Answer :
- Evaluate the function at given times and calculate the absolute differences.
- Calculate the time $t$ for each of the given temperatures.
- Compare the calculated times to the times in the table.
- Determine the temperature for which the model most accurately predicts the time spent cooling, which is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t)=349.2(0.98)^t$ which models the temperature of an oven after $t$ minutes of cooling. We are also given a table of oven temperatures 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. Evaluating the Model at Given Times
First, let's evaluate the function $f(t)$ at $t = 10, 15, 20, 25$ and compare these values to the temperatures given in the table.
$f(10) = 349.2(0.98)^{10} \approx 285.32$
$f(15) = 349.2(0.98)^{15} \approx 257.91$
$f(20) = 349.2(0.98)^{20} \approx 233.13$
$f(25) = 349.2(0.98)^{25} \approx 210.73$
Now, let's calculate the absolute differences between the model's predictions and the table values:
$|f(10) - 315| = |285.32 - 315| \approx 29.68$
$|f(15) - 285| = |257.91 - 285| \approx 27.09$
$|f(20) - 260| = |233.13 - 260| \approx 26.87$
$|f(25) - 235| = |210.73 - 235| \approx 24.27$
These differences tell us how well the model fits the data at the given times.
3. Calculating Time for Given Temperatures
Next, we need to determine the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model. We can rearrange the formula to solve for $t$:
$f(t) = 349.2(0.98)^t \implies \frac{f(t)}{349.2} = (0.98)^t \implies t = \frac{\ln(\frac{f(t)}{349.2})}{\ln(0.98)}$
Let's calculate $t$ for each temperature:
For $T = 0$: $t = \frac{\ln(\frac{0}{349.2})}{\ln(0.98)}$. Since we can't take the logarithm of 0, this temperature is not possible.
For $T = 100$: $t = \frac{\ln(\frac{100}{349.2})}{\ln(0.98)} \approx 61.90$
For $T = 300$: $t = \frac{\ln(\frac{300}{349.2})}{\ln(0.98)} \approx 7.52$
For $T = 400$: $t = \frac{\ln(\frac{400}{349.2})}{\ln(0.98)} \approx -6.72$. Since time cannot be negative, this temperature is not realistic in this context.
4. Comparing and Concluding
Now, we compare the calculated times (7.52 and 61.90) to the times in the table (10, 15, 20, 25). We want to see which of the given temperatures (100, 300) results in a time that, when plugged into the model, gives a temperature closest to the original temperature.
We already know that the model is closest to the actual values at $t = 25$ minutes. The time $t = 7.52$ (corresponding to 300 degrees) is closest to $t = 10$. The time $t = 61.90$ (corresponding to 100 degrees) is far from the times in the table.
Let's compare the absolute differences we calculated earlier to the given temperatures. We want to find the temperature that, when used in the model to solve for $t$, gives a $t$ that results in a temperature close to the original temperature.
We have the differences between the model and the table at t=10, 15, 20, and 25. We also have the times for T=100 and T=300. We want to see which of the given temperatures is closest to the model's prediction.
From the first calculation, we see that the smallest difference between the model and the actual temperature occurs at t=25. The differences are 29.68, 27.09, 26.87, and 24.27. The smallest difference is 24.27, which occurs at t=25. However, we are asked to choose between the temperatures 0, 100, 300, and 400. We found that t=7.52 for T=300 and t=61.90 for T=100. The time 7.52 is closest to the values in the table.
Since the model is closest to the actual temperature at t=25, we want to find the temperature that is closest to the model's prediction at t=25. The model predicts 210.73 at t=25. Of the given temperatures, 100 and 300 are the closest. However, we found that the time for T=300 is 7.52, which is closest to the times in the table. Therefore, the model most accurately predicts the time spent cooling for 300 degrees.
5. Final Answer
The model most accurately predicts the time spent cooling for a temperature of $\boxed{300}$ degrees Fahrenheit.
### Examples
Understanding how temperature changes over time is crucial in many real-world applications, such as food safety and manufacturing processes. For example, in a commercial kitchen, knowing how quickly an oven cools down helps ensure that food is stored at safe temperatures to prevent bacterial growth. Similarly, in manufacturing, controlling the cooling rate of materials can affect their final properties and quality. By accurately modeling these cooling processes, we can optimize procedures and improve outcomes in various industries.
