostly understand what you wrote except 3 things.\n1. Why is the conditional pdf of (z given x,y) equal to dirac's delta function of z-(x+y). Wikipedia just says they are trivially equal. Yes it's true that z needs to be equal to x+y, so you can say there is zero probabiltiy z not equal to x+y. But somehow I feel this is just too much intuition and not rigorous enough.\n2. When the double integral regard to x y become a single integral to x, why does this work out to be fx(x)fy(z-x) ?\n\nYes y=z-x but don't we need\nintegral[fx(x)integral{fy(z-x)d(-x)}dx]. The d(-x) is due to the fact that dy=d(z-x)=-dx\n3. How would we find E(X\|X+Y)\n\nLast edited:\nHomework Helper\nThank you very much for your reply. I mostly understand what you wrote except 3 things.\n1. Why is the conditional pdf of (z given x,y) equal to dirac's delta function of z-(x+y). Wikipedia just says they are trivially equal. Yes it's true that z needs to be equal to x+y, so you can say there is zero probabiltiy z not equal to x+y. But somehow I feel this is just too much intuition and not rigorous enough.\nit can only be that, given X=x & Y=y, Z =X+Y is totally determined to be Z=x+y, hence\n$$f_Z(z=x+y\|X=x,Y=y) = \\delta(z-x+y)$$\nnote this intgrates to one as it must and satifies the constraint\n\n2. When the double integral regard to x y become a single integral to x, why does this work out to be fx(x)fy(z-x) ? Yes y=z-x but don't we need integral[fx(x)integral{fy(z-x)d(-x)}dx]. The d(-x) is due to the fact that dy=d(z-x)=-dx\nthe following integral is contracted by integrating over all y, integrating over the delta function sets y = z-x\n$$= \\int \\int f_{X}(x)f_{Y}(y)\\delta(z-x-y)dxdy = \\int f_{X}(x)f_{Y}(z-x)dx$$\n\nyou could similarly choose to integrate over x first & get\n$$= \\int \\int f_{X}(x)f_{Y}(y)\\delta(z-x-y)dxdy = \\int \\int f_{X}(x)f_{Y}(y)\\delta(z-x-y)dydx = \\int f_{X}(z-y)f_{Y}(y)dy$$\n\nthe results are equivalent as they must be, we choose to integrate over Y first as it makes life easier later on with the constrant on x\n\n3. How would we find E(X\|X+Y)\nthis takes a little more thought as i assume you mean E(X\|Z), with Z = X+Y? you would need to find the conditional probability distribution\n\nfor the previous questions, say you have a random variable X with\n$$E(X) = \\mu$$\n$$VAR(X) = \\sigma^2$$\n\nnote that for a constant a\n$$E(X+a) = \\mu +a$$\n$$VAR(X+a) = \\sigma^2$$\n\ngrossgermany\nI am able to completely understand your post now. Thank you. But I tried to derive the entire thing in your early post but with different objective:\nGive X and Y are indpendent uniform\nFind\nE(X\|Z=X+Y)\nVar(X\|Z=X+Y)\nand I am stuck in the imitation of your proof. Simply because fx(x\|z) is no longer fx(x).\nWould you please show me how to redo your early post in this scenario?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90819323,"math_prob":0.9991825,"size":754,"snap":"2022-27-2022-33","text_gpt3_token_len":226,"char_repetition_ratio":0.12666667,"word_repetition_ratio":0.168,"special_character_ratio":0.25596818,"punctuation_ratio":0.06748466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998534,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T05:17:54Z\",\"WARC-Record-ID\":\"<urn:uuid:10de84ba-5373-46b9-b5ab-a80d8919fed6>\",\"Content-Length\":\"81772\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ea9f33d-f366-4c54-b49c-863d738730b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:9222bdc0-a798-4f54-8873-427fec6e7699>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/sum-of-independent-uniform-distribution-conditional-on-uniform.392882/\",\"WARC-Payload-Digest\":\"sha1:OCBWE2HYGRV7QIM7SG3MACDMXP2V3AVM\",\"WARC-Block-Digest\":\"sha1:ANMTMOBAMP4GYPXOG2E3KCSGNYL54WRA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571147.84_warc_CC-MAIN-20220810040253-20220810070253-00001.warc.gz\"}"}
https://www.studysmarter.us/textbooks/math/essential-calculus-early-transcendentals-2nd/series/q35e-a-use-the-sum-of-the-first-10-terms-and-exercise-33a-to/	[ "", null, "Suggested languages for you:\n\nEurope\n\nAnswers without the blur. Sign up and see all textbooks for free!", null, "Q35E\n\nExpert-verified", null, "Found in: Page 453", null, "### Essential Calculus: Early Transcendentals\n\nBook edition 2nd\nAuthor(s) James Stewart\nPages 830 pages\nISBN 9781133112280", null, "# (a) Use the sum of the first 10 terms and Exercise 33(a) to estimate the sum of the series$$\\sum\\limits_{n = 1}^\\infty {\\frac{1}{{{n^2}}}}$$ . How good is this estimate?(b) Improve this estimate using Exercise 33(b) with n = 10(c) Find a value of n that will ensure that the error in the approximation $$S \\approx {S_n}$$ is less than 0.01\n\n(a) The sum of series $$\\sum\\limits_{n = 1}^\\infty {\\frac{1}{{{n^2}}}}$$ is 1.549768 with error at most 0.1\n\n(b) When n = 10, estimate $$S \\approx 1.64522$$ with error ≤ 0.005\n\n(c) Value of n > 1000 that will ensure the error approx $$S \\approx {S_n}$$ is less than 0.01\n\nSee the step by step solution\n\n## (a) Applying Integral on the Given Function:\n\nLet the associated function $$f(x) = \\frac{1}{{{x^2}}}$$ is positive and continuous then\n\n$$f(x) = \\frac{{ - 2}}{{{x^3}}} < 0{\\rm{ for }}x \\ge 1$$\n\nTherefore on applying integral it gives an upper bound for the error using the mth partial sum as:\n\n$${R_\\infty } = \\sum\\limits_{n \\to 1}^\\infty {\\frac{1}{{{n^2}}} \\le \\int\\limits_1^\\infty {\\frac{1}{{{x^2}}}dx} }$$\n\n$${\\rm{ }} = \\mathop {\\lim }\\limits_{b \\to \\infty } \\left[ {\\frac{{ - 1}}{x}} \\right]_m^b = \\frac{1}{m} - \\mathop {\\lim }\\limits_{b \\to \\infty } \\frac{1}{b} = \\frac{1}{m}$$\n\nThe estimate using the sum of the first ten terms is\n\n$$\\sum\\limits_{n = 1}^\\infty {\\frac{1}{{{n^2}}} \\approx {S_{10}}}$$\n\n$$= 1 + \\frac{1}{{{2^2}}} + \\frac{1}{{{3^2}}} + .... + \\frac{1}{{{{10}^2}}}$$\n\n$$\\approx 1.549768$$\n\nAnd the error of this estimate is $${R_{10}} \\le \\frac{1}{{10}}$$ , so the error is at most 0.1\n\n## (b) Improving Estimate with n = 10\n\nWe have $${S_n} + \\int\\limits_{n + 1}^\\infty {f(x)dx \\le S \\le {S_n} + \\int\\limits_n^\\infty {f(x)dx} }$$\n\nNow with n = 10, we have $${S_{10}} + \\int\\limits_{11}^\\infty {\\frac{1}{{{x^2}}}dx \\( = {S_{10}} + \\frac{1}{{11}} \\le S \\le {S_{10}} + \\frac{1}{{10}}$$\n\n$$= 1.549768 + 0.090909 \\le S \\le 1.549768 + 0.1$$\n\n$$= 1.640677 \\le S \\le 1.6491768$$\n\nSo we get S ≈ 1.64522 which is average of 1.640677 and 1.649768, with error ≤ 0.005 (which is rounded up for half 0.1)\n\n## (c) Finding Value of n\n\nWe have $${S_n} \\le \\int\\limits_n^\\infty {\\frac{1}{{{x^2}}}dx{\\rm{ which is equal to }}\\frac{1}{n}}$$\n\n$${\\rm{That is, }}{{\\rm{S}}_n} \\le \\int\\limits_n^\\infty {\\frac{1}{{{x^2}}}dx} = \\frac{1}{n}$$\n\n$${\\rm{So }}{{\\rm{S}}_n} < 0.01{\\rm{ if }}\\frac{1}{n} < \\frac{1}{{1000}}$$\n\n$${\\rm{Hence }}n > 1000$$", null, "### Want to see more solutions like these?", null, "" ]	[ null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/dist/assets/images/header-logo.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/src/assets/images/ab-test/searching-looking.svg", null, "https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-images/Stewart_essential_calculus.jpeg", null, "https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-images/Stewart_essential_calculus.jpeg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/src/assets/images/ab-test/businessman-superhero.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/img/textbook/banner-top.svg", null, "https://www.studysmarter.us/wp-content/themes/StudySmarter-Theme/img/textbook/cta-icon.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5952772,"math_prob":1.0000001,"size":2327,"snap":"2023-14-2023-23","text_gpt3_token_len":970,"char_repetition_ratio":0.17907877,"word_repetition_ratio":0.032967035,"special_character_ratio":0.5040825,"punctuation_ratio":0.06889353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999981,"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\":\"2023-03-30T02:50:35Z\",\"WARC-Record-ID\":\"<urn:uuid:76575345-8b8e-4814-a366-3f13d8a56ac2>\",\"Content-Length\":\"152637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b94ba16-3e30-4a73-8928-2c80a363fd7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b73a2e0-1cc4-43f1-bbb3-1fca31dff994>\",\"WARC-IP-Address\":\"18.184.139.114\",\"WARC-Target-URI\":\"https://www.studysmarter.us/textbooks/math/essential-calculus-early-transcendentals-2nd/series/q35e-a-use-the-sum-of-the-first-10-terms-and-exercise-33a-to/\",\"WARC-Payload-Digest\":\"sha1:GOHJHA5WKWOZ7RQOV3SZIOOPIY7BUQ2U\",\"WARC-Block-Digest\":\"sha1:JY2BKGZYQGDO5XAB24PNS5L6TKQIFZOS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949093.14_warc_CC-MAIN-20230330004340-20230330034340-00552.warc.gz\"}"}
https://soccerlapeche.ca/scanwise.aspx	[ "Android App Development\n\n# Converting Fractions Improper To Mixed Worksheet\n\n## First form is to fractions mixed\n\nConverting Improper Fractions To Mixed Numbers Worksheet. Apr 22 2013 This free support page will help you understand how to convert improper fraction to mixed fraction and how to convert mixed fractions to improper.\n\nPython\n\n## We count the improper fractions to mixed worksheet\n\nScroll down to a quizizz to this is this is created great way to fractions to improper mixed worksheet maker will need to helping with riddles to practice converting mixed number and improper fractions to solve future problems?\n\nDocuments X\n\n## To create and to worksheet\n\nConverting a decimal to a mixed number and an improper. Worksheet 2 Convert mixed numbers to improper fractions 2 ' This page includes a lesson covering 'Converting from improper fractions to mixed fractions' as.\n\nFrom\n\n## Ready to mixed fractions to improper fractione numerator\n\nClick here for converting fractions improper to mixed worksheet has a different types and publish button in question: convert between different number fraction is required! 10 Ways to Practice Converting Improper Fractions and Mixed.\n\nOf Parts\n\n## The left and converting fractions improper fraction\n\nThe things interesting real life applications as mixed fractions to worksheet, progress so it into an improper fraction worksheets in the number part of the total number. How can you convert an improper fraction to a mixed number. Fraction Worksheets Math is Fun.\n\nManual\n\n## This problem yourself at anytime by converting fractions to improper mixed worksheet onto overhead and draw a means that\n\nHow to Convert Mixed Numbers to Improper Fractions Maths. Create free unlimited worksheets to practice to simplify fractions Set the range for the numerator and denominator and the worksheet is ready in a couple of.\n\nProviders\n\n## Help students start by contrast, to fractions improper mixed numbers are\n\nThe squares to fractions mixed worksheet requires a model of esl math fractions and rectangles with the bottom of the denominator remains the same denominator, you want to? Printable-fraction-worksheets-convert-mixed-numbers-to. Rename Mixed to Fractions PDF4PRO.\n\nSatisfaction\n\n## To all our terms before converting fractions to mixed numbers\n\nIt means that make great deal with rectangles with situation or. Practice writing a fraction as a mixed number and vice versa. Keep the denominator remains the correct and to mixed numbers; give you want to. Converting Improper Fractions to Mixed Fractions A.\n\nDivorce\n\n## Quizizz also mention that\n\nFree Printable Class 4 Converting improper fraction to mixed. Tutorial an animated video mini-lesson and a free worksheet and answer key. Lesson Converting Mixed Numbers to Improper Fractions.\n\nOf System\n\n## Sasha is a mixed number as fractions worksheet has\n\nWhen students have to convert from improper fractions to mixed. Students use themes and whole number together or folding will show you to fractions improper mixed numbers and resources and mixed numbers and more to this.\n\nLaws\n\n## Add a printable converting improper fractions\n\nPlease select the questions for helping with a fraction by your reports, and fill up to join the ten in converting improper fractions can directly from your previous lesson. Convert mixed numbers to improper fractions Fraction and. Improper fractions worksheet.\n\nQuick\n\n## It off a worksheet to fractions improper mixed numbers as well\n\nMixed Fractions to Improper Worksheets Free Printable PDF. Converting Mixed Numbers to Improper Fractions Worksheets. Put a circle around the improper fraction and change them into mixed fractions. Multiplying mixed fractions worksheets.\n\nResume\n\n## Audio instructions for fractions mixed\n\nConvert Improper Fractions and Mixed Numbers Worksheets. Worksheet on Changing Fractions Improper Fraction to a. Also has a mixed fractions can show you can often have permission to worksheet to. Mixed numbers on a number line calculator.\n\nTustin\n\n## Find a fractions mixed numbers\n\nChanging improper fractions to mixed numbers 4th grade. Badges that she practices converting improper fractions to mixed word worksheet has 9 problems multiplying fractions worksheets have an accurate comparison of.\n\nAnd\n\n## This worksheet to fractions improper mixed numbers that\n\nQuiz & Worksheet Changing Between Improper Fraction and. Worksheets Write improper fractions as mixed numbers Below are six versions of our grade 4 fractions worksheet on re-writing improper fractions fractions 1 to.\n\nUnlock", null, "" ]	[ null, "https://soccerlapeche.ca/9dc203.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84989214,"math_prob":0.8154571,"size":4338,"snap":"2022-40-2023-06","text_gpt3_token_len":799,"char_repetition_ratio":0.3077988,"word_repetition_ratio":0.024806201,"special_character_ratio":0.16620563,"punctuation_ratio":0.062849164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98824406,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T16:33:01Z\",\"WARC-Record-ID\":\"<urn:uuid:1ab1d923-627b-45c6-b2fa-ddfaaf1270ed>\",\"Content-Length\":\"122976\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c244110-8328-4b29-8415-331209e0eca8>\",\"WARC-Concurrent-To\":\"<urn:uuid:42e747ec-0979-40f1-baa6-589d696eb126>\",\"WARC-IP-Address\":\"104.21.90.199\",\"WARC-Target-URI\":\"https://soccerlapeche.ca/scanwise.aspx\",\"WARC-Payload-Digest\":\"sha1:NW5FJSOHEWVSS6Z4E7RTL3UOVRFN7N6E\",\"WARC-Block-Digest\":\"sha1:IBQLMDWSXZJ2E5LJY5BVHW6Q3KL5YY6M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00391.warc.gz\"}"}
http://dynamicmathematicslearning.com/thabit-pyth-generalization.html	[ "## Thabit's Generalization of the Theorem of Pythagoras", null, "Let ABC be any triangle as shown below. Construct lines AD and AE as shown in the figure so that angles BAD and CEA are both equal to angle BAC. Then AB2 + AC2 = BC(BD + CE).\n1) Drag any of A, B or C to explore the configuration, and also click on the given button to show the measurements.\n2) An easy way of constructing angles BAD and CEA equal to angle BAC is simply to rotate lines AB and AC around A respectively by the directed angles ACB and ABC.\n3) When the angle at vertex A is a right angle, angles BAD and CEA will also be right angles, and the points D and E will coincide (since the perpendicular from A to BC is unique). In other words, we then obtain the theorem of Pythagoras, namely, AB2 + AC2 = BC2." ]	[ null, "http://dynamicmathematicslearning.com/thabit.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.904388,"math_prob":0.99465686,"size":1285,"snap":"2022-05-2022-21","text_gpt3_token_len":319,"char_repetition_ratio":0.10928962,"word_repetition_ratio":0.0,"special_character_ratio":0.24357976,"punctuation_ratio":0.112840466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99677587,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T16:47:58Z\",\"WARC-Record-ID\":\"<urn:uuid:5f6a342b-37a6-4302-96b0-cb9949b7011f>\",\"Content-Length\":\"3692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3a047df-dab3-44d6-8148-01226f00d397>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e4a5744-e292-47ce-a048-7e15beb14bad>\",\"WARC-IP-Address\":\"66.84.6.23\",\"WARC-Target-URI\":\"http://dynamicmathematicslearning.com/thabit-pyth-generalization.html\",\"WARC-Payload-Digest\":\"sha1:ELKJHYMSGWU5I2MVTQJSOBE2XJNT7RSA\",\"WARC-Block-Digest\":\"sha1:NMQMS6XNWJH4HLFAVJALEGBFBS4HW7PQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662519037.11_warc_CC-MAIN-20220517162558-20220517192558-00772.warc.gz\"}"}
https://www.equationsworksheets.net/translating-words-into-equations-worksheet/	[ "# Translating Words Into Equations Worksheet\n\nTranslating Words Into Equations Worksheet – Expressions and Equations Worksheets are created to assist children in learning faster and more efficiently. These worksheets are interactive and challenges based on sequence of operations. These worksheets will help children can master both basic and complex concepts in a quick amount of period of time. These PDF resources are free to download, and can be used by your child in order to learn math concepts. These resources are helpful for students in 5th-8th grades.", null, "These worksheets can be utilized by students between 5th and 8th grades. These two-step word problems are created using fractions and decimals. Each worksheet contains ten problems. They can be found at any online or print resource. These worksheets are an excellent method to learn how to reorder equations. In addition to allowing students to practice the art of rearranging equations, they assist your student to understand the properties of equality and inverted operations.\n\nThese worksheets are targeted at fifth and eight graders. These are great for students who struggle to calculate percentages. There are three types of problems that you can pick from. You can choose to solve one-step problems that include whole or decimal numbers, or utilize word-based strategies to do fractions or decimals. Each page will have 10 equations. These worksheets for Equations can be used by students in the 5th to 8th grade.", null, "These worksheets are a great source for practicing fractions along with other topics related to algebra. Some of the worksheets let users to select from three types of challenges. You can select one that is word-based, numerical or a combination of both. It is crucial to select the type of problem, as every problem is different. Ten problems are on each page, so they’re fantastic resources for students in the 5th to 8th grades.\n\nThe worksheets will teach students about the relationship between variables and numbers. These worksheets allow students to work on solving polynomial problems and to learn how to apply equations to solve problems in everyday life. If you’re in search of an educational tool that will help you learn about expressions and equations, you can start with these worksheets. These worksheets can teach you about the different kinds of mathematical issues as well as the different symbols that are employed to explain them.", null, "These worksheets are extremely beneficial for students who are in the 1st grade. These worksheets will teach students how to solve equations and graphs. They are great for practice with polynomial variables. These worksheets can help you factor and simplify these variables. You can find a great set of equations, expressions and worksheets designed for children of every grade level. Doing the work yourself is the most efficient way to master equations.\n\nThere are many worksheets to learn about quadratic equations. There are various levels of equations worksheets for each grade. These worksheets are designed in order to assist you in solving problems of the fourth degree. Once you have completed a certain level, you can continue to work on other kinds of equations. You can continue to take on the same problems. For instance, you could discover a problem that uses the same axis, but as an extended number." ]	[ null, "https://www.equationsworksheets.net/wp-content/uploads/2022/10/translating-word-problems-to-equations-word-problems-math-word-2.jpg", null, "https://www.equationsworksheets.net/wp-content/uploads/2022/10/translating-words-into-equations-card-sort-math-words-education-math-scaled.jpg", null, "https://www.equationsworksheets.net/wp-content/uploads/2022/10/translating-words-into-equations-worksheet.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9530452,"math_prob":0.954449,"size":3357,"snap":"2023-40-2023-50","text_gpt3_token_len":628,"char_repetition_ratio":0.18878616,"word_repetition_ratio":0.011173184,"special_character_ratio":0.18111409,"punctuation_ratio":0.082214765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99075955,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T20:01:00Z\",\"WARC-Record-ID\":\"<urn:uuid:037f4631-875a-4644-8afb-0ad09c39b90a>\",\"Content-Length\":\"66422\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61db708c-c2af-4321-bb60-49e63a241906>\",\"WARC-Concurrent-To\":\"<urn:uuid:65b3379d-caaf-4dc5-a50f-64780c30b101>\",\"WARC-IP-Address\":\"104.21.91.193\",\"WARC-Target-URI\":\"https://www.equationsworksheets.net/translating-words-into-equations-worksheet/\",\"WARC-Payload-Digest\":\"sha1:J4OWXEERJ7RGXHS2RKSND5ZFQFHU32WH\",\"WARC-Block-Digest\":\"sha1:4NIY5MQLIM5LNZL7XTC2GKOY67T5ARYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511406.34_warc_CC-MAIN-20231004184208-20231004214208-00363.warc.gz\"}"}
https://math.stackexchange.com/questions/3196064/find-general-formula-for-the-terms	[ "# Find general formula for the terms\n\nFind a general formula for the terms of the sequence\n\n$${a_n}=\\left\\{ \\frac{11}{7},\\frac{107}{49},\\frac{659}{343},\\frac{4883}{2401},\\frac{33371}{16807},\\frac{234569}{117649},\\dots \\right\\}$$\n\nI don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $$7$$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?\n\n• Where does it come from? – Jean-Claude Arbaut Apr 21 at 17:14\n• My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer. – RaV1oLLi Apr 21 at 17:16\n• Have you put it into OEIS? – Dave Apr 21 at 17:22\n• This is in the OEIS, but not much else - oeis.org/A081657 – Peter Foreman Apr 21 at 17:23\n• Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game. – Somos Apr 21 at 18:55\n\n## 3 Answers\n\nThe general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $$n$$ of its (first) members is in principle impossible, as the next (not listed) $$(n+1)^\\mathrm{th}$$ member may be an arbitrary number, and there is still a formula for expressing $$a_1, \\dots, a_n, a_{n+1},$$ e. g. as a polynomial of order $$n$$:\n\n$$a_k = \\sum_{i=0}^nb_ik^i,\\quad k = 1, \\dots,n+1$$\n\nThe process for finding coefficients $$b_0, \\dots, b_n$$ is straightforward enough.\n\nIn other words, if someone will find the formula for your \"sequence\", there is still the infinity number of other formulas, giving different sequences, but all of them producing your \"sequence\", i. e. $$\\frac{11}{7},\\frac{107}{49},\\frac{659}{343},\\frac{4883}{2401},\\frac{33371}{16807},\\frac{234569}{117649}.$$\n\nNote:\n\nIt means that all psychological tests of type\n\nWhat is the next number of the sequence $$1, 2, 3, 4, 5?$$\n\nare in principle meaningless ones, because you may tell \"$$1762$$\", and then show to surprised psychologist a formula supporting the correctness of your answer:\n\n$$a_k = {439\\over30}k^5-{439\\over 2}k^4+{7463\\over 6}k^3-{6585\\over 2}k^2+{60158\\over 15}k-1756$$\n\nIf he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands\n\nsage: var(\"k\")\nsage: a(k) = (439/30)k^5 - (439/2)k^4+(7463/6)k^3-(6585/2)k^2+(60158/15)*k-1756\nsage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)\n\n\nto obtain the result\n\n(1, 2, 3, 4, 5, 1762, 10543, 36884)\n\n\n(and to give the psychologist two more members for free).\n\nNote 2:\n\nIt doesn't mean that there is not a simpler formula - including a recurrent one or other \"recipe\" - for the same (finite) sequence.\n\nFor example, there is so simple one for the rather not so trivial sequence\n\n$$\\color{blue}{1, 11, 21, 1211, 111221}$$\n\nthat even 6-7 year-old child is able to write down the next element ($$\\color{red}{312211}$$) if you tell it the rule, or - perhaps - even without telling it.\n\nNo, you have no chance to discover this simple rule (supposing your age is $$10$$+). Don't waste your time. It's a good advice, believe me.\n\n(Googling for it is a much better approach.)\n\nThe general term seems to be $$a_n=\\frac{2(7^n)+(-3)^n}{7^n}=2+\\left(-\\frac{3}{7}\\right)^n$$ But the last term is given by $$a_6=2-\\left(\\frac{3}{7}\\right)^6$$ so this formula does not always work. A suitable formula could be $$a_n=\\begin{cases}2-\\left(\\frac{3}{7}\\right)^n&n\\equiv0\\mod{6}\\\\ 2+\\left(-\\frac{3}{7}\\right)^n &\\text{otherwise} \\end{cases}$$\n\nThe numerator of the $$n$$-th term seems to be $$2\\cdot 7^n+(-1)^n 3^n.$$\n\n• May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect. – RaV1oLLi Apr 21 at 17:07\n• I was thinking the same, but what about the last term? $234569\\ne 2(7^6)+3^6$ – Peter Foreman Apr 21 at 17:07\n• @PeterForeman Arg, yes, the last is $2(7)^6\\color{red}{-}3^6$! – Jean-Claude Arbaut Apr 21 at 17:10\n• @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms. – Bernard Apr 21 at 17:20" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8821842,"math_prob":0.99608606,"size":2126,"snap":"2019-13-2019-22","text_gpt3_token_len":698,"char_repetition_ratio":0.099434495,"word_repetition_ratio":0.0,"special_character_ratio":0.37770462,"punctuation_ratio":0.16,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994814,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T03:08:48Z\",\"WARC-Record-ID\":\"<urn:uuid:e3407a71-2bb3-42f9-be55-fb04f3dfef7a>\",\"Content-Length\":\"153989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:233dbfdf-4187-466c-9f3f-fdc4ec1e41a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:110703d0-6a74-4f13-8862-1df3c783abc5>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3196064/find-general-formula-for-the-terms\",\"WARC-Payload-Digest\":\"sha1:GU4KKZQ27HEJUYZB6HS5DNIEAGY64ZYE\",\"WARC-Block-Digest\":\"sha1:Y5HSAPQQ2KUIAWA4KEJISYI5B2XRXULC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257847.56_warc_CC-MAIN-20190525024710-20190525050710-00316.warc.gz\"}"}
https://www.convertunits.com/from/decitonne/to/chaldron	[ "## ››Convert decitonne to chaldron\n\n decitonne chaldron\n\nHow many decitonne in 1 chaldron? The answer is 26.9252.\nWe assume you are converting between decitonne and chaldron.\nYou can view more details on each measurement unit:\ndecitonne or chaldron\nThe SI base unit for mass is the kilogram.\n1 kilogram is equal to 0.01 decitonne, or 0.00037139928394218 chaldron.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between decitonnes and chaldrons.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of decitonne to chaldron\n\n1 decitonne to chaldron = 0.03714 chaldron\n\n10 decitonne to chaldron = 0.3714 chaldron\n\n20 decitonne to chaldron = 0.7428 chaldron\n\n30 decitonne to chaldron = 1.1142 chaldron\n\n40 decitonne to chaldron = 1.4856 chaldron\n\n50 decitonne to chaldron = 1.857 chaldron\n\n100 decitonne to chaldron = 3.71399 chaldron\n\n200 decitonne to chaldron = 7.42799 chaldron\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from chaldron to decitonne, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Decitonne\n\nThe SI prefix \"deci\" represents a factor of 10-1, or in exponential notation, 1E-1.\n\nSo 1 decitonne = 10-1 tonnes.\n\nThe definition of a tonne is as follows:\n\nA tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g).\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7544107,"math_prob":0.9467776,"size":1879,"snap":"2019-51-2020-05","text_gpt3_token_len":570,"char_repetition_ratio":0.25226668,"word_repetition_ratio":0.0,"special_character_ratio":0.27142096,"punctuation_ratio":0.15263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9554502,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T05:27:23Z\",\"WARC-Record-ID\":\"<urn:uuid:f36e22b2-2f15-4529-92a1-79a3d7f2fe1e>\",\"Content-Length\":\"31697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c958e22-3b16-48fd-ac8d-a8320d05dd9c>\",\"WARC-Concurrent-To\":\"<urn:uuid:1cdeb375-3950-4b02-902d-4ae8710f8afe>\",\"WARC-IP-Address\":\"54.175.245.234\",\"WARC-Target-URI\":\"https://www.convertunits.com/from/decitonne/to/chaldron\",\"WARC-Payload-Digest\":\"sha1:VNQXGXGHACF77F3BZPMSK5GD2OMOIREG\",\"WARC-Block-Digest\":\"sha1:2WBKR5K4GM76G2O44SAT5R246NDRZOWR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00121.warc.gz\"}"}
