Question

Ill give brainliest!!!

Answer 1

Given:

The coin wrapper volume = 27,480 mm³.

The coin diameter = 21.18 mm.

There are 50 such coins.

To find:

The thickness (height) of 1 coin.

Solution:

The 50 coins form a cylinder with a volume of 27,480 mm³.

If the diameter is 21.18 mm, the radius of a coin will be $\frac{21.18}{2} = 10.59$ mm.

The volume of a cylinder $= \pi r^{2} h.$

The volume of all 50 coins is given as 27,480 mm³. We need to determine the height of a single coin.

$50( \pi r^{2} h)=27,480.$

$50(3.1415)(10.59^{2} )(h) = 27,480.$

$(17,615.662)(h)=27,480.$

$h= \frac{27,480}{17,615.662} = 1.5599$ mm.

So the thickness of 1 coin is 1.5599 mm.