Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
Answer:
6
Step-by-step explanation:
a) 
because it is equal to the area of the shaded region between X=4 and X=6, and the probability that X falls within some interval is given by the area under the PDF.
b) 
because the shaded region is a rectangle of height 1/5 (by virtue of X following a uniform distribution over the interval [2, 7], which has length 5).
If im correct it would be -2(2)+4(-2)-5 which is -4-8-5 which would be -17
Answer:
3,-3,5,-3, the second and fourth are the same
Step-by-step explanation:
