A basketball with a diameter of 9.5 in. is placed in a cubic box with sides 15 in. long. How many cubic inches of packing foam are needed to fill the rest of the box? Round to the nearest tenth.

The volume of foam needed to fill the box is approximately 2926.1 cubic inches.

Step-by-step explanation:

To calculate the amount of foaming that is needed to fill the rest of the box we first need to calculate the volume of the box and the volume of the ball. Since the box is cubic it's volume is given by the formula below, while the formula for the basketball, a sphere, is also shown.

Vcube = a³

Vsphere = (4*pi*r³)/3

Where a is the side of the box and r is the radius of the box. The radius is half of the diameter. Applying the data from the problem to the expressions, we have:

Vcube = 15³ = 3375 cubic inches

Vsphere = (4*pi*(9.5/2)³)/3 = 448.921

The volume of foam there is needed to complete the box is the subtraction between the two volumes above:

Vfoam = Vcube - Vsphere = 3375 - 448.921 = 2926.079 cubic inches

The volume of foam needed to fill the box is approximately 2926.1 cubic inches.

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560