To find the volume of a sphere, we use the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( V \) is the volume and \( r \) is the radius of the sphere. The diameter of the basketball is given as 9.2 inches, so we first need to calculate the radius:
\[ r = \frac{\text{diameter}}{2} = \frac{9.2}{2} = 4.6 \text{ inches} \]
Next, we can substitute the value of \( r \) and \( \pi \approx 3.14 \) into the volume formula:
\[ V = \frac{4}{3} \times 3.14 \times (4.6)^3 \]
Now we calculate \( (4.6)^3 \):
\[ (4.6)^3 = 4.6 \times 4.6 \times 4.6 = 97.336 \]
Now substitute this value back into the volume formula:
\[ V = \frac{4}{3} \times 3.14 \times 97.336 \] \[ V = \frac{4 \times 3.14 \times 97.336}{3} \] \[ V = \frac{1227.1624}{3} \] \[ V \approx 409.05413 \text{ in.}^3 \]
Rounding to the nearest hundredth gives us:
\[ V \approx 409.05 \text{ in.}^3 \]
The closest answer from the options provided is:
\[ \boxed{407.51 \text{ in.}^3} \]
It seems there might be a slight discrepancy in the rounding or values, but this is how the calculation proceeds. So, I would choose 407.51 in.³ from your options, as it's the best match.
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