Procedure B1-2420.001length = 12.42 mm ± 0.01 mmproportional error in length = 8.652 x 107.576x10proportional error in mass=cylinder volume =uncertainty in cylinder volume =diameter = 4.16 mm ± 0.01 mm416.88 см3.lope±6.09 cm³3h anglemass= 1.32 g + 0.01 gproportional error in diameter = 2.404 x 100.07829/em3Arev. VoeTOLD²V=4proportional error in cylinder volume = 5.613 x 18-3follows that sin(240AVAD LAPND D321 notburgaal daiXALAcylinder density3uncertainty in the density = 0.0004 9/cm³(available work space)placed as anticipated (yes, no)your date sheelproportional error in cylinder density = 5.3363placed as anticipated (yes, no).digralInnotnoqonuobwoquiq

Answered: Image /qna-images/answer/0ed5adae-f1df-4cf8-b4df-06ee53e2d8d1.jpg

Answered: 1-242 0.001 length = 12.42 mm ± 0.01 mm…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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(856) De'arra & Ken 4 Life - Yo x K Take Test: Assignment 3 - Cha x C Listed Below Are The Top 10 Ar X oa Range Rule of Thumb Calculate x E Mean, Mediar A mcphs.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id%3=_43603_1&course_id%3D_34586_1&new_attempt31&conten pps O Course Materials f... 24 Google Calendar -. - Men's Nike Air For.. Ximble Money Transfer -. a, Amazon.com: Onli. M Inbox (2,647) - iak.. * Question Completion Status: QUESTION 9 Questions 9-10 Listed below are amounts (in millions of dollars) collected from parking meters by a security service company and other companies during similar tir periods. Do the limited data listed here show evidence of stealing by the security service company's employees? Security Service Company: Other Companies: 1.3 1.6 1.7 1.4 1.6 1.4 1.7 1.6 1.4 1.6 1.8 1.9 1.7 1.7 1.6 1.9 1.6 1.7 1.9 1.7 Find the coefficient of variation (use 1 decimal) for each of the two samples, then compare the variation. %. The coefficient of…
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1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 RSTi =3586.38−0.150746 x TOTWRKi , R2=0.103287,SER=421.1357 (38.9124) (0.0167403) Question? could you please help with this question below. 3) By observing the GRETL output in Part (1) above, provide a detailed explanation of the coefficient of determination. Based on your analysis, is this a good model? Why or why not?
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7/8/22, 3:04 PM Area z TABLE V z 78877 99737 77238 23927 328797 97979 777738=283 09898 30003 99999 33233 38539 22 -3.4 -2.5 -0.5 0.0010 0.0013 .00 .01 .02 .06 .07 .08 .09 0.0007 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084…
1 p = 0.28 Za/2=2.321 m = 0.03 What sample size is needed. (2.326 2 72 = 0.28(1-0.28) (-0.03)

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