Answer:
The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are 0.533 and 0.82
Step-by-step explanation:
According to the given data we have the following:
Total sample of students= 150
80 students preferred to get out 10 minutes early
Therefore, the mean of the sampling distribution of the number of students who preferred to get out early is = 80/150 = 0.533
Therefore, standard deviation of the sampling distribution of the number of students who preferred to get out early= phat - p0/sqrt(p0(1-p)/)
= 0.533-0.5/sqrt(0.5*0.5/15))
= 0.816 = 0.82
Could it possibly be that
Y=69
Z=59
Therefore 69+59=128
Answer:
x=32
Step-by-step explanation:
I believe the angles are supplementary so you could use the equation:
2x+8 + 3x+22 = 180
5x+30=180
5x=160
x=32
4(x+x+7) x=1
4(1+1+7)
4+4+28 = 36
(-2x+7-4) x=2
-2(2)+7-4
-4+7-4 = -1
