Note: The matrix referred to in the question is: ![M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D)
Answer:
a) [5/18, 5/18, 4/9]'
Explanation:
The adjacency matrix is ![M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D)
To start the power iteration, let us start with an initial non zero approximation,
![X_o = \left[\begin{array}{ccc}1\\1\\1\end{array}\right]](https://tex.z-dn.net/?f=X_o%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To get the rank vector for the first Iteration:

![X_1 = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]\left[\begin{array}{ccc}1\\1\\1\end{array}\right] \\\\X_1 = \left[\begin{array}{ccc}5/6\\5/6\\4/3\end{array}\right]\\](https://tex.z-dn.net/?f=X_1%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CX_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F6%5C%5C5%2F6%5C%5C4%2F3%5Cend%7Barray%7D%5Cright%5D%5C%5C)
Multiplying the above matrix by 1/3
![X_1 = \left[\begin{array}{ccc}5/18\\5/18\\4/9\end{array}\right]](https://tex.z-dn.net/?f=X_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F18%5C%5C5%2F18%5C%5C4%2F9%5Cend%7Barray%7D%5Cright%5D)
Using a linear search to find a value that is stored in the last element of an array of 20,000 elements, 20,000 element(s) must be compared.
Answer:
<u> </u><u>Role of websites in internet</u>
- Websites give the online way to the business,it helps the business to grow and reach the higher audience. There was a time,at which there is no apps only HTML/CSS websites are there so it is a building step for reaching the audience.
- Time is a one of the major factors for growth in online sector, and websites provide a interface to use their resources and services at less time ,you don't have to go somewhere now you can order from anywhere and anytime. 24/7 hour services gives a boost to the websites.
- Some of the Applications are Very big but you can access them in very less amount of internet. Apps never replace websites.
- Most of the billionaires are the ones who builds there website like amazon, twitter etc. It is very important to have a website for the business
<u>Importance of internet in our daily life</u>
- Internet provides various resources and services which we need at every time like booking a cab,hotel,flight,trains and tutoring services . It consist a vast variety of information which is required at every moment .
- We really don't need any physical presence of someone to learn something , to understand or for asking it is enough up on internet. Things are changing as technology grows.
- For leisure time, it provide social sites to enjoy and there are much good options to ask the question of any subject.
- It also give option to promote business , it helps to reach to higher audience, many companies give services to us as there business. Digital marketing is also up there to promote on social sites.
Answer:
Security administrator can use power shell command line or the GUI to "Promote a windows server to a domain controller".
Explanation:
A domain controller is a type of server that is used to authenticate the access of the system to the users to resources of windows. Power shell is an important tool that is used to perform different tasks of administration. We can user this tool as security administrator to Promote a Windows server to a domain controller.
integer userInput
integer i
integer mid
integer array(20) number
userInput = 1
for i = 0; userInput >= 0; i = i + 1
if number[i] > -1
userInput = Get next input
number[i] = userInput
i = i - 1
mid = i / 2
if i > 9
Put "Too many inputs" to output
elseif i % 2 == 0
Put number[mid - 1] to output
else
Put number[mid] to output