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garik1379 [7]
3 years ago
5

Find the product and write in lowest term. Give me an integer or simplified fraction. 7/54×27/35

Mathematics
1 answer:
wariber [46]3 years ago
4 0

The product of given expression is \frac{1}{15} \text{ or } 0.0667

<em><u>Solution:</u></em>

Given that we have to find the product and write it in lowest term

Given expression is:

\frac{7}{54} \times \frac{27}{35}

Let us first find common factors of numerator and denominator so, thet simplification becomes easier

54 = 27 \times 3\\\\35 = 7 \times 5

Substituting these in above expression, we get

\rightarrow \frac{7}{27 \times 3} \times \frac{27}{7 \times 5}

Cancel the common factors in numerator and denominator

\rightarrow \frac{1}{15}

In decimal form, we get

\rightarrow \frac{1}{15} = 0.0667

Thus the product of given expression is \frac{1}{15} \text{ or } 0.0667

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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

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

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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

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

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

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

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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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