Answer :
To determine for which temperature the model most accurately predicts the time spent cooling, we will compare the predictions from the function with the actual observed temperatures.
The function given is [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex]. We will calculate the predicted temperatures for the times provided (5, 10, 15, 20, and 25 minutes) and then compare these to the observed temperatures to see which time has the smallest difference between predicted and actual temperatures.
Step-by-step Solution:
1. Predicted Temperatures:
- For [tex]\( t = 5 \)[/tex] minutes: The predicted temperature is approximately 315.65°F.
- For [tex]\( t = 10 \)[/tex] minutes: The predicted temperature is approximately 285.32°F.
- For [tex]\( t = 15 \)[/tex] minutes: The predicted temperature is approximately 257.91°F.
- For [tex]\( t = 20 \)[/tex] minutes: The predicted temperature is approximately 233.13°F.
- For [tex]\( t = 25 \)[/tex] minutes: The predicted temperature is approximately 210.73°F.
2. Observed Temperatures from the Table:
- At [tex]\( t = 5 \)[/tex]: 315°F
- At [tex]\( t = 10 \)[/tex]: 285°F
- At [tex]\( t = 15 \)[/tex]: 260°F
- At [tex]\( t = 20 \)[/tex]: 235°F
- At [tex]\( t = 25 \)[/tex]: 210°F
3. Calculate the Differences:
- Difference at [tex]\( t = 5 \)[/tex]: [tex]\(|315.65 - 315| = 0.65\)[/tex] degrees
- Difference at [tex]\( t = 10 \)[/tex]: [tex]\(|285.32 - 285| = 0.32\)[/tex] degrees
- Difference at [tex]\( t = 15 \)[/tex]: [tex]\(|257.91 - 260| = 2.09\)[/tex] degrees
- Difference at [tex]\( t = 20 \)[/tex]: [tex]\(|233.13 - 235| = 1.87\)[/tex] degrees
- Difference at [tex]\( t = 25 \)[/tex]: [tex]\(|210.73 - 210| = 0.73\)[/tex] degrees
4. Identify the Smallest Difference:
Comparing the differences, the smallest one is 0.32 degrees, which occurs at [tex]\( t = 10 \)[/tex] minutes where the observed temperature was 285°F.
5. Conclusion:
The temperature for which the model most accurately predicts the time spent cooling is approximately 300°F. This corresponds to the predicted temperature closest to the observed temperature at a specific time, in this case at around 10 minutes.
The function given is [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex]. We will calculate the predicted temperatures for the times provided (5, 10, 15, 20, and 25 minutes) and then compare these to the observed temperatures to see which time has the smallest difference between predicted and actual temperatures.
Step-by-step Solution:
1. Predicted Temperatures:
- For [tex]\( t = 5 \)[/tex] minutes: The predicted temperature is approximately 315.65°F.
- For [tex]\( t = 10 \)[/tex] minutes: The predicted temperature is approximately 285.32°F.
- For [tex]\( t = 15 \)[/tex] minutes: The predicted temperature is approximately 257.91°F.
- For [tex]\( t = 20 \)[/tex] minutes: The predicted temperature is approximately 233.13°F.
- For [tex]\( t = 25 \)[/tex] minutes: The predicted temperature is approximately 210.73°F.
2. Observed Temperatures from the Table:
- At [tex]\( t = 5 \)[/tex]: 315°F
- At [tex]\( t = 10 \)[/tex]: 285°F
- At [tex]\( t = 15 \)[/tex]: 260°F
- At [tex]\( t = 20 \)[/tex]: 235°F
- At [tex]\( t = 25 \)[/tex]: 210°F
3. Calculate the Differences:
- Difference at [tex]\( t = 5 \)[/tex]: [tex]\(|315.65 - 315| = 0.65\)[/tex] degrees
- Difference at [tex]\( t = 10 \)[/tex]: [tex]\(|285.32 - 285| = 0.32\)[/tex] degrees
- Difference at [tex]\( t = 15 \)[/tex]: [tex]\(|257.91 - 260| = 2.09\)[/tex] degrees
- Difference at [tex]\( t = 20 \)[/tex]: [tex]\(|233.13 - 235| = 1.87\)[/tex] degrees
- Difference at [tex]\( t = 25 \)[/tex]: [tex]\(|210.73 - 210| = 0.73\)[/tex] degrees
4. Identify the Smallest Difference:
Comparing the differences, the smallest one is 0.32 degrees, which occurs at [tex]\( t = 10 \)[/tex] minutes where the observed temperature was 285°F.
5. Conclusion:
The temperature for which the model most accurately predicts the time spent cooling is approximately 300°F. This corresponds to the predicted temperature closest to the observed temperature at a specific time, in this case at around 10 minutes.
