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Answer :
We are given that the oven cooling behavior is modeled by
[tex]$$
f(t)=349.2(0.98)^t,
$$[/tex]
where [tex]$f(t)$[/tex] is the oven temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes. In addition, experimental data recorded during cooling provides the following approximate values:
[tex]\[
\begin{array}{|c|c|}
\hline
t \text{ (minutes)} & f(t) \text{ (°F)} \\
\hline
5 & 315 \\
10 & 285 \\
15 & 260 \\
20 & 235 \\
25 & 210 \\
\hline
\end{array}
\][/tex]
Because the model was developed using data for temperatures ranging roughly from about 210°F to 315°F, it is reasonable to conclude that the model will be most accurate within this range of temperature values.
Among the answer choices provided ([tex]$0$[/tex], [tex]$100$[/tex], [tex]$300$[/tex], and [tex]$400$[/tex]), the temperature [tex]$300^\circ\text{F}$[/tex] falls within the range of experimentally measured values.
To further justify this, we can determine the time [tex]$t$[/tex] when the model predicts a temperature of [tex]$300^\circ\text{F}$[/tex]. Setting
[tex]$$
349.2(0.98)^t = 300,
$$[/tex]
we solve for [tex]$t$[/tex] by first dividing both sides by [tex]$349.2$[/tex]:
[tex]$$
(0.98)^t = \frac{300}{349.2}.
$$[/tex]
Taking the natural logarithm of both sides gives
[tex]$$
\ln\left((0.98)^t\right) = \ln\left(\frac{300}{349.2}\right).
$$[/tex]
Using the logarithm power rule, this becomes
[tex]$$
t \ln(0.98) = \ln\left(\frac{300}{349.2}\right).
$$[/tex]
Solving for [tex]$t$[/tex] yields
[tex]$$
t = \frac{\ln(300/349.2)}{\ln(0.98)}.
$$[/tex]
Evaluating this expression shows that [tex]$t \approx 7.52$[/tex] minutes. This computed time is within the range of times corresponding to the experimental data (between [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes), giving further confidence in the model's accuracy at or near [tex]$300^\circ\text{F}$[/tex].
Thus, the model most accurately predicts the cooling time for an oven temperature of
[tex]$$
\boxed{300^\circ\text{F}}.
$$[/tex]
[tex]$$
f(t)=349.2(0.98)^t,
$$[/tex]
where [tex]$f(t)$[/tex] is the oven temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes. In addition, experimental data recorded during cooling provides the following approximate values:
[tex]\[
\begin{array}{|c|c|}
\hline
t \text{ (minutes)} & f(t) \text{ (°F)} \\
\hline
5 & 315 \\
10 & 285 \\
15 & 260 \\
20 & 235 \\
25 & 210 \\
\hline
\end{array}
\][/tex]
Because the model was developed using data for temperatures ranging roughly from about 210°F to 315°F, it is reasonable to conclude that the model will be most accurate within this range of temperature values.
Among the answer choices provided ([tex]$0$[/tex], [tex]$100$[/tex], [tex]$300$[/tex], and [tex]$400$[/tex]), the temperature [tex]$300^\circ\text{F}$[/tex] falls within the range of experimentally measured values.
To further justify this, we can determine the time [tex]$t$[/tex] when the model predicts a temperature of [tex]$300^\circ\text{F}$[/tex]. Setting
[tex]$$
349.2(0.98)^t = 300,
$$[/tex]
we solve for [tex]$t$[/tex] by first dividing both sides by [tex]$349.2$[/tex]:
[tex]$$
(0.98)^t = \frac{300}{349.2}.
$$[/tex]
Taking the natural logarithm of both sides gives
[tex]$$
\ln\left((0.98)^t\right) = \ln\left(\frac{300}{349.2}\right).
$$[/tex]
Using the logarithm power rule, this becomes
[tex]$$
t \ln(0.98) = \ln\left(\frac{300}{349.2}\right).
$$[/tex]
Solving for [tex]$t$[/tex] yields
[tex]$$
t = \frac{\ln(300/349.2)}{\ln(0.98)}.
$$[/tex]
Evaluating this expression shows that [tex]$t \approx 7.52$[/tex] minutes. This computed time is within the range of times corresponding to the experimental data (between [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes), giving further confidence in the model's accuracy at or near [tex]$300^\circ\text{F}$[/tex].
Thus, the model most accurately predicts the cooling time for an oven temperature of
[tex]$$
\boxed{300^\circ\text{F}}.
$$[/tex]
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