Please help!!

Mathematics

1 answer:

jenyasd209 [6]3 years ago

8 0

Answer:

x = - 3

y = - 2

Step-by-step explanation:

Start with the second equation. Switch

3y = 2x Divide by 3

3y/3 = 2x/3

y = 2x/3

Now go to the first equation. Put the value for y in the first equation.

x - 3(2x/3) = 3 Cancel the 3s on the left.

x - 2x = 3 Combine

-x = 3 Multiply by -1

-1(-x) = 3(-1)

x = - 3

Now go back to the original equation I have called the second equation

2x = 3y

2(-3) = 3y Put in - 3 for x

- 6 = 3y Divide by 3

-6 / 3 = y

y = - 2

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Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question 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 the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$