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Nonamiya [84]
3 years ago
14

PLEASE HELP AND IF ITS WORNG I WILL BE UPSET BECAUSE THIS QUESTION IS ABOUT WILL I DIE OR LIVE

Mathematics
2 answers:
Evgen [1.6K]3 years ago
8 0

Answer:

  • D. none of these

Step-by-step explanation:

<u>Proper equation would be:</u>

  • 1 - (5/19 + 3/19) = 11/19

This doesn't match any answer options, so the correct option is the last one

IrinaVladis [17]3 years ago
6 0

Answer:

B/2

Step-by-step explanation:

Jeez better not get this wrong... anyway, to start off the total is 19 books. Adding 5 then adding 3, She returned 8 books to the 19 she checked out. Uhhhh no clear answer.

NOT the first one cause 8 isn't the total. Don't think it would be the last one cause that's no fun. So it's either the second or the third one. The third one also makes no sense cause it only includes the returns from yesterday. I'm going with the second one

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Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo
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Answer:

a) P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335

b) P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452

c) P(X \geq 8) = 1-P(X

d) P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda=4  , Var(X)=\lambda=2, Sd(X)=2

a. Compute both P(X≤4) and P(X<4).

P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183

P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733

P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465

P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646

P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692

b. Compute P(4≤X≤ 8).

P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595

P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298

P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452

c. Compute P(8≤ X).

P(X \geq 8) = 1-P(X

P(X \geq 8) = 1-P(X

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

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Answer:

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Step-by-step explanation:

Your Welcome

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Step-by-step explanation:

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