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Let's go through the answers provided for each part of the question, based on the information from the image you uploaded.

### (a) Estimating \( W'(12) \) using data from the table
For part (a), the solution estimates the derivative \( W'(12) \) using the difference quotient with data from \( t = 9 \) and \( t = 15 \). The calculation is:
\[ W'(12) = \frac{W(15) - W(9)}{15 - 9} = \frac{67.9 - 61.8}{6} = \frac{6.1}{6} \approx 1.017 \]
This result indicates that the water temperature is increasing at approximately 1.017°F per minute at \( t = 12 \) minutes.

### (b) Evaluating \( \int_0^{20} W'(t) \, dt \)
In part (b), the solution uses the Fundamental Theorem of Calculus to find the net change in temperature over 20 minutes. It calculates:
\[ \int_0^{20} W'(t) \, dt = W(20) - W(0) = 71.0 - 55.0 = 16 \]
This means the water has warmed by 16°F from the start to 20 minutes.

### (c) Average temperature over the first 20 minutes
For part (c), a left Riemann sum approximation is used to estimate the average temperature over the first 20 minutes:
\[ \frac{1}{20} \int_0^{20} W(t) \, dt \approx \frac{1}{20} \left(4 \cdot 55.0 + 5 \cdot 57.1 + 6 \cdot 61.8 + 5 \cdot 67.9\right) = 60.79 \]
The approximation method chosen is a left Riemann sum, and it's noted that the result is an underestimate since \( W(t) \) is increasing.

### (d) Calculating \( W(25) \) using the first derivative model
Part (d) uses the model \( W'(t) = 0.4 \cdot \cos(0.06 \cdot t) \) to find \( W(25) \):
\[ \int_{20}^{25} W'(t) \, dt = \int_{20}^{25} 0.4 \cdot \cos(0.06 \cdot t) \, dt \approx 0.243155 \]
\[ W(25) = W(20) + \int_{20}^{25} W'(t) \, dt = 71.0 + 0.243155 = 73.043 \]

The results provided in each part are consistent with the mathematical principles and show a thorough understanding of the concepts applied. 

Do you have any specific parts you'd like to discuss further or need clarification on? Here are some related questions we can explore:

1. How would the estimate of \( W'(12) \) change if we used a central difference quotient instead?
2. Can we find a more accurate method to estimate \( \int_0^{20} W(t) \, dt \) than the left Riemann sum?
3. Why does the left Riemann sum provide an underestimate in this scenario?
4. How would the approximation for \( W(25) \) change if we used a different integration method?
5. What are the implications of the derivative model \( W'(t) = 0.4 \cdot \cos(0.06 \cdot t) \) for longer times?

Tip: When estimating derivatives numerically, consider using central difference quotient for a more accurate result when possible.

Explore a detailed solution involving calculus concepts like differentiation, integration, and Riemann sums. This problem examines the change in water temperature over time using mathematical principles, suitable for students in Grades 11-12.

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