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You are working in a primary care office. Flu season is starting. For the sake of public health, it is critical to diagnose peop

Aleksandr-060686 [28]

Answer:

E. 0.11

Step-by-step explanation:

We have these following probabilities:

A 10% probability that a person has the flu.

A 90% probability that a person does not have the flu, just a cold.

If a person has the flu, a 99% probability of having a runny nose.

If a person just has a cold, a 90% probability of having a runny nose.

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have that:

What is the probability that a person has the flu, given that she has a runny nose?

P(B) is the probability that a person has the flu. So P(B) = 0.1.

P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.

P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So

$P(A) = 0.99*0.1 + 0.9*0.9 = 0.909$

What is the probability that this person has the flu?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.1*0.99}{0.909} = 0.1089 = 0.11$

The correct answer is:

E. 0.11

5 0

2 years ago

Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$