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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 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:

A. 4.822 degrees Kelvin because [tex]\hat{y} = -0.4536(\ln 886.7) + 7.9009 = 4.822[/tex].

B. 124.2 degrees Kelvin because [tex]\widehat{\ln y} = -0.4536(\ln 886.7) + 7.9009 = 4.822[/tex] and [tex]e^{4.822} = 124.2[/tex].

C. 709.0 degrees Kelvin because [tex]\widehat{\ln y} = -0.4536(\log 886.7) + 7.9009 = 6.564[/tex] and [tex]\theta^{6.564} = 709.0[/tex].

D. 0.05 degrees Kelvin because [tex]\hat{\ln y} = -0.4536(\ln 886.7) + 0.0706 = -3.008[/tex] and [tex]e^{-3.008} = 0.0494[/tex].



Answer :

GinnyAnswer
To find the mean temperature in Kelvin for Saturn, which is 886.7 million miles away from the sun, we can utilize the relationship between the natural logarithm of the distance from the sun and the natural logarithm of the mean planetary temperature.

1. Step 1: Calculate the natural logarithm of the distance:
- The distance from the sun to Saturn is 886.7 million miles.
- Compute the natural logarithm of [tex]\( 886.7 \)[/tex]:
[tex]\[ \ln(886.7) \approx 6.787 \][/tex]

2. Step 2: Estimate the natural logarithm of the mean temperature:
- Using the given relationship:
[tex]\[ \widehat{\ln y} = -0.4536 \cdot \ln(886.7) + 7.9009 \][/tex]
- Substitute [tex]\( \ln(886.7) = 6.787 \)[/tex]:
[tex]\[ \widehat{\ln y} = -0.4536 \cdot 6.787 + 7.9009 \approx 4.822 \][/tex]

3. Step 3: Find the mean temperature in Kelvin:
- The natural logarithm of the mean temperature [tex]\( \widehat{\ln y} \)[/tex] has been found to be 4.822.
- To find the mean temperature [tex]\( y \)[/tex] in Kelvin, exponentiate [tex]\( \widehat{\ln y} \)[/tex]:
[tex]\[ y = e^{4.822} \approx 124.2 \][/tex]

4. Conclusion:
- The estimated mean temperature for Saturn is approximately [tex]\( 124.2 \)[/tex] degrees Kelvin.

Thus, the correct answer is:

124.2 degrees Kelvin because [tex]\( \widehat{\ln y} = -0.4536(\ln 886.7) + 7.9009 = 4.822 \)[/tex] and [tex]\( e^{4.822} = 124.2 \)[/tex].

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