It is my opinion that it is C) but i am not sure, you might be correct though...
Answer:
x times (x+1).
If it's wrong, then I'm an idiot.
Step-by-step explanation:
Answer:
omg very. bored and ~stressed~
Answer:
The value of the test statistic is 
Step-by-step explanation:
Our test statistic is:

In which X is the sample mean,
is the expected value,
is the standard deviation and n is the size of the sample.
The level of ozone normally found is 7.5 parts/million (ppm).
This means that 
The mean of 24 samples is 7.8 ppm with a standard deviation of 0.7.
This means that 
Test Statistic:



The value of the test statistic is 
Answer:
easy-peasy
Step-by-step explanation:
Perimeter:
P=2(a+b)
P=2(50+70)
P=2(120
P=240
Area:
A=50*70
A=3,500
I really hope it helped:)