Answer :
- Calculate the time $t$ for each temperature (0, 100, 300, 400) using the formula $t = \frac{\ln(f(t)) - \ln(349.2)}{\ln(0.98)}$.
- Calculate the average absolute error between the predicted time and the actual times from the table for each temperature.
- Compare the average absolute errors and identify the temperature with the smallest error.
- The temperature with the smallest average absolute error is 300. Therefore, the final answer is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given a function $f(t)=349.2(0.98)^t$ that models the temperature of an oven as it cools over time $t$. We are also given a table of data points showing the oven temperature at different times. The goal is to determine which of the given temperatures (0, 100, 300, 400) the model most accurately predicts the time spent cooling.
2. Finding the Time for Each Temperature
First, we need to find the time $t$ for each of the given temperatures using the model. We have $f(t) = 349.2(0.98)^t$. We want to solve for $t$ when $f(t)$ is equal to 0, 100, 300, and 400.
3. Calculating the Time
To solve for $t$, we can rearrange the equation: $t = \frac{\ln(f(t)) - \ln(349.2)}{\ln(0.98)}$. Let's calculate the time for each temperature:
For $f(t) = 0$: $t = \frac{\ln(0) - \ln(349.2)}{\ln(0.98)}$. Since $\ln(0)$ is undefined, we can consider the limit as the temperature approaches 0, which would result in $t$ approaching infinity.
For $f(t) = 100$: $t = \frac{\ln(100) - \ln(349.2)}{\ln(0.98)} \approx 61.90$ minutes.
For $f(t) = 300$: $t = \frac{\ln(300) - \ln(349.2)}{\ln(0.98)} \approx 7.52$ minutes.
For $f(t) = 400$: $t = \frac{\ln(400) - \ln(349.2)}{\ln(0.98)} \approx -6.72$ minutes. Since time cannot be negative, this value doesn't make sense in the context of the problem.
4. Calculating Average Absolute Errors
Now, we need to compare these predicted times with the data in the table to see which temperature gives the most accurate prediction. The data points are (5, 315), (10, 285), (15, 260), (20, 235), (25, 210).
For $f(t) = 0$, the predicted time is infinite, so the average error will also be infinite.
For $f(t) = 100$, the predicted time is approximately 61.90 minutes. The absolute differences between this time and the times in the table are |61.90 - 5| = 56.90, |61.90 - 10| = 51.90, |61.90 - 15| = 46.90, |61.90 - 20| = 41.90, |61.90 - 25| = 36.90. The average absolute error is (56.90 + 51.90 + 46.90 + 41.90 + 36.90) / 5 = 46.90.
For $f(t) = 300$, the predicted time is approximately 7.52 minutes. The absolute differences between this time and the times in the table are |7.52 - 5| = 2.52, |7.52 - 10| = 2.48, |7.52 - 15| = 7.48, |7.52 - 20| = 12.48, |7.52 - 25| = 17.48. The average absolute error is (2.52 + 2.48 + 7.48 + 12.48 + 17.48) / 5 = 8.49.
For $f(t) = 400$, the predicted time is approximately -6.72 minutes. Since time cannot be negative, we take the absolute value of the differences: |-6.72 - 5| = 11.72, |-6.72 - 10| = 16.72, |-6.72 - 15| = 21.72, |-6.72 - 20| = 26.72, |-6.72 - 25| = 31.72. The average absolute error is (11.72 + 16.72 + 21.72 + 26.72 + 31.72) / 5 = 21.72.
5. Comparing Errors
Comparing the average absolute errors, we have:
Temperature 0: Average error = infinite
Temperature 100: Average error = 46.90
Temperature 300: Average error = 8.49
Temperature 400: Average error = 21.72
The smallest average absolute error is 8.49, which corresponds to a temperature of 300.
6. Final Answer
Therefore, the model most accurately predicts the time spent cooling for a temperature of 300 degrees Fahrenheit.
### Examples
Understanding how mathematical models predict real-world phenomena, like oven cooling, is crucial in many fields. For example, in manufacturing, predicting the cooling rate of materials is essential for quality control. Similarly, in cooking, knowing how quickly an oven cools helps in planning baking schedules and ensuring consistent results. This type of modeling allows for optimization and efficiency in various processes.
