the answer ot this question is 3/10
Ummmm no.. that's not correct at all.
Each machine is an independent event. so the correct answer is 0.001..
.1 * .1 * .1 = 0.001
or 0.1%
The probability that an adversary will exploit a weakness in your operation, tempered by the impact to your mission, is defined as the:A. Risk
vulnerability
Vulnerability
C(7,2)*(.9)^5*(.1)^2, or about .124 = 12.4% For the desired outcome, considering the seven patients, you need: (Survive,Survive,Survive,Survive,Survive,Die,Die) (Survive,Survive,Survive,Survive,Die,Survive,Die) (Survive,Survive,Survive,Die,Survive,Survive,Die) . . . (Die,Die,Survive,Survive,Survive,Survive,Survive) There are C(7,2) [the number of combinations of 7 things taken 2 at a time] = 21 possible desired outcomes. The probability of each of these outcomes is (.9)*(.9)*(.9)*(.9)*(.9)*(.1)*(.1). Multiplying 21 by (.9)*(.9)*(.9)*(.9)*(.9)*(.1)*(.1) yields the answer.
yes,the histogram equalization operation is idempotent
Probability
The probability that an adversary will exploit a weakness in your operation, tempered by the impact to your mission, is defined as the:A. Risk
vulnerability
Operation Rolling Thunder was the continuous air campaign begun by the US in March 1965.
ORM should only be used when the individual has time to plan an operation or evolution.?
operation measures
12 hrs
Vulnerability
"One go" is a slightly informal phrase meaning "a single action or continuous operation."
oh
It was a continuous large-scale patrolling operation to prevent the South from importing war material.
forcible entry