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AVprozaik [17]
3 years ago
5

The incomplete table below summarizes the number of left-handed students and right-handed students by gender for the eighth-grad

e 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

Mathematics
1 answer:
vova2212 [387]3 years ago
7 0

Answer: A) 0.410

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

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Given parameters;

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Let us represent Simon's money by S

Kande's money by K

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 if Simon gives $20, his money will be  S - 20 lesser;

      When Kande receives $20, his money will increase to K + 20

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Now we have set up two equations, let us solve;

         S - 20  = K + 20  ---- i

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So,      S - 20  = K + 20

          S + 22  = 2K - 44

subtract both equations;

               -20 - 22  = (k -2k)  + 64

                   -42  = -k + 64

                       k  = 106

Using equation i, let us find S;

            S - 20 = K + 20

             S - 20  = 106 + 20

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Therefore, Kande has $106 and Simon has $146

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