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3 years ago

Do the side lengths of 8, 15, and 20 form a triangle? Explain. Can you explain how you did this

Mathematics

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Step-by-step explanation:

8+15=23

will not work it is to long

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Kathy is standing directly between two tall buildings which are 100 feet apart. Her eye level is 5 feet above the ground. Kathy

Zinaida [17]

Angle of taller building Tan 45° = opposite (height of the building)/ adjacent 50 feet (distance between Kathy and the taller building

Tan 45° = 1

Height of taller building from eye level = multiplying adjacent 50 feet * Angle (Tan 45°) 1 = 50 feet opposite

Angle of smaller building Tan 38° = 0.781

Height of smaller building from eye level = Angle * adjacent 50 feet = 39.06 feet

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Taller building height is 50 + 5

Total height of taller building is 55 feet.

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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 $P(X \geq 5)$

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

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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3 years ago