A circle with circumference \blue{6}6start color #6495ed, 6, end color #6495ed has an arc with a 60^\circ60
2 answers:
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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
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(A/B) is the probability that the person being diagnosticated, given that she has cancer. So
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
What is the probability that the person actually does have cancer?
There is a 12.13% probability that the person actually does have cancer.
Two numbers that add up to -19 and multiply to 48 are -16 and -3:
So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.
x+4=0
x = -4
x-4=0
x = 4
x^2-3=0
x^2 = 3
x = +/- √3
So, the solutions are -4, -√3, √3, and 4.
So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.
x+4=0
x = -4
x-4=0
x = 4
x^2-3=0
x^2 = 3
x = +/- √3
So, the solutions are -4, -√3, √3, and 4.