Not sure what quizScore1-5 is, but here it is. Also, I'm not entirely sure what language you're working with here, but null can't be returned as an integer. If it were a string, for example, it'd be possible as string is an object reference type.
public int getData(int dataNumber)
{
if (dataNumber == 1)
return quizScore1;
else if (dataNumber == 2)
return quizScore2;
else if (dataNumber == 3)
return quizScore3;
else if (dataNumber == 4)
return quizScore4;
else if (dataNumber == 5)
return quizScore5;
else
return 0;
}
5 ( x + 4 ) = 4 ( x - 6)
5x + 20 = 4x - 24
5x -4x + 20= 4x - 4x - 24
1x + 20 = -24
1x + 20 - 20 = -24 - 20
1x = -44
check
5 (-44 + 4) = 4 ( -44 - 6)
5 (-40) = 4 (-50)
-200 = -200
Answer:
"Network layer" is the correct answer to the given question .
Explanation:
The network layer is interconnected with different networks by providing permission to the network. This job is done by sending the packets in the network. The main aim of the network layer is to provide the logical addressing, routing, and fragmentation of packets.
The network layer is the layer of the OSI model whose primary function is moving the packets or datagrams in the internetwork which is connected by the routers.
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)
