High School

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

**Oven Cooling Time**

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

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

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

Answer :

To solve the problem, we need to determine which of the given temperatures (0, 100, 300, 400) the model predicts most accurately, based on the provided observed values of the oven's temperature over time. Here's a step-by-step explanation:

1. Understanding the Model: The function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] is given, where [tex]\( t \)[/tex] is the time in minutes and [tex]\( f(t) \)[/tex] is the predicted oven temperature in degrees Fahrenheit.

2. Observed Temperatures and Times:
- At [tex]\( t = 5 \)[/tex] minutes, the observed temperature is 315°F.
- At [tex]\( t = 10 \)[/tex] minutes, the observed temperature is 285°F.
- At [tex]\( t = 15 \)[/tex] minutes, the observed temperature is 260°F.
- At [tex]\( t = 20 \)[/tex] minutes, the observed temperature is 235°F.
- At [tex]\( t = 25 \)[/tex] minutes, the observed temperature is 210°F.

3. Calculate Predicted Temperatures: Using the model function, calculate the predicted temperatures for each time:

- For [tex]\( t = 5 \)[/tex]: [tex]\( 349.2 \times (0.98)^5 = 315.649 \)[/tex]
- For [tex]\( t = 10 \)[/tex]: [tex]\( 349.2 \times (0.98)^{10} = 284.678 \)[/tex]
- For [tex]\( t = 15 \)[/tex]: [tex]\( 349.2 \times (0.98)^{15} = 262.092 \)[/tex]
- For [tex]\( t = 20 \)[/tex]: [tex]\( 349.2 \times (0.98)^{20} = 236.871 \)[/tex]
- For [tex]\( t = 25 \)[/tex]: [tex]\( 349.2 \times (0.98)^{25} = 210.730 \)[/tex]

4. Calculate the Errors: Find the difference between the observed and predicted temperatures for each time to calculate the error:

- Error for [tex]\( t = 5 \)[/tex]: [tex]\( |315 - 315.649| = 0.649 \)[/tex]
- Error for [tex]\( t = 10 \)[/tex]: [tex]\( |285 - 284.678| = 0.322 \)[/tex]
- Error for [tex]\( t = 15 \)[/tex]: [tex]\( |260 - 262.092| = 2.092 \)[/tex]
- Error for [tex]\( t = 20 \)[/tex]: [tex]\( |235 - 236.871| = 1.871 \)[/tex]
- Error for [tex]\( t = 25 \)[/tex]: [tex]\( |210 - 210.730| = 0.730 \)[/tex]

5. Determine the Most Accurate Prediction: Compare the errors to see which prediction has the smallest error. The smallest error is 0.322 at [tex]\( t = 10 \)[/tex] minutes, where the observed temperature is 285°F.

6. Select the Closest Given Temperature: Among the given choices (0, 100, 300, 400), the closest to 285°F is 300°F. Therefore, the model most accurately predicts the temperature for when it approximates 300°F.

Thus, the temperature for which the model most accurately predicts the time spent cooling is closest to 300°F.