Pete runs an ice cream stand that also sells snow cones served in paper cones. %0D%0A%0D%0AThe paper cones he usually uses have a diameter of 3 inches and a height of 4 inches, but his supplier is out of them. %0D%0A%0D%0AAs a replacement, he purchases paper cones with a diameter of 4 inches and a height of 3 inches. How do the volumes of the original and replacement cones compare?%0D%0A%0D%0A(2 points)%0D%0AResponses%0D%0A%0D%0AThe replacement cone and original cone volumes cannot be found from the information.%0D%0AThe replacement cone and original cone volumes cannot be found from the information.%0D%0A%0D%0AThe original and replacement cones have the same volume.%0D%0AThe original and replacement cones have the same volume.%0D%0A%0D%0AThe replacement cone holds more than the original.%0D%0AThe replacement cone holds more than the original.%0D%0A%0D%0AThe original cone holds more than the replacement.

Question

Pete runs an ice cream stand that also sells snow cones served in paper cones. %0D%0A%0D%0AThe paper cones he usually uses have a diameter of 3 inches and a height of 4 inches, but his supplier is out of them. %0D%0A%0D%0AAs a replacement, he purchases paper cones with a diameter of 4 inches and a height of 3 inches. How do the volumes of the original and replacement cones compare?%0D%0A%0D%0A(2 points)%0D%0AResponses%0D%0A%0D%0AThe replacement cone and original cone volumes cannot be found from the information.%0D%0AThe replacement cone and original cone volumes cannot be found from the information.%0D%0A%0D%0AThe original and replacement cones have the same volume.%0D%0AThe original and replacement cones have the same volume.%0D%0A%0D%0AThe replacement cone holds more than the original.%0D%0AThe replacement cone holds more than the original.%0D%0A%0D%0AThe original cone holds more than the replacement.

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1 answer
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Answer 1

Bot GPT 3.5 answered
2 months ago

The original cone holds more than the replacement.

To compare the volumes of the two cones, we can use the formula for the volume of a cone: V = (1/3)πr^2h, where r is the radius and h is the height.

For the original cone with a diameter of 3 inches, the radius is 1.5 inches and the height is 4 inches. Plugging these values into the formula, we get:
V = (1/3)π(1.5)^2(4) = 9π cubic inches

For the replacement cone with a diameter of 4 inches, the radius is 2 inches and the height is 3 inches. Plugging these values into the formula, we get:
V = (1/3)π(2)^2(3) = 4π cubic inches

Since 9π is greater than 4π, the original cone holds more volume than the replacement cone.