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
$$
\boxed{300^\circ\text{F}}.
$$
$$
f(t) = 349.2 \cdot (0.98)^t,
$$
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
$$
\boxed{300^\circ\text{F}}.
$$
