To find the volume of a basketball (which is a sphere) with a diameter of 9.2 inches, we will use the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( V \) is the volume and \( r \) is the radius of the sphere.
- First, we need to find the radius. The radius is half of the diameter:
\[ r = \frac{d}{2} = \frac{9.2}{2} = 4.6 \text{ inches} \]
- Now we can substitute the radius and the value of \(\pi\) (using 3.14) into the volume formula:
\[ V = \frac{4}{3} \times 3.14 \times (4.6)^3 \]
- Calculate \(4.6^3\):
\[ 4.6^3 = 4.6 \times 4.6 \times 4.6 = 97.336 \]
- Now substitute this back into the volume formula:
\[ V = \frac{4}{3} \times 3.14 \times 97.336 \]
- First, calculate \(3.14 \times 97.336\):
\[ 3.14 \times 97.336 \approx 306.554944 \]
- Now calculate \(\frac{4}{3} \times 306.554944\):
\[ V \approx \frac{4 \times 306.554944}{3} \approx \frac{1226.219776}{3} \approx 408.7399253 \]
- Rounding this to the nearest hundredth:
\[ V \approx 408.74 \text{ in}^3 \]
Based on the provided options, the closest response is:
407.51 in.³
It seems there may be a slight discrepancy in the calculations somewhere or in the options provided. However, using the calculated volume of approximately 408.74 in³, none of the responses match precisely. The closest option listed, 407.51 in³, may refer to an approximation or alternate rounding method.