Question

The maximum diameter of a bowling ball is 8.6 inches. What is the volume to the nearest tenth of a bowling ball with this diameter?

Answer 1

The volume of the bowling ball to the nearest tenth is v ≈ 106.0π in³ in terms of π or v≈332.9 in³

Explanation:

To find the volume of the bowling ball, we will follow the steps below;

First write down the formula for calculating the volume of a sphere

v=
`(4)/(3)`πr³

where v = volume of the ball and r = radius of the ball

From the equation given, the diameter of the ball is 8.6 inches but radius is half of diameter, this implies that r= d/2 = 8.6/2 = 4.3, that is; radius=4.3 inches.

Let substitutes the value into the equation

v=
`(4)/(3)`×π×(4.3)³

v =
`(4)/(3)`×π×79.507

v =106.009π

v ≈ 106.0π in³

we can take π=3.14

v = 106.009×3.14

v =332.86826

v≈332.9 in³

The volume of the bowling ball to the nearest tenth is v ≈ 106.0π in³ in terms of π or v≈332.9 in³

Answer 2

333 cubic inches

Explanation:

Hello, I can help you with this.

according to the question the maximum diameter of a bowling ball is 8.6 inches.

`D_(max)=8.6 inches\\`

Step one

using this equation find the radius of the ball

remember

Diameter=2*radius

diameter/2= radius

put the value into the equation

8.6 inches/2= radius

radius=4.3 inches

Step two

using the known radius , find the volume

the volume of a sphere is given by

`V=(4)/(3)\pi *radius^(3)`

put the value of radius into the equation

`V=(4)/(3)\pi *(4.3\ in) ^(3) \\\\V=(4)/(3)\pi *79.507\ in ^(3)\\V=333.03 in ^(3)`

so, the volume to the nearest tenth of a bowling ball with this diameter is 333 cubic inches.

I hope it helps, have a good day.