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Rzqust [24]
3 years ago
8

PLEASE HELP ITS DUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
tia_tia [17]3 years ago
3 0

Answer:

Angle 3 = 103

Angle 4 = 77

Step-by-step explanation:

A)

The sum of interior angles in a triangle is equal to 180 so to find angle 3:

180 - (52+25) = 103

B)

Angle 3 and angle 4 are supplementary so the measure of angle 4:

180 - 103 = 77

C)

Angle 1 and angle 2 and angle 3 are the interior angles of the triangle and their sum is 180 so is the sum of angle 3 and angle 4

D)

<u><em>The sum of two interior angles in a triangle is equal to an exterior angle that is supplementary to the third interior angle so the sum of angle 1 and angle 2 is equal to angle 4</em></u>

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Javier has a job installing windows. He earns a rate of $75 per day plus $8 per window installed. He also receives a stipend of
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4 0
3 years ago
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat
Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 2000, p = 1-0.995 = 0.005

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

P(X < 5) + P(X \geq 5) = 1

We want P(X \geq 5)

So

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044

P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445

P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235

P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480

P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765

P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

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Answer:

first option

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