Question 13 Multiple Choice Worth 1 points) (00.02 LC) Snow's Snow Cones has a special butble gum snow cone on sale. The cone is a regular snow cone that has a spherical plece of bubbie gum nested at the botlom of the cone. The ratis of the snow cone is 4 inches, and the height of the cone is 6 inches. If the diameter of the butbie gum is 0.8 inches, which of the following can be used to calcuite the volume of the cone that can be flled with flavored loe? $$ \frac{1}{3}(3.14)\left( heta^{2} ight)(4)-\frac{4}{3}(3.14)\left(0.4^{3} ight) $$ $$ \frac{1}{3}(3.14)\left(4^{2} imes 6 ight)-\frac{4}{3}(3.14)(0.4) $$ $$ \frac{1}{3}(3.14)\left( heta^{2} ight)(4)-\frac{4}{3}(3.14)\left(0 . .^{3} ight) $$ $$ \frac{1}{3}(3.14)\left(4^{2} ight)(6)-\frac{4}{3}(3.14)\left(0.8^{3} ight) $$

Question

Question 13 Multiple Choice Worth 1 points) (00.02 LC) Snow's Snow Cones has a special butble gum snow cone on sale. The cone is a regular snow cone that has a spherical plece of bubbie gum nested at the botlom of the cone. The ratis of the snow cone is 4 inches, and the height of the cone is 6 inches. If the diameter of the butbie gum is 0.8 inches, which of the following can be used to calcuite the volume of the cone that can be flled with flavored loe? $$ \frac{1}{3}(3.14)\left(	heta^{2}ight)(4)-\frac{4}{3}(3.14)\left(0.4^{3}ight) $$ $$ \frac{1}{3}(3.14)\left(4^{2} 	imes 6ight)-\frac{4}{3}(3.14)(0.4) $$ $$ \frac{1}{3}(3.14)\left(	heta^{2}ight)(4)-\frac{4}{3}(3.14)\left(0 . .^{3}ight) $$ $$ \frac{1}{3}(3.14)\left(4^{2}ight)(6)-\frac{4}{3}(3.14)\left(0.8^{3}ight) $$

UpStudy AI · Accepted Answer

To find the volume of the cone that can be filled with flavored ice, we need to calculate the volume of the cone and subtract the volume of the bubble gum.

The formula for the volume of a cone is given by:
$$ V = \frac{1}{3} \pi r^2 h $$

where:
- $$ V $$ is the volume of the cone,
- $$ r $$ is the radius of the cone,
- $$ h $$ is the height of the cone.

Given:
- Radius of the snow cone ($$ r $$) = 4 inches,
- Height of the cone ($$ h $$) = 6 inches,
- Diameter of the bubble gum = 0.8 inches.

First, let's calculate the volume of the cone:
$$ V_{cone} = \frac{1}{3} \pi (4^2) (6) $$

Next, we need to calculate the volume of the bubble gum. The volume of a sphere is given by:
$$ V_{sphere} = \frac{4}{3} \pi r^3 $$

where:
- $$ r $$ is the radius of the sphere.

Given that the diameter of the bubble gum is 0.8 inches, the radius is half of the diameter:
$$ r_{bubble\ gum} = \frac{0.8}{2} = 0.4 \, \text{inches} $$

Now, we can calculate the volume of the bubble gum:
$$ V_{bubble\ gum} = \frac{4}{3} \pi (0.4)^3 $$

Finally, the volume of the cone that can be filled with flavored ice is the difference between the volume of the cone and the volume of the bubble gum:
$$ V_{flavored\ ice} = V_{cone} - V_{bubble\ gum} $$

Let's calculate the volume of the cone and the bubble gum, and then find the volume of the flavored ice.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1}{3}\times 3.14\times 4^{2}\times 6$$
- step1: Multiply the terms:
  $$\frac{157}{25}\times 4^{2}$$
- step2: Multiply:
  $$\frac{157\times 4^{2}}{25}$$
- step3: Multiply:
  $$\frac{2512}{25}$$
Calculate or simplify the expression $$ \frac{4}{3} * 3.14 * (0.4^3) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{4}{3}\times 3.14\times 0.4^{3}$$
- step1: Convert the expressions:
  $$\frac{4}{3}\times 3.14\left(\frac{2}{5}\right)^{3}$$
- step2: Multiply the terms:
  $$\frac{314}{75}\left(\frac{2}{5}\right)^{3}$$
- step3: Evaluate the power:
  $$\frac{314}{75}\times \frac{2^{3}}{5^{3}}$$
- step4: Multiply the fractions:
  $$\frac{314\times 2^{3}}{75\times 5^{3}}$$
- step5: Multiply:
  $$\frac{2512}{9375}$$
Calculate or simplify the expression $$ 100.48 - 0.26794 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$100.48-0.26794$$
- step1: Subtract the numbers:
  $$100.21206$$
The volume of the cone that can be filled with flavored ice is approximately 100.21 cubic inches.

Therefore, the correct formula to calculate the volume of the cone that can be filled with flavored ice is:
$$ V_{flavored\ ice} = \frac{1}{3} \pi (4^2) (6) - \frac{4}{3} \pi (0.4)^3 $$

This matches option 4: $$ \frac{1}{3}(3.14)(4^{2})(6)-\frac{4}{3}(3.14)(0.8^{3}) $$
The correct formula to calculate the volume of the cone that can be filled with flavored ice is: $$ \frac{1}{3}(3.14)(4^{2})(6)-\frac{4}{3}(3.14)(0.8^{3}) $$ So, the answer is option 4.

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