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
For 27) D
For 28) A
Hope that helps!
Answer:
The percentile is calculated with the following probability:
And using the z score we got:
So then we can conclude that 31 represent approximately the percentile 81 on the distribution given
He scored better than about 80.98 % of all MCAT takers.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
Where 
and 
And for this case we have a score of 31
The percentile is calculated with the following probability:
And using the z score we got:
So then we can conclude that 31 represent approximately the percentile 81 on the distribution given
He scored better than about 80.98 % of all MCAT takers.
Answer:
(x - 1) x (5x + 4)
Step-by-step explanation:
5 x 2 – 4 = 6
5 x 4 – 4 = 16
6 x 2 – 11x – 4 = 56
5 x 2 – x – 4 = (x - 1) x (5x + 4)
