Answer :
Let's analyze the problem step-by-step to find out for which temperature the model most accurately predicts the time spent cooling.
### Step 1: Understanding the model
The model given is [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] where:
- [tex]\( f(t) \)[/tex] represents the temperature of the oven in degrees Fahrenheit.
- [tex]\( t \)[/tex] represents the time in minutes that the oven has been cooling.
### Step 2: Given data
We have a table of times and corresponding actual oven temperatures:
| Time (minutes) | Oven temperature (°F) |
|----------------|------------------------|
| 5 | 315 |
| 10 | 285 |
| 15 | 260 |
| 20 | 235 |
| 25 | 210 |
### Step 3: Calculating predicted temperatures using the model
Using the function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex], we calculate:
- For [tex]\( t = 5 \)[/tex]:
[tex]\( f(5) = 349.2 \times (0.98)^5 \approx 315.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex]:
[tex]\( f(10) = 349.2 \times (0.98)^{10} \approx 285.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex]:
[tex]\( f(15) = 349.2 \times (0.98)^{15} \approx 257.91 \)[/tex]
- For [tex]\( t = 20 \)[/tex]:
[tex]\( f(20) = 349.2 \times (0.98)^{20} \approx 233.13 \)[/tex]
- For [tex]\( t = 25 \)[/tex]:
[tex]\( f(25) = 349.2 \times (0.98)^{25} \approx 210.73 \)[/tex]
### Step 4: Calculating the absolute errors
To find out how accurately the model predicts the time spent cooling, we compare the predicted temperatures with the actual temperatures and calculate the absolute errors:
- For [tex]\( t = 5 \)[/tex]:
Absolute error [tex]\( = |315.65 - 315| = 0.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex]:
Absolute error [tex]\( = |285.32 - 285| = 0.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex]:
Absolute error [tex]\( = |257.91 - 260| = 2.09 \)[/tex]
- For [tex]\( t = 20 \)[/tex]:
Absolute error [tex]\( = |233.13 - 235| = 1.87 \)[/tex]
- For [tex]\( t = 25 \)[/tex]:
Absolute error [tex]\( = |210.73 - 210| = 0.73 \)[/tex]
### Step 5: Finding the smallest error
The smallest absolute error indicates the time at which the model most accurately predicts the temperature:
| Time (minutes) | Absolute error (°F) |
|----------------|----------------------|
| 5 | 0.65 |
| 10 | 0.32 |
| 15 | 2.09 |
| 20 | 1.87 |
| 25 | 0.73 |
The smallest error is 0.32, which corresponds to [tex]\( t = 10 \)[/tex] minutes.
### Conclusion
The temperature for which the model most accurately predicts the cooling time is at [tex]\( t = 10 \)[/tex] minutes. Therefore, the most accurate predicted temperature is [tex]\( 285 \)[/tex] degrees Fahrenheit.
### Step 1: Understanding the model
The model given is [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex] where:
- [tex]\( f(t) \)[/tex] represents the temperature of the oven in degrees Fahrenheit.
- [tex]\( t \)[/tex] represents the time in minutes that the oven has been cooling.
### Step 2: Given data
We have a table of times and corresponding actual oven temperatures:
| Time (minutes) | Oven temperature (°F) |
|----------------|------------------------|
| 5 | 315 |
| 10 | 285 |
| 15 | 260 |
| 20 | 235 |
| 25 | 210 |
### Step 3: Calculating predicted temperatures using the model
Using the function [tex]\( f(t) = 349.2 \times (0.98)^t \)[/tex], we calculate:
- For [tex]\( t = 5 \)[/tex]:
[tex]\( f(5) = 349.2 \times (0.98)^5 \approx 315.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex]:
[tex]\( f(10) = 349.2 \times (0.98)^{10} \approx 285.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex]:
[tex]\( f(15) = 349.2 \times (0.98)^{15} \approx 257.91 \)[/tex]
- For [tex]\( t = 20 \)[/tex]:
[tex]\( f(20) = 349.2 \times (0.98)^{20} \approx 233.13 \)[/tex]
- For [tex]\( t = 25 \)[/tex]:
[tex]\( f(25) = 349.2 \times (0.98)^{25} \approx 210.73 \)[/tex]
### Step 4: Calculating the absolute errors
To find out how accurately the model predicts the time spent cooling, we compare the predicted temperatures with the actual temperatures and calculate the absolute errors:
- For [tex]\( t = 5 \)[/tex]:
Absolute error [tex]\( = |315.65 - 315| = 0.65 \)[/tex]
- For [tex]\( t = 10 \)[/tex]:
Absolute error [tex]\( = |285.32 - 285| = 0.32 \)[/tex]
- For [tex]\( t = 15 \)[/tex]:
Absolute error [tex]\( = |257.91 - 260| = 2.09 \)[/tex]
- For [tex]\( t = 20 \)[/tex]:
Absolute error [tex]\( = |233.13 - 235| = 1.87 \)[/tex]
- For [tex]\( t = 25 \)[/tex]:
Absolute error [tex]\( = |210.73 - 210| = 0.73 \)[/tex]
### Step 5: Finding the smallest error
The smallest absolute error indicates the time at which the model most accurately predicts the temperature:
| Time (minutes) | Absolute error (°F) |
|----------------|----------------------|
| 5 | 0.65 |
| 10 | 0.32 |
| 15 | 2.09 |
| 20 | 1.87 |
| 25 | 0.73 |
The smallest error is 0.32, which corresponds to [tex]\( t = 10 \)[/tex] minutes.
### Conclusion
The temperature for which the model most accurately predicts the cooling time is at [tex]\( t = 10 \)[/tex] minutes. Therefore, the most accurate predicted temperature is [tex]\( 285 \)[/tex] degrees Fahrenheit.
