The total scores on the Medical College Admission Test (MCAT) follow a Normal distribution with mean 25.2 and standard deviation
1 answer:
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Answer:
20. AB = 42
21. BC = 28
22. AC = 70
23. BC = 20.4
24. FH = 48
25. DE = 10, EF = 10, DF = 20
Step-by-step explanation:
✍️Given:
AB = 2x + 7
BC = 28
AC = 4x,
20. Assuming B is between A and C, thus:
AB + BC = AC (Segment Addition Postulate)
2x + 7 + 28 = 4x (substitution)
Collect like terms
2x + 35 = 4x
35 = 4x - 2x
35 = 2x
Divide both side by 2
17.5 = x
AB = 2x + 7
Plug in the value of x
AB = 2(17.5) + 7 = 42
21. BC = 28 (given)
22. AC = 4x
Plug in the value of x
AC = 4(17.5) = 70
✍️Given:
AC = 35 and AB = 14.6.
Assuming B is between A and C, thus:
23. AB + BC = AC (Segment Addition Postulate)
14.6 + BC = 35 (Substitution)
Subtract 14.6 from each side
BC = 35 - 14.6
BC = 20.4
24. FH = 7x + 6
FG = 4x
GH = 24
FG + GH = FH (Segment Addition Postulate)
(substitution)
Collect like terms
Divide both sides by -3
FH = 7x + 6
Plug in the value of x
FH = 7(6) + 6 = 48
25. DE = 5x, EF = 3x + 4
Given that E bisects DF, therefore,
DE = EF
5x = 3x + 4 (substitution)
Subtract 3x from each side
5x - 3x = 4
2x = 4
Divide both sides by 2
x = 2
DE = 5x
Plug in the value of x
DE = 5(2) = 10
EF = 3x + 4
Plug in the value of x
EF = 3(2) + 4 = 10
DF = DE + EF
DE = 10 + 10 (substitution)
DE = 20
