Answer:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Step-by-step explanation:
Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
We find mean = 11.015
Sample std deviation = 3.157
a) 
(Right tailed test)
Mean difference /std error = test statistic

p value =0.174
Since p >0.01, our alpha, fail to reject H0
Conclusion:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
The answer should be 664 in squared
Answer:
10
Step-by-step explanation:
Your answer should be 12.57
You subtract the 11 from both sides, then you get -3x = 33. Next, you divide -3 from both sides, and you get x= -11