The top of the gumball machine on the left is a sphere with a diameter of 20 inches. When the machine is full it can hold 3500 gumballs, leaving 40% of the space in the machine empty. What is the approximate radius of each gumball? Round to the nearest hundredth and use 3.14 for Pi

0.555 in

Step-by-step explanation:

The first thing we must do is calculate the volume of the machine, the volume of a wait is:

v = (4/3) * pi * (r ^ 3)

we know that the diameter is twice the radius, therefore:

r = d / 2 = 20/2

r = 10, replacing:

v = (4/3) * 3.14 * (10 ^ 3)

v = 4186.66 in ^ 3

Now, we know that by leaving 40% of space in the machine there are 3,500 chewing gums, therefore, 60% of that volume is chewing gum.

4186.66 * 0.6 = 2512

to calculate the volume of a chewing gum:

3500 * v = 2512

v = 2512/3500

v = 0.7177 in ^ 2 is the volume of a single chewing gum

if we solve the sphere formula for r, we are left with:

r ^ 3 = v * (3/4) / pi

replacing

r ^ 3 = 0.7177 * (3/4) / 3.14

r ^ 3 = 0.1714

r = 0.1714 ^ (1/3)

r = 0.555

the radius of each gum is 0.555 in