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
113 in.3
113 inches cubed
190.8 in.3
190.8 inches cubed
3,052.1 in.3
3,052.1 inches cubed
381.5 in.3
To find the volume of air needed to fill a basketball, we need to calculate the volume of a sphere, which is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( V \) is the volume, \( \pi \) is the constant (approximately 3.14), and \( r \) is the radius of the sphere.
In this case, the diameter of the basketball is 9 inches, so the radius (\( r \)) is half of that, which is:
\[ r = \frac{9}{2} = 4.5 \text{ inches} \]
Now we can calculate the volume:
\[ V = \frac{4}{3} \times 3.14 \times (4.5)^3 \]
\[ V = \frac{4}{3} \times 3.14 \times 91.125 \]
\[ V = 4.18666667 \times 3.14 \times 91.125 \]
\[ V = 12.566370614359172 \times 91.125 \]
\[ V = 1145.21875 \text{ cubic inches} \]
Rounding to the nearest tenth gives:
\[ V \approx 1145.2 \text{ in}^3 \]
It seems none of the provided options exactly match the calculated value. However, the answer should be rounded to the nearest tenth, so the best approximation based on the given options is:
190.8 in.3 or 190.8 inches cubed
Please note that I found an error in the initial calculation. The correct volume is:
\[ V \approx 381.5 \text{ in}^3 \]
So the correct answer, with rounding to the nearest tenth, is:
381.5 in.3 or 381.5 inches cubed