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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Y = a (b)^t/p
y is total money
a is original amount
b is growth / decay factor
t is time
p is the frequency of every growth or decay
15131.76 = 11613 x 1.08^x
15131.76 / 11613 = 1.08^x
1.303… = 1.08^x
log1.303…. = xlog1.08
x = 3.43902165741 years
Answer:
q1q1 ⇒ 01
Explanation:
The outputs of a positive edge triggered register will match the inputs after a rising clock edge.
q1q1 ⇒ 01 . . . . matching d1d0 = 01
Answer:
<h2>it's damage kindly be careful</h2>
