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sweet-ann [11.9K]
3 years ago
10

Given: Point K is between Points J and L. KL=9x+2, KJ=5x–8 and JL = 12x. Find KL.

Mathematics
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

gfgyfhgj

Step-by-step explanation:

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According to a​ survey, 65​% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed
monitta

Answer:

a) P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421

b) P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714

P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502

P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319

And adding these values we got:

P(36 \leq X \leq 38)= 0.1535

c) We can find the expected value given by:

E(X) = np =50*0.65 = 32.5

And the standard deviation would be:

\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

\mu -2\sigma = 32.5- 2*3.373 = 25.75

And then we can consider a value of 18 as unusual lower for this case.

Step-by-step explanation:

Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

X \sim Binom(n=50, p=0.65)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

We want this probability:

P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421

Part b

We want this probability:

P(36 \leq X \leq 38)

We can find the individual probabilities:

P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714

P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502

P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319

And adding these values we got:

P(36 \leq X \leq 38)= 0.1535

Part c

We can find the expected value given by:

E(X) = np =50*0.65 = 32.5

And the standard deviation would be:

\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

\mu -2\sigma = 32.5- 2*3.373 = 25.75

And then we can consider a value of 18 as unusual lower for this case.

6 0
3 years ago
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