The radius of the cone is 1.25 inches, and its height is 2.75 inches. If the diameter of the bubble gum ball is 0.5 inches, what is the closest approximation of the volume of the cone that can be filled with flavored ice? (4 points)

4.43 in3
0.07 in3
13.50 in3
4.50 in3
https://vvsd6966-vvagpeap-ccl.gradpoint.com/Resource/9331744,0/Assets/74794_53c6cc08/image0034e8484bd.gif

Question

The radius of the cone is 1.25 inches, and its height is 2.75 inches. If the diameter of the bubble gum ball is 0.5 inches, what is the closest approximation of the volume of the cone that can be filled with flavored ice? (4 points)

 	
	4.43 in3
	0.07 in3
	13.50 in3
	4.50 in3
https://vvsd6966-vvagpeap-ccl.gradpoint.com/Resource/9331744,0/Assets/74794_53c6cc08/image0034e8484bd.gif

anonymous · Answer

i got 13.50 in3 is that correct @AihberKhan

aihberkhan · Answer

What you have to do is Volume of the cone - volume of the bubble gum.

This is: $\frac{ 1 }{ 3 }\pi r^2 h - \frac{ 4 }{ 3 } \pi r ^3$

aihberkhan · Answer

Continuing to solve: \($$\frac{ 1 }{ 3 }( \pi )(1.25)^2(2.75) - \frac{ 4 }{ 3 } \pi (.25)^3$$

aihberkhan · Answer

Now using a calulctaor or by hand type 1 divided by 3 times3.14159 times1.25 times 1.25 times2.75  equals

aihberkhan · Answer

Write that number down on a piece of paper.

aihberkhan · Answer

Type the first number you wrote down minus the second number you wrote down equals.

That is your answer

aihberkhan · Answer

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anonymous · Answer

ohhhhh ok makes sense @AihberKhan