How is the distance from the sun for planets in our solar system related to the mean temperature of each planet? To find out, a scatterplot that relates the natural log of the distance of each planet (including Pluto) from the sun in millions of miles and the natural log of the mean planetary temperature in Kelvin was created. 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 ŷ = −0.4536(In 886.7) + 7.9009 = 4.822. 124.2 degrees Kelvin because and e4.822 = 124.2. 709.0 degrees Kelvin because and e6.564 = 709.0. 0.05 degrees Kelvin because and e−3.008 = 0.0494.
How is the distance from the sun for planets in our solar system related to the mean temperature of each planet? To find out, a scatterplot that relates the natural log of the distance of each planet (including Pluto) from the sun in millions of miles and the natural log of the mean planetary temperature in Kelvin was created. 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 ŷ = −0.4536(In 886.7) + 7.9009 = 4.822. 124.2 degrees Kelvin because and e4.822 = 124.2. 709.0 degrees Kelvin because and e6.564 = 709.0. 0.05 degrees Kelvin because and e−3.008 = 0.0494.
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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How is the distance from the sun for planets in our solar system related to the mean temperature of each planet? To find out, a
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 ŷ = −0.4536(In 886.7) + 7.9009 = 4.822.
124.2 degrees Kelvin because and e4.822 = 124.2.
709.0 degrees Kelvin because and e6.564 = 709.0.
0.05 degrees Kelvin because and e−3.008 = 0.0494.
Transcribed Image Text:In(Temperature) vs. In(Distance)
O 124.2 degrees Kelvin because
6.8
6.6
6.4
6.2
In y = - 0.4536(In 886.7) + 7.9009 =4.822 and e4.822 = 124.2.
709.0 degrees Kelvin because
6
5.8
5.6
5.4
5.2
In y = - 0.4536(log 886.7) + 7.9009 =6.564 and e6.564 = 709.0.
O 0.05 degrees Kelvin because
In y = - 0.4536(In 886.7) + 0.0706 = – 3.008 and e-3.008 =
4.8
4.6
4.4
4.2
0.0494.
4
5 6
7
8
9
In(Distance)
SE Coef
0.4381
0.0706
Predictor
Coef
T
P
Constant
7.9009
18.03
0.000
In Distance
-0.4536
-6.42
0.004
S = 0.3446
R-Sq = 85.5
R-Sq (adj) = 83.2%
8642 Ooo4 250o 424
5555
444
In(Temperature)
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