The problem asks us to find the volume of a composite solid consisting of a cone and a hemisphere. The radius of both the cone and hemisphere is 5 inches, and the height of the cone is 14 inches. We need to round the answer to the nearest tenth.

1. Problem Description

The problem asks us to find the volume of a composite solid consisting of a cone and a hemisphere. The radius of both the cone and hemisphere is 5 inches, and the height of the cone is 14 inches. We need to round the answer to the nearest tenth.

2. Solution Steps

First, we calculate the volume of the cone. The formula for the volume of a cone is:

$V_{cone} = \frac{1}{3} \pi r^2 h$

where $r$ is the radius and $h$ is the height.

Substituting the given values, we have:

$V_{cone} = \frac{1}{3} \pi (5^2)(14) = \frac{1}{3} \pi (25)(14) = \frac{350\pi}{3}$

Next, we calculate the volume of the hemisphere. The formula for the volume of a sphere is $\frac{4}{3} \pi r^3$. A hemisphere is half of a sphere, so the volume of a hemisphere is:

$V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$

Substituting the given value for the radius, we have:

$V_{hemisphere} = \frac{2}{3} \pi (5^3) = \frac{2}{3} \pi (125) = \frac{250\pi}{3}$

Now, we add the volume of the cone and the volume of the hemisphere to find the total volume of the composite solid:

$V_{total} = V_{cone} + V_{hemisphere} = \frac{350\pi}{3} + \frac{250\pi}{3} = \frac{600\pi}{3} = 200\pi$

Now, we need to approximate the value using $\pi \approx 3.14159$:

$V_{total} = 200\pi \approx 200 \times 3.14159 = 628.318$

Rounding to the nearest tenth, we get 628.

3.

3. Final Answer

628.3 in$^3$