http://home.ubalt.edu/ntsbarsh/Business-stat/excel/excel.htm	[ "## Excel For Statistical Data Analysis\n\nPara mis visitantes del mundo de habla hispana, este sitio se encuentra disponible en español en:\nSitio Espejo para América Latina Sitio de los E.E.U.U.\n\nExcel is the widely used statistical package, which serves as a tool to understand statistical concepts and computation to check your hand-worked calculation in solving your homework problems. The site provides an introduction to understand the basics of and working with the Excel. Redoing the illustrated numerical examples in this site will help improving your familiarity and as a result increase the effectiveness and efficiency of your process in statistics.\n\nProfessor Hossein Arsham\n\nTo search the site, try Edit \| Find in page [Ctrl + f]. Enter a word or phrase in the dialogue box, e.g. \"variance\" or \"mean\" If the first appearance of the word/phrase is not what you are looking for, try Find Next.\n\nCompanion Sites:\n\n### Introduction\n\nThis site provides illustrative experience in the use of Excel for data summary, presentation, and for other basic statistical analysis. I believe the popular use of Excel is on the areas where Excel really can excel. This includes organizing data, i.e. basic data management, tabulation and graphics. For real statistical analysis on must learn using the professional commercial statistical packages such as SAS, and SPSS.\n\nMicrosoft Excel 2000 (version 9) provides a set of data analysis tools called the Analysis ToolPak which you can use to save steps when you develop complex statistical analyses. You provide the data and parameters for each analysis; the tool uses the appropriate statistical macro functions and then displays the results in an output table. Some tools generate charts in addition to output tables.\n\nIf the Data Analysis command is selectable on the Tools menu, then the Analysis ToolPak is installed on your system. However, if the Data Analysis command is not on the Tools menu, you need to install the Analysis ToolPak by doing the following:\n\nStep 1: On the Tools menu, click Add-Ins.... If Analysis ToolPak is not listed in the Add-Ins dialog box, click Browse and locate the drive, folder name, and file name for the Analysis ToolPak Add-in Analys32.xll usually located in the Program Files\\Microsoft Office\\Office\\Library\\Analysis folder. Once you find the file, select it and click OK.\n\nStep 2: If you don't find the Analys32.xll file, then you must install it.\n\n1. Insert your Microsoft Office 2000 Disk 1 into the CD ROM drive.\n2. Select Run from the Windows Start menu.\n3. Browse and select the drive for your CD. Select Setup.exe, click Open, and click OK.\n4. Click the Add or Remove Features button.\n5. Click the + next to Microsoft Excel for Windows.\n6. Click the + next to Add-ins.\n7. Click the down arrow next to Analysis ToolPak.\n8. Select Run from My Computer.\n9. Select the Update Now button.\n10. Excel will now update your system to include Analysis ToolPak.\n11. Launch Excel.\n12. On the Tools menu, click Add-Ins... - and select the Analysis ToolPak check box.\nStep 3: The Analysis ToolPak Add-In is now installed and Data Analysis... will now be selectable on the Tools menu.\n\nMicrosoft Excel is a powerful spreadsheet package available for Microsoft Windows and the Apple Macintosh. Spreadsheet software is used to store information in columns and rows which can then be organized and/or processed. Spreadsheets are designed to work well with numbers but often include text. Excel organizes your work into workbooks; each workbook can contain many worksheets; worksheets are used to list and analyze data .\n\nExcel is available on all public-access PCs (i.e., those, e.g., in the Library and PC Labs). It can be opened either by selecting Start - Programs - Microsoft Excel or by clicking on the Excel Short Cut which is either on your desktop, or on any PC, or on the Office Tool bar.\n\nOpening a Document:\n\n• Click on File-Open (Ctrl+O) to open/retrieve an existing workbook; change the directory area or drive to look for files in other locations\n• To create a new workbook, click on File-New-Blank Document.\n\nSaving and Closing a Document:\n\nTo save your document with its current filename, location and file format either click on File - Save. If you are saving for the first time, click File-Save; choose/type a name for your document; then click OK. Also use File-Save if you want to save to a different filename/location.\n\nWhen you have finished working on a document you should close it. Go to the File menu and click on Close. If you have made any changes since the file was last saved, you will be asked if you wish to save them.\n\nThe Excel screen", null, "Workbooks and worksheets:\n\nWhen you start Excel, a blank worksheet is displayed which consists of a multiple grid of cells with numbered rows down the page and alphabetically-titled columns across the page. Each cell is referenced by its coordinates (e.g., A3 is used to refer to the cell in column A and row 3; B10:B20 is used to refer to the range of cells in column B and rows 10 through 20).\n\nYour work is stored in an Excel file called a workbook. Each workbook may contain several worksheets and/or charts - the current worksheet is called the active sheet. To view a different worksheet in a workbook click the appropriate Sheet Tab.\n\nYou can access and execute commands directly from the main menu or you can point to one of the toolbar buttons (the display box that appears below the button, when you place the cursor over it, indicates the name/action of the button) and click once.\n\nMoving Around the Worksheet:\n\nIt is important to be able to move around the worksheet effectively because you can only enter or change data at the position of the cursor. You can move the cursor by using the arrow keys or by moving the mouse to the required cell and clicking. Once selected the cell becomes the active cell and is identified by a thick border; only one cell can be active at a time.\n\nTo move from one worksheet to another click the sheet tabs. (If your workbook contains many sheets, right-click the tab scrolling buttons then click the sheet you want.) The name of the active sheet is shown in bold.\n\nMoving Between Cells:\n\nHere is a keyboard shortcuts to move the active cell:\n\n• Home - moves to the first column in the current row\n• Ctrl+Home - moves to the top left corner of the document\n• End then Home - moves to the last cell in the document\n\nTo move between cells on a worksheet, click any cell or use the arrow keys. To see a different area of the sheet, use the scroll bars and click on the arrows or the area above/below the scroll box in either the vertical or horizontal scroll bars.\n\nNote that the size of a scroll box indicates the proportional amount of the used area of the sheet that is visible in the window. The position of a scroll box indicates the relative location of the visible area within the worksheet.\n\n### Entering Data\n\nA new worksheet is a grid of rows and columns. The rows are labeled with numbers, and the columns are labeled with letters. Each intersection of a row and a column is a cell. Each cell has an address, which is the column letter and the row number. The arrow on the worksheet to the right points to cell A1, which is currently highlighted, indicating that it is an active cell. A cell must be active to enter information into it. To highlight (select) a cell, click on it.\n\nTo select more than one cell:\n\n• Click on a cell (e.g. A1), then hold the shift key while you click on another (e.g. D4) to select all cells between and including A1 and D4.\n• Click on a cell (e.g. A1) and drag the mouse across the desired range, unclicking on another cell (e.g. D4) to select all cells between and including A1 and D4.\n• To select several cells which are not adjacent, press \"control\" and click on the cells you want to select. Click a number or letter labeling a row or column to select that entire row or column.\n\nOne worksheet can have up to 256 columns and 65,536 rows, so it'll be a while before you run out of space.\n\nEach cell can contain a label, value, logical value, or formula.\n\n• Labels can contain any combination of letters, numbers, or symbols.\n• Values are numbers. Only values (numbers) can be used in calculations. A value can also be a date or a time\n• Logical values are \"true\" or \"false.\"\n• Formulas automatically do calculations on the values in other specified cells and display the result in the cell in which the formula is entered (for example, you can specify that cell D3 is to contain the sum of the numbers in B3 and C3; the number displayed in D3 will then be a funtion of the numbers entered into B3 and C3).", null, "To enter information into a cell, select the cell and begin typing.\n\nNote that as you type information into the cell, the information you enter also displays in the formula bar. You can also enter information into the formula bar, and the information will appear in the selected cell.\n\nWhen you have finished entering the label or value:\n\n• Press \"Enter\" to move to the next cell below (in this case, A2)\n• Press \"Tab\" to move to the next cell to the right (in this case, B1)\n• Click in any cell to select it\nEntering Labels", null, "Unless the information you enter is formatted as a value or a formula, Excel will interpret it as a label, and defaults to align the text on the left side of the cell.\n\nIf you are creating a long worksheet and you will be repeating the same label information in many different cells, you can use the AutoComplete function. This function will look at other entries in the same column and attempt to match a previous entry with your current entry. For example, if you have already typed \"Wesleyan\" in another cell and you type \"W\" in a new cell, Excel will automatically enter \"Wesleyan.\" If you intended to type \"Wesleyan\" into the cell, your task is done, and you can move on to the next cell. If you intended to type something else, e.g. \"Williams,\" into the cell, just continue typing to enter the term.\n\nTo turn on the AutoComplete funtion, click on \"Tools\" in the menu bar, then select \"Options,\" then select \"Edit,\" and click to put a check in the box beside \"Enable AutoComplete for cell values.\"\n\nAnother way to quickly enter repeated labels is to use the Pick List feature. Right click on a cell, then select \"Pick From List.\" This will give you a menu of all other entries in cells in that column. Click on an item in the menu to enter it into the currently selected cell.\n\nEntering Values\n\nA value is a number, date, or time, plus a few symbols if necessary to further define the numbers [such as: . , + - ( ) % \\$ / ].\n\nNumbers are assumed to be positive; to enter a negative number, use a minus sign \"-\" or enclose the number in parentheses \"()\".\n\nDates are stored as MM/DD/YYYY, but you do not have to enter it precisely in that format. If you enter \"jan 9\" or \"jan-9\", Excel will recognize it at January 9 of the current year, and store it as 1/9/2002. Enter the four-digit year for a year other than the current year (e.g. \"jan 9, 1999\"). To enter the current day's date, press \"control\" and \";\" at the same time.\n\nTimes default to a 24 hour clock. Use \"a\" or \"p\" to indicate \"am\" or \"pm\" if you use a 12 hour clock (e.g. \"8:30 p\" is interpreted as 8:30 PM). To enter the current time, press \"control\" and \":\" (shift-semicolon) at the same time.", null, "An entry interpreted as a value (number, date, or time) is aligned to the right side of the cell, to reformat a value.\n\nRounding Numbers that Meet Specified Criteria: To apply colors to maximum and/or minimum values:\n\n1. Select a cell in the region, and press Ctrl+Shift+* (in Excel 2003, press this or Ctrl+A) to select the Current Region.\n2. From the Format menu, select Conditional Formatting.\n3. In Condition 1, select Formula Is, and type =MAX(\\$F:\\$F) =\\$F1.\n4. Click Format, select the Font tab, select a color, and then click OK.\n5. In Condition 2, select Formula Is, and type =MIN(\\$F:\\$F) =\\$F1.\n6. Repeat step 4, select a different color than you selected for Condition 1, and then click OK.\nNote: Be sure to distinguish between absolute reference and relative reference when entering the formulas.\n\nRounding Numbers that Meet Specified Criteria\n\nProblem: Rounding all the numbers in column A to zero decimal places, except for those that have \"5\" in the first decimal place.\n\nSolution: Use the IF, MOD, and ROUND functions in the following formula: =IF(MOD(A2,1)=0.5,A2,ROUND(A2,0))\n\nTo Copy and Paste All Cells in a Sheet\n\n1. Select the cells in the sheet by pressing Ctrl+A (in Excel 2003, select a cell in a blank area before pressing Ctrl+A, or from a selected cell in a Current Region/List range, press Ctrl+A+A).\nOR\nClick Select All at the top-left intersection of rows and columns.\n2. Press Ctrl+C.\n3. Press Ctrl+Page Down to select another sheet, then select cell A1.\n4. Press Enter.\n\nTo Copy the Entire Sheet\n\nCopying the entire sheet means copying the cells, the page setup parameters, and the defined range Names.\n\nOption 1:\n\n1. Move the mouse pointer to a sheet tab.\n2. Press Ctrl, and hold the mouse to drag the sheet to a different location.\n3. Release the mouse button and the Ctrl key.\n\nOption 2:\n\n1. Right-click the appropriate sheet tab.\n2. From the shortcut menu, select Move or Copy. The Move or Copy dialog box enables one to copy the sheet either to a different location in the current workbook or to a different workbook. Be sure to mark the Create a copy checkbox.\n\nOption 3:\n\n1. From the Window menu, select Arrange.\n2. Select Tiled to tile all open workbooks in the window.\n3. Use Option 1 (dragging the sheet while pressing Ctrl) to copy or move a sheet.\n\nSorting by Columns\n\nThe default setting for sorting in Ascending or Descending order is by row. To sort by columns:\n\n1. From the Data menu, select Sort, and then Options.\n2. Select the Sort left to right option button and click OK.\n3. In the Sort by option of the Sort dialog box, select the row number by which the columns will be sorted and click OK.\n\n### Descriptive Statistics\n\nThe Data Analysis ToolPak has a Descriptive Statistics tool that provides you with an easy way to calculate summary statistics for a set of sample data. Summary statistics includes Mean, Standard Error, Median, Mode, Standard Deviation, Variance, Kurtosis, Skewness, Range, Minimum, Maximum, Sum, and Count. This tool eliminates the need to type indivividual functions to find each of these results. Excel includes elaborate and customisable toolbars, for example the \"standard\" toolbar shown here:", null, "Some of the icons are useful mathematical computation:", null, "is the \"Autosum\" icon, which enters the formula \"=sum()\" to add up a range of cells.", null, "is the \"FunctionWizard\" icon, which gives you access to all the functions available.", null, "is the \"GraphWizard\" icon, giving access to all graph types available, as shown in this display:", null, "Excel can be used to generate measures of location and variability for a variable. Suppose we wish to find descriptive statistics for a sample data: 2, 4, 6, and 8.\n\nStep 1. Select the Tools pull-down menu, if you see data analysis, click on this option, otherwise, click on add-in.. option to install analysis tool pak.\n\nStep 2. Click on the data analysis option.\n\nStep 3. Choose Descriptive Statistics from Analysis Tools list.\n\nStep 4. When the dialog box appears:\n\nEnter A1:A4 in the input range box, A1 is a value in column A and row 1, in this case this value is 2. Using the same technique enter other VALUES until you reach the last one. If a sample consists of 20 numbers, you can select for example A1, A2, A3, etc. as the input range.\n\nStep 5. Select an output range, in this case B1. Click on summary statistics to see the results.\n\nSelect OK.\n\nWhen you click OK, you will see the result in the selected range.\n\nAs you will see, the mean of the sample is 5, the median is 5, the standard deviation is 2.581989, the sample variance is 6.666667,the range is 6 and so on. Each of these factors might be important in your calculation\nof different statistical procedures.\n\n### Normal Distribution\n\nConsider the problem of finding the probability of getting less than a certain value under any normal probability distribution. As an illustrative example, let us suppose the SAT scores nationwide are normally distributed with a mean and standard deviation of 500 and 100, respectively. Answer the following questions based on the given information:\n\nA: What is the probability that a randomly selected student score will be less than 600 points?\nB: What is the probability that a randomly selected student score will exceed 600 points?\nC: What is the probability that a randomly selected student score will be between 400 and 600?\n\nHint: Using Excel you can find the probability of getting a value approximately less than or equal to a given value. In a problem, when the mean and the standard deviation of the population are given, you have to use common sense to find different probabilities based on the question since you know the area under a normal curve is 1.\n\nSolution:\n\nIn the work sheet, select the cell where you want the answer to appear. Suppose, you chose cell number one, A1. From the menus, select \"insert pull-down\".\n\nSteps 2-3 From the menus, select insert, then click on the Function option.\n\nStep 4. After clicking on the Function option, the Paste Function dialog appears from Function Category. Choose Statistical then NORMDIST from the Function Name box; Click OK\n\nStep 5. After clicking on OK, the NORMDIST distribution box appears:\ni. Enter 600 in X (the value box);\nii. Enter 500 in the Mean box;\niii. Enter 100 in the Standard deviation box;\niv. Type \"true\" in the cumulative box, then click OK.\n\nAs you see the value 0.84134474 appears in A1, indicating the probability that a randomly selected student's score is below 600 points. Using common sense we can answer part \"b\" by subtracting 0.84134474 from 1. So the part \"b\" answer is 1- 0.8413474 or 0.158653. This is the probability that a randomly selected student's score is greater than 600 points. To answer part \"c\", use the same techniques to find the probabilities or area in the left sides of values 600 and 400. Since these areas or probabilities overlap each other to answer the question you should subtract the smaller probability from the larger probability. The answer equals 0.84134474 - 0.15865526 that is, 0.68269. The screen shot should look like following:\n\nInverse Case\n\nCalculating the value of a random variable often called the \"x\" value\n\nYou can use NORMINV from the function box to calculate a value for the random variable - if the probability to the left side of this variable is given. Actually, you should use this function to calculate different percentiles. In this problem one could ask what is the score of a student whose percentile is 90? This means approximately 90% of students scores are less than this number. On the other hand if we were asked to do this problem by hand, we would have had to calculate the x value using the normal distribution formula x = m + zd. Now let's use Excel to calculate P90. In the Paste function, dialog click on statistical, then click on NORMINV. The screen shot would look like the following:\n\nWhen you see NORMINV the dialog box appears.\ni. Enter 0.90 for the probability (this means that approximately 90% of students' score is less than the value we are looking for)\nii. Enter 500 for the mean (this is the mean of the normal distribution in our case)\niii. Enter 100 for the standard deviation (this is the standard deviation of the normal distribution in our case)\n\nAt the end of this screen you will see the formula result which is approximately 628 points. This means the top 10% of the students scored better than 628.\n\n### Confidence Interval for the Mean\n\nSuppose we wish for estimating a confidence interval for the mean of a population. Depending on the size of your sample size you may use one of the following cases:\n\nLarge Sample Size (n is larger than, say 30):\n\nThe general formula for developing a confidence interval for a population means is:", null, "In this formula", null, "is the mean of the sample; Z is the interval coefficient, which can be found from the normal distribution table (for example the interval coefficient for a 95% confidence level is 1.96). S is the standard deviation of the sample and n is the sample size.\n\nNow we would like to show how Excel is used to develop a certain confidence interval of a population mean based on a sample information. As you see in order to evaluate this formula you need", null, "\"the mean of the sample\" and the margin of error", null, "Excel will automatically calculate these quantities for you.\n\nThe only things you have to do are:\n\nadd the margin of error", null, "to the mean of the sample,", null, "; Find the upper limit of the interval and subtract the margin of error from the mean to the lower limit of the interval. To demonstrate how Excel finds these quantities we will use the data set, which contains the hourly income of 36 work-study students here, at the University of Baltimore. These numbers appear in cells A1 to A36 on an Excel work sheet.\n\nAfter entering the data, we followed the descriptive statistic procedure to calculate the unknown quantities. The only additional step is to click on the confidence interval in the descriptive statistics dialog box and enter the given confidence level, in this case 95%.\n\nHere is, the above procedures in step-by-step:\n\nStep 1. Enter data in cells A1 to A36 (on the spreadsheet)\nStep 2. From the menus select Tools\nStep 3. Click on Data Analysis then choose the Descriptive Statistics option then click OK.\n\nOn the descriptive statistics dialog, click on Summary Statistic. After you have done that, click on the confidence interval level and type 95% - or in other problems whatever confidence interval you desire. In the Output Range box enter B1 or what ever location you desire.\nNow click on OK. The screen shot would look like the following:", null, "", null, "As you see, the spreadsheet shows that the mean of the sample is", null, "= 6.902777778 and the absolute value of the margin of error", null, "= 0.231678109. This mean is based on this sample information. A 95% confidence interval for the hourly income of the UB work-study students has an upper limit of 6.902777778 + 0.231678109 and a lower limit of 6.902777778 - 0.231678109.\n\nOn the other hand, we can say that of all the intervals formed this way 95% contains the mean of the population. Or, for practical purposes, we can be 95% confident that the mean of the population is between 6.902777778 - 0.231678109 and 6.902777778 + 0.231678109. We can be at least 95% confident that interval [\\$6.68 and \\$7.13] contains the average hourly income of a work-study student.\n\nSmal Sample Size (say less than 30) If the sample n is less than 30 or we must use the small sample procedure to develop a confidence interval for the mean of a population. The general formula for developing confidence intervals for the population mean based on small a sample is:", null, "In this formula", null, "is the mean of the sample.", null, "is the interval coefficient providing an area of", null, "in the upper tail of a t distribution with n-1 degrees of freedom which can be found from a t distribution table (for example the interval coefficient for a 90% confidence level is 1.833 if the sample is 10). S is the standard deviation of the sample and n is the sample size.\n\nNow you would like to see how Excel is used to develop a certain confidence interval of a population mean based on this small sample information.\n\nAs you see, to evaluate this formula you need", null, "\"the mean of the sample\" and the margin of error", null, "Excel will automatically calculate these quantities the way it did for large samples.\n\nAgain, the only things you have to do are: add the margin of error", null, "to the mean of the sample,", null, ", find the upper limit of the interval and to subtract the margin of error from the mean to find the lower limit of the interval.\n\nTo demonstrate how Excel finds these quantities we will use the data set, which contains the hourly incomes of 10 work-study students here, at the University of Baltimore. These numbers appear in cells A1 to A10 on an Excel work sheet.\n\nAfter entering the data we follow the descriptive statistic procedure to calculate the unknown quantities (exactly the way we found quantities for large sample). Here you are with the procedures in step-by-step form:\n\nStep 1. Enter data in cells A1 to A10 on the spreadsheet\nStep 2. From the menus select Tools\nStep 3. Click on Data Analysis then choose the Descriptive Statistics option. Click OK on the descriptive statistics dialog, click on Summary Statistic, click on the confidence interval level and type in 90% or in other problems whichever confidence interval you desire. In the Output Range box, enter B1 or whatever location you desire. Now click on OK. The screen shot will look like the following:", null, "Now, like the calculation of the confidence interval for the large sample, calculate the confidence interval of the population based on this small sample information. The confidence interval is:\n\n6.8 ± 0.414426102\nor\n\\$6.39<===>\\$7.21.\n\nWe can be at least 90% confidant that the interval [\\$6.39 and \\$7.21] contains the true mean of the population.\n\n### Test of Hypothesis Concerning the Population Mean\n\nAgain, we must distinguish two cases with respect to the size of your sample\n\nLarge Sample Size (say, over 30): In this section you wish to know how Excel can be used to conduct a hypothesis test about a population mean. We will use the hourly incomes of different work-study students than those introduced earlier in the confidence interval section. Data are entered in cells A1 to A36. The objective is to test the following Null and Alternative hypothesis:", null, "The null hypothesis indicates that the average hourly income of a work-study student is equal to \\$7 per hour; however, the alternative hypothesis indicates that the average hourly income is not equal to \\$7 per hour.\n\nI will repeat the steps taken in descriptive statistics and at the very end will show how to find the value of the test statistics in this case, z, using a cell formula.\n\nStep 1. Enter data in cells A1 to A36 (on the spreadsheet)\n\nStep 2. From the menus select Tools\n\nStep 3. Click on Data Analysis then choose the Descriptive Statistics option, click OK.