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kompoz [17]
3 years ago
10

List all the prime numbers between 80 and 100

Mathematics
1 answer:
oee [108]3 years ago
4 0

Answer:

81,83,85,87,89,91,93,95,97,99

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Help pls ill mark brainliest!!!!!!!!!!!!!!!!!
ziro4ka [17]

Any points that are less than 6. So 5, 4, 3, 2, etc.

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3 years ago
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How many cubes were used to make this figure?
Paha777 [63]

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I think its 30 I'm not 100% sure tho

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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.

4 0
3 years ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!’
ad-work [718]

Factoring this expression using the identity of a²-b² = (a+b)(a-b)

Given :-

25 - x²

<em>This can be broken down into:- </em>

... ( 5 )² - ( x )²

<em>In this case, a is 5 and b is x </em>

... ( 5 + x ) ( 5 - x ) Is the answer.

Hope it helps!!

7 0
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The table shows the daily time, in minutes, Maggie spent at swim practice and band practice over the last ten days.
jeka94

Answer:

I'm not sure but I think it is the second one.

Step-by-step explanation:

7 0
3 years ago
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