Since the distance a spring stretches varies directly with the weight attached, we can set up a proportion based on the information given.
Let \( d \) be the distance the spring stretches and \( w \) be the weight attached.
From the problem, we know:
- \( d_1 = 11 \) inches when \( w_1 = 75 \) pounds
- We want to find \( d_2 \) when \( w_2 = 65 \) pounds
Using the direct variation, we have the proportion:
\[ \frac{d_1}{w_1} = \frac{d_2}{w_2} \]
Substituting in the known values:
\[ \frac{11}{75} = \frac{d_2}{65} \]
To solve for \( d_2 \), we cross multiply:
\[ 11 \cdot 65 = 75 \cdot d_2 \]
\[ 715 = 75 \cdot d_2 \]
Now, divide both sides by 75:
\[ d_2 = \frac{715}{75} \]
Calculating \( d_2 \):
\[ d_2 \approx 9.4667 \]
Rounding to the nearest tenth:
\[ d_2 \approx 9.5 \text{ inches} \]
Thus, the spring will stretch approximately 9.5 in.