It's fear has a radius of 4 inches in a cube shaped box has an edge length of 7.5 inches JD says the box has greater volume so the sphere will fit in the box is he correct

JD is wrong.

Although, the volume of the cube shaped box V' = 421.88 in³  > volume of the sphere V = 268.1 in³, the sphere cannot fit into the box because its diameter, d = 8 in > side of cube = 7.5 in.

Step-by-step explanation:

The sphere would only fit into the cube shaped box if its diameter is less than the length of side of the cube L = 7.5 inches.

So, diameter of sphere d = 2r where r = radius of sphere = 4 inches. Thus, d = 2r = 2(4 in) = 8 in.

Since the diameter of the sphere d = 8 inches > length of side of the cube = 7.5 inches, the sphere would not fit into the box.

So, JD is wrong.

We calculate both the volume of the sphere and cube.

Volume of a sphere V = 4πr³/3 where r = radius of sphere = 4 inches

So, V = 4πr³/3

= 4π(4 in)³/3

= 4π(64) in³/3

= 256π in³/3

= 804.25 in³/3

= 268.1 in³

Volume of a cube V' = L³ where L = length of side of cube = 7.5 inches

So, V' = L³

= (7.5 in)³

= 421.88 in³  

Although, the volume of the cube shaped box V' = 421.88 in³  > volume of the sphere V = 268.1 in³, the sphere cannot fit into the box because its diameter, d = 8 in > side of cube = 7.5 in.

b is the answer that makes most sense

step-by-step explanation: