A statistics class weighed 20 bags of grapes purchased from the store. The bags are advertised to contain 16 ounces, on average. The class calculated the 90% confidence interval for the true mean weight of bags of grapes from this store to be (15.875, 16.595) ounces. What is the sample mean weight of grapes, and what is the margin of error? O The sample mean weight is 15.875 ounces, and the margin of error is 16.595 ounces.
O The sample mean weight is 16.235 ounces, and the margin of error is 0.360 ounces.
O The sample mean weight is 16.235 ounces, and the margin of error is 0.720 ounces.
O The sample mean weight is 16 ounces, and the margin of error is 0.720 ounces.

The sample mean weight is 16.235 ounces, and the margin of error is 0.360 ounces

Step-by-step explanation:

To find the sample mean, we can find the mean of the confidence interval.

(15.875 + 16.595)/2 = 16.235

To find the margin of error, that is the difference between the mean and one of the edges of the confidence interval. 16.595 - 16.235 = 0.36

The sample mean weight is 16.235 ounces, and the margin of error is 0.360 ounces

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.