Question

1. Ki’von has a sink that is shaped like a half-sphere. The sink has a diameter of 20 inches. One day, his sink clogged. He has to use one of two different cups to scoop the water out of the sink. The sink is completely full when Ki’von begins scooping. (a) What is the exact volume of the sink? Show your work. (3 points) (b) One conical cup has a diameter of 8 in. and a height of 6 in. How many cups of water must Ki’von scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number. Show your work. (6 points) (c) One cylindrical cup has a diameter of 4 in. and a height of 6 in. How many cups of water must he scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number. Show your work. (6 points) Answer:

PLZ HELP WILL GIVE YOU 20 POINTS

Answer 1

A) The exact volume of the sink is (2000π)/3 in³.
B) It would take 21 scoops with the conical cup to empty the sink.
C) It would take 28 scoops with the cylindrical cup to empty the sink.

For A, the volume of a sphere is V=(4/3)πr³. Since the sink is a hemisphere, our volume will be 1/2 of that. The diameter of the sink is 20, so the radius is 10. We have:


`V=(1)/(2)*(4)/(3)\pi(10^3) \\=(1)/(2)*(4)/(3)*(1000)/(1) \pi=(4000\pi)/(6) \\ \\=(2000\pi)/(3)`

For B, we find the volume of the conical cup using the formula V=(1/3)πr²h. Since the diameter of the cup is 8, the radius is 4. We have:


`V=(1)/(3)\pi 4^2*6=(96\pi)/(3)=32\pi`

We divide the volume of the sink by 32π:

(2000π)/3 ÷ 32ππ
= (2000π)/3 ÷ (32π/1)
= (2000π)/3 * 1/(32π)
= (2000π)/(96π) = 2000/96 ≈ 21.

For part C, we first find the volume of the cylindrical cup using the formula V=πr²h. The diameter of the cup is 4, so the radius is 2:

V=π(2²)(6) = 24π

Now we divide the volume of the sink, (2000π)/3, by 24π:

(2000π)/3 ÷24π
= (2000π)/3 ÷ (24π)/1
= (2000π)/3 * 1/(24π)
= (2000π)/(72π) = 2000/72 ≈ 28