The solution for this problem would be:
Given that there is 99.999%.
Let denote n as the network servers and p as the reliability of each server.
So the probability that the network uptime = 1 - (1 - p)^n
Therefore, (1-p) ^n = 0.00001
a. x= log(1-.99999)÷log(1-.97)= 3.2833 is the answer
1-(1-.97)^3= 0.99999 + 0.0001 = 1
b. x = log(1-.99999)÷log(1-.88) = 5.43 is the answer
1-(1-.88)^3= 0.99 + 0.0001 = approx 1
You first have to plug in your given numbers to the equation.
d= rt
d=(52)4.5
Then you solve
52*4.5 = 234
So, your distance or d=234
Answer:
B
Step-by-step explanation:
2/3 ÷ 1/6
=2/3 × 6/1
sorry if I got it wrong
Answer:
70.5
Step-by-step explanation:
Refer to attached image
1.95 = .21(5) + .06m
1.95 = 1.05 + .06m
Solving (if needed)
Subtract 1.05 from each side
1.95 - 1.05 = .06m
.90 = .06m
Divide .06 on both sides
15 = m