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yanalaym [24]
3 years ago
7

**40 POINTS** PLEASE HELP, EXPLAIN WELL IN DETAIL PLEASE DON'T COPY OTHER ANSWERS OFF THE WEB AS NONE OF THEM HAVE MADE SENSE

Mathematics
2 answers:
cestrela7 [59]3 years ago
7 0
Wilson and Alexis can both by correct if the the function is a parabola. It can have zeroes at points x=-1, and x=1. Thus, the average rate of change between the two points would be zero because the y values do not change as they are both zero. In this way, Wilson is correct. Alexis can be correct because the parabola can have it's highest point at its vertex between the two points. It would count as a turning point because the function would increase and then decrease again. 


Hope this helps :)
amm18123 years ago
7 0
If the function is a parabola, with vertex on x= 0, then the average rate of change of the function on {x: -1<x<1} is 0, because the graph is symmetric. There is also a turning point at the vertex.

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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

4 0
3 years ago
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