Question
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)
Responses
3,052.1 in.3
3,052.1 inches cubed
381.5 in.3
381.5 inches cubed
113 in.3
113 inches cubed
190.8 in.3
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)
Responses
3,052.1 in.3
3,052.1 inches cubed
381.5 in.3
381.5 inches cubed
113 in.3
113 inches cubed
190.8 in.3
Answers
GPT-4o mini
To find the volume of a sphere, you can use the formula:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
First, we need to find the radius of the basketball. The diameter is given as 9 inches, so the radius \( r \) is half of the diameter:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
Now, we can substitute \( r \) into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (4.5)^3
\]
Calculating \( (4.5)^3 \):
\[
(4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125
\]
Now, substitute this value back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 91.125
\]
Calculating \( \frac{4}{3} \times 3.14 \):
\[
\frac{4 \times 3.14}{3} = \frac{12.56}{3} \approx 4.1867
\]
Now, multiply this by \( 91.125 \):
\[
V \approx 4.1867 \times 91.125 \approx 381.5
\]
Thus, the volume of the basketball is approximately \( 381.5 \) cubic inches.
Rounding to the nearest tenth, we have:
**381.5 inches cubed**.
So, the correct response is:
**381.5 in.³**.
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
First, we need to find the radius of the basketball. The diameter is given as 9 inches, so the radius \( r \) is half of the diameter:
\[
r = \frac{9}{2} = 4.5 \text{ inches}
\]
Now, we can substitute \( r \) into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (4.5)^3
\]
Calculating \( (4.5)^3 \):
\[
(4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125
\]
Now, substitute this value back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 91.125
\]
Calculating \( \frac{4}{3} \times 3.14 \):
\[
\frac{4 \times 3.14}{3} = \frac{12.56}{3} \approx 4.1867
\]
Now, multiply this by \( 91.125 \):
\[
V \approx 4.1867 \times 91.125 \approx 381.5
\]
Thus, the volume of the basketball is approximately \( 381.5 \) cubic inches.
Rounding to the nearest tenth, we have:
**381.5 inches cubed**.
So, the correct response is:
**381.5 in.³**.