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Answer :
To solve the problem of determining which temperature will most accurately predict the time spent cooling using the model [tex]\( f(t) = 349.2(0.98)^t \)[/tex], we can follow these steps:
1. Understand the Problem:
The function [tex]\( f(t) = 349.2(0.98)^t \)[/tex] models the cooling of an oven over time. The goal is to find which given observed temperature is closest to the temperature predicted by the function for the corresponding times.
2. Identify the Data:
We have observed temperatures at specific times given in the table:
- At 5 minutes: 315°F
- At 10 minutes: 285°F
- At 15 minutes: 260°F
- At 20 minutes: 235°F
- At 25 minutes: 210°F
3. Calculate the Predicted Temperatures:
Using the function [tex]\( f(t) = 349.2(0.98)^t \)[/tex], we calculate the predicted temperatures for each respective time:
- For [tex]\( t = 5 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^5 \)[/tex]
- For [tex]\( t = 10 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{10} \)[/tex]
- For [tex]\( t = 15 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{15} \)[/tex]
- For [tex]\( t = 20 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{20} \)[/tex]
- For [tex]\( t = 25 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{25} \)[/tex]
4. Compare Observed and Predicted Temperatures:
After calculating the predicted temperatures for each time, compare them to the observed temperatures. Determine the difference in each pair to find which observed temperature is closest to its corresponding predicted temperature.
5. Determine the Most Accurate Prediction:
Look for the smallest difference between an observed temperature and its corresponding predicted temperature. The temperature with the smallest difference is considered the most accurately predicted by the model.
In this case, the temperature of 285°F turns out to be the closest to the predicted temperature using the model at 10 minutes. Hence, the model most accurately predicts the time spent cooling for this temperature.
1. Understand the Problem:
The function [tex]\( f(t) = 349.2(0.98)^t \)[/tex] models the cooling of an oven over time. The goal is to find which given observed temperature is closest to the temperature predicted by the function for the corresponding times.
2. Identify the Data:
We have observed temperatures at specific times given in the table:
- At 5 minutes: 315°F
- At 10 minutes: 285°F
- At 15 minutes: 260°F
- At 20 minutes: 235°F
- At 25 minutes: 210°F
3. Calculate the Predicted Temperatures:
Using the function [tex]\( f(t) = 349.2(0.98)^t \)[/tex], we calculate the predicted temperatures for each respective time:
- For [tex]\( t = 5 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^5 \)[/tex]
- For [tex]\( t = 10 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{10} \)[/tex]
- For [tex]\( t = 15 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{15} \)[/tex]
- For [tex]\( t = 20 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{20} \)[/tex]
- For [tex]\( t = 25 \)[/tex]: Calculate [tex]\( 349.2 \times (0.98)^{25} \)[/tex]
4. Compare Observed and Predicted Temperatures:
After calculating the predicted temperatures for each time, compare them to the observed temperatures. Determine the difference in each pair to find which observed temperature is closest to its corresponding predicted temperature.
5. Determine the Most Accurate Prediction:
Look for the smallest difference between an observed temperature and its corresponding predicted temperature. The temperature with the smallest difference is considered the most accurately predicted by the model.
In this case, the temperature of 285°F turns out to be the closest to the predicted temperature using the model at 10 minutes. Hence, the model most accurately predicts the time spent cooling for this temperature.
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