Probability of survival of one disk, p=1-0.05=0.95
If there are n drives, and failures are independent (e.g. mechanical wear and tear, and not a lightning strike or a current spike), then
Probability of survival = p^n
(a) for two disks, P(2) = 0.95^2=0.9025
(b)for three disks P(3) = 0.95^3 = 0.8574
Answer:
1 and 3 I think
Step-by-step explanation:
Answer:
z = -12
Step-by-step explanation:
z + 4 = -8
so, z = -8 - 4
z = -12
I think this is the answer, hope it helps.
Answer:
x=-3/4 X=3
Step-by-step explanation:
Trust me :)
