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

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it