To compare the volumes of the original and replacement cones, 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.
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Original Cone:
- Diameter = 3 inches, so the radius \( r = \frac{3}{2} = 1.5 \) inches.
- Height \( h = 4 \) inches.
Volume of the original cone:
\[ V_{\text{original}} = \frac{1}{3} \pi (1.5)^2 (4) = \frac{1}{3} \pi (2.25)(4) = \frac{1}{3} \pi (9) = 3 \pi \text{ cubic inches} \]
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Replacement Cone:
- Diameter = 4 inches, so the radius \( r = \frac{4}{2} = 2 \) inches.
- Height \( h = 3 \) inches.
Volume of the replacement cone:
\[ V_{\text{replacement}} = \frac{1}{3} \pi (2)^2 (3) = \frac{1}{3} \pi (4)(3) = \frac{1}{3} \pi (12) = 4 \pi \text{ cubic inches} \]
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Comparison:
- Volume of the original cone: \( 3 \pi \)
- Volume of the replacement cone: \( 4 \pi \)
Since \( 4 \pi > 3 \pi \), we conclude that:
The replacement cone holds more than the original.