Step-by-step explanation:
yes that's true
if a^n=b
then
a^n/n=b^1/n
a=b^1/n
6.516 is the number in standard form
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
.93 is far greater than the rest.
