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data were recorded for a cars fuel efficiency, in miles per gallon (mpg…

Question

data were recorded for a cars fuel efficiency, in miles per gallon (mpg), and corresponding speed, in miles per hour (mph). given the least - squares regression line, $ln(\text{fuel efficiency}) = 1.437 + 0.541ln(\text{speed})$, what is the fuel efficiency for a speed of 60 mph?
3.65 mpg
33.90 mpg
36.52 mpg
38.55 mpg

Explanation:

Step1: Substitute speed value into the equation

Given \( \text{Speed} = 60\), substitute into \( \ln(\text{Fuel Efficiency})=1.437 + 0.541\ln(\text{Speed})\). First, calculate \( \ln(60)\). Using a calculator, \( \ln(60)\approx4.0943\). Then, \( 1.437+0.541\times4.0943\).

$$ LATEXBLOCK0 $$

\( 1.437+2.2150163 = 3.6520163\)

Step2: Solve for Fuel Efficiency

Since \( \ln(\text{Fuel Efficiency}) = 3.6520163\), by the property \( y = \ln(x)\Rightarrow x=e^{y}\), then \( \text{Fuel Efficiency}=e^{3.6520163}\). Using a calculator, \( e^{3.6520163}\approx38.55\) (because \( e^{3}=20.0855\), \( e^{4}=54.5982\), and \( 3.6520163\) is closer to \( 3.6\) and \( e^{3.6}\approx36.598\), \( e^{3.7}\approx40.447\)).

Answer:

38.55 mpg