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**

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

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

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

Answer :

The temperature for which the model most accurately predicts the time spent cooling is 300 degrees Fahrenheit. Comparing the given data and modeled values, 300 degrees Fahrenheit closely follows the function's prediction. Therefore, option C is correct.

Let's solve this step-by-step:

  1. We have the function f(t) = 349.2(0.98)^t.
  2. If we look at the given data, we can see the temperatures and the times they occur. By comparing these, we can identify which temperature is closest to our given choices.
  3. For each time t, calculate f(t) and compare with given options:

At t = 5, f(5) = 349.2 * (0.98)^5 ≈ 319 (given as 315)

At t = 10, f(10) = 349.2 * (0.98)^10 ≈ 291 (given as 285)

At t = 15, f(15) = 349.2 * (0.98)^15 ≈ 265 (given as 260)

At t = 20, f(20) = 349.2 * (0.98)^20 ≈ 242 (given as 235)

At t = 25, f(25) = 349.2 * (0.98)^25 ≈ 221 (given as 210)

Thus, the closest modeled temperature from our given options (300, 0, 100, 400) is 300 degrees Fahrenheit.

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