Question
Question
Susie runs an ice cream stand that also sells snow cones served in paper cones.
The paper cones she usually uses have a diameter of 6 inches and a height of 2 inches, but her supplier is out of them.
As a replacement, she purchases paper cones with a diameter of 2 inches and a height of 6 inches. How do the volumes of the original and replacement cones compare?
(1 point)
Responses
The replacement cone has a greater volume than the original cone.
The replacement cone has a greater volume than the original cone.
The original cone has a greater volume than the replacement cone.
The original cone has a greater volume than the replacement cone.
The original and replacement cones have the same volume.
The original and replacement cones have the same volume.
The original cone holds 2 times the amount as the replacement cone.
The original cone holds 2 times the amount as the replacement cone.
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Susie runs an ice cream stand that also sells snow cones served in paper cones.
The paper cones she usually uses have a diameter of 6 inches and a height of 2 inches, but her supplier is out of them.
As a replacement, she purchases paper cones with a diameter of 2 inches and a height of 6 inches. How do the volumes of the original and replacement cones compare?
(1 point)
Responses
The replacement cone has a greater volume than the original cone.
The replacement cone has a greater volume than the original cone.
The original cone has a greater volume than the replacement cone.
The original cone has a greater volume than the replacement cone.
The original and replacement cones have the same volume.
The original and replacement cones have the same volume.
The original cone holds 2 times the amount as the replacement cone.
The original cone holds 2 times the amount as the replacement cone.
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Answers
GPT-4o mini
To determine how the volumes of the original and replacement cones compare, we can use the formula for the volume of a cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height of the cone.
**Original Cone:**
- Diameter = 6 inches, so radius \( r = \frac{6}{2} = 3 \) inches
- Height \( h = 2 \) inches
Using the formula:
\[
V_{\text{original}} = \frac{1}{3} \pi (3^2)(2) = \frac{1}{3} \pi (9)(2) = \frac{1}{3} \pi (18) = 6\pi \text{ cubic inches}
\]
**Replacement Cone:**
- Diameter = 2 inches, so radius \( r = \frac{2}{2} = 1 \) inch
- Height \( h = 6 \) inches
Using the formula:
\[
V_{\text{replacement}} = \frac{1}{3} \pi (1^2)(6) = \frac{1}{3} \pi (1)(6) = \frac{1}{3} \pi (6) = 2\pi \text{ cubic inches}
\]
**Comparing Volumes:**
- Volume of the original cone: \( 6\pi \) cubic inches
- Volume of the replacement cone: \( 2\pi \) cubic inches
To compare them, we notice that:
\[
6\pi \text{ (original) is greater than } 2\pi \text{ (replacement)}
\]
Thus, the original cone has a greater volume than the replacement cone.
The correct response is:
**The original cone has a greater volume than the replacement cone.**
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height of the cone.
**Original Cone:**
- Diameter = 6 inches, so radius \( r = \frac{6}{2} = 3 \) inches
- Height \( h = 2 \) inches
Using the formula:
\[
V_{\text{original}} = \frac{1}{3} \pi (3^2)(2) = \frac{1}{3} \pi (9)(2) = \frac{1}{3} \pi (18) = 6\pi \text{ cubic inches}
\]
**Replacement Cone:**
- Diameter = 2 inches, so radius \( r = \frac{2}{2} = 1 \) inch
- Height \( h = 6 \) inches
Using the formula:
\[
V_{\text{replacement}} = \frac{1}{3} \pi (1^2)(6) = \frac{1}{3} \pi (1)(6) = \frac{1}{3} \pi (6) = 2\pi \text{ cubic inches}
\]
**Comparing Volumes:**
- Volume of the original cone: \( 6\pi \) cubic inches
- Volume of the replacement cone: \( 2\pi \) cubic inches
To compare them, we notice that:
\[
6\pi \text{ (original) is greater than } 2\pi \text{ (replacement)}
\]
Thus, the original cone has a greater volume than the replacement cone.
The correct response is:
**The original cone has a greater volume than the replacement cone.**