We are given a manhole cover modeled as a cylinder with a diameter of 24 inches and a height of 1.25 inches. The density of the steel used to manufacture the cover is 0.282 pounds per cubic inch. We need to find the approximate weight of the manhole cover to the nearest pound.

1. Problem Description

We are given a manhole cover modeled as a cylinder with a diameter of 24 inches and a height of 1.25 inches. The density of the steel used to manufacture the cover is 0.282 pounds per cubic inch. We need to find the approximate weight of the manhole cover to the nearest pound.

2. Solution Steps

First, we need to find the radius of the cylinder. The radius is half of the diameter.

$radius = \frac{diameter}{2}$

$radius = \frac{24}{2} = 12$ inches

Next, we calculate the volume of the cylinder using the formula:

$Volume = \pi * radius^2 * height$

$Volume = \pi * (12^2) * 1.25$

$Volume = \pi * 144 * 1.25$

$Volume = 180\pi$ cubic inches

Now, we can find the weight of the manhole cover using the density and volume.

$Weight = Density * Volume$

$Weight = 0.282 * 180\pi$

$Weight \approx 0.282 * 180 * 3.14159$

$Weight \approx 159.43$ pounds

Finally, we round the weight to the nearest pound.

$Weight \approx 159$ pounds

3. Final Answer

The approximate weight of the manhole cover is 159 pounds.