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.
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.
