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kykrilka [37]
1 year ago
12

Can you please help me out with a question

Mathematics
1 answer:
Finger [1]1 year ago
8 0

The standard equation of a circle is expressed as

(x - h)^2 + (y - k)^2 = r^2

where

h is the x coordinate of the center of the circle

k is the y coordinate of the center of the circle

r is the radius of the circle(the distance from the center of the circle to the circumference

From the graph,

h = - 1

y = 4

r = 5

By substituting these values into the equation, we have

(x - - 1)^2 + (y - 4)^2 = 5^2

(x + 1)^2 + (y - 4)^2 = 25

Thus, the equation of the circle is

(x + 1)^2 + (y - 4)^2 = 25

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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?

So

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

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

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

3 0
3 years ago
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