Answer:
I did not understand a thing from your writing
Answer:
Static
Explanation:
A computer or system administrator who want packets to be filter have to set or configured a filtering rules which will enable and allow the firewall to decide on which of the packets that will be allowed to function and the packets that will be denied not to function to be developed and as well installed on the firewall have to make use of what is called STATIC FILTERING reason been that STATIC FILTERING make it of necessary that the filtering rules should be developed and installed with the firewall.
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
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)
Answer: The amount of interest to be paid is $972
Explanation: This is gotten by firstly getting the amount of money to return per year without the interest which is $114400/10(years)= 11440
11440 is to be paid per year without interest. To get amount of interest is $11440*8.5/100 = $972 this is the amount of interest to be paid per year including in year 2.
Therefore, total amount of money together with interest to be paid each year and in year 2 = $11440+$972=12412.
Answer:
#include <iostream>
using namespace std;
int main()
{
cout << "N\t10*N\t100*N\t1000*N" << endl;
int x=1;
while(x<=5){
cout << x <<"\t"<< x*10 <<"\t"<<x*100 <<"\t"<< x*1000 << endl;
x++;
}
return 0;
}
Explanation:
Print the header, N 10*N 100*N 1000*N, using \t
Initialize the x as 1. It will be used to control the while loop
Create a while loop that iterates while x is smaller than or equal to 5
Inside the loop, print the required values. After printing, increment the value of x by 1
