The coffee and soup machine at the local subway station is supposed to fill cups with 6 ounces of soup. Nine cups of soup are bought with results of a mean of 5.5 ounces and a standard deviation of 0.18 ounces. How large a sample of soups would we need to be 95 percent confident that the sample mean is within 0.03 ounces of the population mean

So the answer for this case would be n=139 rounded up to the nearest integer

Step-by-step explanation:

For this case we have the following info given:

the sample mean

the standard deviation

The margin of error is given by this formula:

   (a)

And on this case we have that ME =0.03 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

  (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution and if we look in the normal standard distirbution we got , replacing into formula (b) we got:

So the answer for this case would be n=139 rounded up to the nearest integer

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement