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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Volume of the original cone:
- Diameter = 6 inches, so radius \( r = \frac{6}{2} = 3 \) inches
- Height \( h = 2 \) inches \[ V_{\text{original}} = \frac{1}{3} \pi (3^2)(2) = \frac{1}{3} \pi (9)(2) = \frac{18}{3} \pi = 6 \pi \text{ cubic inches} \]
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Volume of the replacement cone:
- Diameter = 2 inches, so radius \( r = \frac{2}{2} = 1 \) inch
- Height \( h = 6 \) inches \[ V_{\text{replacement}} = \frac{1}{3} \pi (1^2)(6) = \frac{1}{3} \pi (1)(6) = \frac{6}{3} \pi = 2 \pi \text{ cubic inches} \]
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Comparison of the volumes:
- Original cone volume: \( 6 \pi \)
- Replacement cone volume: \( 2 \pi \)
From this, we see that:
\[ V_{\text{original}} = 6 \pi, \quad V_{\text{replacement}} = 2 \pi \]
To compare:
\[ 6 \pi > 2 \pi \]
This means the original cone has a greater volume than the replacement cone.
Conclusion:
The original cone has a greater volume than the replacement.