The daily revenues of a cafe near the university are approximately normally distributed. the owner recently collected a random sample of 30 daily revenues and found a 95% confidence interval for the average daily revenues in his shop is (966.738, 1033.262). he is unsatisfied by the precision of this confidence interval, however, and wishes to reduce the margin of error by a factor of 3, while retaining the same level of confidence. what sample size do you suggest he use to obtain the desired margin of error? assume the sample standard deviation remains the same as the sample size changes. select one: a. 10b. 90c. 270d. 900

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)

answer: i dont know but if you try hard you will get it right : )

step-by-step explanation: