Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bubble gum nested at the bottom of the cone. The radius of the snow cone is 3 inches, and the height of the cone is 5 inches. If the diameter of the bubble gum is 1.1 inches, which of the following can be used to calculate the volume of the cone that can be filled with flavored ice? (5 points)
Select one:
a. 1 over 3 (3.14)(32)(5) − 4 over 3 (3.14)(1.13)
b. 1 over 3 (3.14)(52)(3) − 4 over 3 (3.14)(1.13)
c. 1 over 3 (3.14)(32)(5) − 4 over 3 (3.14)(0.553)
d. 1 over 3

Question

Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bubble gum nested at the bottom of the cone. The radius of the snow cone is 3 inches, and the height of the cone is 5 inches. If the diameter of the bubble gum is 1.1 inches, which of the following can be used to calculate the volume of the cone that can be filled with flavored ice? (5 points)
Select one:
a. 1 over 3 (3.14)(32)(5) − 4 over 3 (3.14)(1.13)
b. 1 over 3 (3.14)(52)(3) − 4 over 3 (3.14)(1.13)
c. 1 over 3 (3.14)(32)(5) − 4 over 3 (3.14)(0.553)
d. 1 over 3

aakashtomar · Answer

Sorry for the delay. I was on a call, really very sorry.

jmoores · Answer

Don't worry, you're completely fine! :)

jmoores · Answer

I'm just so grateful I can get some kind of help honestly!

aakashtomar · Answer

If you subtract the volume of the sphere from the cone, you'll get the volume you can fill with flavored ice. Right?

jmoores · Answer

I think so

aakashtomar · Answer

Okay, let's make this better through a drawing. 

|dw:1471882967564:dw|
My drawing is really bad, sorry, but I think you can visualize the chewing gum inside the cone.

jmoores · Answer

Ok, I see

aakashtomar · Answer

I mean, that is as far as I can infer from the question. Now, 
$$Volume (sphere) = 4/3 \pi r^3 $$
$$Volume(cone) = 1/3 \pi r^2 \times h$$

aakashtomar · Answer

So that means, The volume left to put in flavored ice is :
$$1/3 \pi r^2 - 4/3 \pi r^3$$

aakashtomar · Answer

Sorry, that is : 
$$1/3 \pi r^2 h - 4/3 \pi r^3$$

aakashtomar · Answer

So putting in the value for radius and height of the cone and radius of the sphere in the 2nd term will do the thing, I think.

jmoores · Answer

Ok, hold one, let me do it and then tell you what I get

jmoores · Answer

on*

aakashtomar · Answer

Sure!

jmoores · Answer

So wait, would I plug it in like this: 1/3(pi)3^2(5)-4/3)(pi)0.55)^2?

aakashtomar · Answer

Just that the radius of sphere is cubed, not squared.

jmoores · Answer

Ok, so switch it to cubed and then I should get my answer?

aakashtomar · Answer

I'm trying to see just that. I seriously think that this is the correct way to solve, but the problem is, the square of the radius of the cone isn't matching any options. It should be 9, but it's given 32 and 52 in the options. Also they haven't cubed the radius of the sphere, which is really wrong.

jmoores · Answer

The course I'm taking is awful, I can say that. Can you type the formula you think I need to use with  the numbers plugged in so I have an idea of where you are getting at?

jmoores · Answer

I'm just confusing myself over here

aakashtomar · Answer

$$1/3 \times 3.14 \times (3)^2 \times 5 - 4/3 \times 3.14 \times (0.55)^3$$

jmoores · Answer

Oh, I see. Hold on, let me calculate it and tell you what I get.

aakashtomar · Answer

Oh!
I See!

aakashtomar · Answer

Okay, I think they didn't raise the power but wrote the exponent beside the number. It must be c. They wrote 32 so as to say 3^2 and wrote 0.553 to say (0.55)^3

jmoores · Answer

Oh my gosh! I think you're right!

aakashtomar · Answer

Yeah, so the answer is c.

jmoores · Answer

That makes more sense now! Thank you so much! You were so so helpful! :)

aakashtomar · Answer

Nah, I'm always so happy to help. :) Welcome!