Question 1205453
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diameter of the ball = 9
radius = diameter/2 = 9/2 = 4.5
A = volume of the ball
A = (4/3)*pi*r^3
A = (4/3)*pi*(4.5)^3
A = 121.5pi


diameter of cylinder container = 12
r = radius = diameter/2 = 12/2 = 6
The water level is h = 9 so that it reaches the top of the 9 cm diameter ball.
B = combined volume of the ball and surrounding water in the cylinder
B = pi*r^2*h
B = pi*6^2*9
B = 324pi


C = volume of water only
C = difference of volumes A and B
C = B - A
C = 324pi - 121.5pi
C = 202.5pi


Use the value of C to compute the new height of the water level after the ball is removed. 
The cylinder's radius r = 6 will stay the same.
V = pi*r^2*h
202.5pi = pi*6^2*h
202.5pi = 36pi*h
h = (202.5pi)/(36pi)
h = 5.625
The water level is 5.625 cm after the ball is removed.
This assumes no water spills out.


The water line started at 9 cm and now is at 5.625 cm
This is a difference of 9 - 5.625 = 3.375 cm which represents how far the water line dropped.
This converts to the mixed number 3 & 3/8 aka {{{matrix(1,2,3,3/8)}}} since 3/8 = 0.375 exactly.



Answer in decimal form <font color=red size=4>3.375 cm</font>
Answer as a mixed number: <font color=red size=4>3 & 3/8 cm</font> or <font color=red size=4>3  3/8 cm</font>
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