Question

A non-scale model of the earth has a diameter of 18 in. Within an outer clay layer, there is a core made of a ball with a diameter of 3 in.
What is the volume of the clay?

Use 3.14 to approximate pi. Round to the nearest hundredth if necessary.
_______cm^3

Answer 1

The answer is 3037.95 in^3

Answer 2

Answer: $3037.95\ cm^3$

Step-by-step explanation:

We know that the volume of sphere is given by :-

$\text{Volume}=\frac{4}{3}\pi r^3$, where r is the radius of sphere.

Given : Diameter of non-scale model of the earth= 18 in.

Radius = $\frac{18}{2}=9\ in.$

We know that the volume of non-scale model of the earthis given by :-

$\text{Volume}=\frac{4}{3}\pi (9)^3\\\\\Rightarrow\ \text{Volume}=\frac{4}{3}(3.14)(729)\\\\\Rightarrow\ \text{Volume}=3052.08\ cm^3$

Diameter of ball= 3 in.

Radius = $\frac{3}{2}\ in.$

We know that the volume of ball is given by :-

$\text{Volume}=\frac{4}{3}\pi (\frac{3}{2})^3\\\\\Rightarrow\ \text{Volume}=\frac{4}{3}(3.14)(\frac{27}{8})\\\\\Rightarrow\ \text{Volume}=14.13\ cm^3$

The volume of the clay= $3052.08\ cm^3-14.13\ cm^3=3037.95\ cm^3$