\nOn the descriptive statistics dialog, click on Summary Statistic. Select the Output Range box, enter B1 or whichever location you desire. Now click OK.\n\n(To calculate the value of the test statistics search for the mean of the sample then the standard error. In this output, these values are in cells C3 and C4.)\n\nStep 4. Select cell D1 and enter the cell formula = (C3 - 7)/C4. The screen shot should look like the following:", null, "The value in cell D1 is the value of the test statistics. Since this value falls in acceptance range of -1.96 to 1.96 (from the normal distribution table), we fail to reject the null hypothesis.\n\nSmall Sample Size (say, less than 30):\n\nUsing steps taken the large sample size case, Excel can be used to conduct a hypothesis for small-sample case. Let's use the hourly income of 10 work-study students at UB to conduct the following hypothesis.", null, "The null hypothesis indicates that average hourly income of a work-study student is equal to \\$7 per hour .The alternative hypothesis indicates that average hourly income is not equal to \\$7 per hour.\n\nI will repeat the steps taken in descriptive statistics and at the very end will show how to find the value of the test statistics in this case \"t\" using a cell formula.\n\nStep 1. Enter data in cells A1 to A10 (on the spreadsheet)\n\nStep 2. From the menus select Tools\n\nStep 3. Click on Data Analysis then choose the Descriptive Statistics option. Click OK.\nOn the descriptive statistics dialog, click on Summary Statistic. Select the Output Range boxes, enter B1 or whatever location you chose. Again, click on OK.\n(To calculate the value of the test statistics search for the mean of the sample then the standard\nerror, in this output these values are in cells C3 and C4.)\n\nStep 4. Select cell D1 and enter the cell formula = (C3 - 7)/C4. The screen shot would look like the following:", null, "Since the value of test statistic t = -0.66896 falls in acceptance range -2.262 to +2.262 (from t table, where", null, "= 0.025 and the degrees of freedom is 9), we fail to reject the null hypothesis.\n\n### Difference Between Mean of Two Populations\n\nIn this section we will show how Excel is used to conduct a hypothesis test about the difference between two population means assuming that populations have equal variances. The data in this case are taken from various offices here at the University of Baltimore. I collected the hourly income data of 36 randomly selected work-study students and 36 student assistants. The hourly income range for work-study students was \\$6 - \\$8 while the hourly income range for student assistants was \\$6-\\$9. The main objective in this hypothesis testing is to see whether there is a significant difference between the means of the two populations. The NULL and the ALTERNATIVE hypothesis is that the means are equal and the means are not equal, respectively.\n\nReferring to the spreadsheet, I chose A1 and A2 as label centers. The work-study students' hourly income for a sample size 36 are shown in cells A2:A37, and the student assistants' hourly income for a sample size 36 is shown in cells B2:B37\n\nData for Work Study Student: 6, 6, 6, 6, 6, 6, 6, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 7, 7, 7, 7, 7, 7, 7, 7.5, 7.5, 7.5, 7.5, 7.5, 7.5, 8, 8, 8, 8, 8, 8, 8, 8, 8.\n\nData for Student Assistant: 6, 6, 6, 6, 6, 6.5, 6.5, 6.5, 6.5, 6.5, 7, 7, 7, 7, 7, 7.5, 7.5, 7.5, 7.5, 7.5, 7.5, 8, 8, 8, 8, 8, 8, 8, 8.5, 8.5, 8.5, 8.5, 8.5, 9, 9, 9, 9.\n\nUse the Descriptive Statistics procedure to calculate the variances of the two samples. The Excel procedure for testing the difference between the two population means will require information on the variances of the two populations. Since the variances of the two populations are unknowns they should be replaced with sample variances. The descriptive for both samples show that the variance of first sample is s12 = 0.55546218, while the variance of the second sample s22 =0.969748.\n\n work-study student student assistant Mean 7.05714286 Mean 7.471429 Standard Error 0.12597757 Standard Error 0.166454 Median 7 Median 7.5 Mode 8 Mode 8 Standard Deviation 0.74529335 Standard Deviation 0.984758 Sample Variance 0.55546218 Sample Variance 0.969748 Kurtosis -1.38870558 Kurtosis -1.192825 Skewness -0.09374375 Skewness -0.013819 Range 2 Range 3 Minimum 6 Minimum 6 Maximum 8 Maximum 9 Sum 247 Sum 261.5 Count 35 Count 35\n\nTo conduct the desired test hypothesis with Excel the following steps can be taken:\n\nStep 1. From the menus select Tools then click on the Data Analysis option.\n\nStep 2. When the Data Analysis dialog box appears:\nChoose z-Test: Two Sample for means then click OK\n\nStep 3. When the z-Test: Two Sample for means dialog box appears:\n\nEnter A1:A36 in the variable 1 range box (work-study students' hourly income)\nEnter B1:B36 in the variable 2 range box (student assistants' hourly income)\nEnter 0 in the Hypothesis Mean Difference box (if you desire to test a mean difference other than 0, enter that value)\nEnter the variance of the first sample in the Variable 1 Variance box\nEnter the variance of the second sample in the Variable 2 Variance box and select Labels\nEnter 0.05 or, whatever level of significance you desire, in the Alpha box\nSelect a suitable Output Range for the results, I chose C19, then click OK.\n\nThe value of test statistic z=-1.9845824 appears in our case in cell D24. The rejection rule for this test is z < -1.96 or z > 1.96 from the normal distribution table. In the Excel output these values for a two-tail test are z<-1.959961082 and z>+1.959961082. Since the value of the test statistic z=-1.9845824 is less than -1.959961082 we reject the null hypothesis. We can also draw this conclusion by comparing the p-value for a two tail -test and the alpha value.\n\nSince p-value 0.047190813 is less than a=0.05 we reject the null hypothesis. Overall we can say, based on the sample results, the two populations' means are different.\n\nSmall Samples: n1 OR n2 are less than 30\n\nIn this section we will show how Excel is used to conduct a hypothesis test about the difference between two population means. - Given that the populations have equal variances when two small independent samples are taken from both populations. Similar to the above case, the data in this case are taken from various offices here at the University of Baltimore. I collected hourly income data of 11 randomly selected work-study students and 11 randomly selected student assistants. The hourly income range for both groups was similar range, \\$6 - \\$8 and \\$6-\\$9. The main objective in this hypothesis testing is similar too, to see whether there is a significant difference between the means of the two populations. The NULL and the ALTERNATIVE hypothesis are that the means are equal and they are not equal, respectively.\n\n work-study student student assistant 6 6 8 9 7.5 8.5 6.5 7 7 6.5 6 7 7.5 7.5 8 6 6 8 6.5 9 7 7.5>\n\nReferring to the spreadsheet, we chose A1 and A2 as label centers. The work-study students' hourly income for a sample size 11 are shown in cells A2:A12, and the student assistants' hourly income for a sample size 11 is shown in cells B2:B12. Unlike previous case, you do not have to calculate the variances of the two samples, Excel will automatically calculate these quantities and use them in the calculation of the value of the test statistic.\n\nSimilar to the previous case, but a bit different in step # 2, to conduct the desired test hypothesis with Excel the following steps can be taken:\n\nStep 1. From the menus select Tools then click on the Data Analysis option.\n\nStep 2. When the Data Analysis dialog box appears:\nChoose t-Test: Two Sample Assuming Equal Variances then click OK\n\nStep 3 When the t-Test: Two Sample Assuming Equal Variances dialog box appears:\n\nEnter A1:A12 in the variable 1 range box (work-study student hourly income)\nEnter B1:B12 in the variable 2 range box (student assistant hourly income)\nEnter 0 in the Hypothesis Mean Difference box(if you desire to test a mean difference other than zero, enter that value) then select Labels\n\nEnter 0.05 or, whatever level of significance you desire, in the Alpha box\n\nSelect a suitable Output Range for the results, I chose C1, then click OK.\n\nThe value of the test statistic t=-1.362229828 appears, in our case, in cell D10. The rejection rule for this test is t<-2.086 or t>+2.086 from the t distribution table where the t value is based on a t distribution with n1-n2-2 degrees of freedom and where the area of the upper one tail is 0.025 ( that is equal to alpha/2).\n\nIn the Excel output the values for a two-tail test are t<-2.085962478 and t>+2.085962478. Since the value of the test statistic t=-1.362229828, is in an acceptance range of t<-2.085962478 and t>+2.085962478, we fail to reject the null hypothesis.\n\nWe can also draw this conclusion by comparing the p-value for a two-tail test and the alpha value.\n\nSince the p-value 0.188271278 is greater than a=0.05 again, we fail to reject the null hypothesis.\n\nOverall we can say, based on sample results, the two populations' means are equal.\n\n work-study student student assistant Mean 6.909090909 7.454545455 Variance 0.590909091 1.172727273 Observations 11 11 Pooled Variance 0.881818182 Hypothesized Mean Difference 0 Df 20 t Stat -1.362229828 P(T<=t) one tail 0.094135639 t Critical one tail 1.724718004 P(T<=t)two tail 0.188271278 t Critical two tail 2.085962478\n\n### ANOVA: Analysis of Variances\n\nIn this section the objective is to see whether or not means of three or more populations based on random samples taken from populations are equal or not. Assuming independents samples are taken from normally distributed populations with equal variances, Excel would do this analysis if you choose one way anova from the menus. We can also choose Anova: two way factor with or without replication option and see whether there is significant difference between means when different factors are involved.\n\nSingle-Factor ANOVA Test\n\nIn this case we were interested to see whether there a significant difference among hourly wages of student assistants in three different service departments here at the University of Baltimore. Six student assistants were randomly were selected from the three departments and their hourly wages were recorded as following:\n\n ARC CSI TCC 10.00 6.50 9.00 8.00 7.00 7.00 7.50 7.00 7.00 8.00 7.50 7.00 7.00 7.00 6.50\n\nEnter data in an Excel work sheet starting with cell A2 and ending with cell C8. The following steps should be taken to find the proper output for interpretation.\n\nStep 1. From the menus select Tools and click on Data Analysis option.\n\nStep 2. When data analysis dialog appears, choose Anova single-factor option; enter A2:C8 in the input range box. Select labels in first row.\n\nStep3.Select any cell as output(in here we selected A11). Click OK.\n\nThe general form of Anova table looks like following:\n Source of Variation SS df MS F P-value F crit Between Groups SSTR K-1 MSTR MST/MSE 0.046725 3.682316674 Within Groups SSE nt-K MSE Total\n\nSuppose the test is done at level of significance a = 0.05, we reject the null hypothesis. This means there is a significant difference between means of hourly incomes of student assistants in these departments.\n\nThe Two-way ANOVA Without Replication\n\nIn this section, the study involves six students who were offered different hourly wages in three different department services here at the University of Baltimore. The objective is to see whether the hourly incomes are the same. Therefore, we can consider the following:\n\nFactor: Department\n\nTreatment: Hourly payments in the three departments\n\nBlocks: Each student is a block since each student has worked in the three different departments\n\n Student ARC CSI TCC 1 10.00 7.50 7.00 2 8.00 7.00 6.00 3 7.00 6.00 6.00 4 8.00 6.50 6.50 5 9.00 8.00 7.00 6 8.00 8.00 6.00\n\nThe general form of Anova table would look like:\n\n Source of Variation Sum of Squares Degrees of freedom Mean Squares F Treatment SST K-1 MST F=MST/MSE Blocks SSB b-1 MSB Error SSE (K-1)(b-1) MSB Total SST nt-1\n\nTo find the Excel output for the above data the following steps can be taken:\n\nStep 1. From the menus select Tools and click on Data Analysis option.\n\nStep2. When data analysis box appears: select Anova two-factor without replication then Enter A2: D8 in the input range. Select labels in first row.\n\nStep3. Select an output range (in here we selected A11) then OK.\n\n SUMMARY COUNT SUM AVERAGE VARIANCE 1 3 24.5 8.166667 2.583333 2 3 21 7 1 3 3 19.5 6.5 0.25 4 3 21.5 7.166667 0.583333 5 3 23 7.666667 2.333333 6 3 22 7.333333 1.333333 ARC 6 50 8.333333 1.066667 CSI 6 43 7.166667 0.666667 TCC 6 38.5 6.416667 0.241667\n\nANOVA\n\n Source of Variation SS df MS F P-value F crit Rows 4.902778 5 0.980556 1.972067 0.168792 3.325837 Columns 11.19444 2 5.597222 11.25698 0.002752 4.102816 Error 4.972222 10 0.497222 Total 21.06944 17\n\nNOTE: F=MST/MSE =0.980556/0.497222 = 1.972067\nF = 3.33 from table (5 numerator DF and 10 denominator DF)\nSince 1.972067<3.33 we fail to reject the null.\n\nConclusion: There is not sufficient evidence to conclude that hourly rates differ for the three departments.\n\nTwo-Way ANOVA with Replication\n\nReferring to the student assistant and the work study hourly wages here at the university of Baltimore the following data shows the hourly wages for the two categories in three different departments:\n\n ARC CSI TCC 6.50 6.10 6.90 Work Study 6.80 6.00 7.20 7.10 6.50 7.10 7.40 6.80 7.50 Student Assistant 7.50 7.00 7.00 8.00 6.60 7.10 Factors\n\nFactor A: Student job category (in here two different job categories exists)\n\nFactor B: Departments (in here we have three departments)\n\nReplication: The number of students in each experimental condition. In this case there are three replications.\n\nInteraction:\n\n ARC CSI TCC 6.50 6.10 6.90 Work Study 6.80 6.00 7.20 7.10 6.50 7.10 7.40 6.80 7.50 Student Assistant 7.50 7.00 7.00 8.00 6.60 7.10\n\n SUMMARY ARC CSI TCC Total Count 3 3 3 9 Sum 20.4 19 21 60.2 Average 6.8 6.2 7.1 6.69 Variance 0.09 0.1 0 0.19\n\n Count 3 3 3 9 Sum 22.9 20 22 64.9 Average 7.63333 6.8 7.2 7.21 Variance 0.10333 0 0.1 0.18 Total\n\n Total Count 6 6 6 Sum 43.3 39 43 Average 7.21667 6.5 7.1 Variance 0.28567 0.2 0\n\nANOVA\n Source of Variation SS df MS F P-value F crit Sample(Factor A) 1.22722 1 1.2 18.6 0.001016557 4.747221 Columns(Factor B) 1.84333 2 0.9 13.9 0.000741998 3.88529 Interaction 0.38111 2 0.2 2.88 0.095003443 3.88529 Within 0.79333 12 0.1 Total 4.245 17\n\nConclusion:\nMean hourly income differ by job category.\nMean hourly income differ by department.\nInteraction is not significant.\n\n### Goodness-of-Fit Test for Discrete Random Variables\n\nThe CHI-SQUARE distribution can be used in a hypothesis test involving a population variance. However, in this section we would like to test and see how close a sample results are to the expected results.\n\nExample: The Multinomial Random Variable\n\nIn this example the objective is to see whether or not based on a randomly selected sample information the standards set for a population is met. There are so many practical examples that can be used in this situation. For example it is assumed the guidelines for hiring people with different ethnic background for the US government is set at 70%(WHITE), 20%(African American) and 10%(others), respectively. A randomly selected sample of 1000 US employees shows the following results that is summarized in a table.\n\n ETHNIC BACKGROUND EXPECTED NUMBER OF EMPLOYEES OBSERVED FROM SAMPLE WHITE 700 =70%OF 1000 750 AFRICAN American 200 =20%OF 1000 170 OTHERS 100 =10%OF 1000 80\n\nAs you see the observed sample numbers for groups two and three are lower than their expected values unlike group one which has a higher expected value. Is this a clear sign of discrimination with respect to ethnic background? Well depends on how much lower the expected values are. The lower amount might not statistically be significant. To see whether these differences are significant we can use Excel and find the value of the CHI-SQUARE. If this value falls within the acceptance region we can assume that the guidelines are met otherwise they are not. Now lets enter these numbers into Excel spread- sheet. We used cells B7-B9 for the expected proportions, C7-C9 for the observed values and D7-D9 for the expected frequency. To calculate the expected frequency for a category, you can multiply the proportion of that category by the sample size (in here 1000). The formula for the first cell of the expected value column, D7 is 1000B7. To find other entries in the expected value column, use the copy and the paste menu as shown in the following picture. These are important values for the chi-square test. The observed range in this case is C7: C9 while the expected range is D7: D9. The null and the alternative hypothesis for this test are as follows:\n\nH0 : PW = 0.70, PA=0.20 and PO =0.10\n\nHA: The population proportions are not PW = 0.70, PA= 0.20 and PO = 0.10\n\nNow lets use Excel to calculate the p-value in a CHI-SQUARE test. Step 1.Select a cell in the work sheet, the location which you like the p value of the CHI-SQUARE to appear. We chose cell D12.\n\nStep 2. From the menus, select insert then click on the Function option, Paste Function dialog box appears.\n\nStep 3.Refer to function category box and choose statistical, from function name box select CHITEST and click on OK.\n\nStep 4.When the CHITEST dialog appears:\nEnter C7: C9 in the actual-range box then enter D7: D9 in the expected-range box, and finally click on OK.\n\nThe p-value will appear in the selected cell, D12.\n\nAs you see the p value is 0.002392 which is less than the value of the level of significance (in this case the level of significance, a= 0.10). Hence the null hypothesis should be rejected. This means based on the sample information the guidelines are not met. Notice if you type \"=CHITEST(C7:C9,D7:D9)\" in the formula bar the p-value will show up in the designated cell.\n\nNOTE: Excel can actually find the value of the CHI-SQUARE. To find this value first select an empty cell on the spread sheet then in the formula bar type \"=CHIINV(D12,2).\" D12 designates the p-Value found previously and 2 is the degrees of freedom (number of rows minus one). The CHI-SQUARE value in this case is 12.07121. If we refer to the CHI-SQUARE table we will see that the cut off is 4.60517 since 12.07121>4.60517 we reject the null. The following screen shot shows you how to the CHI-SQUARE value.\n\n### Test of Independence: Contingency Tables\n\nThe CHI-SQUARE distribution is also used to test and see whether two variables are independent or not. For example based on sample data you might want to see whether smoking and gender are independent events for a certain population. The variables of interest in this case are smoking and the gender of an individual. Another example in this situation could involve the age range of an individual and his or her smoking habit. Similar to case one data may appear in a table but unlike the case one this table may contains several columns in addition to rows. The initial table contains the observed values. To find expected values for this table we set up another table similar to this one. To find the value of each cell in the new table we should multiply the sum of the cell column by the sum of the cell row and divide the results by the grand total. The grand total is the total number of observations in a study. Now based on the following table test whether or not the smoking habit and gender of the population that the following sample taken from are independent. On the other hand is that true that males in this population smoke more than females?\n\nYou could use formula bar to calculate the expected values for the expected range. For example to find the expected value for the cell C5 which is replaced in c11 you could click on the formula bar and enter C6D5/D6 then enter in cell C11.\n\nStep 1. Observed Range b4:c5\n\nSmoking and gender\n\n yes no total male 31 69 100 female 45 122 167 total 76 191 267\n\nStep2. Expected Range b10:c11\n\n 28.4644 71.5356 47.5356 119.464\n\nSo the observed range is b4:c5 and the expected range is b10:c11.\n\nStep 3. Click on fx(paste function)\n\nStep 4. When Paste Function dialog box appears, click on Statistical in function category and CHITEST in the function name then click OK.\n\nWhen the CHITEST box appears, enter b4:c5 for the actual range, then b10:c11 for the expected range.\n\nStep 5. Click on OK (the p-value appears). 0.477395\n\nConclusion: Since p-value is greater than the level of significance (0.05), fails to reject the null. This means smoking and gender are independent events. Based on sample information one can not assure females smoke more than males or the other way around.\n\nStep 6. To find the chi-square value, use CHINV function, when Chinv box appears enter 0.477395 for probability part, then 1 for the degrees of freedom.\n\nDegrees of freedom=(number of columns-1)X(number of rows-1)\n\nCHI-SQUARE=0.504807\n\n### Test Hypothesis Concerning the Variance of Two Populations\n\nIn this section we would like to examine whether or not the variances of two populations are equal. Whenever independent simple random samples of equal or different sizes such as n1 and n2 are taken from two normal distributions with equal variances, the sampling distribution of s12/s22 has F distribution with n1- 1 degrees of freedom for the numerator and n2 - 1 degrees of freedom for the denominator. In the ratio s12/s22 the numerator s12 and the denominator s22 are variances of the first and the second sample, respectively. The following figure shows the graph of an F distribution with 10 degrees of freedom for both the numerator and the denominator. Unlike the normal distribution as you see the F distribution is not symmetric. The shape of an F distribution is positively skewed and depends on the degrees of freedom for the numerator and the denominator. The value of F is always positive.\n\nNow let see whether or not the variances of hourly income of student-assistant and work-study students based on samples taken from populations previously are equal. Assume that the hypothesis test in this case is conducted at a = 0.10. The null and the alternative are:", null, "Rejection Rule: Reject the null hypothesis if F< F0.095 or F> F0.05 where F, the value of the test statistic is equal to s12/s22, with 10 degrees of freedom for both the numerator and the denominator. We can find the value of F.05 from the F distribution table. If s12/s22, we do not need to know the value of F0.095 otherwise, F0.95 = 1/ F0.05 for equal sample sizes.", null, "A survey of eleven student-assistant and eleven work-study students shows the following\ndescriptive statistics. Our objective is to find the value of s12/s22, where s12 is the value of the variance of student assistant sample and s22 is the value of the variance of the work study students sample. As you see these values are in cells F8 and D8 of the descriptive statistic output.", null, "To calculate the value of s12/s22, select a cell such as A16 and enter cell formula = F8/D8 and enter. This is the value of F in our problem. Since this value, F=1.984615385, falls in acceptance area we fail to reject the null hypothesis. Hence, the sample results do support the conclusion that student assistants hourly income variance is equal to the work study students hourly income variance. The following screen shoot shows how to find the F value. We can follow the same format for one tail test(s).", null, "### Linear Correlation and Regression Analysis\n\nIn this section the objective is to see whether there is a correlation between two variables and to find a model that predicts one variable in terms of the other variable. There are so many examples that we could mention but we will mention the popular ones in the world of business. Usually independent variable is presented by the letter x and the dependent variable is presented by the letter y. A business man would like to see whether there is a relationship between the number of cases of sold and the temperature in a hot summer day based on information taken from the past. He also would like to estimate the number cases of soda which will be sold in a particular hot summer day in a ball game. He clearly recorded temperatures and number of cases of soda sold on those particular days. The following table shows the recorded data from June 1 through June 13. The weatherman predicts a 94F degree temperature for June 14. The businessman would like to meet all demands for the cases of sodas ordered by customers on June 14.\n\n DAY Cases of Soda Temperature 1-Jun 57 56 2-Jun 59 58 3-Jun 65 63 4-Jun 67 66 5-Jun 75 73 6-Jun 81 78 7-Jun 86 85 8-Jun 88 85 9-Jun 88 87 10-Jun 84 84 11-Jun 82 88 12-Jun 80 84 13-Jun 83 89\n\nNow lets use Excel to find the linear correlation coefficient and the regression line equation. The linear correlation coefficient is a quantity between -1 and +1. This quantity is denoted by R. The closer R to +1 the stronger positive (direct) correlation and similarly the closer R to -1 the stronger negative (inverse) correlation exists between the two variables. The general form of the regression line is y = mx + b. In this formula, m is the slope of the line and b is the y-intercept. You can find these quantities from the Excel output. In this situation the variable y (the dependent variable) is the number of cases of soda and the x (independent variable) is the temperature. To find the Excel output the following steps can be taken:\n\nStep 1. From the menus choose Tools and click on Data Analysis.\n\nStep 2. When Data Analysis dialog box appears, click on correlation.\n\nStep 3. When correlation dialog box appears, enter B1:C14 in the input range box. Click on Labels in first row and enter a16 in the output range box. Click on OK.\n\n Cases of Soda Temperature Cases of Soda 1 Temperature 0.96659877 1\n\nAs you see the correlation between the number of cases of soda demanded and the temperature is a very strong positive correlation. This means as the temperature increases the demand for cases of soda is also increasing. The linear correlation coefficient is 0.966598577 which is very close to +1.\n\nNow lets follow same steps but a bit different to find the regression equation.\n\nStep 1. From the menus choose Tools and click on Data Analysis\n\nStep 2. When Data Analysis dialog box appears, click on regression.\n\nStep 3. When Regression dialog box appears, enter b1:b14 in the y-range box and c1:c14 in the x-range box. Click on labels.\n\nStep 4. Enter a19 in the output range box.\n\nNote: The regression equation in general should look like Y=m X + b. In this equation m is the slope of the regression line and b is its y-intercept.\n\nSUMMARY OUTPUT\n\n Multiple R 0.966599 R Square 0.934313 Adjusted R Square 0.928341 Standard Error 2.91938 Observations 13\n\nANOVA\n\n df SS MS F Significance F Regression 1 1333.479989 1333.479989 156.4603497 7.58511E-08 Residual 11 93.75078034 8522798213 Total 12 1427.230769\n\n Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 9.17800767 5.445742836 1.685354587 0.120044801 -2.80799756 21.16401 Temperature 0.879202711 0.07028892 12.50841116 7.58511E-08 0.724497763 1.033908\n\nThe relationship between the number of cans of soda and the temperature is: Y = 0.879202711 X + 9.17800767\n\nThe number of cans of soda = 0.879202711(Temperature) + 9.17800767. Referring to this expression we can approximately predict the number of cases of soda needed on June 14. The weather forecast for this is 94 degrees, hence the number of cans of soda needed is equal to; The number of cases of soda=0.879202711*(94) + 9.17800767 = 91.82 or about 92 cases.