Answer :
We are given the model
[tex]$$
f(t) = 349.2 \cdot (0.98)^t,
$$[/tex]
which describes the oven’s temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes of cooling. The cooling data provided includes:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time (minutes)} & \text{Oven Temperature (°F)} \\
\hline
5 & 315 \\
10 & 285 \\
15 & 260 \\
20 & 235 \\
25 & 210 \\
\hline
\end{array}
\][/tex]
### Step 1. Confirm the Model in the Given Range
Let’s verify that the model closely agrees with the observations for times near [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes.
1. For [tex]$t=5$[/tex] minutes, the predicted temperature is
[tex]$$
f(5) = 349.2 \cdot (0.98)^5 \approx 315.65.
$$[/tex]
This value is very close to the observed temperature of 315°F.
2. For [tex]$t=10$[/tex] minutes, the predicted temperature is
[tex]$$
f(10) = 349.2 \cdot (0.98)^{10} \approx 285.32.
$$[/tex]
This is nearly the same as the observed value of 285°F.
Since the outputs for [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes are accurate, the model is trusted to be most reliable in the temperature range approximately between 210°F and 315°F.
### Step 2. Identify the Most Reliable Prediction Temperature
The given answer choices for temperature are 0, 100, 300, and 400°F.
- Temperatures 0°F and 100°F are far outside the interval where the model has been validated.
- 400°F is also outside the available cooling data.
- The temperature 300°F lies very close to where our validation data is located.
### Step 3. Solve for Cooling Time When [tex]$f(t) = 300$[/tex]
To further confirm, we can solve for the time [tex]$t$[/tex] when the oven temperature is 300°F:
Start with the equation
[tex]$$
300 = 349.2 \cdot (0.98)^t.
$$[/tex]
Divide both sides by 349.2:
[tex]$$
(0.98)^t = \frac{300}{349.2} \approx 0.858.
$$[/tex]
Taking the natural logarithm of both sides:
[tex]$$
\ln\big((0.98)^t\big) = \ln(0.858).
$$[/tex]
Using the logarithm power rule, we have:
[tex]$$
t \cdot \ln(0.98) = \ln(0.858).
$$[/tex]
So, solving for [tex]$t$[/tex] gives:
[tex]$$
t = \frac{\ln(0.858)}{\ln(0.98)} \approx 7.52 \text{ minutes}.
$$[/tex]
This result falls between 5 and 10 minutes, consistent with the interval where the model was shown to be accurate.
### Conclusion
Since the predicted temperature of 300°F corresponds to a time within the range verified by the cooling data (approximately 7.52 minutes) and lies within the interval of the measured values, the model will most accurately predict the cooling time at [tex]$\boxed{300}$[/tex]°F.
[tex]$$
f(t) = 349.2 \cdot (0.98)^t,
$$[/tex]
which describes the oven’s temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes of cooling. The cooling data provided includes:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time (minutes)} & \text{Oven Temperature (°F)} \\
\hline
5 & 315 \\
10 & 285 \\
15 & 260 \\
20 & 235 \\
25 & 210 \\
\hline
\end{array}
\][/tex]
### Step 1. Confirm the Model in the Given Range
Let’s verify that the model closely agrees with the observations for times near [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes.
1. For [tex]$t=5$[/tex] minutes, the predicted temperature is
[tex]$$
f(5) = 349.2 \cdot (0.98)^5 \approx 315.65.
$$[/tex]
This value is very close to the observed temperature of 315°F.
2. For [tex]$t=10$[/tex] minutes, the predicted temperature is
[tex]$$
f(10) = 349.2 \cdot (0.98)^{10} \approx 285.32.
$$[/tex]
This is nearly the same as the observed value of 285°F.
Since the outputs for [tex]$t=5$[/tex] and [tex]$t=10$[/tex] minutes are accurate, the model is trusted to be most reliable in the temperature range approximately between 210°F and 315°F.
### Step 2. Identify the Most Reliable Prediction Temperature
The given answer choices for temperature are 0, 100, 300, and 400°F.
- Temperatures 0°F and 100°F are far outside the interval where the model has been validated.
- 400°F is also outside the available cooling data.
- The temperature 300°F lies very close to where our validation data is located.
### Step 3. Solve for Cooling Time When [tex]$f(t) = 300$[/tex]
To further confirm, we can solve for the time [tex]$t$[/tex] when the oven temperature is 300°F:
Start with the equation
[tex]$$
300 = 349.2 \cdot (0.98)^t.
$$[/tex]
Divide both sides by 349.2:
[tex]$$
(0.98)^t = \frac{300}{349.2} \approx 0.858.
$$[/tex]
Taking the natural logarithm of both sides:
[tex]$$
\ln\big((0.98)^t\big) = \ln(0.858).
$$[/tex]
Using the logarithm power rule, we have:
[tex]$$
t \cdot \ln(0.98) = \ln(0.858).
$$[/tex]
So, solving for [tex]$t$[/tex] gives:
[tex]$$
t = \frac{\ln(0.858)}{\ln(0.98)} \approx 7.52 \text{ minutes}.
$$[/tex]
This result falls between 5 and 10 minutes, consistent with the interval where the model was shown to be accurate.
### Conclusion
Since the predicted temperature of 300°F corresponds to a time within the range verified by the cooling data (approximately 7.52 minutes) and lies within the interval of the measured values, the model will most accurately predict the cooling time at [tex]$\boxed{300}$[/tex]°F.
