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
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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.
O 709.0 degrees Kelvin because
In y =-0.4536(log 886.7) +7.9009=6.564 and e6.564 = 709.0.
O 0.05 degrees Kelvin because
5.4
5.2
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.
8.
6.
In(Distance)
redictor
onstant
n Distance
Coef
SE Coef
T
20
0.4381
0.0706
0.000
0.004
7.9009
18.03
-0.4536
-6.42
= 0.3446
R-Sq = 85.54
R-Sq (adj) = 83.2%
CO64268642N5864 24
5555

Transcribed Image Text:Based on the scatterplot and computer output, a
reasonable estimate of mean temperature in Kelvin for
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
Saturn, which is 886.7 million miles away from the sun
is:
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 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
5.8
In y =-0.4536(log 886.7) + 7.909 =6.564 and e6.564 = 709.0.
5.6
5.4
O0.05 degrees Kelvin because
5.2
In y =-0.4536(In 886.7) +0.0706 = - 3.008 and e 3.008 -
E 48
4.6
0.0494.
4.4
4.2
In(Temperature)
6966 5555
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