Question

A spherical fish bowl is half-filled with water. The center of the bowl is C, and the length of segment AB is 16 inches, as shown below. Use Twenty two over seven for pi.

A sphere with diameter 16 inches is drawn.

Which of the following can be used to calculate the volume of water inside the fish bowl?

1 over 24 over 322 over 7(82)(16)
1 over 24 over 322 over 7(83)
1 over 24 over 322 over 7(162)(8)
1 over 24 over 322 over 7(163)

Answer 1

Answer:

The second choice

1 over $2/4$ over $3/22$ over $7/(8^{3})$

Step-by-step explanation:

we know that

The volume of a sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

In this problem we have

$r=16/2=8\ in$

$\pi=22/7$

To find the volume of water calculate the volume of semi-sphere

so

$V=\frac{1}{2}\frac{4}{3}\pi r^{3}$ ------> volume of semi-sphere

substitute the values

$V=\frac{1}{2}\frac{4}{3}(\frac{22}{7})(8)^{3}$      

1 over $2/4$ over $3/22$ over $7/(8^{3})$

Answer 2

Answer:

B. $\frac{1}{2}\times\frac{4}{3}\times \frac{22}{7}\times 8^3$

Step-by-step explanation:

We have been given that a fish bowl has a diameter of 16 inches. We are asked to find the half-volume of the water inside the fish bowl.

We will use volume of sphere formula to solve the given problem.

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

Let us figure out radius of fish bowl by dividing its diameter by 2 as:

$r=\frac{16}{2}=8$

$\text{Volume of half filled fish bowl}=\frac{1}{2}\times\frac{4}{3}\times \frac{22}{7}\times r^3$

$\text{Volume of half filled fish bowl}=\frac{1}{2}\times\frac{4}{3}\times \frac{22}{7}\times 8^3$

Therefore, option B is the correct choice.