Suppose there is a test for esophageal cancer that is 95% accurate both on those that do and those that do not have esophageal cancer. if 0.7 percent of the population has esophageal cancer, compute the probability that a person has this cancer, given that his or her test results indicates so.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

there is nothing below so i cannot

step-by-step explanation: