Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a probability 0.01 of incorrectly identifying an individual as uninfected (False Negative). Assume furthermore that the probability of an individual being infected is 0.000001 (1 in 1 million). If a person tests positive for this disease, what is the probability of actually having the disease

0.00002 = 0.002% probability of actually having the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

In which

P(B|A) is the probability of event B happening, given that A happened.

is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Having the disease

Probability of having a positive test:

0.05 of 1 - 0.000001(false positive)

0.99 of 0.000001 positive. So

Probability of a positive test and having the disease:

0.99 of 0.000001. So

What is the probability of actually having the disease

0.00002 = 0.002% probability of actually having the disease

c: lkllmn.

hope it's you

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.