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leonid [27]
3 years ago
3

21% of the students had a score of 80/100 or higher on both midterms. 25% had a score of 80/100 or higher on midterm 1. what per

centage of the students who had score of 80/100 or higher on midterm 1 had a score of 80/100 or higher on midterm 2
Mathematics
1 answer:
Delvig [45]3 years ago
7 0
Let event A = high score on midterm1.
Let event B = high score on midterm2.

You want P(B|A).
  P(B|A) = P(B&A)/P(A)
  P(B|A) = 0.21/0.25
  P(B|A) = 84%

84% of students who had a high score on midterm1 had a high score on midterm 2.
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The first hexagon: 2, 2, 3, 4, 6, 7

The second hexagon: 3, 3, a, b, c, d

The hexagons are similar. Therefore the sides are in proportion.

\dfrac{a}{3}=\dfrac{3}{2}     multiply both sides by 3

a=\dfrac{3}{2}=4.5

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b=\dfrac{3}{2}\cdot4=6

\dfrac{c}{6}=\dfrac{3}{2}     multiply both sides by 6

c=\dfrac{3}{2}\cdot6=9

\dfrac{d}{7}=\dfrac{3}{2}    multiply both sides by 7

d=\dfrac{21}{2}=10.5

The perimeter:

P=4.5+6+9+10.5=30

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