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11 June, 02:17

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

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  1. 11 June, 04:01
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    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
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