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{array}{|c|c|}

\hline

\text{Time (minutes) } t & \text{Oven temperature (degrees Fahrenheit) } f(t) \\

\hline

5 & 315 \\

\hline

10 & 285 \\

\hline

15 & 260 \\

\hline

20 & 235 \\

\hline

25 & 210 \\

\hline

\end{array}

\][/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 of determining for which temperature the model most accurately predicts the time spent cooling, we should follow these steps:

1. Understand the Function: The function given is [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex], which models the oven's cooling based on time [tex]\( t \)[/tex] in minutes.

2. Calculate Predicted Temperatures:
We need to calculate the predicted temperatures using the model for the given times:
- For [tex]\( t = 5 \)[/tex]: [tex]\( f(5) = 349.2 \times (0.98)^5 \approx 315.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex]: [tex]\( f(10) = 349.2 \times (0.98)^{10} \approx 285.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex]: [tex]\( f(15) = 349.2 \times (0.98)^{15} \approx 257.91 \)[/tex]
- For [tex]\( t = 20 \)[/tex]: [tex]\( f(20) = 349.2 \times (0.98)^{20} \approx 233.13 \)[/tex]
- For [tex]\( t = 25 \)[/tex]: [tex]\( f(25) = 349.2 \times (0.98)^{25} \approx 210.73 \)[/tex]

3. Compare with Given Data:
We are given specific temperatures for these times. We need to compare the predicted temperatures with the actual temperatures provided:
- For [tex]\( t = 5 \)[/tex], given [tex]\( f(t) = 315 \)[/tex]: Difference [tex]\( |315.65 - 315| = 0.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex], given [tex]\( f(t) = 285 \)[/tex]: Difference [tex]\( |285.32 - 285| = 0.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex], given [tex]\( f(t) = 260 \)[/tex]: Difference [tex]\( |257.91 - 260| = 2.09 \)[/tex]
- For [tex]\( t = 20 \)[/tex], given [tex]\( f(t) = 235 \)[/tex]: Difference [tex]\( |233.13 - 235| = 1.87 \)[/tex]
- For [tex]\( t = 25 \)[/tex], given [tex]\( f(t) = 210 \)[/tex]: Difference [tex]\( |210.73 - 210| = 0.73 \)[/tex]

4. Find Minimum Difference:
The smallest difference indicates the best match between the model prediction and actual data. The smallest difference is:
- Time [tex]\( t = 10 \)[/tex] minutes, difference [tex]\( 0.32 \)[/tex]

5. Evaluate Options:
Among the temperature options given (0, 100, 300, 400), determine which temperature the model predicts most accurately. From the numerical results we have:
- The model prediction most closely matches when the temperature equals 300 degrees.

Therefore, the temperature for which the model most accurately predicts the time spent cooling is 300 degrees Fahrenheit.