In(Temperature) vs. In(Distance)O 124.2 degrees Kelvin because6.86.66.46.2In y =-0.4536(In 886.7) +7.9009 =4.822 and e4.822 = 124.2.O 709.0 degrees Kelvin becauseIn y =-0.4536(log 886.7) +7.9009=6.564 and e6.564 = 709.0.O 0.05 degrees Kelvin because5.45.2In y =-0.4536(In 886.7) +0.0706=- 3.008 and e-3.008 =4.84.64.44.20.0494.4.8.6.In(Distance)redictoronstantn DistanceCoefSE CoefT200.43810.07060.0000.0047.900918.03-0.4536-6.42= 0.3446R-Sq = 85.54R-Sq (adj) = 83.2%CO64268642N5864 245555 Based on the scatterplot and computer output, areasonable estimate of mean temperature in Kelvin forHow is the distance from the sun for planets in our solarsystem related to the mean temperature of each planet?To find out, a scatterplot that relates the natural log of thedistance of each planet (including Pluto) from the sun inmillions of miles and the natural log of the meanSaturn, which is 886.7 million miles away from the sunis:O 4.822 degrees Kelvin because ý = -0.4536(In 886.7)+ 7.9009 = 4.822.planetary temperature in Kelvin was created.In(Temperature) vs. In(Distance)O124.2 degrees Kelvin because6.86.66.46.2In y = -0.4536(In 886.7) + 7.9009=4.822 and e4.822 = 124.2.709.0 degrees Kelvin because5.8In y =-0.4536(log 886.7) + 7.909 =6.564 and e6.564 = 709.0.5.65.4O0.05 degrees Kelvin because5.2In y =-0.4536(In 886.7) +0.0706 = - 3.008 and e 3.008 -E 484.60.0494.4.44.2In(Temperature)6966 5555

Name: In(Temperature) vs. In(Distance)O 124.2 degrees Kelvin because6.86.66
Uploaded: 2024-03-20T07:07:12.000Z
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Description: In(Temperature) vs. In(Distance)O 124.2 degrees Kelvin because6.86.66.46.2In y =-0.4536(In 886.7) +7.9009 =4.822 and e4.822 = 124.2.O 709.0 degrees Kelvin becauseIn y =-0.4536(log 886.7) +7.9009=6.564 and e6.564 = 709.0.O 0.05 degrees Kelvin because

From the output, the regression equation is

Math

Statistics

Based on the scatterplot and computer output, a reasonable estimate of mean temperature in Kelvin for Saturn, which is 886.7 million miles away from the sun is: 4.822 degrees Kelvin because ý = -0.4536(In 886.7) + 7.9009 = 4.822. %3D 124.2 degrees Kelvin because In y = -0.4536(In 886.7) +7.9009 =4.822 and e4.822 = 124.2. %3D %3D

Based on the scatterplot and computer output, a reasonable estimate of mean temperature in Kelvin for Saturn, which is 886.7 million miles away from the sun is: 4.822 degrees Kelvin because ý = -0.4536(In 886.7) + 7.9009 = 4.822. %3D 124.2 degrees Kelvin because In y = -0.4536(In 886.7) +7.9009 =4.822 and e4.822 = 124.2. %3D %3D

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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