Answer :
To solve this problem, we want to determine which temperature the model most accurately predicts the time spent cooling. The model is expressed by the function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex], which gives the temperature of the oven, [tex]\( f(t) \)[/tex], at a given time [tex]\( t \)[/tex] in minutes.
Here’s the step-by-step solution:
1. Understand the problem: We need to compare the model's predictions with the actual temperatures given in the provided data at specific times.
2. Review the provided table: The table lists times in minutes (5, 10, 15, 20, and 25) and their corresponding actual recorded temperatures (315°F, 285°F, 260°F, 235°F, and 210°F).
3. Use the model to predict temperatures: Apply the function for each of the given time samples to predict the temperatures:
- At [tex]\( t = 5 \)[/tex]: [tex]\( f(5) = 349.2 \times (0.98)^5 \)[/tex]
- At [tex]\( t = 10 \)[/tex]: [tex]\( f(10) = 349.2 \times (0.98)^{10} \)[/tex]
- At [tex]\( t = 15 \)[/tex]: [tex]\( f(15) = 349.2 \times (0.98)^{15} \)[/tex]
- At [tex]\( t = 20 \)[/tex]: [tex]\( f(20) = 349.2 \times (0.98)^{20} \)[/tex]
- At [tex]\( t = 25 \)[/tex]: [tex]\( f(25) = 349.2 \times (0.98)^{25} \)[/tex]
The predictions from these calculations are:
- [tex]\( f(5) \approx 315.65 \)[/tex]
- [tex]\( f(10) \approx 285.32 \)[/tex]
- [tex]\( f(15) \approx 257.91 \)[/tex]
- [tex]\( f(20) \approx 233.13 \)[/tex]
- [tex]\( f(25) \approx 210.73 \)[/tex]
4. Calculate the differences: For each time point, find the absolute difference between the predicted temperature and the actual recorded temperature:
- At [tex]\( t = 5 \)[/tex]: Difference is [tex]\( |315 - 315.65| = 0.65 \)[/tex]
- At [tex]\( t = 10 \)[/tex]: Difference is [tex]\( |285 - 285.32| = 0.32 \)[/tex]
- At [tex]\( t = 15 \)[/tex]: Difference is [tex]\( |260 - 257.91| = 2.09 \)[/tex]
- At [tex]\( t = 20 \)[/tex]: Difference is [tex]\( |235 - 233.13| = 1.87 \)[/tex]
- At [tex]\( t = 25 \)[/tex]: Difference is [tex]\( |210 - 210.73| = 0.73 \)[/tex]
5. Determine the smallest difference: Identify which temperature has the smallest difference, indicating that the model prediction was closest to the actual temperature. The smallest difference here is 0.32, corresponding to the temperature 285°F at [tex]\( t = 10 \)[/tex] minutes.
6. Conclusion: The model most accurately predicts the temperature of 285°F as the time spent cooling. Therefore, with all considerations and analysis made, the most accurate prediction was made for the temperature of 285 degrees Fahrenheit.
I hope this helps clear up how to verify the model's accuracy at different temperatures! If you have any more questions or need further clarification, feel free to ask.
Here’s the step-by-step solution:
1. Understand the problem: We need to compare the model's predictions with the actual temperatures given in the provided data at specific times.
2. Review the provided table: The table lists times in minutes (5, 10, 15, 20, and 25) and their corresponding actual recorded temperatures (315°F, 285°F, 260°F, 235°F, and 210°F).
3. Use the model to predict temperatures: Apply the function for each of the given time samples to predict the temperatures:
- At [tex]\( t = 5 \)[/tex]: [tex]\( f(5) = 349.2 \times (0.98)^5 \)[/tex]
- At [tex]\( t = 10 \)[/tex]: [tex]\( f(10) = 349.2 \times (0.98)^{10} \)[/tex]
- At [tex]\( t = 15 \)[/tex]: [tex]\( f(15) = 349.2 \times (0.98)^{15} \)[/tex]
- At [tex]\( t = 20 \)[/tex]: [tex]\( f(20) = 349.2 \times (0.98)^{20} \)[/tex]
- At [tex]\( t = 25 \)[/tex]: [tex]\( f(25) = 349.2 \times (0.98)^{25} \)[/tex]
The predictions from these calculations are:
- [tex]\( f(5) \approx 315.65 \)[/tex]
- [tex]\( f(10) \approx 285.32 \)[/tex]
- [tex]\( f(15) \approx 257.91 \)[/tex]
- [tex]\( f(20) \approx 233.13 \)[/tex]
- [tex]\( f(25) \approx 210.73 \)[/tex]
4. Calculate the differences: For each time point, find the absolute difference between the predicted temperature and the actual recorded temperature:
- At [tex]\( t = 5 \)[/tex]: Difference is [tex]\( |315 - 315.65| = 0.65 \)[/tex]
- At [tex]\( t = 10 \)[/tex]: Difference is [tex]\( |285 - 285.32| = 0.32 \)[/tex]
- At [tex]\( t = 15 \)[/tex]: Difference is [tex]\( |260 - 257.91| = 2.09 \)[/tex]
- At [tex]\( t = 20 \)[/tex]: Difference is [tex]\( |235 - 233.13| = 1.87 \)[/tex]
- At [tex]\( t = 25 \)[/tex]: Difference is [tex]\( |210 - 210.73| = 0.73 \)[/tex]
5. Determine the smallest difference: Identify which temperature has the smallest difference, indicating that the model prediction was closest to the actual temperature. The smallest difference here is 0.32, corresponding to the temperature 285°F at [tex]\( t = 10 \)[/tex] minutes.
6. Conclusion: The model most accurately predicts the temperature of 285°F as the time spent cooling. Therefore, with all considerations and analysis made, the most accurate prediction was made for the temperature of 285 degrees Fahrenheit.
I hope this helps clear up how to verify the model's accuracy at different temperatures! If you have any more questions or need further clarification, feel free to ask.
