Answer :
We are given the model
[tex]$$
f(t) = 349.2(0.98)^t,
$$[/tex]
which represents the oven's temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes. To decide for which temperature the model most accurately predicts the cooling time, we consider solving for [tex]$t$[/tex] when [tex]$f(t)$[/tex] equals the stated temperatures.
Since solving for [tex]$t$[/tex] is easier when taking the logarithm, we rearrange the equation. For a general temperature [tex]$T$[/tex], we have:
[tex]$$
349.2(0.98)^t = T.
$$[/tex]
Dividing both sides by [tex]$349.2$[/tex] yields:
[tex]$$
(0.98)^t = \frac{T}{349.2}.
$$[/tex]
Taking the natural logarithm of both sides:
[tex]$$
\ln\left((0.98)^t\right) = \ln\left(\frac{T}{349.2}\right).
$$[/tex]
Using the logarithm power rule, this becomes:
[tex]$$
t \cdot \ln(0.98) = \ln\left(\frac{T}{349.2}\right).
$$[/tex]
Thus, solving for [tex]$t$[/tex] gives:
[tex]$$
t = \frac{\ln\left(\frac{T}{349.2}\right)}{\ln(0.98)}.
$$[/tex]
Now, let’s analyze the given temperature choices:
1. For [tex]$T = 0$[/tex]:
The expression [tex]$\ln(0)$[/tex] is undefined. Thus, [tex]$T = 0$[/tex] is not valid.
2. For [tex]$T = 100$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{100}{349.2}\right)}{\ln(0.98)} \approx 61.9 \text{ minutes}.
$$[/tex]
This cooling time is far outside the range of our measured cooling times (from 5 to 25 minutes).
3. For [tex]$T = 300$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{300}{349.2}\right)}{\ln(0.98)} \approx 7.52 \text{ minutes}.
$$[/tex]
This cooling time is within the range of times used in the data table, making it a reasonable value for applying the model.
4. For [tex]$T = 400$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{400}{349.2}\right)}{\ln(0.98)} \approx -6.72 \text{ minutes}.
$$[/tex]
A negative time is not physically meaningful, so [tex]$T = 400$[/tex] is not applicable.
Since the model gives [tex]$t \approx 7.52$[/tex] minutes for [tex]$T = 300$[/tex], which is consistent with the measured cooling times presented, the temperature at which the model most accurately predicts the cooling time is
[tex]$$
\boxed{300}.
$$[/tex]
[tex]$$
f(t) = 349.2(0.98)^t,
$$[/tex]
which represents the oven's temperature (in degrees Fahrenheit) after [tex]$t$[/tex] minutes. To decide for which temperature the model most accurately predicts the cooling time, we consider solving for [tex]$t$[/tex] when [tex]$f(t)$[/tex] equals the stated temperatures.
Since solving for [tex]$t$[/tex] is easier when taking the logarithm, we rearrange the equation. For a general temperature [tex]$T$[/tex], we have:
[tex]$$
349.2(0.98)^t = T.
$$[/tex]
Dividing both sides by [tex]$349.2$[/tex] yields:
[tex]$$
(0.98)^t = \frac{T}{349.2}.
$$[/tex]
Taking the natural logarithm of both sides:
[tex]$$
\ln\left((0.98)^t\right) = \ln\left(\frac{T}{349.2}\right).
$$[/tex]
Using the logarithm power rule, this becomes:
[tex]$$
t \cdot \ln(0.98) = \ln\left(\frac{T}{349.2}\right).
$$[/tex]
Thus, solving for [tex]$t$[/tex] gives:
[tex]$$
t = \frac{\ln\left(\frac{T}{349.2}\right)}{\ln(0.98)}.
$$[/tex]
Now, let’s analyze the given temperature choices:
1. For [tex]$T = 0$[/tex]:
The expression [tex]$\ln(0)$[/tex] is undefined. Thus, [tex]$T = 0$[/tex] is not valid.
2. For [tex]$T = 100$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{100}{349.2}\right)}{\ln(0.98)} \approx 61.9 \text{ minutes}.
$$[/tex]
This cooling time is far outside the range of our measured cooling times (from 5 to 25 minutes).
3. For [tex]$T = 300$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{300}{349.2}\right)}{\ln(0.98)} \approx 7.52 \text{ minutes}.
$$[/tex]
This cooling time is within the range of times used in the data table, making it a reasonable value for applying the model.
4. For [tex]$T = 400$[/tex]:
[tex]$$
t = \frac{\ln\left(\frac{400}{349.2}\right)}{\ln(0.98)} \approx -6.72 \text{ minutes}.
$$[/tex]
A negative time is not physically meaningful, so [tex]$T = 400$[/tex] is not applicable.
Since the model gives [tex]$t \approx 7.52$[/tex] minutes for [tex]$T = 300$[/tex], which is consistent with the measured cooling times presented, the temperature at which the model most accurately predicts the cooling time is
[tex]$$
\boxed{300}.
$$[/tex]
