Answer:
The probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
Step-by-step explanation:
We are given that . At Johnson University, the mean time is 5 hrs with a standard deviation of 1.2 hrs.
Mean = 
Standard deviation = 
We are supposed to find the probability that the average time 100 random students on campus will spend more than 5 hours on the internet i.e. P(X>5)


Z=0
P(X>5)=1-P(X<5)=1-P(Z<0)=1-0.5=0.5
Hence the probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
Answer:
1(left) goes with 1
2(left) goes with 3
3(left) goes with 2
4(left) goes with 4
Step-by-step explanation:
2(5+6) = 22
3(5+3) = 24
2(4+5) = 18
2(3+2) = 10
10 + 12 = 22
8+ 10 = 18
15 + 9 = 24
6 + 4 = 10
1(left) goes with 1
2(left) goes with 3
3(left) goes with 2
4(left) goes with 4
#13
Letter D
We don't know if 5 = 6 or
6 = 8
#14
You need slope and y- intercept. It is a negative slope because the line is going down from left to right. Count rise over run.
M = - 4/6 = -2/3
If the line continued to the y axis it would cross at 10.
y = -2/3x + 10
Answer:
the answerrrrrrrrrrr is 7x^2-9x+1