Answer:
B only
Explanation:
Squeeze-type resistance spot welding (STRSW)is a type of electric resistance welding that brings about the weld on interfacing sheet metal pieces through which heat generated from electric resistance bring about fusion and welding of the two pieces together
Therefore, it is not meant for opening but joints but it can be used for making replacement spot welds adjacent to the original spot weld due to the smaller heat affected zone (HAZ) created by the STRSW process.
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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