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3 years ago

If U = { 1 , 2 , 3 , 4 , 5 ,6 , 7 , 8 , 9 }. A={1 , 2 , 3 , 4 } , B= {2 , 4 , 6 , 8 } .Find

Mathematics

1 answer:

Zina [86]3 years ago

4 0

Answer:

(A - B)' = {2, 4, 5, 6, 7, 8, 9}

(A n B) = {2, 4}

Step-by-step explanation:

Given the set :

U = { 1 , 2 , 3 , 4 , 5 ,6 , 7 , 8 , 9 }

A={1 , 2 , 3 , 4 } , B= {2 , 4 , 6 , 8 }

(A - B) = elements of A which are not in B

(A - B) = {1, 3}

(A - B)' = the complement of A - B ; element in the universal set which are not in A - B

(A - B)' = {2, 4, 5, 6, 7, 8, 9}

(A n B) = elements in both set A and set B

(A n B) = {2, 4}

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QUESTION 1

We want to find the digit that should fill the blank space to make

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QUESTION 2

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QUESTION 3.

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QUESTION 4.

We want to find the greatest common factor of

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(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

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(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

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$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let X = results of a certain blood test.

It is provided that the random variable X follows a Normal distribution with parameters $\mu = 60$ and $s = 18$ .

The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.

The z-scores follows a Standard normal distribution, N (0, 1).

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Compute the probability that the results are more than 45 as follows:

$P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z$

The percentage of results more than 45 is: $0.7967\times100=79.67\%$

Thus, the percentage of results more than 45 is 79.67%.

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Compute the probability that the results are less than 85 as follows:

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The percentage of results less than 85 is: $0.9177\times100=91.77\%$

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

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The percentage of results are between 75 and 90 is: $0.1558\times100=15.58\%$

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

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Then the probability that the results outside the range 20 to 100 is: $1-0.9736=0.0264$ .

The percentage of results outside the range 20 to 100 is: $0.0264\times100=2.64\%$

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

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