g ) If it is raining, a home security system detects an intruder with probability 0.70. If it is NOT raining, the probability becomes 0.92. The probability of rain on any 2 given day is 0.25. To test the system on a randomly chosen day, the system technician pretends to be an intruder. Given that the technician will NOT be detected, what is the probability that it is NOT raining

0.4444 = 44.44% probability that it is NOT raining

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

In which

P(B|A) is the probability of event B happening, given that A happened.

is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Technician not detected.

Event B: Not raining.

Probability the technician is not detected:

0.3 of 0.25(raining).

0.08 of 0.75(not raining). So

Probability the technician is not detected and it is not raining:

0.08 of 0.75. So

Given that the technician will NOT be detected, what is the probability that it is NOT raining?

0.4444 = 44.44% probability that it is NOT raining

answer:

c) the x-coordinate of the center of the circle increases by 3

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.