\n\n### Moving Average and Exponential Smoothing\n\nMoving Average Models: Use the Add Trendline option to analyze a moving average forecasting model in Excel. You must first create a graph of the time series you want to analyze. Select the range that contains your data and make a scatter plot of the data. Once the chart is created, follow these steps:\n1. Click on the chart to select it, and click on any point on the line to select the data series. When you click on the chart to select it, a new option, Chart, s added to the menu bar.\n2. From the Chart menu, select Add Trendline.\n\nThe following is the moving average of order 4 for weekly sales:", null, "Exponential Smoothing Models: The simplest way to analyze a timer series using an Exponential Smoothing model in Excel is to use the data analysis tool. This tool works almost exactly like the one for Moving Average, except that you will need to input the value of a instead of the number of periods, k. Once you have entered the data range and the damping factor, 1- a , and indicated what output you want and a location, the analysis is the same as the one for the Moving Average model.\n\n### Applications and Numerical Examples\n\nDescriptive Statistics: Suppose you have the following, n = 10, data:\n\n1.2, 1.5, 2.6, 3.8, 2.4, 1.9, 3.5, 2.5, 2.4, 3.0\n\n1. Type your n data points into the cells A1 through An.\n2. Click on the \"Tools\" menu. (At the bottom of the \"Tools\" menu will be a submenu \"Data Analysis...\", if the Analysis Tool Pack has been properly installed.)\n3. Clicking on \"Data Analysis...\" will lead to a menu from which \"Descriptive Statistics\" is to be selected.\n4. Select \"Descriptive Statistics\" by pointing at it and clicking twice, or by highlighting it and clicking on the \"Okay\" button.\n5. Within the Descriptive Statistics submenu,\na. for the \"input range\" enter \"A1:Dn\", assuming you typed the data into cells A1 to An.\n\nb. click on the \"output range\" button and enter the output range \"C1:C16\".\n\nc. click on the Summary Statistics box\n\nd. finally, click on \"Okay.\"\n\nThe Central Tendency: The data can be sorted in ascending order:\n\n1.2, 1.5, 1.9, 2.4, 2.4, 2.5, 2.6, 3.0, 3.5, 3.8\n\nThe mean, median and mode are computed as follows:\n\n(1.2 1.5 2.6 3.8 2.4 1.9 3.5 2.5 2.4 3.0) / 10 = 2.48\n\n(2.4 + 2.5) / 2 = 2.45\n\nThe mode is 2.4, since it is the only value that occurs twice.\n\nThe midrange is (1.2+ 3.8) / 2 = 2.5.\n\nNote that the mean, median and mode of this set of data are very close to each other. This suggests that the data is very symmetrically distributed.\n\nVariance: The variance of a set of data is the average of the cumulative measure of the squares of the difference of all the data values from the mean.\n\nThe sample variance-based estimation for the population variance are computed differently. The sample variance is simply the arithmetic mean of the squares of the difference between each data value in the sample and the mean of the sample. On the other hand, the formula for an estimate for the variance in the population is similar to the formula for the sample variance, except that the denominator in the fraction is (n-1) instead of n. However, you should not worry about this difference if the sample size is large, say over 30. Compute an estimate for the variance of the population, given the following sorted data:\n\n1.2, 1.5, 1.9, 2.4, 2.4, 2.5, 2.6, 3.0, 3.5, 3.8 mean = 2.48 as computed earlier. An estimate for the population variance is: s2 = 1 / (10-1) [ (1.2 - 2.48)2 + (1.5 - 2.48)2 + (1.9 - 2.48)2 + (2.4 -2.48)2 + (2.4 - 2.48)2 + (2.5 - 2.48)2 + (2.6 - 2.48)2 + (3.0 - 2.48)2 + (3.5 -2.48)2 + (3.8 - 2.48)2 ]\n= (1 / 9) (1.6384 + 0.9604 + 0.3364 + 0.0064 + 0.0064 + 0.0004 + 0.0144 + 0.2704 + 1.0404 + 1.7424) = 0.6684\n\nTherefore, the standard deviation is s = ( 0.6684 )1/2 = 0.8176\n\nProbability and Expected Values: Newsweek reported that \"average take\" for bank robberies was \\$3,244 but 85 percent of the robbers were caught. Assuming 60 percent of those caught lose their entire take and 40 percent lose half, graph the probability mass function using EXCEL. Calculate the expected take from a bank robbery. Does it pay to be a bank robber?\n\nTo construct the probability function for bank robberies, first define the random variable x, bank robbery take. If the robber is not caught, x = \\$3,244. If the robber is caught and manages to keep half, x = \\$1,622. If the robber is caught and loses it all, then x = 0. The associated probabilities for these x values are 0.15 = (1 - 0.85), 0.34 = (0.85)(0.4), and 0.51 = (0.85)(0.6). After entering the x values in cells A1, A2 and A3 and after entering the associated probabilities in B1, B2, and B3, the following steps lead to the probability mass function:\n\n1. Click on ChartWizard. The \"ChartWizard Step 1 of 4\" screen will appear.\n2. Highlight \"Column\" at \"ChartWizard Step 1 of 4\" and click \"Next.\"\n3. At \"ChartWizard Step 2 of 4 Chart Source Data,\" enter \"=B1:B3\" for \"Data range,\" and click \"column\" button for \"Series in.\" A graph will appear. Click on \"series\" toward the top of the screen to get a new page.\n4. At the bottom of the \"Series\" page, is a rectangle for \"Category (X) axis labels:\" Click on this rectangle and then highlight A1:A3.\n5. At \"Step 3 of 4\"; move on by clicking on \"Next,\" and at \"Step 4 of 4\", click on \"Finish.\"\n\nThe expected value of a robbery is \\$1,038.08.\n\nE(X) = (0)(0.51)+(1622)(0.34) + (3244)(0.15) = 0 + 551.48 + 486.60 = 1038.08\n\nThe expected return on a bank robbery is positive. On average, bank robbers get \\$1,038.08 per heist. If criminals make their decisions strictly on this expected value, then it pays to rob banks. A decision rule based only on an expected value, however, ignores the risks or variability in the returns. In addition, our expected value calculations do not include the cost of jail time, which could be viewed by criminals as substantial.\n\nDiscrete & Continuous Random Variables:\n\nBinomial Distribution Application: A multiple choice test has four unrelated questions. Each question has five possible choices but only one is correct. Thus, a person who guesses randomly has a probability of 0.2 of guessing correctly. Draw a tree diagram showing the different ways in which a test taker could get 0, 1, 2, 3 and 4 correct answers. Sketch the probability mass function for this test. What is the probability a person who guesses will get two or more correct?\n\nSolution: Letting Y stand for a correct answer and N a wrong answer, where the probability of Y is 0.2 and the probability of N is 0.8 for each of the four questions, the probability tree diagram is shown in the textbook on page 182. This probability tree diagram shows the \"branches\" that must be followed to show the calculations captured in the binomial mass function for n = 4 and = 0.2. For example, the tree diagram shows the six different branch systems that yield two correct and two wrong answers (which corresponds to 4!/(2!2!) = 6. The binomial mass function shows the probability of two correct answers as\n\nP(x = 2 \| n = 4, p = 0.2) = 6(.2)2(.8)2 = 6(0.0256) = 0.1536 = P(2)\n\nWhich is obtained from excel by using the \"BINOMDIST\" Command, where the first entry is x, the second is n, and the third is mass (0) or cumulative (1); that is, entering\n\n=BINOMDIST(2,4,0.2,0) IN ANY EXCEL CELL YIELDS 0.1536 AND\n=BINOMDIST(3,4,0.2,0) YIELDS P(x=3\|n=4, p = 0.2) = 0.0256\n=BINOMDIST(4,4,0.2,0) YIELDS P(x=4\|n=4, p = 0.2) = 0.0016\n=1-BINOMDIST(1,4,0.2,1) YIELDS P(x ³ 2 \| n = 4, p = 0.2) = 0.1808\n\nNormal Example: If the time required to complete an examination by those with a certain learning disability is believed to be distributed normally, with mean of 65 minutes and a standard deviation of 15 minutes, then when can the exam be terminated so that 99 percent of those with the disability can finish?\n\nSolution: Because the average and standard deviation are known, what needs to be established is the amount of time, above the mean time, such that 99 percent of the distribution is lower. This is a distance that is measured in standard deviations as given by the Z value corresponding to the 0.99 probability found in the body of Appendix B, Table 5,as shown in the textbook OR the commands entered into any cell of Excel to find this Z value is =NORMINV(0.99,0,1) for 2.326342.\n\nThe closest cumulative probability that can be found is 0.9901, in the row labeled 2.3 and column headed by .03, Z = 2.33, which is only an approximation for the more exact 2.326342 found in Excel. Using this more exact value the calculation with mean m and standard deviation s in the following formula would be\n\nZ = ( X - m ) / s\nThat is, Z = ( x - 65)/15\nThus, x = 65 + 15(2.32634) = 99.9 minutes.\n\nAlternatively, instead of standardizing with the Z distribution using Excel we can simply work directly with the normal distribution with a mean of 65 and standard deviation of 15 and enter \"=NORMINV(0.99,65,15)\". In general to obtain the x value for which alpha percent of a normal random variable's values are lower, the following \"NORMINV\" command may be used, where the first entry is a, the second is m , and the third is s.\n\nAnother Example: In the early 1980s, the Toro Company of Minneapolis, Minnesota, advertised that it would refund the purchase price of a snow blower if the following winter's snowfall was less than 21 percent of the local average. If the average snowfall is 45.25 inches, with a standard deviation of 12.2 inches, what is the likelihood that Toro will have to make refunds?\n\nSolution: Within limits, snowfall is a continuous random variable that can be expected to vary symmetrically around its mean, with values closer to the mean occurring most often. Thus, it seems reasonable to assume that snowfall (x) is approximately normally distributed with a mean of 45.25 inches and standard deviation of 12.2 inches. Nine and one half inches is 21 percent of the mean snowfall of 45.25 inches and, with a standard deviation of 12.2 inches, the number of standard deviations between 45.25 inches and 9.5 inches is Z:\n\nZ = ( x - m ) / s = (9.50 - 45.25)/12.2 = -2.93\n\nUsing Appendix B, Table 5, the textbook demonstrates the determination of P(x £ 9.50) = P(z £ -2.93) = 0.17, the probability of snowfall less than 9.5 inches. Using Excel, this normal probability is obtained with the \"NORMDIST\" command, where the first entry is x, the second is mean m , the third is standard deviation s, and the fourth is CUMULATIVE (1). Entering\n\n=NORMDIST(9.5,45.25,12.2,1), Gives P( x £ 9.50) = 0.001693.\n\nSampling Distribution and the Central Limit Theorem : A bakery sells an average of 24 loaves of bread per day. Sales (x) are normally distributed with a standard deviation of 4.\n\nIf a random sample of size n = 1 (day) is selected, what is the probability this x value will exceed 28?\n\nIf a random sample of size n = 4 (days) is selected, what is theprobability that xbar ³ 28?\n\nWhy does the answer in part 1 differ from that in part 2?\n\nSolutions:\n\n1. The sampling distribution of the sample mean xbar is normal with a mean of 24 and a standard error of the mean of 4. Thus, using Excel, 0.15866 =1-NORMDIST(28,24,4,1).\n\n2. The sampling distribution of the sample mean xbar is normal with a mean of 24 and a standard error of the mean of 2 using Excel, 0.02275 =1-NORMDIST(28,24,2,1).\n\nRegression Analysis: The highway deaths per 100 million vehicle miles and highway speed limits for 10 countries, are given below:\n\n(Death, Speed) = (3.0, 55), (3.3, 55), (3.4, 55), (3.5, 70), (4.1, 55), (4.3, 60), (4.7, 55), (4.9, 60), (5.1, 60), and (6.1, 75).\n\nFrom this we can see that five countries with the same speed limit have very different positions on the safety list. For example, Britain ... with a speed limit of 70 is demonstrably safer than Japan, at 55. Can we argue that, speed has little to do with safety. Use regression analysis to answer this question.\n\nSolution: Enter the ten paired y and x data into cells A2 to A11 and B2 to B11, with the \"death\" rate label in A1 and \"speed\" limits label in B1, the following steps produce the regression output.\n\nChoose \"Regression\" from \"Data Analysis\" in the \"Tools\" menu. The Regression dialog box will will appear.\n\nNote: Use the mouse to move between the boxes and buttons. Click on the desired box or button. The large rectangular boxes require a range from the worksheet. A range may be typed in or selected by highlighting the cells with the mouse after clicking on the box. If the dialog box blocks the data, it can be moved on the screen by clicking on the title bar and dragging.\n\nFor the \"Input Y Range,\" enter A1 to A11, and for the \"Input X Range\" enter B1 to B11.\n\nBecause the Y and X ranges include the \"Death\" and \"Speed\" labels in A1 and B1, select the \"Labels\" box with a click.\n\nClick the \"Output Range\" button and type reference cell, which in this demonstration is A13.\n\nTo get the predicted values of Y (Death rates) and residuals select the \"Residuals\" box with a click.\n\nYour screen display should show a Table, clicking \"OK\" will give the \"SUMMARY OUTPUT,\" \"ANOVA\" AND RESIDUAL OUTPUT\"\n\nThe first section of the EXCEL printout gives \"SUMMARY OUTPUT.\" The \"Multiple R\" is the square root of the \"R Square;\" the computation and interpretation of which we have already discussed. The \"Standard Error\" of estimate (which will be discussed in the next chapter) is s = 0.86423, which is the square root of \"Residual SS\" = 5.97511 divided by its degrees of freedom, df = 8, as given in the \"ANOVA\" section. We will also discuss the adjusted R-square of 0.21325 in the following chapters.\n\nUnder the \"ANOVA\" section are the estimated regression coefficients and related statistics that will be discussed in detail in the next chapter. For now it is sufficient to recognize that the calculated coefficient values for the slope and y intercept are provided (b = 0.07556 and a = -0.29333). Next to these coefficient estimates is information on the variability in the distribution of the least-squares estimators from which these specific estimates were drawn: the column titled \"Std. Error\" contains the standard deviations (standard errors) of the intercept and slope distributions; the \"t-ratio\" and \"p\" columns give the calculated values of the t statistics and associated p-values. As shown in Chapter 13, the t statistic of 1.85458 and p-value of 0.10077, for example, indicates that the sample slope (0.07556) is sufficiently different from zero, at even the 0.10 two-tail Type I error level, to conclude that there is a significant relationship between deaths and speed limits in the population. This conclusion is contrary to assertion that \"speed has little to do with safety.\"\n\nSUMMARY OUTPUT: Multiple R = 0.54833, R Square = 0.30067, Adjusted R Square = 0.21325, Standard Error = 0.86423, Observations = 10\n\nANOVA df SS MS F P-value\nRegression 1 2.56889 2.56889 3.43945 0.10077\nResidual 8 5.97511 0.74689\nTotal 9 8.54400\n\nCoeffs. Estimate Std. Error T Stat P-value Lower 95% Upper 95%\nIntercept -0.29333 2.45963 -0.11926 0.90801 -5.96526 5.37860\nSpeed 0.07556 0.04074 1.85458 0.10077 -0.01839 0.16950\n\nResidual Output:\n\nPredicted Residuals\n3.86222 -0.86222\n3.86222 -0.56222\n3.86222 -0.46222\n4.99556 -1.49556\n3.86222 0.23778\n4.24000 0.06000\n3.86222 0.83778\n4.24000 0.66000\n4.24000 0.86000\n5.37333 0.72667\n\n### Microsoft Excel Add-Ins\n\nForecasting with regression requires the Excel add-in called \"Analysis ToolPak,\" and linear programming requires the Excel add-in called \"Solver.\" How you check to see if these are activated on your computer, and how to activate them if they are not active, varies with Excel version. Here are instructions for the most common versions. If Excel will not let you activate Data Analysis and Solver, you must use a different computer.\n\nExcel 2002/2003:\nStart Excel, then click Tools and look for Data Analysis and for Solver. If both are there, press Esc (escape) and continue with the respective assignment. Otherwise click Tools, Add-Ins, and check the boxes for Analysis ToolPak and for Solver, then click OK. Click Tools again, and both tools should be there.\n\nExcel 2007:\nStart Excel 2007 and click the Data tab at the top. Look to see if Data Analysis and Solver show in the Analysis section at the far right. If both are there, continue with the respective assignment. Otherwise, do the following steps exactly as indicated:\n-click the “Office Button” at top left\n-click the Excel Options button near the bottom of the resulting window\n-click the Add-ins button on the left of the next screen\n-near the bottom at Manage Excel Add-ins, click Go\n-check the boxes for Analysis ToolPak and Solver Add-in if they are not already checked, then click OK\n-click the Data tab as above and verify that the add-ins show.\n\nExcel 2010:\nStart Excel 2010 and click the Data tab at the top. Look to see if Data Analysis and Solver show in the Analysis section at the far right. If both are there, continue with the respective assignment. Otherwise, do the following steps exactly as indicated:\n-click the File tab at top left\n-click the Options button near the bottom of the left side\n-click the Add-ins button near the bottom left of the next screen\n-near the bottom at Manage Excel Add-ins, click Go\n-check the boxes for Analysis ToolPak and Solver Add-in if they are not already checked, then click OK\n-click the Data tab as above and verify that the add-ins show.\n\n### Computer-assisted Learning: E-Labs and Computational Tools\n\nMy teaching style deprecates the 'plug the numbers into the software and let the magic box work it out' approach. Personal computers, spreadsheets, e.g., Excel, professional statistical packages (e.g., such as SPSS), and other information technologies are now ubiquitous in statistical data analysis. Without using these tools, one cannot perform any realistic statistical data analysis on large data sets.\n\nThe appearance of other computer software, JavaScript Applets, Statistical Demonstrations Applets, and Online Computation are the most important events in the process of teaching and learning concepts in model-based statistical decision making courses. These tools allow you to construct numerical examples to understand the concepts, and to find their significance for yourself.\n\nUse any or online interactive tools available on the WWW to perform statistical experiments (with the same purpose, as you used to do experiments in physics labs to learn physics) to understand statistical concepts such as Central Limit Theorem are entertaining and educating.\n\nComputer-assisted learning is similar to the experiential model of learning. The adherents of experiential learning are fairly adamant about how we learn. Learning seldom takes place by rote. Learning occurs because we immerse ourselves in a situation in which we are forced to perform and think. You get feedback from the computer output and then adjust your thinking-process if needed.\n\nProfessional Software:\n\n• A SPSS-Example, SPSS-Examples, SPSS-More Examples, (Statistical Package for the Social Sciences) is a data management and analysis product. It can perform a variety of data analysis and presentation functions, including statistical analyses and graphical presentation of data.\n\nAvailable at: SPSS Package on Citrix (Installing and Accessing)\n\n• SAS (Statistical Analysis System) is a system of software packages; some of its basic functions and uses are: database management inputting, cleaning and manipulating data, statistical analysis, calculating simple statistics such as means, variances, correlations; running standard routines such as regressions.\n\nAvailable at: SPSS/SAS Packages on Citrix (Installing and Accessing) Use your email ID and Password: Technical Difficulties OTS Call Center (401) 837-6262\n\n• Excel Examples, Excel More Examples It is Excellent for Descriptive Statistics, and getting acceptance is improving, as computational tool for Inferential Statistics.\n\nThe Value of Performing Experiment: If the learning environment is focused on background information, knowledge of terms and new concepts, the learner is likely to learn that basic information successfully. However, this basic knowledge may not be sufficient to enable the learner to carry out successfully the on-the-job tasks that require more than basic knowledge. Thus, the probability of making real errors in the business environment is high. On the other hand, if the learning environment allows the learner to experience and learn from failures within a variety of situations similar to what they would experience in the \"real world\" of their job, the probability of having similar failures in their business environment is low. This is the realm of simulations-a safe place to fail.\n\nThe appearance of statistical software is one of the most important events in the process of decision making under uncertainty. Statistical software systems are used to construct examples, to understand the existing concepts, and to find new statistical properties. On the other hand, new developments in the process of decision making under uncertainty often motivate developments of new approaches and revision of the existing software systems. Statistical software systems rely on a cooperation of statisticians, and software developers.\n\nBeside the professional statistical software Online statistical computation, and the use of a scientific calculator is required for the course. A Scientific Calculator is the one, which has capability to give you, say, the result of square root of 5. Any calculator that goes beyond the 4 operations is fine for this course. These calculators allow you to perform simple calculations you need in this course, for example, enabling you to take square root, to raise e to the power of say, 0.36. and so on. These types of calculators are called general Scientific Calculators. There are also more specific and advanced calculators for mathematical computations in other areas such as Finance, Accounting, and even Statistics. The last one, for example, computes mean, variance, skewness, and kurtosis of a sample by simply entering all data one-by-one and then pressing any of the mean, variance, skewness, and kurtosis keys.\n\nWithout a computer one cannot perform any realistic statistical data analysis. Students who are signing up for the course are expected to know the basics of Excel.\n\nAs a starting point, you need visiting the Excel Web site created for this course. If you are challenged by or unfamiliar with Excel, you may seek tutorial help from the Academic Resource Center at 410-837-5385, E-mail.\n\nLeaning Tools for Statistical Concepts is a part of the E-labs learning objects for decision making.\n\nWhat and How to Hand-in My Computer Assignment? For the computer assignment I do recommend in checking your hand computation homework, and checking some of the numerical examples from your textbook. As part of your homework assignment you don not have to hand in the printout of the computer assisted learning, however, you must include within your handing homework a paragraph entitled \"Computer Implementation\" describing your (positive or negative) experience.\n\n### Interesting and Useful Sites\n\nBack to\n\nThe Copyright Statement: The fair use, according to the 1996 Fair Use Guidelines for Educational Multimedia, of materials presented on this Web site is permitted for non-commercial and classroom purposes only.\nThis site may be mirrored intact (including these notices), on any server with public access. All files are available at http://www.mirrorservice.org/sites/home.ubalt.edu/ntsbarsh/Business-stat for mirroring.\n\nKindly e-mail me your comments, suggestions, and concerns. 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http://www.nanodic.com/general/Heisenberg_Uncertainty_Principle.htm	[ " Dictionary of nanotechnology - Heisenberg Uncertainty Principle\nHeisenberg Uncertainty Principle\n\n A quantum-mechanical principle with the consequence that the position and momentum of an object cannot be precisely determined. The Heisenberg principle helps determine the size of electron clouds, and hence the size of atoms. Source A consequence from quantum mechanics: the position and the momentum of a particle can not be determined with the same accuracy at the same time. When for example the speed of an electron is well known, it's position is unpredictable within a certain volume. This effect is negligible in macroscopic dimensions. Source A quantum-mechanical principle with the consequence that the position and momentum of an object cannot be precisely determined. The Heisenberg principle helps determine the size of electron clouds, and hence the size of atoms. [NTN] \"The more precisely the POSITION is determined, the less precisely the MOMENTUM is known\" [Werner Heisenberg] Source A quantum-mechanical principle with the consequence that the position and momentum of an object cannot be precisely determined. The Heisenberg principle helps determine the size of electron clouds, and hence the size of atoms. [NTN] \"The more precisely the POSITION is determined, the less precisely the MOMENTUM is known\" [Werner Heisenberg] Source", null, "", null, "", null, "", null, "", null, "", null, "", null, "___________________ Refer to this page: ___________________ Related Terms:", null, "", null, "Note: If a company/institute/site doesn't want to present its own information in nanodic.com, it can sent one e-mail to [email protected]." ]	[ null, "http://www.nanodic.com/images/top02.gif", null, "http://www.nanodic.com/images/right_space.gif", null, "http://www.nanodic.com/tell/Image/button_black2.gif", null, "http://www.nanodic.com/images/favorites.png", null, "http://www.nanodic.com/images/print.png", null, "http://www.nanodic.com/images/SendResponse.png", null, "http://www.nanodic.com/images/saveaspdf.jpg", null, "http://www.nanodic.com/images/bottom_01.gif", null, "http://www.nanodic.com/images/copyright.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8384806,"math_prob":0.90661967,"size":1223,"snap":"2022-40-2023-06","text_gpt3_token_len":274,"char_repetition_ratio":0.14438064,"word_repetition_ratio":0.6775956,"special_character_ratio":0.17825021,"punctuation_ratio":0.0776699,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9535616,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,8,null,8,null,8,null,8,null,8,null,8,null,8,null,8,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T00:08:59Z\",\"WARC-Record-ID\":\"<urn:uuid:96852b5a-6ac1-4052-8766-d68df70c3982>\",\"Content-Length\":\"16783\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b4ce9d60-be33-4030-82dd-78ff33870d5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:700b8fca-1be0-43b2-9209-e8b07adea1b9>\",\"WARC-IP-Address\":\"185.55.224.160\",\"WARC-Target-URI\":\"http://www.nanodic.com/general/Heisenberg_Uncertainty_Principle.htm\",\"WARC-Payload-Digest\":\"sha1:AH6XYHGDYRTTCKL2D4DSBZEXYULXRN4O\",\"WARC-Block-Digest\":\"sha1:WCGCTOLADJD7OKFBLKVFKZFRWBM7IN3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337889.44_warc_CC-MAIN-20221006222634-20221007012634-00008.warc.gz\"}"}
