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Valentin
22 days ago
13

Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A1 be the ev

ent that the receiver functions properly throughout the warranty period, A2 be the event that the speakers function properly throughout the warranty period, and A3 be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A1) 5 .95, P(A2) 5 .98, and P(A3) 5 .80.(a) What is the probability that all three components function properly throughout the warranty period?(b) What is the probability that at least one component needs service during the warranty period?
Mathematics
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Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Svet_ta [12734]

Response:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Detailed explanation:

In Problem 8-4, the computer time-sharing system experiences teleport inquiries at an average rate of 0.1 per millisecond. We are tasked with determining the probabilities of the inquiries over a specific period of 50 milliseconds:

Given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Applying the Poisson process, we find that

(a) at most 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) exactly 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) more than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) exactly 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) within the range of 10 to 15, inclusive

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Thus, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
2 months ago
On the average, 1 1/3 bushels of seed are needed to plant 1 acre of wheat. How many bushels of seed would be required to plant 3
lawyer [12517]

Answer: You would need 40 bushels of seed to plant 30 acres of wheat.

Step-by-step explanation:

<pTypically, 1 1/3 bushels of seed are necessary to cultivate 1 acre of wheat. In improper fraction form, 1 1/3 bushels becomes 4/3 bushels.

Let x denote the bushels of seed required for 30 acres. Thus, we have

4/3 bushels = 1 acre

x = 30 acres

Cross multiplying gives us

x × 1 = 4/3 × 30

x = 40 bushels of seed

6 0
3 months ago
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