Answer :
To determine which temperature the model most accurately predicts based on the data given, we compare the predicted temperatures from the model with the observed temperatures at different times. Here is the step-by-step process:
1. Understand the Model:
The function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] models how the oven's temperature changes over time, [tex]\( t \)[/tex], in minutes.
2. Observed Variables:
We have a table showing observed oven temperatures at different times:
- 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. Calculate Predicted Temperatures:
According to the function, we calculate the predicted temperatures:
- For 5 minutes: [tex]\( 349.2 \times (0.98)^5 \)[/tex]
- For 10 minutes: [tex]\( 349.2 \times (0.98)^{10} \)[/tex]
- For 15 minutes: [tex]\( 349.2 \times (0.98)^{15} \)[/tex]
- For 20 minutes: [tex]\( 349.2 \times (0.98)^{20} \)[/tex]
- For 25 minutes: [tex]\( 349.2 \times (0.98)^{25} \)[/tex]
4. Compare with Observed Temperatures:
We calculate the differences between the predicted and observed temperatures for each time value:
- Difference at 5 minutes: very close to 0.649
- Difference at 10 minutes: very close to 0.322
- Difference at 15 minutes: very close to 2.092
- Difference at 20 minutes: very close to 1.871
- Difference at 25 minutes: very close to 0.730
5. Identify the Most Accurate Prediction:
Check which time has the smallest difference between predicted and observed temperatures. The smallest difference is approximately 0.322 at 10 minutes, when the observed temperature is 285°F.
6. Conclusion:
The model most accurately predicts the time spent cooling when the observed temperature is 285°F.
1. Understand the Model:
The function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] models how the oven's temperature changes over time, [tex]\( t \)[/tex], in minutes.
2. Observed Variables:
We have a table showing observed oven temperatures at different times:
- 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. Calculate Predicted Temperatures:
According to the function, we calculate the predicted temperatures:
- For 5 minutes: [tex]\( 349.2 \times (0.98)^5 \)[/tex]
- For 10 minutes: [tex]\( 349.2 \times (0.98)^{10} \)[/tex]
- For 15 minutes: [tex]\( 349.2 \times (0.98)^{15} \)[/tex]
- For 20 minutes: [tex]\( 349.2 \times (0.98)^{20} \)[/tex]
- For 25 minutes: [tex]\( 349.2 \times (0.98)^{25} \)[/tex]
4. Compare with Observed Temperatures:
We calculate the differences between the predicted and observed temperatures for each time value:
- Difference at 5 minutes: very close to 0.649
- Difference at 10 minutes: very close to 0.322
- Difference at 15 minutes: very close to 2.092
- Difference at 20 minutes: very close to 1.871
- Difference at 25 minutes: very close to 0.730
5. Identify the Most Accurate Prediction:
Check which time has the smallest difference between predicted and observed temperatures. The smallest difference is approximately 0.322 at 10 minutes, when the observed temperature is 285°F.
6. Conclusion:
The model most accurately predicts the time spent cooling when the observed temperature is 285°F.
