Question

A snow cone is a tasty treat with flavored ice and a spherical bubble gum ball at the bottom.

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?

4.43
0.07
13.50
4.50

Answer 1

Correct answer is A. 4.43 in^3

Explanation:

The previous answer listed an incorrect answer to due leaving out the 1/3 calculation in the Volume of the Cone which is as follows:

V=1/3(pi)(r^2)(h) so 1/3 (3.14) (1.25)^2(2.75) then Vcone=1/3(13.492187) then Vcone =4.497. Volume of Sphere is V(Bubble gum)=4/3(pi)(r)^3 so 4/3 (3.14)(0.25)^3 then V(Bubble gum)=4/3(0.0490625) then Volume of (Bubble gum sphere)=0.0654. In order to determine the volume of the cone that an be filled with favored ice, you have to deduct the volume of gum from the volume of the cone as V(Cone)-V(Bubble gum)=4.997 - 0.0654=4.43 inches cubed.

Answer 2

So here is how we are going to answer this question. Firstly, we are going to find volume of the cone and the volume of the sphere. So for the volume of the cone, the formula would be πr2h/3 and the volume of the sphere would be 4/3πr3. Now let us solve.
V(C)=(3.14)(1.25)^2(2.75)
V(C)= (3.14)(1.5625)(2.75)
V(C) = 13.492 cubic inches

Next, let us solve for the sphere
V(B) = 4/3(3.14)(0.25)^3
V(B) = 1.33(3.14)(0.015625)
V(B) = 1.33 (0.491)
V(B) = 0.0653 cubic inches

Now in order to find the volume of the cone that can be filled with ice cream, deduct the volume of the gum from the volume of the cone.
V(C) - V(B) = V (Ice Cream)
13.492 - 0.0653 = 13.43 cubic inches