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
3.3
-1/8 - 55/8 = -56/8 = -7
3.4
(-4x7)/10 = -28/10 = -14/5
3.5
-19/2 ÷ 38/7 <=> -19/2 x 7/38 and simplify = -7/4
Answer:
2 5/14
Step-by-step explanation:
Answer:
36.58
Step-by-step explanation:
38.3-1.72=36.58