- Calculate the time $t$ for each of the given temperatures.
- Compare the calculated times to the times in the table.
- Determine the temperature for which the model most accurately predicts the time spent cooling, which is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t)=349.2(0.98)^t$ which models the temperature of an oven after $t$ minutes of cooling. We are also given a table of oven temperatures 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. Evaluating the Model at Given Times
First, let's evaluate the function $f(t)$ at $t = 10, 15, 20, 25$ and compare these values to the temperatures given in the table.
$f(10) = 349.2(0.98)^{10} \approx 285.32$
$f(15) = 349.2(0.98)^{15} \approx 257.91$
$f(20) = 349.2(0.98)^{20} \approx 233.13$
$f(25) = 349.2(0.98)^{25} \approx 210.73$
Now, let's calculate the absolute differences between the model's predictions and the table values:
$|f(10) - 315| = |285.32 - 315| \approx 29.68$
$|f(15) - 285| = |257.91 - 285| \approx 27.09$
$|f(20) - 260| = |233.13 - 260| \approx 26.87$
$|f(25) - 235| = |210.73 - 235| \approx 24.27$
These differences tell us how well the model fits the data at the given times.
3. Calculating Time for Given Temperatures
Next, we need to determine the time $t$ for each of the given temperatures (0, 100, 300, 400) using the model. We can rearrange the formula to solve for $t$:
$f(t) = 349.2(0.98)^t \implies \frac{f(t)}{349.2} = (0.98)^t \implies t = \frac{\ln(\frac{f(t)}{349.2})}{\ln(0.98)}$
Let's calculate $t$ for each temperature:
For $T = 0$: $t = \frac{\ln(\frac{0}{349.2})}{\ln(0.98)}$. Since we can't take the logarithm of 0, this temperature is not possible.
For $T = 100$: $t = \frac{\ln(\frac{100}{349.2})}{\ln(0.98)} \approx 61.90$
For $T = 300$: $t = \frac{\ln(\frac{300}{349.2})}{\ln(0.98)} \approx 7.52$
For $T = 400$: $t = \frac{\ln(\frac{400}{349.2})}{\ln(0.98)} \approx -6.72$. Since time cannot be negative, this temperature is not realistic in this context.
4. Comparing and Concluding
Now, we compare the calculated times (7.52 and 61.90) to the times in the table (10, 15, 20, 25). We want to see which of the given temperatures (100, 300) results in a time that, when plugged into the model, gives a temperature closest to the original temperature.
We already know that the model is closest to the actual values at $t = 25$ minutes. The time $t = 7.52$ (corresponding to 300 degrees) is closest to $t = 10$. The time $t = 61.90$ (corresponding to 100 degrees) is far from the times in the table.
Let's compare the absolute differences we calculated earlier to the given temperatures. We want to find the temperature that, when used in the model to solve for $t$, gives a $t$ that results in a temperature close to the original temperature.
We have the differences between the model and the table at t=10, 15, 20, and 25. We also have the times for T=100 and T=300. We want to see which of the given temperatures is closest to the model's prediction.
From the first calculation, we see that the smallest difference between the model and the actual temperature occurs at t=25. The differences are 29.68, 27.09, 26.87, and 24.27. The smallest difference is 24.27, which occurs at t=25. However, we are asked to choose between the temperatures 0, 100, 300, and 400. We found that t=7.52 for T=300 and t=61.90 for T=100. The time 7.52 is closest to the values in the table.
Since the model is closest to the actual temperature at t=25, we want to find the temperature that is closest to the model's prediction at t=25. The model predicts 210.73 at t=25. Of the given temperatures, 100 and 300 are the closest. However, we found that the time for T=300 is 7.52, which is closest to the times in the table. Therefore, the model most accurately predicts the time spent cooling for 300 degrees.
5. Final Answer
The model most accurately predicts the time spent cooling for a temperature of $\boxed{300}$ degrees Fahrenheit.
### Examples
Understanding how temperature changes over time is crucial in many real-world applications, such as food safety and manufacturing processes. For example, in a commercial kitchen, knowing how quickly an oven cools down helps ensure that food is stored at safe temperatures to prevent bacterial growth. Similarly, in manufacturing, controlling the cooling rate of materials can affect their final properties and quality. By accurately modeling these cooling processes, we can optimize procedures and improve outcomes in various industries.
