Question

Use the formula for the volume of a sphere to solve the following problem. How much air is needed to fill a basketball with a diameter of 9 inches? Use 3.14 as an approximation for pi. Round your answer to the nearest tenth. (1 point)

113 in.3 113 inches cubed 190.8 in.3 190.8 inches cubed 381.5 in.3 381.5 inches cubed 3,052.1 in.3

Answer 1

To find the volume of a basketball using the formula for the volume of a sphere, we can use the formula:

\[
V = \frac{4}{3} \pi r^3
\]

where \( V \) is the volume and \( r \) is the radius of the sphere.

1. First, we need to find the radius of the basketball. The diameter of the basketball is 9 inches, so the radius \( r \) is half of the diameter:

\[
r = \frac{9 \text{ in}}{2} = 4.5 \text{ in}
\]

2. Now we can plug the radius into the volume formula. Using \( \pi \approx 3.14 \):

\[
V = \frac{4}{3} \cdot 3.14 \cdot (4.5)^3
\]

3. First, calculate \( (4.5)^3 \):

\[
(4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125
\]

4. Now substitute \( (4.5)^3 \) back into the volume formula:

\[
V = \frac{4}{3} \cdot 3.14 \cdot 91.125
\]

5. Calculate \( \frac{4}{3} \cdot 3.14 \):

\[
\frac{4 \cdot 3.14}{3} \approx \frac{12.56}{3} \approx 4.18667
\]

6. Now multiply this value by 91.125:

\[
V \approx 4.18667 \cdot 91.125 \approx 381.5
\]

Thus, the volume of the basketball is approximately:

\[
\boxed{381.5 \text{ in}^3}
\]