Math, asked by MrDestruction, 9 months ago

The incomplete table above summarizes the number of left-handed students and right-handed students by gender for the eighth grade students at Keisel Middle School. There are 5 times as many right-handed female students as there are left-handed female students, and there are 9 times as many right-handed male students as there are left-handed male students. if there is a total of 18 left-handed students and 122 right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.) A) 0.410 B) 0.357 C) 0.333 D) 0.250

Answers

Answered by Anonymous
0

In order to solve this problem, you should create two equations using two variables (x and y) and the information you're given. Let x be the number of left-handed female students and let y be the number of left-handed male students. Using the information given in the problem, the number of right-handed female students will be 5x and the number of right-handed male students will be 9y. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

x+y=18

5x+9y=122

When you solve this system of equations, you get x=10 and y=8. Thus, 5*10, or 50, of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is  

50 /122 , which to the nearest thousandth is 0.410.

The final answer is A.

\rule{200}{2}

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