Answer:
A) 0.005
B) 0.001
C)0.0495
Step-by-step explanation:
Let A be the event that an aircraft is present and let B be the event the radar registers its presence.. Thus;
P(A) = Probability that an aircraft is present
P(A') = Probability that an aircraft is not present
P(B) = Probability that the radar generates an alarm
P(B') = Probability that the radar doesn't generate an alarm
Thus from what we are given, we have;
P(A) = 0.05
P(A') = 0.95
P(B) = 0.99
P(B') = 0.01
P(B|A') = 0.1
A) Probability of a false alarm will be;
P(A' ∩ B) = P(A) × P(B|A')
P(A' ∩ B) = 0.05 × 0.1 = 0.005
B) probability of missed detection is;
0.1 × (1 - 0.99) = 0.001
C) probability that an aircraft is present given that the radar registers a presence will be;
P(A ∩ B) = P(A) × P(B)
P(A ∩ B) = 0.05 × 0.99
P(A ∩ B) = 0.0495
Answer:9090
it is simple you just need to multiply by 2
Answer:
√2 < 1.8
Step-by-step explanation:
Square of 2 is 1.41421356237309 and that is less than 1.8 continued.
Answer:
its A
Step-by-step explanation:
