Question

data were recorded for the temperature, in degrees celsius, of a cup of coffee over a 30 - minute period. given the regression equation, $ln(\text{temp}) = 4.20 - 0.023(\text{time})$, what is the predicted temperature after 5 minutes?

$59.44 ^{circ}c$

$64.26 ^{circ}c$

$65.17 ^{circ}c$

$66.69 ^{circ}c$

Explanation:

Step1: Substitute Time = 5 into the regression equation

Given the regression equation \(\ln(\text{Temp})=4.20 - 0.023(\text{Time})\). Substitute \(\text{Time}=5\) into the equation:
\(\ln(\text{Temp})=4.20-0.023\times5\)
\(\ln(\text{Temp})=4.20 - 0.115\)
\(\ln(\text{Temp})=4.085\)

Step2: Solve for Temp using the exponential function

Since \(y = \ln(x)\) implies \(x=e^{y}\), we have \(\text{Temp}=e^{4.085}\)
Using a calculator, \(e^{4.085}\approx59.44\)

Answer:

\(59.44^{\circ}\text{C}\) (First option)