Answer :
- Calculate $f(5)$ using the given function $f(t)=349.2(0.98)^t$, which is approximately 315.7.
- Calculate the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model $f(t)$.
- Compare the model's predictions with the data in the table by calculating the time $t$ for each temperature in the table (315, 285, 260, 235) using the model and then find the absolute difference between the predicted time and the given time.
- Determine that the model most accurately predicts the time spent cooling for the temperature of $\boxed{300}$ degrees Fahrenheit.
### Explanation
1. Problem Analysis
We are given a function that models the temperature of an oven as it cools over time, and we want to determine which of the given temperatures (0, 100, 300, or 400) the model predicts most accurately. To do this, we will compare the model's predictions to the data provided in the table.
2. Calculate f(5)
First, let's calculate $f(5)$ using the given function $f(t) = 349.2(0.98)^t$. This will give us the temperature at $t=5$ minutes according to the model.
3. Calculating f(5)
Now, we will calculate $f(5) = 349.2(0.98)^5$. The result of this calculation is approximately 315.7.
4. Solving for t
Next, we want to find the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model $f(t) = 349.2(0.98)^t$. We will solve for $t$ in the equation $temperature = 349.2(0.98)^t$. This can be rewritten as $t = \frac{\log(\frac{temperature}{349.2})}{\log(0.98)}$.
5. Calculating t for each temperature
Let's calculate the time $t$ for each temperature:
For temperature 0: $t = \frac{\log(\frac{0}{349.2})}{\log(0.98)}$. Since the logarithm of 0 is undefined, we cannot use this temperature.
For temperature 100: $t = \frac{\log(\frac{100}{349.2})}{\log(0.98)} = \frac{\log(0.2864)}{\log(0.98)} \approx \frac{-0.5428}{-0.00877} \approx 61.9$ minutes.
For temperature 300: $t = \frac{\log(\frac{300}{349.2})}{\log(0.98)} = \frac{\log(0.8591)}{\log(0.98)} \approx \frac{-0.0657}{-0.00877} \approx 7.5$ minutes.
For temperature 400: $t = \frac{\log(\frac{400}{349.2})}{\log(0.98)} = \frac{\log(1.1455)}{\log(0.98)} \approx \frac{0.0590}{-0.00877} \approx -6.7$ minutes. Since time cannot be negative, we discard this temperature.
6. Comparing Model Predictions with Table Data
Now, let's compare the model's predictions with the data in the table. We will calculate the time $t$ for each temperature in the table (315, 285, 260, 235) using the model and then find the absolute difference between the predicted time and the given time.
For 315: $t = \frac{\log(\frac{315}{349.2})}{\log(0.98)} = \frac{\log(0.9021)}{\log(0.98)} \approx \frac{-0.0448}{-0.00877} \approx 5.1$ minutes. The table gives 10 minutes, so the error is $|10 - 5.1| = 4.9$ minutes.
For 285: $t = \frac{\log(\frac{285}{349.2})}{\log(0.98)} = \frac{\log(0.8161)}{\log(0.98)} \approx \frac{-0.0882}{-0.00877} \approx 10.1$ minutes. The table gives 15 minutes, so the error is $|15 - 10.1| = 4.9$ minutes.
For 260: $t = \frac{\log(\frac{260}{349.2})}{\log(0.98)} = \frac{\log(0.7446)}{\log(0.98)} \approx \frac{-0.1280}{-0.00877} \approx 14.6$ minutes. The table gives 20 minutes, so the error is $|20 - 14.6| = 5.4$ minutes.
For 235: $t = \frac{\log(\frac{235}{349.2})}{\log(0.98)} = \frac{\log(0.6730)}{\log(0.98)} \approx \frac{-0.1720}{-0.00877} \approx 19.6$ minutes. The table gives 25 minutes, so the error is $|25 - 19.6| = 5.4$ minutes.
7. Determining the Most Accurate Temperature
Now, let's consider the temperature 300. We found that $t \approx 7.5$ minutes. We can compare this to the data points in the table. The closest temperatures are 315 at 10 minutes and $f(5) \approx 315.7$ at 5 minutes. The average of these times is 7.5 minutes, which matches the model's prediction. Therefore, the model most accurately predicts the time spent cooling for a temperature of 300 degrees.
8. Final Answer
The model most accurately predicts the time spent cooling for the temperature of 300 degrees Fahrenheit.
### Examples
Understanding how temperature changes over time is crucial in many real-world applications, such as food safety. For example, knowing how quickly an oven cools helps determine how long food remains at a temperature where bacteria can grow. This model can be used to predict cooling times and ensure food is cooled safely, preventing foodborne illnesses. Similarly, in manufacturing, controlling the cooling rate of materials can affect their final properties, making accurate temperature models essential for quality control.
- Calculate the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model $f(t)$.
- Compare the model's predictions with the data in the table by calculating the time $t$ for each temperature in the table (315, 285, 260, 235) using the model and then find the absolute difference between the predicted time and the given time.
- Determine that the model most accurately predicts the time spent cooling for the temperature of $\boxed{300}$ degrees Fahrenheit.
### Explanation
1. Problem Analysis
We are given a function that models the temperature of an oven as it cools over time, and we want to determine which of the given temperatures (0, 100, 300, or 400) the model predicts most accurately. To do this, we will compare the model's predictions to the data provided in the table.
2. Calculate f(5)
First, let's calculate $f(5)$ using the given function $f(t) = 349.2(0.98)^t$. This will give us the temperature at $t=5$ minutes according to the model.
3. Calculating f(5)
Now, we will calculate $f(5) = 349.2(0.98)^5$. The result of this calculation is approximately 315.7.
4. Solving for t
Next, we want to find the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model $f(t) = 349.2(0.98)^t$. We will solve for $t$ in the equation $temperature = 349.2(0.98)^t$. This can be rewritten as $t = \frac{\log(\frac{temperature}{349.2})}{\log(0.98)}$.
5. Calculating t for each temperature
Let's calculate the time $t$ for each temperature:
For temperature 0: $t = \frac{\log(\frac{0}{349.2})}{\log(0.98)}$. Since the logarithm of 0 is undefined, we cannot use this temperature.
For temperature 100: $t = \frac{\log(\frac{100}{349.2})}{\log(0.98)} = \frac{\log(0.2864)}{\log(0.98)} \approx \frac{-0.5428}{-0.00877} \approx 61.9$ minutes.
For temperature 300: $t = \frac{\log(\frac{300}{349.2})}{\log(0.98)} = \frac{\log(0.8591)}{\log(0.98)} \approx \frac{-0.0657}{-0.00877} \approx 7.5$ minutes.
For temperature 400: $t = \frac{\log(\frac{400}{349.2})}{\log(0.98)} = \frac{\log(1.1455)}{\log(0.98)} \approx \frac{0.0590}{-0.00877} \approx -6.7$ minutes. Since time cannot be negative, we discard this temperature.
6. Comparing Model Predictions with Table Data
Now, let's compare the model's predictions with the data in the table. We will calculate the time $t$ for each temperature in the table (315, 285, 260, 235) using the model and then find the absolute difference between the predicted time and the given time.
For 315: $t = \frac{\log(\frac{315}{349.2})}{\log(0.98)} = \frac{\log(0.9021)}{\log(0.98)} \approx \frac{-0.0448}{-0.00877} \approx 5.1$ minutes. The table gives 10 minutes, so the error is $|10 - 5.1| = 4.9$ minutes.
For 285: $t = \frac{\log(\frac{285}{349.2})}{\log(0.98)} = \frac{\log(0.8161)}{\log(0.98)} \approx \frac{-0.0882}{-0.00877} \approx 10.1$ minutes. The table gives 15 minutes, so the error is $|15 - 10.1| = 4.9$ minutes.
For 260: $t = \frac{\log(\frac{260}{349.2})}{\log(0.98)} = \frac{\log(0.7446)}{\log(0.98)} \approx \frac{-0.1280}{-0.00877} \approx 14.6$ minutes. The table gives 20 minutes, so the error is $|20 - 14.6| = 5.4$ minutes.
For 235: $t = \frac{\log(\frac{235}{349.2})}{\log(0.98)} = \frac{\log(0.6730)}{\log(0.98)} \approx \frac{-0.1720}{-0.00877} \approx 19.6$ minutes. The table gives 25 minutes, so the error is $|25 - 19.6| = 5.4$ minutes.
7. Determining the Most Accurate Temperature
Now, let's consider the temperature 300. We found that $t \approx 7.5$ minutes. We can compare this to the data points in the table. The closest temperatures are 315 at 10 minutes and $f(5) \approx 315.7$ at 5 minutes. The average of these times is 7.5 minutes, which matches the model's prediction. Therefore, the model most accurately predicts the time spent cooling for a temperature of 300 degrees.
8. Final Answer
The model most accurately predicts the time spent cooling for the temperature of 300 degrees Fahrenheit.
### Examples
Understanding how temperature changes over time is crucial in many real-world applications, such as food safety. For example, knowing how quickly an oven cools helps determine how long food remains at a temperature where bacteria can grow. This model can be used to predict cooling times and ensure food is cooled safely, preventing foodborne illnesses. Similarly, in manufacturing, controlling the cooling rate of materials can affect their final properties, making accurate temperature models essential for quality control.
