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Tcecarenko [31]
3 years ago
14

Cot? 2x + cos2x + sin’ 2x = csc? 2x Need help with 40

Mathematics
1 answer:
kkurt [141]3 years ago
4 0

Answer:

\rule{100mm}{0.1mm}

    \cot^2(2x)+\cos^2(2x)+\sin^2(2x)=\csc^2(2x)

\dfrac{\cos^2(2x)}{\sin^2(2x)} +[(\cos^2(2x)+\sin^2(2x)]=\csc^2(2x)

                               \dfrac{\cos^2(2x)}{\sin^2(2x)} +1=\csc^2(2x)

                    \dfrac{\cos^2(2x)+\sin^2(2x)}{\sin^2(2x)}=\csc^2(2x)

                                    \dfrac{1}{sin^2(2x)} =\csc^2(2x)

                                     \boxed{\csc^2(2x)=\csc^2(2x)} ✔

\rule{100mm}{0.1mm}

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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

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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

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

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b. Compute P(4≤X≤ 8).

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

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

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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

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

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

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