All bags entering a research facility are screened. ninety-seven percent of the bags that contain forbidden material trigger an alarm. fifteen percent of the bags that do not contain forbidden material also trigger the alarm. if 1 out of every 1,000bags entering the building contains forbidden material, what is the probabilitythat a bag that triggers the alarm will actually contain forbidden material? (a) 0.00097(b) 0.00640(c) 0.03000(d) 0.14550(e) 0.97000

(b) 0.00640

Step-by-step explanation:

Let

P(A|F) = The probability of bags that contain forbidden material trigger an alarm = 0.97 or 97%

P(A| not F)= The probability of bags that do not contain forbidden material also trigger the alarm= 0.15 or 15%

P(F) = the probability that a bag contain forbidden material = 0.001

P(not F) = the probability that a bag does not contain forbidden material = 0.999

Then P(A)= P(F) × P(A|F)  + P(not F)  × P(A| not F) = 0.001×0.97 + 0.999×0.15 = 0.15082

Then the probability that a bag that triggers the alarm will actually contain forbidden material =

= ≈ 0.0064

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)

mia khalifa

step-by-step explanation: