Answer :
To identify the temperature for which the model [tex]\( f(t) = 349.2 \cdot (0.98)^t \)[/tex] most accurately predicts the time spent cooling, we need to compare the predicted temperatures with the actual temperatures at given time points.
Given data points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes) } t & \text{Oven temperature (degrees Fahrenheit)} \\ \hline 5 & 315 \\ \hline 10 & 285 \\ \hline 15 & 260 \\ \hline 20 & 235 \\ \hline 25 & 210 \\ \hline \end{array} \][/tex]
First, we calculate the predicted temperatures using the model [tex]\( f(t) = 349.2 \cdot (0.98)^t \)[/tex] and evaluate the absolute differences between the predicted and actual temperatures at each time point.
1. At [tex]\( t = 5 \)[/tex]:
- Predicted temperature [tex]\( f(5) = 349.2 \cdot (0.98)^5 \approx 314.35085775744005 \)[/tex]
- Actual temperature = 315
- Difference = [tex]\( |314.35085775744005 - 315| \approx 0.6491422425599467 \)[/tex]
2. At [tex]\( t = 10 \)[/tex]:
- Predicted temperature [tex]\( f(10) = 349.2 \cdot (0.98)^{10} \approx 284.6781758348687 \)[/tex]
- Actual temperature = 285
- Difference = [tex]\( |284.6781758348687 - 285| \approx 0.32182416513131784 \)[/tex]
3. At [tex]\( t = 15 \)[/tex]:
- Predicted temperature [tex]\( f(15) = 349.2 \cdot (0.98)^{15} \approx 257.908330643775 \)[/tex]
- Actual temperature = 260
- Difference = [tex]\( |257.908330643775 - 260| \approx 2.0916693562249975 \)[/tex]
4. At [tex]\( t = 20 \)[/tex]:
- Predicted temperature [tex]\( f(20) = 349.2 \cdot (0.98)^{20} \approx 233.1287037368789 \)[/tex]
- Actual temperature = 235
- Difference = [tex]\( |233.1287037368789 - 235| \approx 1.8712962631211099 \)[/tex]
5. At [tex]\( t = 25 \)[/tex]:
- Predicted temperature [tex]\( f(25) = 349.2 \cdot (0.98)^{25} \approx 209.2701163612093 \)[/tex]
- Actual temperature = 210
- Difference = [tex]\( |209.2701163612093 - 210| \approx 0.7298836387907102 \)[/tex]
Next, we compare these differences to determine the smallest difference, which signifies the most accurate prediction by the model.
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes)} & \text{Absolute Difference} \\ \hline 5 & 0.6491422425599467 \\ \hline 10 & 0.32182416513131784 \\ \hline 15 & 2.0916693562249975 \\ \hline 20 & 1.8712962631211099 \\ \hline 25 & 0.7298836387907102 \\ \hline \end{array} \][/tex]
The smallest absolute difference is at [tex]\( t = 10 \)[/tex] minutes, with a difference of approximately [tex]\( 0.32182416513131784 \)[/tex]. The actual temperature at this time is [tex]\( 285 \)[/tex] degrees Fahrenheit.
Thus, the temperature for which the model most accurately predicts the time spent cooling is [tex]\( 285 \)[/tex] degrees Fahrenheit.
Given data points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes) } t & \text{Oven temperature (degrees Fahrenheit)} \\ \hline 5 & 315 \\ \hline 10 & 285 \\ \hline 15 & 260 \\ \hline 20 & 235 \\ \hline 25 & 210 \\ \hline \end{array} \][/tex]
First, we calculate the predicted temperatures using the model [tex]\( f(t) = 349.2 \cdot (0.98)^t \)[/tex] and evaluate the absolute differences between the predicted and actual temperatures at each time point.
1. At [tex]\( t = 5 \)[/tex]:
- Predicted temperature [tex]\( f(5) = 349.2 \cdot (0.98)^5 \approx 314.35085775744005 \)[/tex]
- Actual temperature = 315
- Difference = [tex]\( |314.35085775744005 - 315| \approx 0.6491422425599467 \)[/tex]
2. At [tex]\( t = 10 \)[/tex]:
- Predicted temperature [tex]\( f(10) = 349.2 \cdot (0.98)^{10} \approx 284.6781758348687 \)[/tex]
- Actual temperature = 285
- Difference = [tex]\( |284.6781758348687 - 285| \approx 0.32182416513131784 \)[/tex]
3. At [tex]\( t = 15 \)[/tex]:
- Predicted temperature [tex]\( f(15) = 349.2 \cdot (0.98)^{15} \approx 257.908330643775 \)[/tex]
- Actual temperature = 260
- Difference = [tex]\( |257.908330643775 - 260| \approx 2.0916693562249975 \)[/tex]
4. At [tex]\( t = 20 \)[/tex]:
- Predicted temperature [tex]\( f(20) = 349.2 \cdot (0.98)^{20} \approx 233.1287037368789 \)[/tex]
- Actual temperature = 235
- Difference = [tex]\( |233.1287037368789 - 235| \approx 1.8712962631211099 \)[/tex]
5. At [tex]\( t = 25 \)[/tex]:
- Predicted temperature [tex]\( f(25) = 349.2 \cdot (0.98)^{25} \approx 209.2701163612093 \)[/tex]
- Actual temperature = 210
- Difference = [tex]\( |209.2701163612093 - 210| \approx 0.7298836387907102 \)[/tex]
Next, we compare these differences to determine the smallest difference, which signifies the most accurate prediction by the model.
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (minutes)} & \text{Absolute Difference} \\ \hline 5 & 0.6491422425599467 \\ \hline 10 & 0.32182416513131784 \\ \hline 15 & 2.0916693562249975 \\ \hline 20 & 1.8712962631211099 \\ \hline 25 & 0.7298836387907102 \\ \hline \end{array} \][/tex]
The smallest absolute difference is at [tex]\( t = 10 \)[/tex] minutes, with a difference of approximately [tex]\( 0.32182416513131784 \)[/tex]. The actual temperature at this time is [tex]\( 285 \)[/tex] degrees Fahrenheit.
Thus, the temperature for which the model most accurately predicts the time spent cooling is [tex]\( 285 \)[/tex] degrees Fahrenheit.
