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Mila [183]
2 years ago
10

consider there are three machines. All of the machines can produce 1000 pins at a time. Therate of producing a faulty pin from M

achine 1 be 10%, from Machine 2 be 20% andfrom Machine3 be 5%. What is the probability that a produced pin will be faulty and it will be from the firstmachin?
Mathematics
1 answer:
blondinia [14]2 years ago
6 0

Answer:

0.03333

Step-by-step explanation:

Given that there are three machines. All of the machines can produce 1000 pins at a time.

Chance for any machine is thus equaly likely = 1/3

The rate of producing a faulty pin from Machine 1 be 10%, from Machine 2 be 20% andfrom Machine3 be 5%.

Machine             I                     II              III                  total

Faulty              0.10                0.20        0.05

Prob                0.3333           0.3333    0.3334             1

Faulty*prob    0.03333         0.06666  0.16665         0.26664

Thus probability  that a produced pin will be faulty and it will be from the firstmachine

= 0.03333

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Simplify the expression to a + bi form:<br> (-2 - 6i)?
mixer [17]

The expression (-2 - 6i)-(-2-4i) to a + bi form is 0 - 2i.

Complete question.

Simplify the expression to a + bi form:

(-2 - 6i)-(-2-4i)

Square root of any negative number are expressed as a complex number. For example i = √-1

Complex numbers are generally written in the format z = x+iy

Given the expression (-2 - 6i)-(-2-4i)), in expansion:

(-2 - 6i)-(-2-4i)

= -2 - 6i + 2+4i

Collect the like terms

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Therefore the expression (-2 - 6i)-(-2-4i) to a + bi form is 0 - 2i.

Learn more on complex number here: brainly.com/question/12375854

6 0
2 years ago
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2 years ago
The probability that a lab specimen contains high levels of contamination is 0.10. A group of 4 independent samples are checked.
galben [10]

Answer:

a) 0.6561

b) 0.2916

c) 0.3439    

Step-by-step explanation:

We are given the following information:

Let us treat high level of contamination as our success.

p = P(High level of contamination) = P(success) = 0.10

n = 4

The, by binomial distribution:

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}\\\text{where x is the number of success}

a) P(No high level of contamination)

We put x = 0, in the formula.

P(X=0) = \binom{4}{0}.(0.10)^0.(1-0.10)^{4} =0.6561

Probability that no lab specimen contain high level of contamination is 0.6561

b) P(Exactly one high level of contamination)

We put x = 1, in the formula.

P(X=1) = \binom{4}{1}.(0.10)^1.(1-0.10)^{3} =0.2916

Probability that no lab specimen contain high level of contamination is 0.6561

c) P(At least one contains high level of contamination)

p(x \geq 1) = 1 - p( x = 0) = 1 - 0.6561 = 0.3439

Probability that at least 1 lab specimen contain high level of contamination is 0.3439

8 0
3 years ago
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astra-53 [7]

Answer:

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Step-by-step explanation:

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