48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
54: 1, 2, 3, 6, 9, 18, 27, 54
gcf: 6
Notice the grid, the points are (-5, -2) and (4, -1), thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 4}} &,&{{ -1}}~) % (c,d) &&(~{{ -5}} &,&{{ -2}}~) \end{array}~~ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[-5-4]^2+[-2-(-1)]^2}\implies d=\sqrt{(-5-4)^2+(-2+1)^2} \\\\\\ d=\sqrt{(-9)^2+(-1)^2}\implies d=\sqrt{81+1}\implies d=\sqrt{82}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%20-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%7B%7B%20-5%7D%7D%20%26%2C%26%7B%7B%20-2%7D%7D~%29%0A%5Cend%7Barray%7D~~%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-5-4%5D%5E2%2B%5B-2-%28-1%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5-4%29%5E2%2B%28-2%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-9%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B81%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B82%7D)
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
Answer:
10.8 hours
Step-by-step explanation:
