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mote1985 [20]
3 years ago
11

What is the mathematical shape made by each net

Mathematics
1 answer:
Bumek [7]3 years ago
5 0
A) cylinder
b) triangular pyramid
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hsuaiabshsisnsbkjwnsjns

Step-by-step explanation:

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At a concert, 20% of the audience members were teenagers. if the number of teenagers at a concert was 360, what was the total nu
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The total number of audience members is 1800.

Step-by-step explanation:

In order to find this, we need to divide the number of teenagers by the percentage share.

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If 6 times a certain number is added to 8, the result is 32.. . Which of the following equations could be used to solve the prob
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9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat
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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.

4 0
3 years ago
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