A factory produces weighted balls to use for exercise, by filling spherical rubber shells of different sizes with a sand-like material. The material's density is 1.5 grams per cubic centimeter.
Assuming the shell weighs 10 grams, what should be the ball's radius so, when full, it weighs 1 kilogram (or 1000 grams)?
Round your answer, if necessary, to the nearest tenth.

r=5.4

As mentioned above, the actual \goldE{\text{total quantity}}total quantitystart color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, end color #a75a05 is 1000-10=\goldE{990}1000−10=9901000, minus, 10, equals, start color #a75a05, 990, end color #a75a05 grams, since we are trying to find the volume of the air within the shell.

Let's denote the ball's radius as rrr. Then, the \maroonD{\text{volume}}volumestart color #ca337c, start text, v, o, l, u, m, e, end text, end color #ca337c is \maroonD{\dfrac43\pi r^3}  

3

4

πr  

3

start color #ca337c, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c.

Now we can plug \blueE{\text{density}=1.5}density=1.5start color #0c7f99, start text, d, e, n, s, i, t, y, end text, equals, 1, point, 5, end color #0c7f99, \goldE{\text{total quantity}=990}total quantity=990start color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, equals, 990, end color #a75a05, and \maroonD{\text{volume}=\dfrac43\pi r^3}volume=  

3

4

πr  

3

r=5.4

As mentioned above, the actual \goldE{\text{total quantity}}total quantitystart color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, end color #a75a05 is 1000-10=\goldE{990}1000−10=9901000, minus, 10, equals, start color #a75a05, 990, end color #a75a05 grams, since we are trying to find the volume of the air within the shell.

Let's denote the ball's radius as rrr. Then, the \maroonD{\text{volume}}volumestart color #ca337c, start text, v, o, l, u, m, e, end text, end color #ca337c is \maroonD{\dfrac43\pi r^3}  

3

4

πr  

3

start color #ca337c, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c.

Now we can plug \blueE{\text{density}=1.5}density=1.5start color #0c7f99, start text, d, e, n, s, i, t, y, end text, equals, 1, point, 5, end color #0c7f99, \goldE{\text{total quantity}=990}total quantity=990start color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, equals, 990, end color #a75a05, and \maroonD{\text{volume}=\dfrac43\pi r^3}volume=  

3

4

πr  

3

start color #ca337c, start text, v, o, l, u, m, e, end text, equals, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c in the equation.

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Step-by-step explanation:

r=5.4

As mentioned above, the actual \goldE{\text{total quantity}}total quantitystart color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, end color #a75a05 is 1000-10=\goldE{990}1000−10=9901000, minus, 10, equals, start color #a75a05, 990, end color #a75a05 grams, since we are trying to find the volume of the air within the shell.

Let's denote the ball's radius as rrr. Then, the \maroonD{\text{volume}}volumestart color #ca337c, start text, v, o, l, u, m, e, end text, end color #ca337c is \maroonD{\dfrac43\pi r^3}  

3

4

πr  

3

start color #ca337c, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c.

Now we can plug \blueE{\text{density}=1.5}density=1.5start color #0c7f99, start text, d, e, n, s, i, t, y, end text, equals, 1, point, 5, end color #0c7f99, \goldE{\text{total quantity}=990}total quantity=990start color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, equals, 990, end color #a75a05, and \maroonD{\text{volume}=\dfrac43\pi r^3}volume=  

3

4

πr  

3

start color #ca337c, start text, v, o, l, u, m, e, end text, equals, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c in the equation.

answer: 19 is   2

step-by-step explanation: i just looked at your answer and wrote it down

probably the last option

step-by-step explanation: