The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays
<h3>How to utilize z-score statistics?</h3>
We are given;
Mean; μ = 15
Standard Deviation; σ = 5
We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.
Formula for z-score is;
z = (x' - μ)/σ
We want to find the value of x such that the probability is 0.95;
P(X > x) = P(z > (x - 15)/5) = 0.95
⇒ 1 - P(z ≤ (x - 15)/5) = 0.95
Thus;
P(z ≤ (x - 15)/5) = 1 - 0.95
P(z ≤ (x - 15)/5) = 0.05
The value of z from the z-table of 0.05 is -1.645
Thus;
(x - 15)/5 = -1.645
x ≈ 7
Complete Question is;
A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. 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.
Read more about Z-score at; brainly.com/question/25638875
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John Smeatom, U.K. 18th century, was the first self-proclaimed, civil engineer in the 18th century and IS considered “the father of modern, civil engineering”.
hoped this helped! :)
Answer:
Q = 424523.22 kw
Explanation:
k = 48.9 W/m - K
c = 0.115 KJ/kg- K
T_∞ = 35 degree celcius
velocity of air stream = 15 m/s
D = 40 cm
L = 200 cm
mass flow rate
solving for h
h = 675.6 kw/m^2K
Q = 675.6*2.513*(285-35)
Q = 424523.22 kw
Answer:
cant see the moon sorry dude