https://piccard.info/matpower-41-28/	[ "Matpower is designed to give the best performance possible while keeping the code simple to understand and modify. Most of the formulation More information. The solution values are stored as shown in Table Solution of Linear Systems Chapter 3 Solution of Linear Systems In this chapter we study algorithms for possibly the most commonly occurring problem in scientific computing, the solution of linear systems of equations. This paper is extracted from the M. Each Newton step involves computing the mismatch g x , forming the Jacobian based on the sensitivities of these mismatches to changes in x and solving for an updated value of x by factorizing this Jacobian. The initial need for Matlab-based power flow and optimal power flow code was born out of the computational requirements of the PowerWeb project 2.", null, "", null, "", null, "An n-dimensional vector is a row or a column. An optimization problem usually has three essential ingredients: These options are passed to the simulation routines as a Matpower options vector.\n\nAs a result, for the DC case 5. Matlab is a registered trademark of The MathWorks, Inc.\n\nThe standard version of each takes the following form: April 12, Contents 2. Systems of Linear Equations Definition. The function g is convex if either of the following two conditions.", null, "The notation in this section, as well as Sections 4 and 5, is based on this internal numbering, with all generators and branches assumed to be in-service. All of Matpower s solvers exploit the sparsity of the problem and, except for Gauss-Seidel, scale well to very large systems. W03 Analysis of DC Circuits.\n\nmatpowe", null, "Enhanced Fritz John Optimality Conditions In this example, it allows us to access the real power demand column of the bus matrix using the name PD without having to remember that it is the 3 rd column. The bus data includes the voltage, angle and total generation mqtpower load at each bus. An Example Consider the following linear program: Algebra I Part IV: Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear More information.\n\nIntroduction to PowerWorld Simulator: In the first form, the objective is to maximize, the material.\n\n## Matpower 4.1 User s Manual\n\nChristopher Lum lum u. The parameters r s, x s, b c, and shift are specified directly in columns 3, 4, 5, 9 and 10, respectively, of the corresponding row of the branch matrix.\n\nFurthermore, the n l n b sparse connection matrices C f and C t used in building matpowfr system admittance matrices can be defined as follows. Follow the download instructions on the Matpower home page 6. Familiarization with the Network Analyzer Measurements to characterize networks at high frequencies RF and microwave frequencies are usually done in terms of scattering parameters Matpoder parameters.\n\nAlgorithms and Applications Matrix Math Review The purpose of this document is to give a brief review of selected linear algebra concepts that will be useful for the course and to develop More information." ]	[ null, "https://www.researchgate.net/profile/Ray_Zimmerman/publication/265542880/figure/tbl2/AS:669378353836047@1536603644478/2-Power-Flow-Options_Q320.jpg", null, "https://piccard.info/download.png", null, "https://imgv2-1-f.scribdassets.com/img/document/314185815/149x198/ffc361076f/1464551614", null, "https://docplayer.net/docs-images/40/13628725/images/page_4.jpg", null, "https://image.slidesharecdn.com/matpowermanual-151121185035-lva1-app6892/95/mat-power-manual-41-638.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8530879,"math_prob":0.9352683,"size":4268,"snap":"2020-34-2020-40","text_gpt3_token_len":899,"char_repetition_ratio":0.104831144,"word_repetition_ratio":0.038235296,"special_character_ratio":0.19868791,"punctuation_ratio":0.102827765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928527,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,5,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-24T02:17:53Z\",\"WARC-Record-ID\":\"<urn:uuid:565858c9-7c7e-4bb6-b6ab-4ce6698af751>\",\"Content-Length\":\"36562\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ca56992-e53a-4aa7-8ce0-4ec68c913d1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:0228bcc5-7b37-4369-a1ad-28fcce395c3e>\",\"WARC-IP-Address\":\"172.67.212.5\",\"WARC-Target-URI\":\"https://piccard.info/matpower-41-28/\",\"WARC-Payload-Digest\":\"sha1:OB3LHXZNPB5AEQ6ZGJ7VCIGZS2GTAIWI\",\"WARC-Block-Digest\":\"sha1:EKXN47QTQ5SAFV7Q3HHVOEM4WQPPSWRA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400213006.47_warc_CC-MAIN-20200924002749-20200924032749-00388.warc.gz\"}"}
https://zoidberg.ukp.informatik.tu-darmstadt.de/jenkins/job/DKPro%20Text%20Classification%20Framework%20(GitHub)/javadoc/org/dkpro/tc/ml/weka/evaluation/MulanEvaluationWrapper.html	[ "org.dkpro.tc.ml.weka.evaluation\n\n## Class MulanEvaluationWrapper\n\n• ```public class MulanEvaluationWrapper\nextends Object```\nA wrapper for evaluation measures calculated by the Mulan framework for multi-label classification.\n• ### Constructor Summary\n\nConstructors\nConstructor and Description\n`MulanEvaluationWrapper()`\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`static boolean[][]` `getBooleanMatrix(int[][] actuals)`\nConverts a list of {0,1}-integer arrays into a boolean-matrix.\n`static List<mulan.evaluation.measure.Measure>` ```getMeasures(mulan.classifier.MultiLabelOutput prediction, int numOfLabels, boolean strict)```\n`static List<mulan.evaluation.measure.Measure>` ```getMulanEvals(double[][] predictions, boolean[][] actuals, double threshold)```\nRetrieves evaluation measures calculated by the Mulan framework for multi-label classification\n`static mulan.evaluation.measure.Measure` ```getMulanMeasure(double[][] predictions, boolean[][] actuals, double[] thresholds, mulan.evaluation.measure.Measure m)```\nRetrieves a single evaluation measure calculated by the Mulan framework for multi-label classification\n• ### Methods inherited from class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Constructor Detail\n\n• #### MulanEvaluationWrapper\n\n`public MulanEvaluationWrapper()`\n• ### Method Detail\n\n• #### getMulanEvals\n\n```public static List<mulan.evaluation.measure.Measure> getMulanEvals(double[][] predictions,\nboolean[][] actuals,\ndouble threshold)```\nRetrieves evaluation measures calculated by the Mulan framework for multi-label classification\nParameters:\n`predictions` - predictions by the classifier\n`actuals` - gold standard\n`threshold` - a threshold to create bipartitions from rankings\nReturns:\nmeasures as defined in `getMeasures(MultiLabelOutput, int, boolean)`\n• #### getMeasures\n\n```public static List<mulan.evaluation.measure.Measure> getMeasures(mulan.classifier.MultiLabelOutput prediction,\nint numOfLabels,\nboolean strict)```\n• #### getBooleanMatrix\n\n`public static boolean[][] getBooleanMatrix(int[][] actuals)`\nConverts a list of {0,1}-integer arrays into a boolean-matrix.\nParameters:\n`actuals` - a list of {0,1}-integer arrays\nReturns:\na matrix holding only boolean values\n• #### getMulanMeasure\n\n```public static mulan.evaluation.measure.Measure getMulanMeasure(double[][] predictions,\nboolean[][] actuals,\ndouble[] thresholds,\nmulan.evaluation.measure.Measure m)\nthrows IOException```\nRetrieves a single evaluation measure calculated by the Mulan framework for multi-label classification\nParameters:\n`predictions` - predictions by the classifier\n`actuals` - gold standard\n`thresholds` - a threshold to create bipartitions from rankings (one per instance)\n`m` - the measure\nReturns:\nthe updated measure\nThrows:\n`IOException` - an exception\n\nCopyright © 2013–2019 Ubiquitous Knowledge Processing (UKP) Lab. All rights reserved." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5639317,"math_prob":0.91989803,"size":1898,"snap":"2019-13-2019-22","text_gpt3_token_len":406,"char_repetition_ratio":0.17001057,"word_repetition_ratio":0.22222222,"special_character_ratio":0.18967333,"punctuation_ratio":0.18560606,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98103815,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T16:01:09Z\",\"WARC-Record-ID\":\"<urn:uuid:3668e8e7-eb10-46fb-bb5a-253edc385f82>\",\"Content-Length\":\"19068\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e542a2f-207b-48e5-99de-cd87c78e1eb3>\",\"WARC-Concurrent-To\":\"<urn:uuid:03a8d2cc-ac9c-4847-ab02-c64696da8474>\",\"WARC-IP-Address\":\"130.83.167.140\",\"WARC-Target-URI\":\"https://zoidberg.ukp.informatik.tu-darmstadt.de/jenkins/job/DKPro%20Text%20Classification%20Framework%20(GitHub)/javadoc/org/dkpro/tc/ml/weka/evaluation/MulanEvaluationWrapper.html\",\"WARC-Payload-Digest\":\"sha1:UJ5YELZ5ESMV7VXJDUWFACE3IIS25WQ7\",\"WARC-Block-Digest\":\"sha1:HGFXH22I4AZJBHA6GODVU3QR4DQPJSJP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256858.44_warc_CC-MAIN-20190522143218-20190522165218-00339.warc.gz\"}"}
https://www.financehomeworkhelp.org/how-to-calculate-capitalization-rate/	[ "# Find out How to Calculate Capitalization Rate\n\n## Capitalization Rate: How to Calculate", null, "Assignment:\n\nXYZ Ltd., expects a net income of Rs. 1,50,000. The company has 10% of\n5,00,000 Debentures. The equity capitalization rate of the company is 10%.\n(a) Calculate the value of the firm and overall capitalization rate according to\nthe net income approach (ignoring income tax).\n(b) If the debenture debt is increased to Rs. 7,50,000 and interest of debt is change\nto 9%. What is the value of the firm and overall capitalization rate?a. Calculation the firm value and overall capitalization rate according to the net income approach.\n\nAccording to net income approach, the firm value is the summary its debt and equity. The cost of debt and cost of equity are independent. The main purpose of any company is raising its value using the most optimal mix of debt and equity. While the firm will achieve to maximize its value, the overall capitalization rate will be declined. So how to calculate capitalization rate? The overall capitalization rate means how much net operating income is generated by the dollar of market value its assets. And what is the value of the firm and how do business analysis and valuation work? The value of the company (V) is defined the following formulas, where S is the sum of market value of its equity and B is the market value of its debt.", null, "Using NI approach, the market value of equity(S) is calculated as earning available to equity shareholders (NI) divided by the equity capitalization rate (k e).", null, "The earnings available to equity shareholders (NI) are determined as net operating income (EBIT) less interest on debenture (I).", null, "The overall capitalization rate (ko) is defined as net operating income (EBIT) divided by the value of the company (V).", null, "Particulars Rs. Net operating income (EBIT) 1,50,000 Less: Interest of 10% on debenture of 5,00,000 (I) 50,000 Earnings available to equity shareholders (NI) 1,00,000 Divided by: Equity capitalization rate (K e)% 10% Market value of equity (S= Nix100/Ke) 10,00,000 Plus: Market value of debt (B) 5,00,000 Value of the firm (V=S+B) 15,00,000 Overall capitalization rate (Ko=EBIT/V)% 10.00%\n\nSo, the value of the firm will be equal Rs. 15, 00,000, while the overall capitalization rate will be the same as cost of equity and debt. Typically, the cost of debt is lower than cost of equity as the lenders require lower rate of return than equity shareholders. So, the company is more interested in attraction additional debt than equity for its operating activity. In the above case, the firm has as the equal finance structure, as the same cost of its sources. This is the main reason why the overall capitalization rate is the same.\n\nb. Calculation the firm value and overall capitalization rate if the debt will be increased.\n\nAccording to the above calculation in requirement #a, if the debenture debt is increased to Rs. 7, 50,000 while the cost of debt is reduced to 9%, the calculation the firms’ value will be following. As the cost of debt is decreased, but the proportion in capital structure ups the value of the company will be increased as shown in the table below. As the result of this shift in capital structure the overall capitalization rate is decreased from 10 percent to 9.52 percent.\n\n Particulars Rs Net operating income (EBIT) 1,50,000 Less: Interest of 9% on debenture of 7,50,000 (I) 67,500 Earnings available to equity shareholders (NI) 82,500 Divided by: Equity capitalization rate (K e)% 10% Market value of equity (S= Nix100/K e) 825,000 Plus: Market value of debt (B) 7,50,000 Value of the firm (V=S+B) 15,75,000 Overall capitalization rate (K o=EBIT/V)% 9.52%" ]	[ null, "https://www.financehomeworkhelp.org/wp-content/uploads/2017/10/how-to-calculate-capitalization-rate-solution.png", null, 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https://www.isprimenumber.com/prime/26927	[ "# Is 26927 a Prime Number?\n\nYes, 26927 is a prime number. 26927 is divisible by 1 and itself.\n\nYes, 26927 is a prime number.\n\n### Why 26927 is a prime number?\n\nBecause 26927 has no positive divisors rather than 1 and itself.\n\nThe next prime number of 26927 is 26947\nThe previous prime number of 26927 is 26921.\n\n### Related Prime Numbers\n\nBiggest 10 prime numbers smaller than 26927\n\nSmallest 10 prime numbers bigger than 26927\n\n### Related Non-Prime Numbers\n\nBiggest 10 composite numbers smaller than 26927\n\nSmallest 10 composite numbers bigger than 26927" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84833604,"math_prob":0.9205378,"size":531,"snap":"2019-13-2019-22","text_gpt3_token_len":147,"char_repetition_ratio":0.25426945,"word_repetition_ratio":0.11494253,"special_character_ratio":0.33898306,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95886827,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T08:21:10Z\",\"WARC-Record-ID\":\"<urn:uuid:6d4a1473-37c4-4d39-a350-ab25eb7adb1e>\",\"Content-Length\":\"12629\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c13bcd2-8523-4e6d-8e16-c453d82d1fb9>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8bdcc9e-49fe-4717-bfa3-7dffb86e743e>\",\"WARC-IP-Address\":\"104.197.145.71\",\"WARC-Target-URI\":\"https://www.isprimenumber.com/prime/26927\",\"WARC-Payload-Digest\":\"sha1:RGNGB2JAKPHKAKPLPT6SIA2AFVTZ724A\",\"WARC-Block-Digest\":\"sha1:N2LL3JNPDZQAVYETHCCVN2HZN2HPSHAD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202781.33_warc_CC-MAIN-20190323080959-20190323102959-00117.warc.gz\"}"}
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Newton1/ObjCh3Inertia.html	[ "# Physics - Chapter 4 Newtons First Law of Motion - Inertia Terms & Objectives", null, "", null, "", null, "", null, "", null, "BHS -> Mr. Stanbrough -> Physics -> Mechanics -> Newton's Laws -> Newton's First Law -> this page\n\nNote: The following terms and objectives are based on the Indiana Standards 2000 for Physics 1. You may download the complete physics standards by clicking here, or view the standards relevant to Chapter 4 - Newton's First Law here.\n\n## New Terms in Chapter 4:\n\n accelerating reference frame inertia Newton (unit) fictitious force kilogram Newton's First Law force Law of Inertia violent motion friction equilibrium static equilibrium inertial reference frame natural motion support force heliocentric solar system geocentric solar system normal force net force\n\n## Chapter 4 Objectives:\n\nAfter completing Chapter 4, you should be able to:\n\n1. ... define force.\n2. ...find the net force acting on an object if:\n1. no forces push/pull on it.\n2. 1 force pushes/pulls on it.\n3. 2 forces push/pull on it in the same direction.\n4. 2 forces push/pull on it in opposite directions.\n3. ... state Newton's First Law.\n4. ... define inertia.\n5. ... identify the conditions necessary for a body to be in equilibrium, and describe all of the forces acting on the body.\n6. ... correctly apply Newton's First Law and the concept of inertia to explain various motions, such as the motion of a person inside an accelerating car.\n\n## Possible Preconceptions to Correct:\n\nDo you believe that any of the following statements are true? They AREN'T! When you finish Chapter 4, you should understand that each of these statements is FALSE, and WHY it is false.", null, "", null, "", null, "", null, "", null, "BHS -> Mr. Stanbrough -> Physics -> Mechanics -> Newton's Laws -> Newton's First Law -> this page\n\nlast update October 15, 2009 by JL Stanbrough" ]	[ null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Prev.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Next.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Up.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Home.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Help.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Prev.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Next.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Up.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Home.GIF", null, "http://www.batesville.k12.in.us/physics/PhyNet/NavIcons/Help.GIF", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8913592,"math_prob":0.7428063,"size":706,"snap":"2021-43-2021-49","text_gpt3_token_len":161,"char_repetition_ratio":0.13105413,"word_repetition_ratio":0.23140496,"special_character_ratio":0.25354108,"punctuation_ratio":0.10569106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9838202,"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,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\":\"2021-11-29T13:48:42Z\",\"WARC-Record-ID\":\"<urn:uuid:7c639215-f5f4-4899-8066-fe4abe6adfa5>\",\"Content-Length\":\"9038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:174bb6d5-b863-400f-9550-4048a692ef12>\",\"WARC-Concurrent-To\":\"<urn:uuid:1be6eb43-9325-4fce-9ba9-2d852ab3c721>\",\"WARC-IP-Address\":\"69.16.250.147\",\"WARC-Target-URI\":\"http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Newton1/ObjCh3Inertia.html\",\"WARC-Payload-Digest\":\"sha1:CU5AGJ6ZB6IILLS3TSQA7YPQODIPFQ4U\",\"WARC-Block-Digest\":\"sha1:TFTBU5KUFPMMY5XW2L5SUL4CG5G2NVJI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358774.44_warc_CC-MAIN-20211129134323-20211129164323-00113.warc.gz\"}"}
https://www.journaldev.com/26700/python-and-operator	[ "# Python AND Operator\n\nFiled Under: Python\n\nPython operators can be divided into multiple categories. Two of them are – bitwise operators and logical operators. Bitwise operator performs operations on integers in binary format whereas logical operators perform operations on boolean values.\n\n## Python AND Operator\n\nThere are two types of Python AND Operators.\n\n1. Bitwise AND Operator: It’s denoted by `&` and work with integers. The numbers are converted into binary format and bitwise AND operation is performed. Finally, the output is returned in the decimal format.\n2. Logical AND Operator: It’s denoted by `and` and work with boolean values. The output is boolean – either `True` or `False`.\n\n## Python Bitwise AND Operator Example\n\nLet’s look at an example of bitwise and operator. We will ask the user to enter two numbers and print their binary and operation output.\n\n``````\na = int(input('Please enter an integer:\\n'))\nb = int(input('Please enter another integer:\\n'))\n\nprint(f'{a} in binary is {str(bin(a))[2:]}')\nprint(f'{b} in binary is {str(bin(b))[2:]}')\nprint(f'Binary AND of {a} and {b} is {a&b}')\n``````\n\nOutput:", null, "Python And Operator – Bitwise\n\n## Python Logical AND Operator Example\n\nLet’s look at an example of logical and operator. We will ask the user to enter a single digit number and print if it’s positive or negative. If the user doesn’t enter single digit integer, then we will print it too.\n\n``````\nx = int(input('Please enter a single digit integer:\\n'))\n\nif x > 0 and x < 10:\nprint('You entered positive single digit number')\nelif x < 0 and x > -10:\nprint('You entered negative single digit number')\nelse:\nprint('You did not entered single digit integer')\n``````\n\nOutput:", null, "Python Logical And Operator\n\n## Summary\n\nPython provides a lot of operators to work with different kinds of data types. Here we learned how to use Python and operators with boolean and bitwise binary integers.\n\nYou can checkout complete python script and more Python examples from our GitHub Repository.\nclose\nGeneric selectors\nExact matches only\nSearch in title\nSearch in content\nSearch in posts\nSearch in pages" ]	[ null, "https://cdn.journaldev.com/wp-content/uploads/2019/02/python-and-operator-bitwise-binary-integers.png", null, "https://cdn.journaldev.com/wp-content/uploads/2019/02/python-logical-and-operator.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81557685,"math_prob":0.94199026,"size":1761,"snap":"2019-51-2020-05","text_gpt3_token_len":397,"char_repetition_ratio":0.15082528,"word_repetition_ratio":0.07801419,"special_character_ratio":0.2350937,"punctuation_ratio":0.08928572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98841876,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T12:45:01Z\",\"WARC-Record-ID\":\"<urn:uuid:ee3af733-add5-4d7f-8507-f3f1df43faa3>\",\"Content-Length\":\"164856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3579680-0e6b-42e5-94cb-5243420f2d89>\",\"WARC-Concurrent-To\":\"<urn:uuid:05ea91fe-3709-479b-afd4-09769eec5df6>\",\"WARC-IP-Address\":\"3.95.122.236\",\"WARC-Target-URI\":\"https://www.journaldev.com/26700/python-and-operator\",\"WARC-Payload-Digest\":\"sha1:557DPHHTZL227F6SIUOVOZSV6KWM2RP2\",\"WARC-Block-Digest\":\"sha1:GSIDYS3HGPT44YNCHSNEMW2STGIHBIYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00216.warc.gz\"}"}
https://www.colorhexa.com/0c9a75	[ "# #0c9a75 Color Information\n\nIn a RGB color space, hex #0c9a75 is composed of 4.7% red, 60.4% green and 45.9% blue. Whereas in a CMYK color space, it is composed of 92.2% cyan, 0% magenta, 24% yellow and 39.6% black. It has a hue angle of 164.4 degrees, a saturation of 85.5% and a lightness of 32.5%. #0c9a75 color hex could be obtained by blending #18ffea with #003500. Closest websafe color is: #009966.\n\n• R 5\n• G 60\n• B 46\nRGB color chart\n• C 92\n• M 0\n• Y 24\n• K 40\nCMYK color chart\n\n#0c9a75 color description : Dark cyan - lime green.\n\n# #0c9a75 Color Conversion\n\nThe hexadecimal color #0c9a75 has RGB values of R:12, G:154, B:117 and CMYK values of C:0.92, M:0, Y:0.24, K:0.4. Its decimal value is 825973.\n\nHex triplet RGB Decimal 0c9a75 `#0c9a75` 12, 154, 117 `rgb(12,154,117)` 4.7, 60.4, 45.9 `rgb(4.7%,60.4%,45.9%)` 92, 0, 24, 40 164.4°, 85.5, 32.5 `hsl(164.4,85.5%,32.5%)` 164.4°, 92.2, 60.4 009966 `#009966`\nCIE-LAB 56.557, -43.047, 9.976 14.917, 24.472, 20.766 0.248, 0.407, 24.472 56.557, 44.188, 166.952 56.557, -46.721, 20.138 49.469, -32.747, 9.74 00001100, 10011010, 01110101\n\n# Color Schemes with #0c9a75\n\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #9a0c31\n``#9a0c31` `rgb(154,12,49)``\nComplementary Color\n• #0c9a2e\n``#0c9a2e` `rgb(12,154,46)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #0c789a\n``#0c789a` `rgb(12,120,154)``\nAnalogous Color\n• #9a2e0c\n``#9a2e0c` `rgb(154,46,12)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #9a0c78\n``#9a0c78` `rgb(154,12,120)``\nSplit Complementary Color\n• #9a750c\n``#9a750c` `rgb(154,117,12)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #750c9a\n``#750c9a` `rgb(117,12,154)``\nTriadic Color\n• #319a0c\n``#319a0c` `rgb(49,154,12)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #750c9a\n``#750c9a` `rgb(117,12,154)``\n• #9a0c31\n``#9a0c31` `rgb(154,12,49)``\nTetradic Color\n• #06533f\n``#06533f` `rgb(6,83,63)``\n• #086b51\n``#086b51` `rgb(8,107,81)``\n• #0a8263\n``#0a8263` `rgb(10,130,99)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #0eb287\n``#0eb287` `rgb(14,178,135)``\n• #10c999\n``#10c999` `rgb(16,201,153)``\n• #12e1ab\n``#12e1ab` `rgb(18,225,171)``\nMonochromatic Color\n\n# Alternatives to #0c9a75\n\nBelow, you can see some colors close to #0c9a75. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0c9a52\n``#0c9a52` `rgb(12,154,82)``\n• #0c9a5d\n``#0c9a5d` `rgb(12,154,93)``\n• #0c9a69\n``#0c9a69` `rgb(12,154,105)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #0c9a81\n``#0c9a81` `rgb(12,154,129)``\n• #0c9a8d\n``#0c9a8d` `rgb(12,154,141)``\n• #0c9a99\n``#0c9a99` `rgb(12,154,153)``\nSimilar Colors\n\n# #0c9a75 Preview\n\nText with hexadecimal color #0c9a75\n\nThis text has a font color of #0c9a75.\n\n``<span style=\"color:#0c9a75;\">Text here</span>``\n#0c9a75 background color\n\nThis paragraph has a background color of #0c9a75.\n\n``<p style=\"background-color:#0c9a75;\">Content here</p>``\n#0c9a75 border color\n\nThis element has a border color of #0c9a75.\n\n``<div style=\"border:1px solid #0c9a75;\">Content here</div>``\nCSS codes\n``.text {color:#0c9a75;}``\n``.background {background-color:#0c9a75;}``\n``.border {border:1px solid #0c9a75;}``\n\n# Shades and Tints of #0c9a75\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010806 is the darkest color, while #f5fefc is the lightest one.\n\n• #010806\n``#010806` `rgb(1,8,6)``\n• #021b14\n``#021b14` `rgb(2,27,20)``\n• #032d22\n``#032d22` `rgb(3,45,34)``\n• #053f30\n``#053f30` `rgb(5,63,48)``\n• #06513e\n``#06513e` `rgb(6,81,62)``\n• #08634c\n``#08634c` `rgb(8,99,76)``\n• #097659\n``#097659` `rgb(9,118,89)``\n• #0b8867\n``#0b8867` `rgb(11,136,103)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #0dac83\n``#0dac83` `rgb(13,172,131)``\n• #0fbe91\n``#0fbe91` `rgb(15,190,145)``\n• #10d19e\n``#10d19e` `rgb(16,209,158)``\n• #12e3ac\n``#12e3ac` `rgb(18,227,172)``\nShade Color Variation\n• #1bedb6\n``#1bedb6` `rgb(27,237,182)``\n• #2defbc\n``#2defbc` `rgb(45,239,188)``\n• #3ff0c2\n``#3ff0c2` `rgb(63,240,194)``\n• #51f1c8\n``#51f1c8` `rgb(81,241,200)``\n• #64f3ce\n``#64f3ce` `rgb(100,243,206)``\n• #76f4d3\n``#76f4d3` `rgb(118,244,211)``\n• #88f6d9\n``#88f6d9` `rgb(136,246,217)``\n• #9af7df\n``#9af7df` `rgb(154,247,223)``\n• #acf9e5\n``#acf9e5` `rgb(172,249,229)``\n• #bffaeb\n``#bffaeb` `rgb(191,250,235)``\n• #d1fbf0\n``#d1fbf0` `rgb(209,251,240)``\n• #e3fdf6\n``#e3fdf6` `rgb(227,253,246)``\n• #f5fefc\n``#f5fefc` `rgb(245,254,252)``\nTint Color Variation\n\n# Tones of #0c9a75\n\nA tone is produced by adding gray to any pure hue. In this case, #525453 is the less saturated color, while #06a078 is the most saturated one.\n\n• #525453\n``#525453` `rgb(82,84,83)``\n• #4c5a56\n``#4c5a56` `rgb(76,90,86)``\n• #456159\n``#456159` `rgb(69,97,89)``\n• #3f675d\n``#3f675d` `rgb(63,103,93)``\n• #396d60\n``#396d60` `rgb(57,109,96)``\n• #327463\n``#327463` `rgb(50,116,99)``\n• #2c7a66\n``#2c7a66` `rgb(44,122,102)``\n• #268069\n``#268069` `rgb(38,128,105)``\n• #1f876c\n``#1f876c` `rgb(31,135,108)``\n• #198d6f\n``#198d6f` `rgb(25,141,111)``\n• #129472\n``#129472` `rgb(18,148,114)``\n• #0c9a75\n``#0c9a75` `rgb(12,154,117)``\n• #06a078\n``#06a078` `rgb(6,160,120)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c9a75 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.58931506,"math_prob":0.6653302,"size":3700,"snap":"2019-13-2019-22","text_gpt3_token_len":1685,"char_repetition_ratio":0.122835495,"word_repetition_ratio":0.011049724,"special_character_ratio":0.56027025,"punctuation_ratio":0.23698781,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811403,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T00:58:33Z\",\"WARC-Record-ID\":\"<urn:uuid:a6b02cdc-78d1-43d1-8543-4881b42ec688>\",\"Content-Length\":\"36427\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:380e3d4b-0fad-4bac-b016-d3f2f61a6c57>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ef73f56-e14b-42fc-a619-4a6fb87d74ce>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0c9a75\",\"WARC-Payload-Digest\":\"sha1:FTBJTYRBKJSJHHQW6AFBR546WZAVYIIY\",\"WARC-Block-Digest\":\"sha1:FYZORGKC3GJNZREZMKZCOYF6NXS4TUER\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256600.32_warc_CC-MAIN-20190522002845-20190522024845-00243.warc.gz\"}"}
http://www.usaco.org/current/data/sol_necklace.html	[ "Analysis: Necklace by Jakub Pachocki\n\nAs it is often the case when counting the number of subsequences fulfilling some criteria, the solution to this problem uses dynamic programming. Assume the initial string representing Bessie's necklace is S, and that the other cow's name is T. For every i, we would like to compute the number of subsequences of S[1..i] that do not contain T. This information is not enough to maintain by itself, so we need to split our state some more. One possibility would be remembering the last \|T\| - 1 letters of our chosen subsequence; that is something we would be able to update upon adding one letter, and it would also allow us to determine whether we contain T as a substring.\n\nUnfortunately, the number of states in the proposed approach is exponential in \|T\|, and thus far too large for our needs. Maybe we do not really care what the last \|T\| - 1 letters are exactly? In fact, it would be enough to know how 'similar' they are to T. The correct measure of similarity here is the longest suffix of these letters that is also some prefix of T. This is also information that we can maintain upon state transition, and it also allows us to determine whether the chosen subsequence contains T as a substring.\n\nTo speed up the state transitions, it may be useful to precompute which prefix of T we contain after matching T[1..j] and adding any letter c. This gives us A * \|T\| state transitions, where A = 26 is the size of the alphabet. The total running of our solution is thus O(A * \|T\| * \|T\| + \|S\| * \|T\|), if we just precompute the transitions naively. Although that is fast enough, it is actually just as easy to precompute the transitions with DP, reducing the complexity to O(A * \|T\| + \|T\| * \|S\|). A sample implementation follows.\n\n```/\nLANG: C++\n/\n\n#include <iostream>\n#include <fstream>\n#include <algorithm>\n#include <vector>\n#include <string>\n\nusing namespace std;\n\nint main() {\nifstream in;\nofstream out;\nin.open(\"necklace.in\");\nout.open(\"necklace.out\");\n\nstring text, pattern;\nin >> text >> pattern;\n\nint n = text.size();\nint m = pattern.size();\n\nfor (int i = 0; i < n; ++i) {\ntext[i] -= 'a';\n}\n\nfor (int i = 0; i < m; ++i) {\npattern[i] -= 'a';\n}\n\nvector<vector<int> > next(m, vector<int>(26, 0));\n\nfor (int i = 1; i < m; ++i) {\nint prev = next[i - 1][pattern[i - 1]];\nnext[i - 1][pattern[i - 1]] = i;\nfor (int j = 0; j < 26; ++j) {\nnext[i][j] = next[prev][j];\n}\n}\nnext[m - 1][pattern[m - 1]] = m;\n\nvector<int> max_taken(m, -n - 1);\nmax_taken = 0;\nfor (int i = 0; i < n; ++i) {\nvector<int> next_taken = max_taken;\nfor (int j = 0; j < m; ++j) {\nint cur = next[j][text[i]];\nif (cur < m) {\nnext_taken[cur] = max(next_taken[cur], max_taken[j] + 1);\n}\n}\nmax_taken = next_taken;\n}\n\nint result = 0;\nfor (int i = 0; i < m; ++i) {\nresult = max(result, max_taken[i]);\n}\n\nout << n - result << endl;\n\nin.close();\nout.close();\n}\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83428794,"math_prob":0.9882266,"size":2843,"snap":"2019-13-2019-22","text_gpt3_token_len":800,"char_repetition_ratio":0.10003522,"word_repetition_ratio":0.07471264,"special_character_ratio":0.31902918,"punctuation_ratio":0.1539763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99476177,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-26T03:55:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a9b30324-8a42-4e09-8f7c-feca72431e7a>\",\"Content-Length\":\"4027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69b5182b-8222-4bd4-bb12-1a1183c0647f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f2808374-a8c8-44c4-9249-87207ef1b565>\",\"WARC-IP-Address\":\"130.127.215.7\",\"WARC-Target-URI\":\"http://www.usaco.org/current/data/sol_necklace.html\",\"WARC-Payload-Digest\":\"sha1:UZQITXXXKE543LMRTOYFJHJVPFURRLEC\",\"WARC-Block-Digest\":\"sha1:RELG7JRUXOIR5EKWMO7BLQFPBWD4SG54\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204790.78_warc_CC-MAIN-20190326034712-20190326060712-00407.warc.gz\"}"}
https://effort.academickids.com/encyclopedia/index.php/Gravity	[ "# Gravity\n\nThe popular model of general relativity, as causing a flat surface like a rubber sheet to curve into a manifold is unhelpful to further progress in unifying quantum space with gravitation, since physical space fills volume, not surface area. This obvious fact is obfuscated by jargon in physics! Instead of 11 dimensional and multiple universe speculation (http://www.math.columbia.edu/~woit/blog/), consider what physics can do with experimentally-proved properties of space: real space is in physics as electric impedance (mc = 377 ohms for a vacuum, allowing radio transmission), electric permittivity, magnetic permeability (m), it forms a vacuum foam and causes particle-wave duality (Dirac space sea).\n\nFact 1: the big bang is receding galaxies with observable speed increasing linearly with time past, v/t = constant acceleration\n\nFact 2: by Newtons 2nd empirical law, F = ma, the mass of the galaxies times acceleration is outward force of the big bang\n\nFact 3: by Newtons 3rd empirical law, forces have equal and opposite reaction: so equal inward force is proved, thus gravity\n\nThis inward force is carried by the fabric of space around accelerating matter, so the inward pressure (force/area) gives rise to the inverse-square law of gravity (see illustrations), causing gravitation where shielded by mass (http://cdsweb.cern.ch/search.py?recid=706468&ln=en).\n\nIn many interesting situations the source of the gravitational field can be taken to be a perfect fluid. A fluid is a continuum that flows... A perfect fluid is defined as one in which all antislipping forces are zero, and the only force between neighboring fluid elements is pressure. Bernard Schutz, General Relativity, Cambridge University Press, 1986, pp. 89-90.\n\nThe correct force of gravity is proved below: F = mMG/r2, where G = (3/4) H2/( p r local e3 ) = 0.0119H2/r local = 6.7 x 10-11 Nm2kg-2. This is not speculative: all 3 facts above are hard experimental results. Electromagnetism and nuclear forces are well-known energy exchange processes, but gravity is not.\n\n• Art and Cultures\n• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)\n• Space and Astronomy" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8796769,"math_prob":0.96803707,"size":2085,"snap":"2021-21-2021-25","text_gpt3_token_len":493,"char_repetition_ratio":0.086496875,"word_repetition_ratio":0.0,"special_character_ratio":0.2321343,"punctuation_ratio":0.13636364,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.982163,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T01:21:46Z\",\"WARC-Record-ID\":\"<urn:uuid:4a309e62-7724-48f2-b68e-6ece0352f9dd>\",\"Content-Length\":\"24375\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c8aea76-f5ee-4897-aca9-1816d9af05e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:51abbd61-c742-499a-850d-35eea2662195>\",\"WARC-IP-Address\":\"108.59.0.28\",\"WARC-Target-URI\":\"https://effort.academickids.com/encyclopedia/index.php/Gravity\",\"WARC-Payload-Digest\":\"sha1:OAE3TFKNZQOXHKFXWO7UFFSLL5SUW4HB\",\"WARC-Block-Digest\":\"sha1:UYZZBZGVS2FSC4L2OMLY4FTOGUMRVTEP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991553.4_warc_CC-MAIN-20210510235021-20210511025021-00354.warc.gz\"}"}
https://ftp.aimsciences.org/journal/1930-5311/2017/11/0	[ "", null, "", null, "", null, "", null, "", null, "ISSN:\n1930-5311\n\neISSN:\n1930-532X\n\n## Journal Home\n\n• Open Access Articles", null, "All Issues\n\n## Journal of Modern Dynamics\n\n2017 , Volume 11\n\nSelect all articles\n\nExport/Reference:\n\n2017, 11: 1-16 doi: 10.3934/jmd.2017001 +[Abstract](5670) +[HTML](134) +[PDF](186.6KB)\nAbstract:\n\nWe prove analogs of the logarithm laws of Sullivan and KleinbockMargulis in the context of unipotent flows. In particular, we prove results for horospherical actions on homogeneous spaces G/Γ.\n\n2017, 11: 17-41 doi: 10.3934/jmd.2017002 +[Abstract](4071) +[HTML](117) +[PDF](255.8KB)\nAbstract:\n\nWe compute the algebraic equation of the universal family over the Kenyon-Smillie (2, 3, 4)-Teichmüller curve, and we prove that the equation is correct in two different ways. Firstly, we prove it in a constructive way via linear conditions imposed by three special points of the Teichmüller curve. Secondly, we verify that the equation is correct by computing its associated Picard-Fuchs equation. We also notice that each point of the Teichmüller curve has a hyperflex and we see that the torsion map is a central projection from this point.\n\n2017, 11: 43-56 doi: 10.3934/jmd.2017003 +[Abstract](4098) +[HTML](81) +[PDF](196.7KB)\nAbstract:\n\nThe celebrated KAM theory says that if one makes a small perturbation of a non-degenerate completely integrable system, we still see a huge measure of invariant tori with quasi-periodic dynamics in the perturbed system. These invariant tori are known as KAM tori. What happens outside KAM tori draws a lot of attention. In this paper we present a Lagrangian perturbation of the geodesic flow on a flat 3-torus. The perturbation is $C^\\infty$ small but the flow has a positive measure of trajectories with positive Lyapunov exponent. The measure of this set is of course extremely small. Still, the flow has positive metric entropy. From this result we get positive metric entropy outside some KAM tori.\n\n2017, 11: 57-98 doi: 10.3934/jmd.2017004 +[Abstract](5314) +[HTML](101) +[PDF](388.2KB)\nAbstract:\n\nIn the moduli space of degree $d$ polynomials, we prove the equidistribution of postcritically finite polynomials toward the bifurcation measure. More precisely, using complex analytic arguments and pluripotential theory, we prove the exponential speed of convergence for $\\mathscr{C}^2$-observables. This improves results obtained with arithmetic methods by Favre and Rivera-Letellier in the unicritical family and Favre and the first author in the space of degree $d$ polynomials.\n\nWe deduce from that the equidistribution of hyperbolic parameters with $(d-1)$ distinct attracting cycles of given multipliers toward the bifurcation measure with exponential speed for $\\mathscr{C}^1$-observables. As an application, we prove the equidistribution (up to an explicit extraction) of parameters with $(d-1)$ distinct cycles with prescribed multiplier toward the bifurcation measure for any $(d-1)$ multipliers outside a pluripolar set.\n\n2017, 11: 99-123 doi: 10.3934/jmd.2017005 +[Abstract](3964) +[HTML](92) +[PDF](282.7KB)\nAbstract:\n\nIn this paper we investigate relations between Koopman, groupoid and quasi-regular representations of countable groups. We show that for an ergodic measure class preserving action of a countable group G on a standard Borel space the associated groupoid and quasi-regular representations are weakly equivalent and weakly contained in the Koopman representation. Moreover, if the action is hyperfinite then the Koopman representation is weakly equivalent to the groupoid. As a corollary of our results we obtain a continuum of pairwise disjoint pairwise equivalent irreducible representations of weakly branch groups. As an illustration we calculate spectra of regular, Koopman and groupoid representations associated to the action of the 2-group of intermediate growth constructed by the second author in 1980.\n\n2017, 11: 125-142 doi: 10.3934/jmd.2017006 +[Abstract](3854) +[HTML](84) +[PDF](223.8KB)\nAbstract:\n\nWe study close-to-constants quasiperiodic cocycles in \\begin{document} $\\mathbb{T} ^{d} \\times G$ \\end{document}, where \\begin{document} $d \\in \\mathbb{N} ^{*}$ \\end{document} and \\begin{document} $G$ \\end{document} is a compact Lie group, under the assumption that the rotation in the basis satisfies a Diophantine condition. We prove differentiable rigidity for such cocycles: if such a cocycle is measurably conjugate to a constant one satisfying a Diophantine condition with respect to the rotation, then it is \\begin{document} $C^{\\infty}$ \\end{document}-conjugate to it, and the KAM scheme actually produces a conjugation. We also derive a global differentiable rigidity theorem, assuming the convergence of the renormalization scheme for such dynamical systems.\n\n2017, 11: 143-153 doi: 10.3934/jmd.2017007 +[Abstract](5012) +[HTML](117) +[PDF](183.1KB)\nAbstract:\n\nWe give upper and lower bounds for Diophantine exponents measuring how well a point in the plane can be approximated by points in the orbit of a lattice \\begin{document}$\\Gamma < {\\rm{S}}{{\\rm{L}}_2}\\left( {\\mathbb{R}} \\right)$\\end{document} acting linearly on \\begin{document}${\\mathbb{R}^2}$\\end{document}. Our method gives bounds that are uniform for almost all orbits.\n\n2017, 11: 155-188 doi: 10.3934/jmd.2017008 +[Abstract](2572) +[HTML](94) +[PDF](326.2KB)\nAbstract:\n\nLet \\begin{document}$\\Gamma$\\end{document} be a lattice in \\begin{document}$G=\\mathrm{SL}(2, \\mathbb{C})$\\end{document}. We give an effective equidistribution result with precise error terms for expanding translates of pieces of horospherical orbits in \\begin{document}$\\Gamma\\backslash G$\\end{document}. Our method of proof relies on the theory of unitary representations.\n\n2017, 11: 189-217 doi: 10.3934/jmd.2017009 +[Abstract](5312) +[HTML](122) +[PDF](386.4KB)\nAbstract:\n\nConsider a general circle packing \\begin{document}$\\mathscr{P}$\\end{document} in the complex plane \\begin{document}$\\mathbb{C}$\\end{document} invariant under a Kleinian group \\begin{document}$\\Gamma$\\end{document}. When \\begin{document}$\\Gamma$\\end{document} is convex cocompact or its critical exponent is greater than 1, we obtain an effective equidistribution for small circles in \\begin{document}$\\mathscr{P}$\\end{document} intersecting any bounded connected regular set in \\begin{document}$\\mathbb{C}$\\end{document}; this provides an effective version of an earlier work of Oh-Shah . In view of the recent result of McMullen-Mohammadi-Oh , our effective circle counting theorem applies to the circles contained in the limit set of a convex cocompact but non-cocompact Kleinian group whose limit set contains at least one circle. Moreover, consider the circle packing \\begin{document}$\\mathscr{P}(\\mathscr{T})$\\end{document} of the ideal triangle attained by filling in largest inner circles. We give an effective estimate to the number of disks whose hyperbolic areas are greater than \\begin{document}$t$\\end{document}, as \\begin{document}$t\\to0$\\end{document}, effectivizing the work of Oh .\n\n2017, 11: 219-248 doi: 10.3934/jmd.2017010 +[Abstract](4820) +[HTML](96) +[PDF](685.9KB)\nAbstract:\n\nIt is known since a 40-year-old paper by M.Keane that minimality is a generic (i.e., holding with probability one) property of an irreducible interval exchange transformation. If one puts some integral linear restrictions on the parameters of the interval exchange transformation, then minimality may become an \"exotic\" property. We conjecture in this paper that this occurs if and only if the linear restrictions contain a Lagrangian subspace of the first homology of the suspension surface. We partially prove it in the `only if' direction and provide a series of examples to support the converse one. We show that the unique ergodicity remains a generic property if the restrictions on the parameters do not contain a Lagrangian subspace (this result is due to Barak Weiss).\n\n2017, 11: 249-262 doi: 10.3934/jmd.2017011 +[Abstract](3523) +[HTML](83) +[PDF](183.8KB)\nAbstract:\n\nLet \\begin{document}$T$\\end{document} be an \\begin{document}$m$\\end{document}-interval exchange transformation. By the rank of \\begin{document}$T$\\end{document} we mean the dimension of the \\begin{document}$\\mathbb{Q}$\\end{document}-vector space spanned by the lengths of the exchanged intervals. We prove that if \\begin{document}$T$\\end{document} is minimal and the rank of \\begin{document}$T$\\end{document} is greater than \\begin{document}$1+\\lfloor m/2 \\rfloor$\\end{document}, then \\begin{document}$T$\\end{document} cannot be written as a power of another interval exchange. We also demonstrate that this estimate on the rank cannot be improved.\n\nIn the case that \\begin{document}$T$\\end{document} is a minimal 3-interval exchange transformation, we prove a stronger result: \\begin{document}$T$\\end{document} cannot be written as a power of another interval exchange if and only if \\begin{document}$T$\\end{document} satisfies Keane's infinite distinct orbit condition. In the course of proving this result, we give a classification (up to conjugacy) of those minimal interval exchange transformations whose discontinuities all belong to a single orbit.\n\n2017, 11: 263-312 doi: 10.3934/jmd.2017012 +[Abstract](4112) +[HTML](89) +[PDF](456.5KB)\nAbstract:\n\nWe show that genuinely higher rank expanding actions of abelian semigroups on compact manifolds are \\begin{document}$C.{\\infty}$\\end{document}-conjugate to affine actions on infra-nilmanifolds. This is based on the classification of expanding diffeomorphisms up to Hölder conjugacy by Gromov and Shub, and is similar to recent work on smooth classification of higher rank Anosov actions on tori and nilmanifolds. To prove regularity of the conjugacy in the higher rank setting, we establish exponential mixing of solenoid actions induced from semigroup actions by nilmanifold endomorphisms, a result of independent interest. We then proceed similar to the case of higher rank Anosov actions.\n\n2017, 11: 313-339 doi: 10.3934/jmd.2017013 +[Abstract](5570) +[HTML](107) +[PDF](425.1KB)\nAbstract:\n\nThe first author introduced a relative symplectic capacity \\begin{document}$C$\\end{document} for a symplectic manifold \\begin{document}$(N,\\omega_N)$\\end{document} and its subset \\begin{document}$X$\\end{document} which measures the existence of non-contractible periodic trajectories of Hamiltonian isotopies on the product of \\begin{document}$N$\\end{document} with the annulus \\begin{document}$A_R=(-R,R)\\times\\mathbb{R}/\\mathbb{Z}$\\end{document}. In the present paper, we give an exact computation of the capacity \\begin{document}$C$\\end{document} of the \\begin{document}$2n$\\end{document}-torus \\begin{document}$\\mathbb{T}^{2n}$\\end{document} relative to a Lagrangian submanifold \\begin{document}$\\mathbb{T}^n$\\end{document} which implies the existence of non-contractible Hamiltonian periodic trajectories on \\begin{document}$A_R\\times\\mathbb{T}^{2n}$\\end{document}. Moreover, we give a lower bound on the number of such trajectories.\n\n2017, 11: 341-368 doi: 10.3934/jmd.2017014 +[Abstract](4008) +[HTML](82) +[PDF](287.0KB)\nAbstract:\n\nLet \\begin{document}$f$\\end{document} be a measure-preserving transformation of a Lebesgue space \\begin{document}$(X,\\mu)$\\end{document} and let \\begin{document}${\\mathscr{F}}$\\end{document} be its extension to a bundle \\begin{document}$\\mathscr{E} = X \\times {\\mathbb{R}}^m$\\end{document} by smooth fiber maps \\begin{document}${\\mathscr{F}}_x : {\\mathscr{E}}_x \\to {\\mathscr{E}}_{fx}$\\end{document} so that the derivative of \\begin{document}${\\mathscr{F}}$\\end{document} at the zero section has negative Lyapunov exponents. We construct a measurable system of smooth coordinate changes \\begin{document}${\\mathscr{H}}_x$\\end{document} on \\begin{document}${\\mathscr{E}}_x$\\end{document} for \\begin{document}$\\mu$\\end{document}-a.e. \\begin{document}$x$\\end{document} so that the maps \\begin{document}${\\mathscr{P}}_x ={\\mathscr{H}}_{fx} \\circ {\\mathscr{F}}_x \\circ {\\mathscr{H}}_x^{-1}$\\end{document} are sub-resonance polynomials in a finite dimensional Lie group. Our construction shows that such \\begin{document}${\\mathscr{H}}_x$\\end{document} and \\begin{document}${\\mathscr{P}}_x$\\end{document} are unique up to a sub-resonance polynomial. As a consequence, we obtain the centralizer theorem that the coordinate change \\begin{document}$\\mathscr{H}$\\end{document} also conjugates any commuting extension to a polynomial extension of the same type. We apply our results to a measure-preserving diffeomorphism \\begin{document}$f$\\end{document} with a non-uniformly contracting invariant foliation \\begin{document}$W$\\end{document}. We construct a measurable system of smooth coordinate changes \\begin{document}${\\mathscr{H}}_x: W_x \\to T_xW$\\end{document} such that the maps \\begin{document}${\\mathscr{H}}_{fx} \\circ f \\circ {\\mathscr{H}}_x^{-1}$\\end{document} are polynomials of sub-resonance type. Moreover, we show that for almost every leaf the coordinate changes exist at each point on the leaf and give a coherent atlas with transition maps in a finite dimensional Lie group.\n\n2017, 11: 369-407 doi: 10.3934/jmd.2017015 +[Abstract](3869) +[HTML](99) +[PDF](1611.2KB)\nAbstract:\n\nWe show that in positive characteristic the homogeneous probability measure supported on a periodic orbit of the diagonal group in the space of \\begin{document}$2$\\end{document}-lattices, when varied along rays of Hecke trees, may behave in sharp contrast to the zero characteristic analogue: For a large set of rays, the measures fail to converge to the uniform probability measure on the space of \\begin{document}$2$\\end{document}-lattices. More precisely, we prove that when the ray is rational there is uniform escape of mass, that there are uncountably many rays giving rise to escape of mass, and that there are rays along which the measures accumulate on measures which are not absolutely continuous with respect to the uniform measure on the space of \\begin{document}$2$\\end{document}-lattices.\n\n2017, 11: 409-423 doi: 10.3934/jmd.2017016 +[Abstract](3414) +[HTML](168) +[PDF](358.22KB)\nAbstract:\n\nThe deformation space of a branched cover \\begin{document}$f:(S^2,A)\\to (S^2,B)$\\end{document} is a complex submanifold of a certain Teichmüller space, which consists of classes of marked rational maps \\begin{document}$F:(\\mathbb{P}^1,A')\\to (\\mathbb{P}^1,B')$\\end{document} that are combinatorially equivalent to \\begin{document}$f$\\end{document}. In the case \\begin{document}$A=B$\\end{document}, under a mild assumption on \\begin{document}$f$\\end{document}, William Thurston gave a topological criterion for which the deformation space of \\begin{document}$f:(S^2,A)\\to (S^2,B)$\\end{document} is nonempty, and he proved that it is always connected. We show that if \\begin{document}$A\\subsetneq B$\\end{document}, then the deformation space need not be connected. We exhibit a family of quadratic rational maps for which the associated deformation spaces are disconnected; in fact, each has infinitely many components.\n\n2017, 11: 425-445 doi: 10.3934/jmd.2017017 +[Abstract](3699) +[HTML](173) +[PDF](264.84KB)\nAbstract:\n\nFor area preserving C2 surface diffeomorphisms, we give an explicit finite information condition on the exponential growth of the number of Bowen's (n, δ)-balls needed to cover a positive proportion of the space, that is sufficient to guarantee positive topological entropy. This can be seen as an effective version of Katok's horseshoe theorem in the conservative setting. We also show that the analogous result is false in dimension larger than 3.\n\n2017, 11: 447-476 doi: 10.3934/jmd.2017018 +[Abstract](3488) +[HTML](131) +[PDF](312.47KB)\nAbstract:\n\nWe prove logarithm laws for unipotent flows on non-compact finite-volume hyperbolic manifolds. Our method depends on the estimate of norms of certain incomplete Eisenstein series.\n\n2017, 11: 477-499 doi: 10.3934/jmd.2017019 +[Abstract](3285) +[HTML](165) +[PDF](632.75KB)\nAbstract:\n\nWe prove the existence and some properties of the limiting gap distribution for the directions of some Schottky group orbits in the Poincaré disk. A key feature is that the fundamental domains for these groups have infinite area.\n\n2017, 11: 501-550 doi: 10.3934/jmd.2017020 +[Abstract](3140) +[HTML](148) +[PDF](464.61KB)\nAbstract:\n\nWe prove an analogue of a theorem of Eskin-Margulis-Mozes . Suppose we are given a finite set of places \\begin{document} $S$ \\end{document} over \\begin{document} ${\\mathbb{Q}}$ \\end{document} containing the Archimedean place and excluding the prime \\begin{document} $2$ \\end{document}, an irrational isotropic form \\begin{document} ${\\mathbf q}$ \\end{document} of rank \\begin{document} $n\\geq 4$ \\end{document} on \\begin{document} ${\\mathbb{Q}}_S$ \\end{document}, a product of \\begin{document} $p$ \\end{document}-adic intervals \\begin{document} $\\mathsf{I}_p$ \\end{document}, and a product \\begin{document} $\\Omega$ \\end{document} of star-shaped sets. We show that unless \\begin{document} $n=4$ \\end{document} and \\begin{document} ${\\mathbf q}$ \\end{document} is split in at least one place, the number of \\begin{document} $S$ \\end{document}-integral vectors \\begin{document} $\\mathbf v \\in {\\mathsf{T}} \\Omega$ \\end{document} satisfying simultaneously \\begin{document} ${\\mathbf q}(\\mathbf v) \\in I_p$ \\end{document} for \\begin{document} $p \\in S$ \\end{document} is asymptotically given by\n\n\\begin{document}$\\begin{split}\\lambda({\\mathbf q}, \\Omega) \|\\,\\mathsf{I}\\,\| \\cdot \\\| {\\mathsf{T}} \\\|^{n-2}\\end{split}$ \\end{document}\n\nas \\begin{document} ${\\mathsf{T}}$ \\end{document} goes to infinity, where \\begin{document} $\|\\,\\mathsf{I}\\,\|$ \\end{document} is the product of Haar measures of the \\begin{document} $p$ \\end{document}-adic intervals \\begin{document} $I_p$ \\end{document}. The proof uses dynamics of unipotent flows on \\begin{document} $S$ \\end{document}-arithmetic homogeneous spaces; in particular, it relies on an equidistribution result for certain translates of orbits applied to test functions with a controlled growth at infinity, specified by an \\begin{document} $S$ \\end{document}-arithmetic variant of the \\begin{document} $\\alpha$ \\end{document}-function introduced in , and an \\begin{document} $S$ \\end{document}-arithemtic version of a theorem of Dani-Margulis .\n\n2017, 11: 551-562 doi: 10.3934/jmd.2017021 +[Abstract](3369) +[HTML](208) +[PDF](163.67KB)\nAbstract:\n\nWe consider densities \\begin{document}$D_\\Sigma(A)$\\end{document}, \\begin{document}$\\overline{D}_\\Sigma(A)$\\end{document} and \\begin{document}$\\underline{D}_\\Sigma(A)$\\end{document} for a subset \\begin{document}$A$\\end{document} of \\begin{document}$\\mathbb{N}$\\end{document} with respect to a sequence \\begin{document}$\\Sigma$\\end{document} of finite subsets of \\begin{document}$\\mathbb{N}$\\end{document} and study Fourier coefficients of ergodic, weakly mixing and strongly mixing \\begin{document}$\\times p$\\end{document}-invariant measures on the unit circle \\begin{document}$\\mathbb{T}$\\end{document}. Combining these, we prove the following measure rigidity results: on \\begin{document}$\\mathbb{T}$\\end{document}, the Lebesgue measure is the only non-atomic \\begin{document}$\\times p$\\end{document}-invariant measure satisfying one of the following: (1) \\begin{document}$\\mu$\\end{document} is ergodic and there exist a Følner sequence \\begin{document}$\\Sigma$\\end{document} in \\begin{document}$\\mathbb{N}$\\end{document} and a nonzero integer \\begin{document}$l$\\end{document} such that \\begin{document}$\\mu$\\end{document} is \\begin{document}$\\times (p^j+l)$\\end{document}-invariant for all \\begin{document}$j$\\end{document} in a subset \\begin{document}$A$\\end{document} of \\begin{document}$\\mathbb{N}$\\end{document} with \\begin{document}$D_\\Sigma(A)=1$\\end{document}; (2) \\begin{document}$\\mu$\\end{document} is weakly mixing and there exist a Følner sequence \\begin{document}$\\Sigma$\\end{document} in \\begin{document}$\\mathbb{N}$\\end{document} and a nonzero integer \\begin{document}$l$\\end{document} such that \\begin{document}$\\mu$\\end{document} is \\begin{document}$\\times (p^j+l)$\\end{document}-invariant for all \\begin{document}$j$\\end{document} in a subset \\begin{document}$A$\\end{document} of \\begin{document}$\\mathbb{N}$\\end{document} with \\begin{document}$\\overline{D}_\\Sigma(A)>0$\\end{document}; (3) \\begin{document}$\\mu$\\end{document} is strongly mixing and there exists a nonzero integer \\begin{document}$l$\\end{document} such that \\begin{document}$\\mu$\\end{document} is \\begin{document}$\\times (p^j+l)$\\end{document}-invariant for infinitely many \\begin{document}$j$\\end{document}. Moreover, a \\begin{document}$\\times p$\\end{document}-invariant measure satisfying (2) or (3) is either a Dirac measure or the Lebesgue measure.\n\nAs an application we prove that for every increasing function \\begin{document}$\\tau$\\end{document} defined on positive integers with \\begin{document}$\\lim_{n\\to\\infty}\\tau(n)=\\infty$\\end{document}, there exists a multiplicative semigroup \\begin{document}$S_\\tau$\\end{document} of \\begin{document}$\\mathbb{Z}^+$\\end{document} containing \\begin{document}$p$\\end{document} such that \\begin{document}$\|S_\\tau\\cap[1,n]\|\\leq (\\log_p n)^{\\tau(n)}$\\end{document} and the Lebesgue measure is the only non-atomic ergodic \\begin{document}$\\!\\times \\!p$\\end{document}-invariant measure which is \\begin{document}$\\times q$\\end{document}-invariant for all \\begin{document}$q$\\end{document} in \\begin{document}$S_\\tau$\\end{document}.\n\n2017, 11: 563-588 doi: 10.3934/jmd.2017022 +[Abstract](4919) +[HTML](172) +[PDF](277.58KB)\nAbstract:\n\nWe show that Masur's logarithmic law of geodesics in the moduli space of translation surfaces does not imply unique ergodicity of the translation flow, but that a similar law involving the flat systole of a Teichmüller geodesic does imply unique ergodicity. It shows that the flat geometry has a better control on ergodic properties of translation flow than hyperbolic geometry.\n\n2020 Impact Factor: 0.848\n5 Year Impact Factor: 0.815\n2020 CiteScore: 0.9" ]	[ null, "https://ftp.aimsciences.org:443/style/web/images/white_google.png", null, "https://ftp.aimsciences.org:443/style/web/images/white_facebook.png", null, "https://ftp.aimsciences.org:443/style/web/images/white_twitter.png", null, "https://ftp.aimsciences.org:443/style/web/images/white_linkedin.png", null, "https://ftp.aimsciences.org/fileAIMS/journal/img/cover/coverJMD_ad.png", null, "https://ftp.aimsciences.org:443/style/web/images/OA.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8785081,"math_prob":0.9865323,"size":6261,"snap":"2021-43-2021-49","text_gpt3_token_len":1379,"char_repetition_ratio":0.11539076,"word_repetition_ratio":0.0020811656,"special_character_ratio":0.18798915,"punctuation_ratio":0.0679702,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986996,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T12:19:42Z\",\"WARC-Record-ID\":\"<urn:uuid:11e63fc3-e924-452c-8c6f-f59897554bfe>\",\"Content-Length\":\"155897\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e88f0978-5c7e-46dd-a5ca-9b35946d9058>\",\"WARC-Concurrent-To\":\"<urn:uuid:fec0dce5-ac79-4844-97d6-a641961ff84c>\",\"WARC-IP-Address\":\"107.161.80.18\",\"WARC-Target-URI\":\"https://ftp.aimsciences.org/journal/1930-5311/2017/11/0\",\"WARC-Payload-Digest\":\"sha1:CTFGYIUFKYYLT2T6YSLZSEU53KKXRQGV\",\"WARC-Block-Digest\":\"sha1:VE2SOGAYOO77BLDY4WMZCCCKJSEWAWVV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585405.74_warc_CC-MAIN-20211021102435-20211021132435-00202.warc.gz\"}"}
