Answer :
Let's analyze the data and the given function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] to determine which temperature will be most accurately predicted by the model.
First, we will organize the necessary information and identify the key points to solve this problem:
1. Given Data Points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes)} & \text{Temperature (°F)} \\ \hline 5 & 315 \\ \hline 10 & 285 \\ \hline 15 & 260 \\ \hline 20 & 235 \\ \hline 25 & 210 \\ \hline \end{array} \][/tex]
2. Function for Cooling Temperature:
[tex]\[ f(t) = 349.2 \times (0.98)^t \][/tex]
3. Temperatures to Check:
- 0°F
- 100°F
- 300°F
- 400°F
Let's proceed by computing the predicted times for the given temperatures using the logarithmic transformation of the function [tex]\( f(t) \)[/tex]. Here's a step-by-step breakdown:
### Step-by-Step Analysis
1. Calculate Predicted Times:
To find the time [tex]\( t \)[/tex] corresponding to each temperature [tex]\( T \)[/tex]:
[tex]\[ T = 349.2 \times (0.98)^t \implies t = \frac{\log(T / 349.2)}{\log(0.98)} \][/tex]
Applying this formula:
- For 100°F:
[tex]\[ t = \frac{\log(100 / 349.2)}{\log(0.98)} \approx 61.90 \][/tex]
- For 300°F:
[tex]\[ t = \frac{\log(300 / 349.2)}{\log(0.98)} \approx 7.52 \][/tex]
- For 400°F:
[tex]\[ t = \frac{\log(400 / 349.2)}{\log(0.98)} \approx -6.72 \quad (\text{Not possible, as time cannot be negative}) \][/tex]
2. Calculate Errors for Given Data Points:
The error for each time is calculated as the absolute difference between the actual temperature and the modeled temperature [tex]\( f(t) \)[/tex]:
- At [tex]\( t = 5 \)[/tex]:
[tex]\[ \text{Error} = |315 - f(5)| \approx 0.65 \][/tex]
- At [tex]\( t = 10 \)[/tex]:
[tex]\[ \text{Error} = |285 - f(10)| \approx 0.32 \][/tex]
- At [tex]\( t = 15 \)[/tex]:
[tex]\[ \text{Error} = |260 - f(15)| \approx 2.09 \][/tex]
- At [tex]\( t = 20 \)[/tex]:
[tex]\[ \text{Error} = |235 - f(20)| \approx 1.87 \][/tex]
- At [tex]\( t = 25 \)[/tex]:
[tex]\[ \text{Error} = |210 - f(25)| \approx 0.73 \][/tex]
3. Identify Minimum Error:
The minimum error occurs at [tex]\( t = 10 \)[/tex]:
[tex]\[ \text{Error} \approx 0.32 \][/tex]
### Conclusion:
Among the temperatures provided (0°F, 100°F, 300°F, 400°F), and given the minimum error calculation for the cooling times, the model most accurately predicts the time spent cooling for the temperature [tex]\(285°F\)[/tex].
This corresponds to [tex]\( t = 10 \)[/tex] minutes, which had the minimum prediction error. Thus, the model most accurately predicts the time spent cooling for the temperature:
[tex]\[ \boxed{285 \text{°F}} \][/tex]
First, we will organize the necessary information and identify the key points to solve this problem:
1. Given Data Points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes)} & \text{Temperature (°F)} \\ \hline 5 & 315 \\ \hline 10 & 285 \\ \hline 15 & 260 \\ \hline 20 & 235 \\ \hline 25 & 210 \\ \hline \end{array} \][/tex]
2. Function for Cooling Temperature:
[tex]\[ f(t) = 349.2 \times (0.98)^t \][/tex]
3. Temperatures to Check:
- 0°F
- 100°F
- 300°F
- 400°F
Let's proceed by computing the predicted times for the given temperatures using the logarithmic transformation of the function [tex]\( f(t) \)[/tex]. Here's a step-by-step breakdown:
### Step-by-Step Analysis
1. Calculate Predicted Times:
To find the time [tex]\( t \)[/tex] corresponding to each temperature [tex]\( T \)[/tex]:
[tex]\[ T = 349.2 \times (0.98)^t \implies t = \frac{\log(T / 349.2)}{\log(0.98)} \][/tex]
Applying this formula:
- For 100°F:
[tex]\[ t = \frac{\log(100 / 349.2)}{\log(0.98)} \approx 61.90 \][/tex]
- For 300°F:
[tex]\[ t = \frac{\log(300 / 349.2)}{\log(0.98)} \approx 7.52 \][/tex]
- For 400°F:
[tex]\[ t = \frac{\log(400 / 349.2)}{\log(0.98)} \approx -6.72 \quad (\text{Not possible, as time cannot be negative}) \][/tex]
2. Calculate Errors for Given Data Points:
The error for each time is calculated as the absolute difference between the actual temperature and the modeled temperature [tex]\( f(t) \)[/tex]:
- At [tex]\( t = 5 \)[/tex]:
[tex]\[ \text{Error} = |315 - f(5)| \approx 0.65 \][/tex]
- At [tex]\( t = 10 \)[/tex]:
[tex]\[ \text{Error} = |285 - f(10)| \approx 0.32 \][/tex]
- At [tex]\( t = 15 \)[/tex]:
[tex]\[ \text{Error} = |260 - f(15)| \approx 2.09 \][/tex]
- At [tex]\( t = 20 \)[/tex]:
[tex]\[ \text{Error} = |235 - f(20)| \approx 1.87 \][/tex]
- At [tex]\( t = 25 \)[/tex]:
[tex]\[ \text{Error} = |210 - f(25)| \approx 0.73 \][/tex]
3. Identify Minimum Error:
The minimum error occurs at [tex]\( t = 10 \)[/tex]:
[tex]\[ \text{Error} \approx 0.32 \][/tex]
### Conclusion:
Among the temperatures provided (0°F, 100°F, 300°F, 400°F), and given the minimum error calculation for the cooling times, the model most accurately predicts the time spent cooling for the temperature [tex]\(285°F\)[/tex].
This corresponds to [tex]\( t = 10 \)[/tex] minutes, which had the minimum prediction error. Thus, the model most accurately predicts the time spent cooling for the temperature:
[tex]\[ \boxed{285 \text{°F}} \][/tex]
