Answer:
The margin of error for this estimate is of 14.79 yards per game.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.093
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.
This means that 
What is the margin of error for this estimate?



The margin of error for this estimate is of 14.79 yards per game.
Answer:
80.95
Step-by-step explanation:
Percentage increase =
*100
=> (38-21)* 100/21 = 80.9523%
The answer round it to the nearest 100th = 80.95%
Answer: X- 0,1,2,3,4 and Y-0,6,12,18,24
Step-by-step explanation: i had to take this test ): luck
Suppose:
C=cost
c=cost per kilo-watt hour
w=number of watts
h=number of hours
thus
C=kwhc
where:
k=constant of proportionality:
k=C/(whc)
but
C=15 when w=6000, h=10/60=1/6 hours, c=7.5 cents per kw/h
thus
k=15/(6000*1/6*7.5)
k=0.002
thus:
C=0.002(whc)
hence:
when
h=70*2=140=140/6=70/3 hours the cost will be:
C=0.002(6000*70/3*7.5)
C=2100 cents
C=$21