The bulls-eye on a target has a diameter of 3 inches. the whole target has a diameter of 15 inches. what part of the whole target is the bulls-eye? explain.

Since a target is a circle and the bulls-eye is also a circle, the percent of the circle that is bulls-eye would be (Area of the bulls eye)/(Area of the target)

[tex] A = \pi r^{2} \\
d = 2r \\ r = \frac{d}{2} \\\\
\frac{ \pi ( \frac{d}{2})^{2}}{ \pi ( \frac{d}{2})^{2} }= \frac{ \pi ( \frac{3}{2})^{2}}{ \pi ( \frac{15}{2})^{2} }\\
\frac{ \pi ( \frac{3}{2})^{2} }{ \pi ( \frac{15}{2})^{2} } = \frac{ \pi (1.5)^{2} }{ \pi (7.5)^{2} } \\
\frac{ \pi (1.5)}{ \pi (7.5) } = \frac{ \pi (2.25)}{ \pi (56.25)}\\
\frac{ \pi (2.25)}{ \pi (56.25)}=\frac{2.25}{56.25}= 0.04 [tex]

So the bulls-eye takes up 4% of the target.

The answer is 109 times per second