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Answer :
To determine which temperature the given model most accurately predicts the cooling time, we can analyze the differences between the actual oven temperatures provided and those predicted by the model for specific times. Here's how we can do it:
1. Understand the Function:
The function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] models the temperature of the oven after it has been cooling for [tex]\( t \)[/tex] minutes.
2. Provided Data:
We have a table with given times and corresponding oven temperatures:
- At 5 minutes, the temperature is 315°F.
- At 10 minutes, the temperature is 285°F.
- At 15 minutes, the temperature is 260°F.
- At 20 minutes, the temperature is 235°F.
- At 25 minutes, the temperature is 210°F.
3. Predict Temperatures using the Model:
For each time given in the table, use the model to calculate the predicted temperature:
- At 5 minutes, [tex]\( f(5) = 315.65 \)[/tex] (approximately).
- At 10 minutes, [tex]\( f(10) = 285.32 \)[/tex] (approximately).
- At 15 minutes, [tex]\( f(15) = 257.91 \)[/tex] (approximately).
- At 20 minutes, [tex]\( f(20) = 233.13 \)[/tex] (approximately).
- At 25 minutes, [tex]\( f(25) = 210.73 \)[/tex] (approximately).
4. Calculate the Differences:
Determine the absolute differences between the actual and predicted temperatures:
- Difference at 5 minutes: [tex]\( |315 - 315.65| \approx 0.65 \)[/tex]
- Difference at 10 minutes: [tex]\( |285 - 285.32| \approx 0.32 \)[/tex]
- Difference at 15 minutes: [tex]\( |260 - 257.91| \approx 2.09 \)[/tex]
- Difference at 20 minutes: [tex]\( |235 - 233.13| \approx 1.87 \)[/tex]
- Difference at 25 minutes: [tex]\( |210 - 210.73| \approx 0.73 \)[/tex]
5. Find the Most Accurate Prediction:
The smallest difference shows where the model's prediction is most accurate. In this case, the smallest difference is approximately 0.32 at 10 minutes, corresponding to a temperature of 285°F.
Therefore, the model most accurately predicts the oven's cooling time when the temperature is close to 285°F.
1. Understand the Function:
The function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] models the temperature of the oven after it has been cooling for [tex]\( t \)[/tex] minutes.
2. Provided Data:
We have a table with given times and corresponding oven temperatures:
- At 5 minutes, the temperature is 315°F.
- At 10 minutes, the temperature is 285°F.
- At 15 minutes, the temperature is 260°F.
- At 20 minutes, the temperature is 235°F.
- At 25 minutes, the temperature is 210°F.
3. Predict Temperatures using the Model:
For each time given in the table, use the model to calculate the predicted temperature:
- At 5 minutes, [tex]\( f(5) = 315.65 \)[/tex] (approximately).
- At 10 minutes, [tex]\( f(10) = 285.32 \)[/tex] (approximately).
- At 15 minutes, [tex]\( f(15) = 257.91 \)[/tex] (approximately).
- At 20 minutes, [tex]\( f(20) = 233.13 \)[/tex] (approximately).
- At 25 minutes, [tex]\( f(25) = 210.73 \)[/tex] (approximately).
4. Calculate the Differences:
Determine the absolute differences between the actual and predicted temperatures:
- Difference at 5 minutes: [tex]\( |315 - 315.65| \approx 0.65 \)[/tex]
- Difference at 10 minutes: [tex]\( |285 - 285.32| \approx 0.32 \)[/tex]
- Difference at 15 minutes: [tex]\( |260 - 257.91| \approx 2.09 \)[/tex]
- Difference at 20 minutes: [tex]\( |235 - 233.13| \approx 1.87 \)[/tex]
- Difference at 25 minutes: [tex]\( |210 - 210.73| \approx 0.73 \)[/tex]
5. Find the Most Accurate Prediction:
The smallest difference shows where the model's prediction is most accurate. In this case, the smallest difference is approximately 0.32 at 10 minutes, corresponding to a temperature of 285°F.
Therefore, the model most accurately predicts the oven's cooling time when the temperature is close to 285°F.
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