Abakery wants to determine how many trays of doughnuts it should prepare each day. demand is normal with a mean of 5 trays and standard deviation of 1 tray. if the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

4 trays should he prepared, if the owner wants a service level of at least 95%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5

Standard Deviation, σ = 1

We are given that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula:

P(X > x) = 0.95

We have to find the value of x such that the probability is 0.95

P(X > x)  

Calculation the value from standard normal z table, we have,  

Hence,  4 trays should he prepared, if the owner wants a service level of at least 95%.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet