Question

a fishbowl shaped like a sphere is filled with water. The fishbowl has a diameter of 16 inches. Which measurement is closest to the volume of water in the fishbowl in cubic inches?

Answer 1

Answer:

2,144.66 in³

Step-by-step explanation:

∵ The volume of a sphere,

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

Where,

r = radius of the sphere,

If Diameter of a sphere = 16 inches,

Then the radius of the sphere, r = $\frac{16}{2}$ = 8 inches

So, the volume of the sphere,

$V=\frac{4}{3}\pi (8)^3$

$=\frac{4}{3}\times 512$

$=2144.66058485$

$\approx 2144.66\text{ square inches }$

∵ the  fishbowl shaped like a sphere having diameter 16 in is filled with water.

Thus, the volume of the water in the fish bowl = volume of sphere having diameter 16 inches

=  2,144.66 in³

Answer 2

Answer:

The volume of the water in the fishbowl is equal to

$(682\frac{2}{3})\pi\ in^{3}$  or  $2,144.6\ in^{3}$

Step-by-step explanation:

we know that

The volume of the sphere (a fishbowl) is equal to

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

In this problem we have

$r=16/2=8\ in$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (8^{3})=\frac{2,048}{3}\pi\ in^{3}$

convert to mixed number

$\frac{2,048}{3}\pi\ in^{3}=\pi (\frac{2,046}{3}+\frac{2}{3})=(682\frac{2}{3})\pi\ in^{3}$

$(3.14156)*(682\frac{2}{3})=2,144.6\ in^{3}$