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2 years ago

14x25x4 breakdown using subraction distributive properties

Mathematics

2 answers:

Novosadov [1.4K]2 years ago

6 0

(10)+(4) x(20)+(5) x(4)
=1,400

WINSTONCH [101]2 years ago

3 0

5x5x2x7x2x2 is the one i got

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Help pls!!!

expeople1 [14]

That is a good definition

5 0

2 years ago

A football is made of 12 black panels and 20 white panels. There is a joint whenever two panels meet along the edge how many joi

oee [108]

Answer:

12+20=32

Step-by-step explanation:

None Hope this helps

8 0

2 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 $P(X \geq 5)$

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

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

2 years ago

Deondra has $12 to spend on a mixture of green and red grapes. What equation can she use to graph a line showing the different a

Sonbull [250]

Answer:

It depends on the cost of grapes.

Step-by-step explanation:

Half of 12 is 6, so if each pack of grapes cost 6 or less, she can buy a pack of green and red grapes.

6 0

3 years ago

Evaluate 5.45×1.5/0.025 leave your answer in standard form

malfutka [58]

= 5.45 × 1.5/0.025
= 5.45 x 60
= 327

Answer= 327

Hope this helps! :)

3 0

2 years ago