Answer :
Sure! Let's go through the problem step-by-step to understand for which temperature the model will most accurately predict the time spent cooling.
We are given a function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] that models the relationship between [tex]\( t \)[/tex], the time an oven spends cooling, and the oven temperature [tex]\( f(t) \)[/tex]. We also have the following table of observed values:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time (minutes), } t & \text{Oven temperature (°F), } f(t) \\
\hline
5 & 315 \\
\hline
10 & 285 \\
\hline
15 & 260 \\
\hline
20 & 235 \\
\hline
25 & 210 \\
\hline
\end{array}
\][/tex]
We need to determine for which temperature the model most accurately predicts the cooling time.
### Step-by-Step Solution:
1. Calculate the predicted temperatures using the model:
We use the given function to calculate the temperatures at [tex]\( t = 5, 10, 15, 20, 25 \)[/tex] minutes.
[tex]\[
f(5) = 349.2 \times (0.98)^5 \approx 315.65
\][/tex]
[tex]\[
f(10) = 349.2 \times (0.98)^{10} \approx 285.32
\][/tex]
[tex]\[
f(15) = 349.2 \times (0.98)^{15} \approx 257.91
\][/tex]
[tex]\[
f(20) = 349.2 \times (0.98)^{20} \approx 233.13
\][/tex]
[tex]\[
f(25) = 349.2 \times (0.98)^{25} \approx 210.73
\][/tex]
2. Compare the predicted temperatures with the actual observed temperatures to find the differences:
[tex]\[
\text{Difference at } t=5: |315.65 - 315| \approx 0.65
\][/tex]
[tex]\[
\text{Difference at } t=10: |285.32 - 285| \approx 0.32
\][/tex]
[tex]\[
\text{Difference at } t=15: |257.91 - 260| \approx 2.09
\][/tex]
[tex]\[
\text{Difference at } t=20: |233.13 - 235| \approx 1.87
\][/tex]
[tex]\[
\text{Difference at } t=25: |210.73 - 210| \approx 0.73
\][/tex]
3. Identify the smallest difference:
The differences are:
- [tex]\( \approx 0.65 \)[/tex] at [tex]\( t=5 \)[/tex]
- [tex]\( \approx 0.32 \)[/tex] at [tex]\( t=10 \)[/tex]
- [tex]\( \approx 2.09 \)[/tex] at [tex]\( t=15 \)[/tex]
- [tex]\( \approx 1.87 \)[/tex] at [tex]\( t=20 \)[/tex]
- [tex]\( \approx 0.73 \)[/tex] at [tex]\( t=25 \)[/tex]
The smallest difference is [tex]\( \approx 0.32 \)[/tex].
4. Conclusion:
The model most accurately predicts the temperature at [tex]\( t = 10 \)[/tex] minutes, with an observed temperature of 285°F.
Therefore, the temperature 285°F is where the model most accurately predicts the time spent cooling.
We are given a function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] that models the relationship between [tex]\( t \)[/tex], the time an oven spends cooling, and the oven temperature [tex]\( f(t) \)[/tex]. We also have the following table of observed values:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time (minutes), } t & \text{Oven temperature (°F), } f(t) \\
\hline
5 & 315 \\
\hline
10 & 285 \\
\hline
15 & 260 \\
\hline
20 & 235 \\
\hline
25 & 210 \\
\hline
\end{array}
\][/tex]
We need to determine for which temperature the model most accurately predicts the cooling time.
### Step-by-Step Solution:
1. Calculate the predicted temperatures using the model:
We use the given function to calculate the temperatures at [tex]\( t = 5, 10, 15, 20, 25 \)[/tex] minutes.
[tex]\[
f(5) = 349.2 \times (0.98)^5 \approx 315.65
\][/tex]
[tex]\[
f(10) = 349.2 \times (0.98)^{10} \approx 285.32
\][/tex]
[tex]\[
f(15) = 349.2 \times (0.98)^{15} \approx 257.91
\][/tex]
[tex]\[
f(20) = 349.2 \times (0.98)^{20} \approx 233.13
\][/tex]
[tex]\[
f(25) = 349.2 \times (0.98)^{25} \approx 210.73
\][/tex]
2. Compare the predicted temperatures with the actual observed temperatures to find the differences:
[tex]\[
\text{Difference at } t=5: |315.65 - 315| \approx 0.65
\][/tex]
[tex]\[
\text{Difference at } t=10: |285.32 - 285| \approx 0.32
\][/tex]
[tex]\[
\text{Difference at } t=15: |257.91 - 260| \approx 2.09
\][/tex]
[tex]\[
\text{Difference at } t=20: |233.13 - 235| \approx 1.87
\][/tex]
[tex]\[
\text{Difference at } t=25: |210.73 - 210| \approx 0.73
\][/tex]
3. Identify the smallest difference:
The differences are:
- [tex]\( \approx 0.65 \)[/tex] at [tex]\( t=5 \)[/tex]
- [tex]\( \approx 0.32 \)[/tex] at [tex]\( t=10 \)[/tex]
- [tex]\( \approx 2.09 \)[/tex] at [tex]\( t=15 \)[/tex]
- [tex]\( \approx 1.87 \)[/tex] at [tex]\( t=20 \)[/tex]
- [tex]\( \approx 0.73 \)[/tex] at [tex]\( t=25 \)[/tex]
The smallest difference is [tex]\( \approx 0.32 \)[/tex].
4. Conclusion:
The model most accurately predicts the temperature at [tex]\( t = 10 \)[/tex] minutes, with an observed temperature of 285°F.
Therefore, the temperature 285°F is where the model most accurately predicts the time spent cooling.