https://www.prepbytes.com/blog/cpp-programming/factorial-program-in-cpp/	[ "# Factorial Program in C++", null, "In this article, we will discuss what is a factorial number, and what is its logic to implement it in C++. We will look and see multiple approaches to implement factorial programs.\nFirstly, let’s see what a Factorial Number is?\n\n## Factorial Number\n\nCalculation of factorial numbers is simple mathematics. The factorial of a number is 1 2 3 number.\nFor Example, factorial 6 is\n6! = 1 2 3 4 5 * 6 = 720.\n\n## Explanation of Program\n\nLet’s create a factorial program using recursive functions. Until the value is not equal to zero, the recursive function will call itself.\nFactorial can be calculated using the following recursive formula:\nn! = n * (n – 1)!\nn! = 1 if n = 0 or n = 1\n\n### Approach 1: (Using Recursion)\n\n1. Build a recursive function factorial which takes an integer n as a parameter and returns an unsigned integer as the output.\n2. If n ==0 or n==1, then return 1.\n3. Else return n * factorial(n – 1).\n\nCode Implementation (Using Recursion)\n\n```#include <iostream>\nusing namespace std;\n\nunsigned int factorial(unsigned int n)\n{\nif (n == 0 \|\| n == 1)\nreturn 1;\nreturn n * factorial(n - 1);\n}\n\nint main()\n{\nint num = 5;\ncout << \"Factorial of \"\n<< num << \" is: \" << factorial(num) << endl;\nreturn 0;\n}\n```\n\nTime complexity: O(n) will be the time complexity for implementing a factorial program in C++ using recursion.\n\nAuxiliary Space: O(n) will be the space complexity for implementing a factorial program in C++ using recursion.\n\n### Approach 2: (Using For loop)\n\nFollow the steps to solve the problem:\n\n1. Using a for loop, we will write a program for finding the factorial of a number.\n2. An integer variable with a value of 1 will be used in the program.\n3. With each iteration, the value will increase by 1 until it equals the value entered by the user.\n4. The factorial of the number entered by the user will be the final value in the fact variable.\n\nCode Implementation (Using For Loop)\n\n```#include <iostream>\nusing namespace std;\n\nunsigned int factorial(unsigned int n)\n{\nint res = 1, i;\nfor (i = 2; i <= n; i++)\nres = i;\nreturn res;\n}\n\nint main()\n{\nint num = 5;\ncout << \"Factorial of \"\n<< num << \" is \"\n<< factorial(num) << endl;\nreturn 0;\n}\n```\n\nTime complexity: O(n) will be the time complexity for implementing a factorial program in C++ using a for loop.\n\nAuxiliary Space: O(1) will be the space complexity for implementing a factorial program in C++ using a for loop.\n\nApproach 3: In this approach, we will see how we can use the while loop for implementing the factorial of a number.\n\nCode Implementation (Using While Loop)\n\n```#include <iostream>\nusing namespace std;\nunsigned int factorial(unsigned int n)\n{\nif(n == 0)\nreturn 1;\nint i = n, fact = 1;\nwhile (n / i != n) {\nfact = fact i;\ni--;\n}\nreturn fact;\n}\nint main()\n{\nint num = 5;\ncout << \"Factorial of \"\n<< num << \" is \"\n<< factorial(num) << endl;\nreturn 0;\n}\n```\n\nTime complexity: O(n) will be the time complexity for implementing a factorial program in C++ using a while loop.\n\nAuxiliary Space: O(1) will be the space complexity for implementing a factorial program in C++ using a while loop.\n\nApproach 4: A ternary operator can be thought of as a shorthand for an if…else statement. The conditions are provided, along with statements to be executed based on them. Here’s the program for factorial using a ternary operator.\n\nCode Implementation (Using Ternary Operators)\n\n```#include <iostream>\nusing namespace std;\n\nint factorial(int n)\n{\nreturn (n == 1 \|\| n == 0) ? 1 : n * factorial(n - 1);\n}\n\nint main()\n{\nint num = 5;\ncout << \"Factorial of \" << num << \" is \"<< factorial(num);\nreturn 0;\n}\n```\n\nTime complexity: O(n) will be the time complexity for implementing a factorial program in C++ using a ternary operator for the recursive approach.\n\nAuxiliary Space: O(n) will be the space complexity for implementing a factorial program in C++ using a ternary operator for the recursive approach.\n\nTips\nVarious service based companies like TCS, Wipro, Accenture and Capgemini have asked for a factorial program for their technical interviews.\n\nConclusion\nThis article tried its best to explain the various approaches to implement factorial programs in C++. Having knowledge about mathematical topics like factorial numbers will always help in the logic building. And for beginners, this problem must be solved to improve their coding as well as problem solving skills.\n\nOther C++ Programs" ]	[ null, "https://prepbytes-misc-images.s3.ap-south-1.amazonaws.com/assets/1670564301916-Factorial%20Program%20in%20Cpp.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76338345,"math_prob":0.99248004,"size":4485,"snap":"2022-40-2023-06","text_gpt3_token_len":1096,"char_repetition_ratio":0.18946664,"word_repetition_ratio":0.30920246,"special_character_ratio":0.2717949,"punctuation_ratio":0.11452184,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99955285,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T02:24:53Z\",\"WARC-Record-ID\":\"<urn:uuid:614cb14e-5de7-41aa-b0fd-5138227a6b20>\",\"Content-Length\":\"148359\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae31d988-7e11-43ba-bd8b-5b0237d7f673>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d366151-509f-4df2-8090-1123a34d90b2>\",\"WARC-IP-Address\":\"18.160.18.47\",\"WARC-Target-URI\":\"https://www.prepbytes.com/blog/cpp-programming/factorial-program-in-cpp/\",\"WARC-Payload-Digest\":\"sha1:XBWU2G5HT6E5BB73GIK4HUHCRINBPVHS\",\"WARC-Block-Digest\":\"sha1:ZOCBO2AQ6ZXGE7BGVMCHDBZTOH7YEIRR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500080.82_warc_CC-MAIN-20230204012622-20230204042622-00251.warc.gz\"}"}
https://alpha-brains.com/Courses/magic-math/	[ "3000 PKR\n\n# Magic Maths Cultivates an interest in numbers and eliminates the maths phobia\n\nGet the benefit from the 20 years of\nMr. Sohaib Intesar\n\nFirst Certified Abacus Trainer in Pakistan.", null, "### Magic Math\n\nAlpha Brains rearranged Magic Maths course for kids and launched this course in Pakistan, UAE & UK. Magic maths Course has been introduced as a system of ancient times for Mathematics which was introduced by Late S. K Theta. He has discovered and re-organized the techniques in the form of 16 main techniques and 13 sub techniques. Alpha brains provides excellent opportunity to learn the Magic maths course by online or offline and get certification. It is simple and more interesting than the regular way of learning conventional maths.\nThe course is based on the natural way of learning; which helps to learn and perform the calculations at a much faster rate with smart efficiency that can save the time to solve mathematical problems. This works as a system which helps in checking the independent set of methods which can be useful in cross-checking the answers that can reduce the possibility of doing careless mistakes while solving the problems.\nThe tricks and formulas applied are direct, straightforward, and simple which reduces the stress work of finger counting and scratch work. Thus, it is helpful in developing your creativity, intelligence by improving the clarity in mind. It is very helpful in preparing for competitive exams like SAT, CSS, MCAT Engineering and many more.\nMagic Maths Course is for teachers, students, homemaker, and businessman for doing calculations and even teaching students as a mentor. With the help of its learning, one can perform the calculations 10-15 times faster than the conventional maths methods.\n\n### Benefits of Magic Math\n\n⦁ Magic Mathematics cultivates an interest in numbers and eliminates the maths-phobia\n⦁ Magic Maths is very easy to understand and practice.\n⦁ Any type of complex and critical multiplication or division can be done with simplicity\n⦁ High-Speed Magic Mathematics is 10 to 15 times faster than conventional Maths\n⦁ Magic Maths not only simplifies and eases the toughness of mathematics but also boosts the brain power\n⦁ Kids will improve their results in Schools\n⦁ Helps in Simplifying Calculations\n⦁ Enhances Ability for Numbers\n⦁ Improves Focus, Memory & Concentration\n⦁ Better Performances in Competitive Exams\n\n# Who Can Learn Magic Math?", null, "", null, "", null, "", null, "", null, "", null, "# Course Outline\n\n## M1\n\n2 Months\n• Multiplication by 11\n• Multiplication by 22-99\n• Multiplication by 12 to 19\n• Multiplication by 111\n• Multiplication by 222 to 999\n• If the sum of the units digits is 10 & tens place digits are same\n• If the sum of the tens place digits is 10 & units place digits are same\n• Multiplication by 9\n• Multiplication by number ending with 9\n• General Method Multiplication 2D x 2D & 3D x 3D\n• Multiplication (Below Base 10-90)\n• Multiplication (Below Base 100-900)\n• Multiplication Below Base Method\u0002Practice (100-10000)\n• Multiplication (Above Base 10-90)\n• Multiplication (Above Base 100-900)\n• Multiplication Above Base Method\u0002Practice (1000-10000)\n• Subtraction (All from 9 & last from 10)\n• Vinculum\n• Devinculum\n• Subtraction using Vinculum\n• Subtraction Base Method\n• Division by 9\n• Division by 8\n• Division by 11 & 12\n• Division by Below base 100\n• Division by Above base 100\n• Division Base Method (Above)\n• Division Base Method (Below) by other divisors\n• Squares (Above the Base & Below the Base 10 to 900)\n• Squares of number starting with 5 & number ending with 5\n• Mock Papers\n\n## M2\n\n\\$ 2 Months\n• Tables using vinculum\n• Multiplication by number of nines\n• Multiplication (General Method) 2D x 2D, 3D x3D, 4Dx4D\n• High Speed Multiplications\n• Division (General Method) by 2 digit divisor (flag method)\n• Division (General Method) by 3 digit divisor (flag method)\n• Division (General Method)by 4 digit divisor (flag method)\n• Many More...\n\n## M3\n\n\\$ 2 Months\n• Auxiliary Fraction (Divisor Ending with 9, 8, 7, 6 & 1)\n• Multiplication (General Method) 2Dx2Dx2D ( 3 rows )\n• Multiplication (Base Method), 3 rows\n• Magic Squares\n• Algebraic Multiplication\n• Algebraic Division\n• Divisibility (by 2 to 23)\n• Many More...\n\n## M4\n\n\\$ 2 Months\n• Product of two numbers ending with 5\n• Revision Multiplication Base Method (Below Base)\n• Revision Multiplication Base Method (Above Base)\n• Revision Multiplication Base Method (Mixed Base)\n• (one number below and other Above - Same base)\n• Base Method multiplication\n• Many More..." ]	[ null, "https://alpha-brains.com/wp-content/uploads/2022/03/upaa2.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg1-1.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg2.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg3.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg4.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg5.png", null, "http://alpha-brains.com/wp-content/uploads/2022/03/bg6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9401106,"math_prob":0.888705,"size":2372,"snap":"2022-40-2023-06","text_gpt3_token_len":482,"char_repetition_ratio":0.11697635,"word_repetition_ratio":0.015789473,"special_character_ratio":0.18254638,"punctuation_ratio":0.071428575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811901,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T15:42:32Z\",\"WARC-Record-ID\":\"<urn:uuid:c5a374cf-8138-42c5-a509-254e98e43cac>\",\"Content-Length\":\"100734\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb4f0013-d381-4028-817d-c6f554d63abc>\",\"WARC-Concurrent-To\":\"<urn:uuid:e3bbb124-ca77-4c5b-8323-7cd41cbbc61c>\",\"WARC-IP-Address\":\"185.212.71.38\",\"WARC-Target-URI\":\"https://alpha-brains.com/Courses/magic-math/\",\"WARC-Payload-Digest\":\"sha1:6MXNXYTRPJMEKBUUVHU7MDTRXNQMDSCR\",\"WARC-Block-Digest\":\"sha1:G3EW56CP37MHTYFRM6LPCKCIZII6AV2N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00608.warc.gz\"}"}
https://simbody.github.io/simbody-3.6-doxygen/api/structSimTK_1_1Narrowest.html	[ "Simbody 3.6\nSimTK::Narrowest< R1, R2 > Struct Template Reference\n\nThis class is specialized for all 16 combinations of standard types (that is, real and complex types in each of two precisions) and has typedefs \"Type\" which is the appropriate \"narrowed\" type for use when R1 & R2 appear in an operation together where the result must be of the narrower precision, and \"Precision\" which is the expected precision of the result (float, double). More...\n\n## Detailed Description\n\n### template<class R1, class R2> struct SimTK::Narrowest< R1, R2 >\n\nThis class is specialized for all 16 combinations of standard types (that is, real and complex types in each of two precisions) and has typedefs \"Type\" which is the appropriate \"narrowed\" type for use when R1 & R2 appear in an operation together where the result must be of the narrower precision, and \"Precision\" which is the expected precision of the result (float, double).\n\nFor example, if R1=complex< double > and R2=float, Narrowest<R1,R2>::Type is complex< float > and Narrowest<R1,R2>::Precision is float.\n\nThe documentation for this struct was generated from the following file:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90888035,"math_prob":0.8856204,"size":911,"snap":"2022-40-2023-06","text_gpt3_token_len":219,"char_repetition_ratio":0.13781698,"word_repetition_ratio":0.82758623,"special_character_ratio":0.23929748,"punctuation_ratio":0.12068965,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9690087,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T22:54:53Z\",\"WARC-Record-ID\":\"<urn:uuid:6f89780a-ff03-429f-aa28-72be1afbe214>\",\"Content-Length\":\"5884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfca9f70-8cb0-413d-b26c-bb04f0766c6c>\",\"WARC-Concurrent-To\":\"<urn:uuid:29fb62e0-6968-4871-9466-1673ff4c535a>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://simbody.github.io/simbody-3.6-doxygen/api/structSimTK_1_1Narrowest.html\",\"WARC-Payload-Digest\":\"sha1:VXTXZRP55J7CZ7SHGYZQX5HUNVK4NA5H\",\"WARC-Block-Digest\":\"sha1:FRYAXSDLGNICQH423VC6XHJXUKGTRCHI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334942.88_warc_CC-MAIN-20220926211042-20220927001042-00322.warc.gz\"}"}
https://www.midasbridge.com/en/blog/bridge-insight/wood-armer-moments-for-skew-slabs	[ "BLOG BRIDGE INSIGHT\n\n# Wood Armer Moments for Skew Slabs\n\n## 1. Introduction\n\nA bridge deck configuration depends upon the site conditions and road alignment. This article will discuss how to calculate the design moments along the reinforcement directions in a skew slab. Slab bridges tend to span in a direction running from support to support. As skew increases, the moments vary towards edges, maximum at the obtuse end and minimum towards the acute end. These can be seen from the moment contours, as shown in Figure 1. Various parametric study results have indicated that shear forces and bending moments increase with increasing skew angle.", null, "Figure 1. Image of trajectories for different skew angles.\n\nThere are many theoretical approaches to obtain optimum reinforcement angles for the most economical design for different skew angles in a slab. However, such theoretical optimization techniques are usually not practical and may not be economical to execute. A few of the different possible layouts for a skew bridge are shown in Figure 2. One can create the meshing as per different configurations such as parallel to skew, transversely normal, and longitudinally normal type of meshing. Each has some variations depending upon the loadings.", null, "Figure 2. Different skew layouts.\n\nIt is observed that as the skew increases:\n\n1. The deflection reduces\n2. The longitudinal moment reduces\n3. The point of maximum longitudinal moment shift towards the obtuse corner\n4. The shear force concentrate at the obtuse corner\n5. The transverse moment increases\n\nThe layout with straight bars in a skew slab is not the most practical. Research on skew deck slab has shown that an orthogonal layout is optimized most of the time and deviates to provide the higher amount of reinforcement.\n\nSeveral parameters can affect the magnitude of moments at the obtuse corner\n\n1. structural continuity\n2. geometry of the slab\n3. Cracked conditions\n\n## 2. Wood Armer Moments\n\nThe design moments along the reinforcement for the skew slab are a combination of (a) normal bending moments and a torsional component. These are called wood armer moments. The Wood-Armer equations were derived for the design of skew as well as curved reinforced concrete slabs.\n\n## 3. Midas Civil Features for Wood Armer Moments\n\nTo calculate the wood armer moments in midas Civil, we need to generate the slabs as plate elements. From there, we need to define the domain and sub-domain to provide the reinforcement directions along to which these forces need to arrive. A Sub-Domain can be defined while a mesh is generated using the Auto mesh feature in midas Civil or it can also be defined from the Define Sub-Domain.", null, "Figure 3. Domain definitions in midas Civil", null, "Figure 4. Sub-domain definition in midas Civil\n\nThe reference axis used to define the rebar direction can be: (Refer to Figure 3&4)\n\n1. Local\n2. UCS\n3. Reference Axis\n\n1. Local: We can define the rebar direction along the Local Coordinate System of the plate elements.\n\n2. UCS: We can define the rebar direction along the User-defined Coordinate System. If user-defined coordinates are not specified, the Global Coordinate System can be used.\n\n3. Reference Axis: We can define the rebar direction along the user-defined Reference Axis. Here we need two nodes for V1 to be specified to define the x-direction from the origin of the coordinate system. Then V2 is selected to decide the plane, and z-direction is selected as per ‘Right-Hand’ rule.\n\n## 4. Moment Calculations in midas Civil\n\nFrom the midas Civil analysis results, the following plate moments about the local axis are obtained:\n\n• mxx\n• myy\n• mxy\n\nIn order to calculate design forces or wood armer moments in the reinforcement direction, angle α and φ will be taken as shown in Figure 5.", null, "Figure 5. Domain and sub-domain definitions in midas Civil.\n\nwhere,\n\nx, y: the local axis of the plate element.\n\n1, 2: the reinforcement direction.\n\nα: the angle between local x-direction and rebar direction 1.\n\nφ: angle between rebar direction 1 and rebar direction 2.\n\nFirst, the internal forces (mxx, myy and mxy) are calculated into a-b coordinate system as shown in fig. 5.", null, "Then, Wood-Armer moments are calculated as follows:", null, "", null, "where,\n\nmud1 : is the initial moments in direction 1\n\nmud2 : is the initial moments in direction 2\n\nm’ud1 : is the wood armer moments in direction 1\n\nm’ud2 : is the wood armer moments in direction 2\n\n## 5. Verification of Wood Armer Moments\n\nA verification of Wood Armer Moment with manual calculations is explained here.\nConsider case 3 for verification purposes. To understand the calculations, we will consider node number 96, which has a maximum moment; refer to fig.6.", null, "Figure 7. Reference element for verification.\n\n### Manual Calculations for Wood Armer Moments\n\nLocal forces from the analysis are listed here: (Refer to Figure 8)\n\nElement number = 388;\n\nNode Number = 96;\n\nMxx = -30 kNm;\n\nMyy = -3.398 kNm;\n\nMxy = -12.694 kNm;\n\nSkew angle = 45;\n\nReinforcement Direction 1 = 0 from local axis x;\n\nReinforcement Direction 2 = 90 from local axis x", null, "Figure 8. Plate local forces in midas Civil\n\nReinforcement Angle α = 0 = 0 x (π/180) = 0 rad;\n\nand φ = 90 = 90 x (π/180) = 1.5708 rad;", null, "", null, "", null, "Top Design Force\n\nAs, M’ud1 < 0 and M’ud2 < 0,\n\nM’ud1 = 42.694 kNm\n\nM’ud2 = 16.092 kNm\n\nBottom Design Force\n\nAs M’ud1 > 0 and M’ud2 > 0,\n\nM’ud1 = 0 kNm", null, "M’ud2 = (-3.398) – l (-12.694 – (-3.398) x cot2 (1.5708)) / (-30 – 2 x -12.694 x cot (1.5708)) + (-3.398 x cot2 (1.5708)) l = 1.97325 kNm\n\n Wood Armer Moments Manual Calculations Midas Civil Direction Direction 1 Direction 2 Direction 1 Direction 2 Top 42.694 16.093 42.694 16.093 Bottom 0 1.973 0 1.973\n\nTable 1. Comparison of Wood Armer Moments from manual calculations and midas Civil.\n\nThe Wood Armer Moments in each directions calculated by from the Empirical formulae can be validated with midas Civil output as shown in the table 1.", null, "", null, "" ]	[ null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%201.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%202.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%203-1.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%203-2.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%204.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%205-1.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/image-png-Oct-20-2021-06-25-44-96-AM.png", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%205-3.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/Fig%206.jpg", null, "https://www.midasbridge.com/hubfs/image-png-Oct-20-2021-06-36-14-39-AM.png", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/1.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/2.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/3.jpg", null, "https://www.midasbridge.com/hs-fs/hubfs/International_Bridge/midasBridge_Blog/2021%20midasBridge%20contents/Wood%20Armer%20Moments%20for%20Skew%20Slabs/4.jpg", null, "https://www.midasbridge.com/hubfs/International_Bridge/midasBridge_Blog/2020%20midasBridge%20Contents%20(2020.10.26~2020.12.31)/Photos%20of%20specialists/Pratap-1.png", null, "https://www.midasbridge.com/hubfs/International_Bridge/email/midas%20newsletter%20MIAC/20210913/front-page.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8328241,"math_prob":0.9393348,"size":5742,"snap":"2021-43-2021-49","text_gpt3_token_len":1449,"char_repetition_ratio":0.15144649,"word_repetition_ratio":0.049196787,"special_character_ratio":0.25322187,"punctuation_ratio":0.11879433,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9701723,"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],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,3,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T10:59:26Z\",\"WARC-Record-ID\":\"<urn:uuid:8bf15a8b-d9f1-4cf4-8e34-5940cbb9dec7>\",\"Content-Length\":\"198909\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0a118f6-f500-4da3-9ba1-73dffb377eb9>\",\"WARC-Concurrent-To\":\"<urn:uuid:79e609fd-0c5a-48d3-ade8-95b148c6dc5b>\",\"WARC-IP-Address\":\"199.60.103.226\",\"WARC-Target-URI\":\"https://www.midasbridge.com/en/blog/bridge-insight/wood-armer-moments-for-skew-slabs\",\"WARC-Payload-Digest\":\"sha1:NZNYAH6AUPRQFN5NWKE6XMCKPEBOMM2N\",\"WARC-Block-Digest\":\"sha1:ZEAQJ7HVNDEVWYWFC37UJCBO5G7MEFAM\",\"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-20211127103444-20211127133444-00260.warc.gz\"}"}
