Question

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

Answer 1

$A) \: \: 0.410 \: \: is \: \: answer$

Step-by-step explanation:

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

$\frac{50}{122} , which \: \: to \: \: the \: \: nearest \: \: thousandth \: \: is \: \: 0.410.$

The final answer is A.