The relationship between the distance a spring stretches and the weight attached can be represented using the concept of direct variation. This means 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. We know from the problem that:
\[ d_1 = 11 \text{ inches} \text{ when } w_1 = 75 \text{ pounds} \]
We want to find \( d_2 \) when \( w_2 = 65 \text{ pounds} \). Since the stretching distance varies directly with weight, we can set up the following proportion:
\[ \frac{d_1}{w_1} = \frac{d_2}{w_2} \]
Substituting in the known values:
\[ \frac{11}{75} = \frac{d_2}{65} \]
To find \( d_2 \), we can cross-multiply:
\[ 11 \times 65 = 75 \times d_2 \]
Calculating \( 11 \times 65 \):
\[ 11 \times 65 = 715 \]
So we have:
\[ 715 = 75 \times d_2 \]
Now, we solve for \( d_2 \):
\[ d_2 = \frac{715}{75} \]
Calculating the division:
\[ d_2 \approx 9.4667 \]
Rounding to the nearest tenth of an inch:
\[ d_2 \approx 9.5 \]
Thus, the spring will stretch approximately 9.5 inches with 65 pounds attached.