http://slideplayer.com/slide/6196369/	[ "", null, "Holt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson.\n\nPresentation on theme: \"Holt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson.\"— Presentation transcript:\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson PresentationPreview\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Warm Up A rectangle has sides lengths of 12 ft and 20 ft. 2. Find the area. 1. Find the perimeter. 64 ft 240 ft 2\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids MG2.1 Use formulas routinely for finding the perimeter and area of basic two- dimensional figures and the surface area and volume of basic three-dimensional figures, including rectangles, parallelograms, trapezoids, squares, triangles, circles, prisms, and cylinders. Also covered: MG3.2 California Standards\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids DO NOT COPY A triangle or trapezoid can be thought of as half of a parallelogram. Therefore, the formulas for the area of a triangle or trapezoid have as a factor. 1212\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Additional Example 1: Using Perimeter 71 = 55 + d d = 16 in. P = 22 + 15 + 18 + d 16 = d Substitute 71 for P. Find the missing measurement when the perimeter is 71 in. 18 in. 22 in. 15 in. Subtract 55 from both sides. 55 d\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Check It Out! Example 1 58 = 49 + d d = 9 in. P = 28 + 7 + 14 + d 9 = d Substitute 58 for P. Find the missing measurement when the perimeter is 58 in. 14 in. 28 in. 7 in. Subtract 49 from both sides. 49 d\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Additional Example 2: Multi-Step Application Step 1: Find the length of the third side. Substitute 12 for a and 20 for c. A homeowner wants to plant a border of shrubs around her yard that is in the shape of a right triangle. She knows that the length of the shortest side of the yard is 12 feet and the length of the longest side is 20 feet. How long will the border be? b = 16 a 2 + b 2 = c 2 12 2 + b 2 = 20 2 144 + b 2 = 400 b 2 = 256 Use the Pythagorean Theorem. √256 = 16.\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Additional Example 3 Continued Step 2: Find the perimeter of the yard. Add all sides. P = a + b + c = 12 + 20 + 16 = 48 The border will be 48 feet long.\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Check It Out! Example 2 Step 1: Find the length of the third side. Substitute 38 for a and 32 for b. A gardener wants to plant a border of flowers around the building that is in the shape of a right triangle. He knows that the length of the shortest sides of the building are 38 feet and 32 feet. How long will the border be? c ≈ 49.68 a 2 + b 2 = c 2 38 2 + 32 2 = c 2 1444 + 1024 = c 2 2468 = c 2 Use the Pythagorean Theorem. √2468  49.68.\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Lesson Quiz Use the figure to find the following measurements. 1. the perimeter of the triangle 2. the perimeter of the trapezoid 3. the perimeter of the combined figure 4. the area of the triangle 5. the area of the trapezoid 36 cm 44 cm 54 cm 2 64 cm 104 cm 2\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Exit ticket Page 441 - 442 Problems 1 – 4, 17 -22\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Check It Out! Example 2 Continued Step 2: Find the perimeter of the yard. Add all sides. P = a + b + c  38 + 32 + 49.68  119.68 The border will be about 119.68 feet long.\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids A. (–2, 2), (4, 2), (0, 5) x y 6 3 = 9 units 2 A = bh 1212 Additional Example 3: Finding the Area of Triangles and Trapezoids Graph and find the area of the figure with the given vertices. (4, 2) (0, 5) (–2, 2) = 6 3 1212 Area of a triangle Substitute for b and h.\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids B. (–1, 1), (4, 1), (4, 4), (0, 4) = 13.5 units 2 A = h(b 1 + b 2 ) 1212 = 3(5 + 4) 1212 Area of a trapezoid Substitute for h, b 1, and b 2. (–1, 1) (0, 4) (4, 4) x y (4, 1) Graph and find the area of the figure with the given vertices. Additional Example 3: Finding the Area of Triangles and Trapezoids 5 3 4\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids A. (–1, –2), (5, –2), (5, 2), (–1, 6) Check It Out! Example 3 Graph and find the area of the figure with the given vertices. = 36 units 2 A = h(b 1 + b 2 ) 1212 = 6(8 + 4) 1212 Area of a trapezoid Substitute for h, b 1, and b 2. (–1, –2) (–1, 6) (5, 2) x y (5, –2) 48 6\n\nHolt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids B. (–1, 1), (5, 1), (1, 5) x y 6 4 = 12 units 2 A = bh 1212 Check It Out! Example 3 Graph and find the area of the figure with the given vertices. (5, 1) (1, 5) (–1, 1) = 6 4 1212 Area of a triangle Substitute for b and h.\n\nDownload ppt \"Holt CA Course 1 9-2 Perimeter and Area of Triangles and Trapezoids Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson.\"\n\nSimilar presentations" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81523144,"math_prob":0.9893209,"size":5128,"snap":"2019-43-2019-47","text_gpt3_token_len":1722,"char_repetition_ratio":0.20745511,"word_repetition_ratio":0.34427768,"special_character_ratio":0.3521841,"punctuation_ratio":0.118297875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99478465,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T12:19:53Z\",\"WARC-Record-ID\":\"<urn:uuid:5dc6f1a4-db94-491b-b018-80f7b2f4e30f>\",\"Content-Length\":\"167138\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1836cc26-d528-4751-ae08-4ce760d9603b>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e41c558-8354-45f1-a4bd-25e19c772754>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/6196369/\",\"WARC-Payload-Digest\":\"sha1:2IJR6MZLT6YN65U6W4E2P4EKZAAPMHDV\",\"WARC-Block-Digest\":\"sha1:3IQNKUIP55R5HPXOHSI5SRNFTT7VHDUS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986693979.65_warc_CC-MAIN-20191019114429-20191019141929-00314.warc.gz\"}"}
http://www.nabla.hr/Z_MemoHU-010.htm	[ "10\n\nALGEBRA\nExponential and logarithmic functions and equations\n:: Exponential and logarithmic functions (are mutually) inverse functions\n The inverse function, usually written f -1, is the function whose domain and the range are respectively the range and domain of a given function f, that is f -1(x) = y if and only if f (y) = x. Thus, the composition of the inverse function and the given function returns x, which is called the identity function, i.e., f -1(ƒ(x)) = x and f (f -1(x)) = x. The inverse of a function undoes the procedure (or function) of the given function. Therefore, to obtain the inverse of a function y = f (x), exchange the variables x and y writing x = f (y) and solve for y. That is, form the composition f (f -1(x)) = x and solve for f -1.\n Example: Given y = f (x) = log2 x determine f -1(x). Recall, a logarithm is the exponent (the power) to which a base must be raised to yield a given number that is, y = loga x if x = a y. Solution: Rewrite y = f (x) = log2 x to x = log2 y and solve for y, which gives y = f -1(x) = 2x. or, form f (f -1(x)) = x that is, log2 (f -1(x)) = x and solve for f -1, which gives y = f -1(x) = 2x.\n Graphs of inverse functions The graphs of a pair of inverse functions are symmetrical with respect to the line y = x.", null, "Exponential functions\n f (x) = y = ex <=> x = ln y, e = 2.718281828..., where e is the base of the natural logarithm. The exponential function is inverse of the natural logarithm function, so that e ln x = x.\nf (x) = y = ax <=> x = loga y, where a > 0 and a is not 1.\n The exponential function with base a is inverse of the logarithmic function, so that", null, "The graph of the exponential function y = ax = ebx, where a > 0 and b = ln a", null, "The graph of the logarithmic function y = logaxa > 0 and for a = ey = logex = ln x", null, "The logarithmic function is inverse of the f (x) = ax since its domain and the range are respectively the range and domain of the exponential function, so that", null, "The domain of f (x) = loga x is the set of all positive real numbers. The range of f (x) = loga x is the set of all real numbers. If a > 1 then f is an increasing function and if 0 < a < 1 then f is a decreasing function. The graph of the logarithmic function passes through the point (1, 0). The y-axis is the vertical asymptote to the graph, as shows the above picture.\nTranslated logarithmic and exponential functions\n Note that y - y0 = loga (x - x0) represents translated logarithmic function to base a and y - y0 = a (x - x0) represents translated exponential function with base a where, x0 and y0 are the coordinates of translations of the graph in the direction of the coordinate axes.\n Example: Given translated logarithmic function y - 2 = log3 (x + 1), find its inverse and draw their graphs. Solution: Exchange variables and solve for y, that is x - 2 = log3 (y + 1) which gives y + 1 = 3x - 2.", null, "", null, "", null, "", null, "Contents A", null, "" ]	[ null, "http://www.nabla.hr/MeInvLogExB2.GIF", null, "http://www.nabla.hr/Me83.GIF", null, "http://www.nabla.hr/MeExpFunct.GIF", null, "http://www.nabla.hr/MeLogFun.GIF", null, "http://www.nabla.hr/Me84.GIF", null, "http://www.nabla.hr/MeTranslLE.GIF", null, "http://www.nabla.hr/Previous.gif", null, "http://www.nabla.hr/TOP.gif", null, "http://www.nabla.hr/Next.gif", null, "http://www.nabla.hr/HOME1.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7292481,"math_prob":0.9998567,"size":3066,"snap":"2022-05-2022-21","text_gpt3_token_len":1013,"char_repetition_ratio":0.18451992,"word_repetition_ratio":0.13213213,"special_character_ratio":0.3330072,"punctuation_ratio":0.10696921,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000007,"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,2,null,2,null,2,null,2,null,2,null,2,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T18:27:58Z\",\"WARC-Record-ID\":\"<urn:uuid:aa7be2fe-c52b-4844-9b72-10773af0b921>\",\"Content-Length\":\"50037\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:158c572c-7e15-44c2-8d19-5937530bcdac>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4f43827-f75f-45f2-9749-6df44b171c67>\",\"WARC-IP-Address\":\"213.202.101.150\",\"WARC-Target-URI\":\"http://www.nabla.hr/Z_MemoHU-010.htm\",\"WARC-Payload-Digest\":\"sha1:B75G53KNLK7RDJEM4TDTIW6ICPHFHS7J\",\"WARC-Block-Digest\":\"sha1:KFFUYDTPGCE5A2QUQCHQOQ5E56PBFPGV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662619221.81_warc_CC-MAIN-20220526162749-20220526192749-00756.warc.gz\"}"}
https://blog.cropbytes.com/deflation-protocol/	[ "# Game Coins and Deflation Protocol\n\nCropBytes is the top crypto game since 2018. Since the start, the development and growth of CropBytes has been constant. CropBytes has grown from a 2D game with basic game economics to a full-fledged multi-platform game with unmatched game economics and the world’s most advanced trading platform for games.\n\nThe Crypto market is soaring right now. To keep up we are bringing in some major improvements in the game and the economy. Two of these are deflation protocol and Game Coin.\n\nLet’s take a look at these in detail.\n\n## Game Coins\n\n#### What are Game Coins?\n\nGame Cards will now be called Game Coins. Game Coins are an in-game currency used in the CropBytes ecosystem. These coins can only be minted by players in the game.\n\n#### How are they created?\n\nMinting of a Game Coin requires a player to maintain a farm and complete game activities to collect produce from animals. Animal products can then be burnt to create Game Coins.\n\n#### Applications\n\n##### Current\n• Mining: Game Coins are mined by players in the CropBytes ecosystem.\n• Melting: Game Coins can be converted to Pro assets.\n• Trade: Game Coins are trade-able with Tron (TRX)\n##### Future\n• Trading platforms: Game Coins (GC) will be tradable with BTC, ETH, USDT on major trading platforms.\n• Mini-Games: Used as tokens in various mini-games.\n• Alliances: Will be used by alliances and partners of CropBytes.\n• Game transactions: Trading fee and other purchases in the CropBytes ecosystem.\n\n## Deflation protocol\n\nAs more and more players start playing, game activity increases, the production of extracts also increases, and so does the minting of GC. This eventually will lead to an unlimited supply of Game Coins and a reduction in their value.Game Coins are based on the Deflation Protocol, which means the rate of production slows down as the supply increases.\n\nGame Coins can also be melted to acquire pro assets, thus reducing the supply.\n\n#### How does it work?\n\nTo mine GC, players need to convert (burn) their produced goods. The factor of conversion (Difficulty) is based on the total GC in circulation.\n\nFormula for GC Generation:\n\n##### (Extract x Conversion rate ) x Difficulty = GC\n\nExample: Where conversion rate is 4 milk = 1 GC\n\nWhen Difficulty = 1\n(Milk x Conversion rate = 4) x Difficulty (1) ~ 4×1= 4\n∴ 4 Milk = 1 GC\n\nWhen Difficulty = 0.5\n(Milk x Conversion rate = 4) x Difficulty (0.5) ~ 4×0.5 = 2\n∴ 2 Milk = 1 GC\n\nWhen Difficulty = 1.5\n(Milk x Conversion rate = 4) x Difficulty (1.5) ~ 4×1.5= 6\n∴ 6 Milk = 1 GC\n\nExample: Where conversion rate is 8 milk = 1 GC\n\n• When Difficulty = 1\n(Milk x Conversion rate = 8) x Difficulty (1) ~ 8×1= 8\n∴ 8 Milk = 1 GC\n• When Difficulty = 0.5\n(Milk x Conversion rate = 8) x Difficulty (0.5) ~ 8×0.5= 4\n∴ 4 Milk = 1 GC\n• When Difficulty = 1.5\n(Milk x Conversion rate = 8) x Difficulty (1.5) ~ 8×1.5= 12\n∴ 12 Milk = 1 GC\n\nFormula for difficulty:\n\n##### Difficulty = ( Actual produced GC / Estimated Gc) ^ 5\n\nExample:\n\n• If estimated GC supply by X weeks is 1,42,000 and actually GC produced is 1,52,000.\nDifficulty = (1,52,000/ 1,42,000)⁵ = 1.145\nMilk required = 4 * 1.145 = 4.58\n∴ 4.58 Milk = 1 GC\n• If estimated GC supply by X weeks is 1,42,000, and actually GC produced is 1,35,000.\nDifficulty = ( 135000 / 142000)⁵ = 0.77\nMilk required = 4 * 0.77 = 3.08\n∴ 3.08 Milk = 1 GC\n\nFor example:\nMax 200000 GC is estimated to be generated per week. The count of estimated GC minted will be reduced by 50% every 10 weeks.\n\nTherefore:\n\n• Week 10–20: Estimated GC per week = 200,000\n• Week 21–30: Estimated GC per week= 200,000 / 2 = 100,000\n\nAs GC is produced and burned, the difficulty factor fluctuates. If the Total GC generated is more or less than the estimated GC by that particular time period, then it will affect the difficulty.\n\n#### Conversion rate\n\nAfter every period, the standard conversion rate for extracts to GC increases.\n\nExample:\n\n• Week 10- 20: 4 milk = 1 GC\n• Week 21–30: 8 milk = 1 GC\n• Week 31–40: 16 milk = 1 GC\n\n### Supply\n\n• Current Circulation: 2,892,893 \| 16.17%\n• Pre Minted (team): 3,000,000 \| 16.77%\n• Pre Minted (operations): 2,000,000 \| 11.18%\n• Mintable GC: 10,000,000 \| 55.89%\n##### Total Supply: 17,892,893.00\n\nAll the users can see the total supply in circulation & total GC produced per day on the GC Conversion Page.\n\n#### Unlocking Process\n\nTotal GC allocated for team & CropBytes will be locked for the first year. After which, 20% of the total will be unlocked each year.\n\n#### TRC20 Token\n\nSoon, GC will be converted to TRC20 tokens. The distribution of tokens will remain the same as current GC distribution." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9016742,"math_prob":0.97987336,"size":4455,"snap":"2021-21-2021-25","text_gpt3_token_len":1274,"char_repetition_ratio":0.14895529,"word_repetition_ratio":0.11965812,"special_character_ratio":0.30774412,"punctuation_ratio":0.13796791,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95745486,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T06:57:18Z\",\"WARC-Record-ID\":\"<urn:uuid:6fa19d24-881d-46fd-993b-a7a581b9b1f1>\",\"Content-Length\":\"76077\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09418f68-3ec2-432a-bc37-cc7f86a2c4bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5569d9f-1eab-4f46-a21d-f2eb2ccee4df>\",\"WARC-IP-Address\":\"45.93.101.7\",\"WARC-Target-URI\":\"https://blog.cropbytes.com/deflation-protocol/\",\"WARC-Payload-Digest\":\"sha1:OAHYIMD3PPAOIFZCZ7YG55KEPUNMZKOJ\",\"WARC-Block-Digest\":\"sha1:RWA5EAPB5626XKKHAEDKJ7ONBBALDXAV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487643703.56_warc_CC-MAIN-20210619051239-20210619081239-00187.warc.gz\"}"}
https://www.percentagecal.com/answer/what-is-%2010-percent-of-%2024500	[ "#### Solution for What is 10 percent of 24500:\n\n10 percent * 24500 =\n\n( 10:100)* 24500 =\n\n( 10* 24500):100 =\n\n245000:100 = 2450\n\nNow we have: 10 percent of 24500 = 2450\n\nQuestion: What is 10 percent of 24500?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 24500.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{ 24500}={100\\%}.\n\nStep 4: Similarly, {x}={ 10\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{ 24500}={100\\%}(1).\n\n{x}={ 10\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{ 24500}{x}=\\frac{100\\%}{ 10\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{ 24500}=\\frac{ 10}{100}\n\n\\Rightarrow{x} = {2450}\n\nTherefore, { 10\\%} of { 24500} is {2450}\n\n#### Solution for What is 24500 percent of 10:\n\n24500 percent * 10 =\n\n( 24500:100)* 10 =\n\n( 24500* 10):100 =\n\n245000:100 = 2450\n\nNow we have: 24500 percent of 10 = 2450\n\nQuestion: What is 24500 percent of 10?\n\nPercentage solution with steps:\n\nStep 1: Our output value is 10.\n\nStep 2: We represent the unknown value with {x}.\n\nStep 3: From step 1 above,{ 10}={100\\%}.\n\nStep 4: Similarly, {x}={ 24500\\%}.\n\nStep 5: This results in a pair of simple equations:\n\n{ 10}={100\\%}(1).\n\n{x}={ 24500\\%}(2).\n\nStep 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both\nequations have the same unit (%); we have\n\n\\frac{ 10}{x}=\\frac{100\\%}{ 24500\\%}\n\nStep 7: Again, the reciprocal of both sides gives\n\n\\frac{x}{ 10}=\\frac{ 24500}{100}\n\n\\Rightarrow{x} = {2450}\n\nTherefore, { 24500\\%} of { 10} is {2450}\n\nCalculation Samples" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.764586,"math_prob":0.99928135,"size":1984,"snap":"2020-45-2020-50","text_gpt3_token_len":729,"char_repetition_ratio":0.19040404,"word_repetition_ratio":0.3547486,"special_character_ratio":0.500504,"punctuation_ratio":0.15690866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999455,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T14:46:43Z\",\"WARC-Record-ID\":\"<urn:uuid:6d1b5425-b4ee-45cf-927c-7d53c9e76645>\",\"Content-Length\":\"10307\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:619fc5d6-e4c4-426a-8ad9-42b10424aa7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:cef2e4b1-73c4-4b9e-99fb-ecf222fe0848>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"https://www.percentagecal.com/answer/what-is-%2010-percent-of-%2024500\",\"WARC-Payload-Digest\":\"sha1:7T5EL6SGGB24VANIVRZTZ4AUGRAEG6IW\",\"WARC-Block-Digest\":\"sha1:VMOF3OFZEHYBHWEABOW72A5PX5VECWUA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107904287.88_warc_CC-MAIN-20201029124628-20201029154628-00119.warc.gz\"}"}
https://crypto.stackexchange.com/questions/68859/what-is-the-difference-between-discrete-then-gaussian-and-gaussian-then-discrete	[ "# What is the difference between discrete-then-gaussian and gaussian-then-discrete?\n\nIn lattice cryptography, we always face the probem of discrete gaussian sampling. To the beginners, it is a bit complex. However, gaussian sampling from a continous space is much easier to understand, and a lot of tools are available. Say, we can use MATLAB to do gaussian sampling very efficiently. So, I want to know what is the difference between the following to process:\n\n(1) discrete-then-gaussian: Just as required is many lattice cryptography papers.\n\n(2) gaussian-then-discrete: At first, get continous gaussian samples, say by using MATLAB, and then perform nearest rounding operations, i.e., discrete to the nearest integers.\n\n• The resulting distributions of (1) and (2) are not quite the same, and in some applications even a tiny difference in these distributions can be fatal for proving/guaranteeing security. – TMM Apr 18 at 1:11\n• @TMM Thanks! Can you give me more references or explanations on the differences? – Licheng Wang Apr 18 at 6:10\n• Simply from looking at both definitions you should be able to see there is no immediate reason for the two distributions to be equivalent - sure, they both mimic a continuous Gaussian on a discrete set, but the probability mass function is different. – TMM Apr 18 at 23:53" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95144534,"math_prob":0.93686575,"size":634,"snap":"2019-51-2020-05","text_gpt3_token_len":134,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.0,"special_character_ratio":0.20347004,"punctuation_ratio":0.176,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935185,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T01:18:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9f1c5801-4b81-40b4-bfce-6031a894c225>\",\"Content-Length\":\"128038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f9c3ae7-2122-43c1-b688-94183ffeff43>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed3cdbab-fc28-4e2c-8df9-f9cf1fbef927>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://crypto.stackexchange.com/questions/68859/what-is-the-difference-between-discrete-then-gaussian-and-gaussian-then-discrete\",\"WARC-Payload-Digest\":\"sha1:6KUHEZNZ7FUEF6BUSCPBRRZK5MJOGQT2\",\"WARC-Block-Digest\":\"sha1:Y5Q5BZAMJVLCMTIJROPEDTGL6XBBXUXG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540503656.42_warc_CC-MAIN-20191207233943-20191208021943-00553.warc.gz\"}"}
https://www.pustudy.com/calculators/calc/math/Arctan_Calculator.html	[ "# Arctan Calculator\n\nFor example, If the tangent of 45° is 1:\n\ntan(45°) = 1\n\nThen the arctangent of 1 is 45°:\n\narctan(1) = tan-1(1) = 45°\n\n## Arctangent table\n\ny x = arctan(y)\n-1.732050808 -60° -π/3\n-1 -45° -π/4\n-0.577350269 -30° -π/6\n0 0\n0.577350269 30° π/6\n1 45° π/4\n1.732050808 60° π/3\n\nHere we discuss list of online math calculators that help us for doing calculation easily.\n\n### Arctan Calculator\n\nThe arctan is the inverse of the tangent. It is normally represented by arctan(θ) or tan-1(θ).Use arctan calculator to easily calculate the arctan of a given number. Online arctangent calculation tool to compute the arcus tangens function in degrees or radians. Supports input of decimal numbers (0.5, 6, -1, etc.) and fractions (1/3, 3/4, 1/6, -4/3 etc.).\n\nArctangent definition\n\nThe arctangent function is the inverse function of y = tan(x).\n\narctan(y) = tan-1(y) = x+ kπ\n\nFor every\n\nk = {...,-2,-1,0,1,2,...}\n\nFor example, If the tangent of 45° is 1:\n\ntan(45°) = 1\n\nThen the arctangent of 1 is 45°:\n\narctan(1) = tan-1(1) = 45°\n\nHow to Use the Arctan Calculator?\n\nThe procedure to use the arctan calculator is as follows:\n\n1. Enter the tangent value in the respective input field\n\n2. Now click the button “Calculate Arctan” to get the angle\n\n3. Finally, the angle in both degree and radian measure will be displayed in the output field\n\nWhat is arctan?\n\nArctangent is the inverse of the tangent function. Simply speaking, we use arctan when we want to find an angle for which we know the tangent value.\n\nHowever, in the strictest sense, because the tangent is a periodic trigonometric function, it doesn't have an inverse function. Still, we can define an inverse function if we restrict the domain to the interval where the function is monotonic. Using the tan-1x convention may lead to confusion about the difference between arctangent and cotangent. It turns out that arctan and cot are really separate things:\n\n1. cot(x) = 1/tan(x), so cotangent is basically the reciprocal of a tangent, or, in other words, the multiplicative inverse\n\n2. arctan(x) is the angle whose tangent is x\n\nWe hope that now you do not doubt that arctan and cotan are different. To avoid any further misunderstandings, you may want to use the arctan(x) rather than tan-1x notation\n\nArctan function :- The arctan (a.k.a. arcus tangens) is one of the inverse trigonometric functions (antitrigonometric functions) and is the inverse of the tangent function. It is sometimes written as tan-1(x), but this notation should be avoided as it can cause confusion with an exponent notation. The arctan is used to obtain an angle from the tangent trigonometric ratio, which is the ratio between the side opposite to the angle and the adjacent side of the triangle.\n\nThe function spans all real numbers (-∞ - +∞) and so do the results from our calculator. The range of the angle values is usually between -90° and 90°. There are a number of arctan rules, like that tan(arctan(x)) = x, or that arctanα + arctanβ = arctan((α + β) / (1-αβ)), as well as sine of the arctangent: sin(arctan(x)) = x / √(1+x2), which can help you in trigonometry calculations." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8200155,"math_prob":0.99828005,"size":3308,"snap":"2023-14-2023-23","text_gpt3_token_len":975,"char_repetition_ratio":0.1822034,"word_repetition_ratio":0.11270125,"special_character_ratio":0.286578,"punctuation_ratio":0.12071535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99936527,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-31T20:26:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1faf278f-c439-4d84-ab22-e43de4f1b28d>\",\"Content-Length\":\"23388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b348e786-ce94-4c6c-924e-d2793442079d>\",\"WARC-Concurrent-To\":\"<urn:uuid:efdef45a-50c9-497f-988e-62dc0602e273>\",\"WARC-IP-Address\":\"103.93.17.55\",\"WARC-Target-URI\":\"https://www.pustudy.com/calculators/calc/math/Arctan_Calculator.html\",\"WARC-Payload-Digest\":\"sha1:HK2OFJY35ZWWSMHANP2OSYVX7WXFOUVY\",\"WARC-Block-Digest\":\"sha1:QG3SNTJ536WIDF7X5376GNSU2Y5SHLXS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647409.17_warc_CC-MAIN-20230531182033-20230531212033-00497.warc.gz\"}"}
https://featheressays.com/let-professional-handle-your-assignment-get-an-a-cloud-architectures-application-121/	[ "Referencing Styles : APA Part A Circuit Option 1: Part A Only (35 marks):The implementation for this part must use only the three basic logic gates (AND, OR, NOT) with maximum 2 inputs.You are required to implement a circuit where the user (you) can input a key value (K1, K2, and K3) and a lock value (L1, L2, and L3) and the circuit decodes the K1, K2, K3 and L1, L2, L3 values using a decoder (see lecture notes) as well as other permitted logic gates to determine if the key matches the lock.The output will be a single L … View More Part A Circuit Option 1: Part A Only (35 marks): The implementation for this part must use only the three basic logic gates (AND, OR, NOT) with maximum 2 inputs. You are required to implement a circuit where the user (you) can input a key value (K1, K2, and K3) and a lock value (L1, L2, and L3) and the circuit decodes the K1, K2, K3 and L1, L2, L3 values using a decoder (see lecture notes) as well as other permitted logic gates to determine if the key matches the lock. The output will be a single LED labelled OPEN which is lit if the key matches the lock, and is not lit if the key and lock are different. Part B Circuit Option 2: Part A and Part B (60 marks): For this part, the lock allows a certain number (n) of incorrect attempts. The number of times an invalid key value can be input ranges from 0 to 7 and should be set via a combination of three separate inputs: X1, X2 and X3 Note: You can combine the 3 inputs into a single 3 bit input. With an n value of 0 you can make 0 incorrect key attempts (ie. The first attempt has to work and the OPEN LED has to light up and be set to 1). Using the same circuit as Part A, add additional circuitry to count how many incorrect attempts have been made. Read Less" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86670923,"math_prob":0.9841381,"size":1757,"snap":"2020-24-2020-29","text_gpt3_token_len":459,"char_repetition_ratio":0.12093554,"word_repetition_ratio":0.44606414,"special_character_ratio":0.26465565,"punctuation_ratio":0.10997442,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97470886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T09:47:58Z\",\"WARC-Record-ID\":\"<urn:uuid:78963a6f-4c9c-4c9d-a331-27892484d02d>\",\"Content-Length\":\"50151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab6aef0c-3e98-4b0f-b5c7-33cba52a4971>\",\"WARC-Concurrent-To\":\"<urn:uuid:53c1a2df-8268-4649-920f-bd52ace7d160>\",\"WARC-IP-Address\":\"198.54.114.164\",\"WARC-Target-URI\":\"https://featheressays.com/let-professional-handle-your-assignment-get-an-a-cloud-architectures-application-121/\",\"WARC-Payload-Digest\":\"sha1:MZDNAJYL27AQSUTHFM6VYX5OQ4SEV3RU\",\"WARC-Block-Digest\":\"sha1:VXLVNUJXLU3ZNKQ2RRHSDCX2LGYJEZJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348513230.90_warc_CC-MAIN-20200606093706-20200606123706-00180.warc.gz\"}"}