The daily revenue at a university snack bar has been recorded for the past five years. records indicate that the mean daily revenue is $2700 and the standard deviation is $400. suppose that 100 days are randomly selected. what is the probability that the average daily revenue of the sample is higher than $2600?

0.9938

Step-by-step explanation:

Let x be the random variable that represents the daily revenue at a university snack bar.

As per given , we have

, and n= 100

Using formula ,

z-score for x= 2600

The probability that the average daily revenue of the sample is higher than $2600 :

  [P(Z>-z)=P(Z<z)]

Therefore, the probability that the average daily revenue of the sample is higher than $2600 = 0.9938

nada

step-by-step explanation:

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.