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3 years ago

Help plz help plz help plz help plz

Mathematics

2 answers:

Valentin [98]3 years ago

6 0

Answer:

25. x = 17°

26. x = 50°

Step-by-step explanation:

25. There is a straight line which means :

108 +4x = 180

solving the obtained equation

108+4x = 180

4x = 180 - 108

4x = 72

x = 72/4

x = 17°

26. Again we can see that the angles in which 101 and x + 29 exists is a straight line . which means :

101 + x + 29 = 180

101 + 29 + x = 180

130 + x = 180

x = 180 - 130

x = 50°

pogonyaev3 years ago

3 0

So the answer of question no. 25 is 18 degree.

The answer of question no 26 is 50 degree.

Look at the attached picture

hope it will help you

good luck on your assignment 

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Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backu

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Answer:

a) There is a 10.24% probability that both generators fail during a power outage.

b) There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

Step-by-step explanation:

For each emergency backup generator, there are only two possible outcomes. Either they work correctly, or they fail. So we use the binomial probability distribution to solve this problem.

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:

Assume that emergency backup generators fail 32% of the times when they are needed. So they work correctly 100-32 = 68% of the time. So $p = 0.68$

There are two generators, so $n = 2$

a. Find the probability that both generators fail during a power outage

This is $P(X = 0)$

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

$P(X = 0) = C_{2,0}.(0.68)^{0}.(0.32)^{2} = 0.1024$

There is a 10.24% probability that both generators fail during a power outage.

b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital? Assume the hospital needs both generators to fail less than 1% of the time when needed.

Either there are no working generators, or there is at least one working generator. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$