- Calculate the average absolute error between the predicted time and the actual times from the table for each temperature.
- Compare the average absolute errors and identify the temperature with the smallest error.
- The temperature with the smallest average absolute error is 300. Therefore, the final answer is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given a function $f(t)=349.2(0.98)^t$ that models the temperature of an oven as it cools over time $t$. We are also given a table of data points showing the oven temperature at different times. The goal is to determine which of the given temperatures (0, 100, 300, 400) the model most accurately predicts the time spent cooling.
2. Finding the Time for Each Temperature
First, we need to find the time $t$ for each of the given temperatures using the model. We have $f(t) = 349.2(0.98)^t$. We want to solve for $t$ when $f(t)$ is equal to 0, 100, 300, and 400.
3. Calculating the Time
To solve for $t$, we can rearrange the equation: $t = \frac{\ln(f(t)) - \ln(349.2)}{\ln(0.98)}$. Let's calculate the time for each temperature:
For $f(t) = 0$: $t = \frac{\ln(0) - \ln(349.2)}{\ln(0.98)}$. Since $\ln(0)$ is undefined, we can consider the limit as the temperature approaches 0, which would result in $t$ approaching infinity.
For $f(t) = 100$: $t = \frac{\ln(100) - \ln(349.2)}{\ln(0.98)} \approx 61.90$ minutes.
For $f(t) = 300$: $t = \frac{\ln(300) - \ln(349.2)}{\ln(0.98)} \approx 7.52$ minutes.
For $f(t) = 400$: $t = \frac{\ln(400) - \ln(349.2)}{\ln(0.98)} \approx -6.72$ minutes. Since time cannot be negative, this value doesn't make sense in the context of the problem.
4. Calculating Average Absolute Errors
Now, we need to compare these predicted times with the data in the table to see which temperature gives the most accurate prediction. The data points are (5, 315), (10, 285), (15, 260), (20, 235), (25, 210).
For $f(t) = 0$, the predicted time is infinite, so the average error will also be infinite.
For $f(t) = 100$, the predicted time is approximately 61.90 minutes. The absolute differences between this time and the times in the table are |61.90 - 5| = 56.90, |61.90 - 10| = 51.90, |61.90 - 15| = 46.90, |61.90 - 20| = 41.90, |61.90 - 25| = 36.90. The average absolute error is (56.90 + 51.90 + 46.90 + 41.90 + 36.90) / 5 = 46.90.
For $f(t) = 300$, the predicted time is approximately 7.52 minutes. The absolute differences between this time and the times in the table are |7.52 - 5| = 2.52, |7.52 - 10| = 2.48, |7.52 - 15| = 7.48, |7.52 - 20| = 12.48, |7.52 - 25| = 17.48. The average absolute error is (2.52 + 2.48 + 7.48 + 12.48 + 17.48) / 5 = 8.49.
For $f(t) = 400$, the predicted time is approximately -6.72 minutes. Since time cannot be negative, we take the absolute value of the differences: |-6.72 - 5| = 11.72, |-6.72 - 10| = 16.72, |-6.72 - 15| = 21.72, |-6.72 - 20| = 26.72, |-6.72 - 25| = 31.72. The average absolute error is (11.72 + 16.72 + 21.72 + 26.72 + 31.72) / 5 = 21.72.
5. Comparing Errors
Comparing the average absolute errors, we have:
Temperature 0: Average error = infinite
Temperature 100: Average error = 46.90
Temperature 300: Average error = 8.49
Temperature 400: Average error = 21.72
The smallest average absolute error is 8.49, which corresponds to a temperature of 300.
6. Final Answer
Therefore, the model most accurately predicts the time spent cooling for a temperature of 300 degrees Fahrenheit.
### Examples
Understanding how mathematical models predict real-world phenomena, like oven cooling, is crucial in many fields. For example, in manufacturing, predicting the cooling rate of materials is essential for quality control. Similarly, in cooking, knowing how quickly an oven cools helps in planning baking schedules and ensuring consistent results. This type of modeling allows for optimization and efficiency in various processes.
