A research company desires to know the mean consumption of meat per week among people over age 29. A sample of 2092 people over age 29 was drawn and the mean meet consumption was 2.9 pounds. Assume that the standard deviation is known to be 1.4 pounds. Construct a 95% confidence interval for the mean consumption of meat among people over age 29. Round your answer to one decimal place.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 2.9 - 0.1 = 2.8 pounds

The upper end of the interval is the sample mean added to M. So it is 2.9 + 0.1 = 3 pounds.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

tigers

step-by-step explanation:

cougars 12/10 (12-10 = 2)

bobcats 20/10 (20-10 =10)

bulldogs 8/5 (8-5 =3)

tigers 3/5 (3-5 =-2)

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1