Mathematics, 26.12.2019 19:31 reinasuarez964

There are two traffic lights on the route used by a certain individual to go from home to work. let e denote the event that the individual must stop at the first light, and define the event f in a similar manner for the second light. suppose that p(e) = .4, p(f) = .2 and p(e intersect f) = .15.
(a) what is the probability that the individual must stop at at least one light; that is, what is the probability of the event p(e union f)?

(b) what is the probability that the individual needn't stop at either light?

(c) what is the probability that the individual must stop at exactly one of the two lights?
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(d) what is the probability that the individual must stop just at the first light? (hint: how is the probability of this event related to p(e) and p(e intersect f)? a venn diagram might .)
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Answer from: hellicuh

a) There is a 45% probability that the individual must stop at at least one light.

b) 55% probability that the individual needn't stop at either light.

c) 30% probability that the individual must stop at exactly one of the two lights.

d) 25% probability that the individual must stop just at the first light.

Step-by-step explanation:

The problem says that:

P(E) is the probability that a person has to stop at the first light.

P(F) is the probability that a person has to stop at the second light.

We have that:

$P(E) = P(e) + P(E \cap F)$

In which P(e) is the probability that a person has to stop at the first light but not the second and $P(E \cap F)$ is the probability that a person has to stop at both these lights

By the same logic, we have that:

$P(F) = P(f) + P(E \cap F)$

Lets find each value:

$P(E \cap F) = 0.15, P(F) = 0.2, P(E) = 0.4$

$P(F) = P(f) + P(E \cap F)$

0.2 = P(f) + 0.15

P(f) = 0.05

---------

$P(E) = P(e) + P(E \cap F)$

0.4 = P(e) + 0.15

P(e) = 0.25

(a) What is the probability that the individual must stop at at least one light; that is, what is the probability of the event P(E union F)?

This is

$P = P(e) + P(f) + P(E \cap F) = 0.05 + 0.25 + 0.15 = 0.45$

There is a 45% probability that the individual must stop at at least one light.

(b) What is the probability that the individual needn't stop at either light?

Either a person has to stop in at least one light, or the person does not have to stop at any.

The sum of these probabilities is decimal 1. So this probability is 1 subtracted by the probability found in a).

So 1-0.45 = 0.55 = 55% probability that the individual needn't stop at either light.

This is P(e) + P(f) = 0.25 + 0.05 = 0.30 = 30% probability that the individual must stop at exactly one of the two lights.

(d) What is the probability that the individual must stop just at the first light?

This is P(e) = 0.25.

So there is a 25% probability that the individual must stop just at the first light.

Answer from: Quest

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function $y=f(x)$ the average rate of change from point $a(a,f(a))$ to point $b(b,f(b))$ can be calculated as

$\text{average rate of change}=\dfrac{f(b)-f(a)}{b-a}.$

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,

$\text{average rate of change}=\dfrac{2-8}{10-2}=\dfrac{-6}{8}=-\dfrac{3}{4}.$

The graph below shows the number of houses sold over x days. what is the average rate of change from

Answer from: Quest

The answer is 30 minutes to an hour

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