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brilliants [131]
3 years ago
10

Which of the following keys moves the insertion point to the end of data in a cell?

Computers and Technology
1 answer:
IgorC [24]3 years ago
5 0
C) End is the key that moves the insertion point to the end of data in a cell. 
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A network administrator is implementing multifactor authentication for employees who travel and use company devices remotely by
devlian [24]

Answer:

A networking administrator is a multi-factor authentication for employees who use company devices by using a VPN. The level of authentication is 802.1X and OTP.

Explanation:

Network administration is the computer network and uses multiple computers. A company and organization uses multiple computers and need a network admin to connect with a different system. Network administrators depend upon the organization. Some duties of the network administration are split and clearly defined such as

Configure the network hardware, servers, and routers

Upgrade the computer networks

Troubleshooting network issues

Assigning the network models

Deploying the software

Managing the servers

Implementing the security and measure

Network administration requires technical competence and will benefit as a network administrator.

For the level of authentication 802.1, X and OTP are suitable for network administration.

5 0
4 years ago
There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou
Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:-  n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex  u ∈ V

\rightarrow  for each vertex v ∈ V

\rightarrow\rightarrowif( d(pu, pj) > k )

\rightarrow\rightarrow\rightarrow add vertex u to Adj[v]   // Adj[v] represents adjacency list of v

\rightarrow\rightarrow\rightarrow add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3.  bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

\rightarrowvisited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph  

6. for each vertex u ∈ V  

\rightarrow if ( visited[u] == false)

\rightarrow\rightarrow color[u] = 0;

\rightarrow\rightarrow isbipartite = BFS(G,u,color,visited)  // if the vertices reachable from u form a bipartite graph, it returns true

\rightarrow\rightarrow if (isbipartite == false)

\rightarrow\rightarrow\rightarrow print " No solution exists "

\rightarrow\rightarrow\rightarrow exit(0)

7.  for each vertex u ∈V

\rightarrow if (color[u] == 0 )

\rightarrow\rightarrow print " Student u is assigned Bus 1"

\rightarrowelse

\rightarrow\rightarrow print " Student v is assigned Bus 2"

BFS(G,s,color,visited)  

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

\rightarrow u = Q.pop()

\rightarrow for each vertex v ∈ Adj[u]

\rightarrow\rightarrow if (visited[v] == false)

\rightarrow\rightarrow\rightarrow color[v] = (color[u] + 1) % 2

\rightarrow\rightarrow\rightarrow visited[v] = true

\rightarrow\rightarrow\rightarrow Q.push(v)

\rightarrow\rightarrow else

\rightarrow\rightarrow\rightarrow if (color[u] == color[v])

\rightarrow\rightarrow\rightarrow\rightarrow return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0
3 years ago
The minimum recommended standards for the operating system, processor, primary memory (RAM), and storage capacity for certain so
timurjin [86]

Answer:

They are called System Requirements.

Explanation:

3 0
2 years ago
When you cut and then paste a file, what are you doing?
frutty [35]

Answer:

removing it from it's current location and placing it in another

4 0
3 years ago
What is the return data type of the following function?
Olin [163]

Answer: the floor

Explanation:

8 0
3 years ago
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