Certain virus infects one in every 300 people. a test used to detect the virus in a person is positive 80% of the time if the person has the virus and 8% of the time if the person does not have the virus. (this 8% result is called a false positive.) let a be the event "the person is infected" and b be the event "the person tests positive". hint: make a tree diagram a) find the probability that a person has the virus given that they have tested positive, i. e. find p(a|b). round your answer to the nearest tenth of a percent and do not include a percent sign. p(a|b)= % b) find the probability that a person does not have the virus given that they test negative, i. e. find p(a'|b'). round your answer to the nearest tenth of a percent and do not include a percent sign. p(a'|b') =

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

Step-by-step explanation:

a) The probability of B: "the person tests positive" is the sum of:

- The probability that a person has the virus and the test is positive.

- The probability that a person don't have the virus but the test is positive.

Then, the probability of B: "the person tests positive" is

We have

Then

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

b) The probability P(A'|B') is

We have

Then

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

366

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2