If an aircraft is present in a certain area, a radar correctly registers its presence with probability 0.99. if it is not present, the radar falsely registers an aircraft presence with probability 0.10. we assume that an aircraft is present with probability 0.05. what is the probability of false alarm (a false indication of aircraft presence), and the probability of missed detection (notng registers, even though an aircraft is present)? what is the probability that an aircraft is present given that the radar registers a presence?

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

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

v = 1/3 a^2h

step-by-step explanation:

the volume is 1/3 a^2h.