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2 years ago

7

Determine if the statement is always, sometimes or never true. The measure of an exterior angle at the vertex of a polygon equal

s the measure of the adjacent interior angle. always

1 answer:

jeka942 years ago

6 0

Answer:

I believe it is never

Step-by-step explanation:

Sorry if I'm wrong ...

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2 years ago

If 20% of the people in a community use the emergency room at a hospital in one year, find

Pie

Answer:

a) 87.91% probability that at most three used the emergency room

b) 20.13% probability that exactly three used the emergency room.

c) 3.28% probability that at least five used the emergency room

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use the emergency room, or they do not. The probability of a person using the emergency room is independent of any other person. 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.

Sample of 10 people:

This means that $n = 10$

20% of the people in a community use the emergency room at a hospital in one year

This means that $p = 0.2$

a) At most three used the emergency room

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$

87.91% probability that at most three used the emergency room

b) Exactly three used the emergency room

$P(X = 3) = C_{10,3}.(0.2)^{3}.(0.8)^{7} = 0.2013$

20.13% probability that exactly three used the emergency room.

c) At least five used the emergency room

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

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

From 0 to 3, we already have in a).

$P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1074 + 0.2684 + 0.3020 + 0.2013 + 0.0881 = 0.9672$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9672 = 0.0328$

3.28% probability that at least five used the emergency room

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3 years ago