In baseball, each time a player attempts to hit the ball, it is recorded. The ratio of hits compared to total attempts is their batting average. Each player on the team wants to have the highest batting average to help their team the most. For the season so far, Jana has hit the ball 8 times out of 10 attempts, while Tasha has hit the ball 9 times out of 12 attempts. Which player has a better batting average?

A. Tasha, because she has the highest ratio since [tex]$0.75 \ \textless \ 0.8$[/tex]

B. Tasha, because she has the highest ratio since [tex]$\frac{48}{60} \ \textgreater \ \frac{45}{60}$[/tex]

C. Jana, because she has the highest ratio since [tex]$0.8 \ \textgreater \ 0.75$[/tex]

D. Jana, because she has the highest ratio since [tex]$\frac{48}{60} \ \textgreater \ \frac{45}{60}$[/tex]

First, we calculate the batting average for each player by dividing the number of hits by the number of attempts.

For Jana:
[tex]$$\text{Batting average} = \frac{8}{10} = 0.8$$[/tex]

For Tasha:
[tex]$$\text{Batting average} = \frac{9}{12} = 0.75$$[/tex]

Since a higher batting average is better, we compare the two:
[tex]$$0.8 > 0.75$$[/tex]

We can also express these averages with a common denominator. If we choose 60 as the common denominator, then:
- Jana's average becomes:
[tex]$$\frac{8}{10} = \frac{8 \times 6}{10 \times 6} = \frac{48}{60}$$[/tex]
- Tasha's average becomes:
[tex]$$\frac{9}{12} = \frac{9 \times 5}{12 \times 5} = \frac{45}{60}$$[/tex]

Clearly, since
[tex]$$\frac{48}{60} > \frac{45}{60},$$[/tex]
Jana has the better batting average.

Thus, the correct answer is:

d Jana, because she has the highest ratio since [tex]$\frac{48}{60} > \frac{45}{60}$[/tex].