High School

The function [tex]f(t) = 349.2(0.98)^t[/tex] models the relationship between the time [tex]t[/tex] an oven spends cooling and the temperature of the oven.

Oven Cooling Time:

\[
\begin{tabular}{|l|l|}
\hline
Time (minutes) [tex]t[/tex] & Oven Temperature (degrees Fahrenheit) [tex]f(t)[/tex] \\
\hline
5 & 315 \\
\hline
10 & 285 \\
\hline
15 & 260 \\
\hline
20 & 235 \\
\hline
25 & 210 \\
\hline
\end{tabular}
\]

For which temperature will the model most accurately predict the time spent cooling?

A. 0
B. 100
C. 300
D. 400

Answer :

We start with the model

$$
f(t) = 349.2 \cdot (0.98)^t,
$$

which gives the temperature of the oven after $t$ minutes of cooling. The table of measured values shows that the oven temperature ranges from about 315°F to 210°F as the cooling time increases from 5 to 25 minutes. Out of the given options (0, 100, 300, 400°F), the temperature of 300°F falls within the range of the observed data, so we expect the model to be most reliable there.

To confirm this, we can determine the cooling time $t$ when the temperature is 300°F by solving

$$
349.2 \cdot (0.98)^t = 300.
$$

Divide both sides by 349.2:

$$
(0.98)^t = \frac{300}{349.2}.
$$

Taking the natural logarithm on both sides gives

$$
\ln\bigl((0.98)^t\bigr) = \ln\left(\frac{300}{349.2}\right).
$$

By the logarithmic property $\ln(a^b) = b \ln(a)$, this becomes

$$
t \cdot \ln(0.98) = \ln\left(\frac{300}{349.2}\right).
$$

Solving for $t$, we obtain

$$
t = \frac{\ln\left(\frac{300}{349.2}\right)}{\ln(0.98)}.
$$

Evaluating this expression gives a cooling time of approximately $7.52$ minutes when the temperature is 300°F. Since this temperature lies within the range of the measured data, we conclude that the model most accurately predicts the cooling time for an oven temperature of

$$
\boxed{300^\circ\text{F}}.
$$