How is the distance from the sun for planets in oursolar system related to the mean temperature ofeach planet? To find out, a scatterplot that relates thenatural log of the distance of each planet (includingPluto) from the sun in millions of miles and thenatural log of the mean planetary temperature inKelvin was created.In(Temperature) vs. In(Distance)6.86.66.46.265.85.65.45.24.84.64.44.2446 78.In(Distance)PredictorCoef7.9009SE CoefPConatant0.438118.030.000In Distance-0.45360.0706-6.420.004s = 0.3446R-Sq = 85.5R-8q (adj) = 83.2%Based on the scatterplot and computer output, areasonable estimate of mean temperature in Kelvinfor Saturn, which is 886.7 million miles away fromthe sun is:O 4.822 degrees Kelvin because ý = -0.4536(In886.7) + 7.9009 = 4.822.O 124.2 degrees Kelvin becausein y = -0,4536(in 886,7) + 7.9009 =4,822and e4.822 = 124.2.O 709.0 degrees Kelvin becauseIn y = - 0.4536(log 886.7) + 7.9009 =6.564and e6.564 = 709.0.O 0.05 degrees Kelvin becauseIn y = -0,4536(in 886.7) + 0.0706 = - 3.008and e-3.008 = 0.0494.In(Temperature)3.

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. In(Temperature) vs. In(Distance) 6.8 6.6 6.2 E 5.8 5.6 5.4 5.2 4.6 4.4 4.2 4 In(Distance) Predictor Constant Coef 7.9009 -0.4536 SE Coef 0.4381 18.03 0.000 -6.42 0.004 In Distance 0.0706 S= 0.3446 R-8g - 85.54 R-8q (adj) - 83.24 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: O 4.822 degrees Kelvin because ý = -0.4536(In 886.7) + 7.9009 = 4.822. O 124.2 degrees Kelvin because iny = -0,4536(mn 86.7) +7.9009 =4,822 and et 82 = 124.2. O 709.0 degrees Kelvin because iny = -0,4536(kog 886,7) + 7.9KK9 = 6,564 and ef 564 = 709.0. O 0.05 degrees Kelvin because In y = -0,4536(in 886.7) + 0.0706 = – 3.008 and e3008 = 0.0494. 889426D O642 58642 6666 555 4444 In(Temperature)

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. In(Temperature) vs. In(Distance) 6.8 6.6 6.2 E 5.8 5.6 5.4 5.2 4.6 4.4 4.2 4 In(Distance) Predictor Constant Coef 7.9009 -0.4536 SE Coef 0.4381 18.03 0.000 -6.42 0.004 In Distance 0.0706 S= 0.3446 R-8g - 85.54 R-8q (adj) - 83.24 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: O 4.822 degrees Kelvin because ý = -0.4536(In 886.7) + 7.9009 = 4.822. O 124.2 degrees Kelvin because iny = -0,4536(mn 86.7) +7.9009 =4,822 and et 82 = 124.2. O 709.0 degrees Kelvin because iny = -0,4536(kog 886,7) + 7.9KK9 = 6,564 and ef 564 = 709.0. O 0.05 degrees Kelvin because In y = -0,4536(in 886.7) + 0.0706 = – 3.008 and e3008 = 0.0494. 889426D O642 58642 6666 555 4444 In(Temperature)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Red Shift

It is an astronomical phenomenon. In this phenomenon, increase in wavelength with corresponding decrease in photon energy and frequency of radiation of light. It is the displacement of spectrum of any kind of astronomical object to the longer wavelengths (red) side.

Question

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.
In(Temperature) vs. In(Distance)
6.8
6.6
6.4
6.2
6
5.8
5.6
5.4
5.2
4.8
4.6
4.4
4.2
4
4
6 7
8.
In(Distance)
Predictor
Coef
7.9009
SE Coef
P
Conatant
0.4381
18.03
0.000
In Distance
-0.4536
0.0706
-6.42
0.004
s = 0.3446
R-Sq = 85.5
R-8q (adj) = 83.2%
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:
O 4.822 degrees Kelvin because ý = -0.4536(In
886.7) + 7.9009 = 4.822.
O 124.2 degrees Kelvin because
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
In y = -0,4536(in 886.7) + 0.0706 = - 3.008
and e-3.008 = 0.0494.
In(Temperature)
3.

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