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.
Answer:
y=42
Step-by-step explanation:
y+48+90=180
y=180-138
y=42
1 1/2 servings are in 3/4 bc 2/4= 1/2 and theres 1/2 a cookie left and if u take half of that it gets to 3/4
Since, population of species A is represented by : 
Let us find the population of species A, at the end of week 1:
i.e., x = 1
i.e., 
i.e., 
i.e., 
Also, since population of species B is represented by : 
Let us find the population of species B, at the end of week 1:
i.e., x = 1
i.e., 
i.e., 
i.e.,
Thus, at the end of 1 week, species A and species B will have the same population.
Hence, option D is correct.
<span>the value of the digit in the tenths place is 5/10 or 0.5
</span>
the value of the digit in the thousandth place
is 5/1000 = 0.005
0.5 = 100 x 0.005
Therefore, the value of the digit in the tenths place is 100 times as much as the value of the digit in the thousandth place
