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Aleksandr [31]
3 years ago
6

The weight of an elephant is 1000 times the weight of a cat. If the elephant weighs 14,000 pounds how many pounds does the cat w

eigh
Mathematics
2 answers:
ohaa [14]3 years ago
5 0
To solve this problem, all you have to do is 14,000/1,000, which is 14, because the elephant weighs 14,000 lbs and the cat is 1/1000 of that weight. The cat weighs 14 pounds.
Ad libitum [116K]3 years ago
4 0
Weight of the elephant= 1,000x
x= weight of the cat
14,000=1,000x
x=14 pounds is what the cat weighs. 
Check: 14*1,000= 14,000
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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.

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

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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

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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

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

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=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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