Telephone calls arrive at the Global Airline reservation office in Louisville according to a Poisson distribution with a mean of
1 answer:
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The weight of first shipment = 1 ton 115 pounds
As we know, 1 ton = 2000 pounds, so
So, the weight of first shipment = pounds
The weight of second shipment = 1723 pounds 7 ounces
The weight of third shipment = 957 pounds 11 ounces
So, adding all three weights =
So, all three weights = 4795 pounds 18 ounces
As we know, 16 ounces = 1 pound, so
all three weights = 4795 pound 1 pound 2 ounces
All three weights = 4796 pounds 2 ounces
4796 could be wriiten as 4000 + 796
As we know, 2000 pound = 1 ton so
All three weight = 2 tons 796 pounds 2 ounces
So, the total weight of three shipments = 2 tons 796 pounds 2 ounces : Answer
Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire
On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters
Answer:
Step-by-step explanation:
The step by step calculation is as shown in the attached file.
