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?
Accepted Solution
A:
Answer:The volume of the water in the fishbowl is equal to [tex](682\frac{2}{3})\pi\ in^{3}[/tex] or [tex]2,144.6\ in^{3}[/tex]Step-by-step explanation:we know thatThe volume of the sphere (a fishbowl) is equal to[tex]V=\frac{4}{3}\pi r^{3}[/tex]In this problem we have[tex]r=16/2=8\ in[/tex] ----> the radius is half the diametersubstitute[tex]V=\frac{4}{3}\pi (8^{3})=\frac{2,048}{3}\pi\ in^{3}[/tex]convert to mixed number[tex]\frac{2,048}{3}\pi\ in^{3}=\pi (\frac{2,046}{3}+\frac{2}{3})=(682\frac{2}{3})\pi\ in^{3}[/tex][tex](3.14156)*(682\frac{2}{3})=2,144.6\ in^{3}[/tex]