Acompany has 2 photocopy machines. the probability that machine a will break down on a given day is 2%. the probability that machine b will break down is 2.5%. what is the probability that exactly one machine will break down on a given day?

 4.44 %

Step-by-step explanation:

Let A is the event that machine a will break down,

A' is the event that machine a will not break down,

B is the event that machine b will break down,

And, B' is the event that machine will not break down.

According to the question,

P(A) = 2%

⇒ P(A) = 0.02

P(A') = 1 - 0.02 = 0.98

Similarly, P(B) = 0.025

⇒ P(B') = 1 - 0.025 = 0.975

Thus, the probability that exactly one machine will break down on a given day = P(A) × P(B') + P(A') × P(B)

= 0.02 × 0.975 + 0.98 × 0.025

= 0.044

= 4.44 %

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

52 \43 i don't know this i tried