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suter [353]
3 years ago
7

Which is the term for a computer typically located in an area with limited security and loaded with software and data files that

appear to be authentic, yet they are imitations of real data files?
Computers and Technology
1 answer:
otez555 [7]3 years ago
8 0
Sorry..................
You might be interested in
For this assignment: Analyze and describe the network infrastructure. Describe and explain the various policies that will be nee
mamaluj [8]

Answer:

Explanation:

The Network infrastructure shown here are LAN and WAN. Wired and wireless communications.

The Various policies are:

1. Group related items together, for instance, grouping all Windows servers, into one virtual LAN (VLAN). Other asset groups might include infrastructure (routers, switches, VPNs and VoIP) in one VLAN and security assets (IDS, firewalls, web filters and scanners) may be grouped in another.

2. In general, it is good to adopt a default deny access posture for each VLAN.

3. Network segmentation is a very significant, long-term project, but each step along the way increases security. Log all traffic between segments to determine what is normal and needed for effective functioning.

4. Network segmentation is undeniably and unquestionably an effective component in a defense in depth strategy. Organizations that implement it must be prepared to manage scores of firewalls, switches and routers, each with hundreds of rules, all of which may be affected by the network segmentation process and potentially by updates and changes, even after it is in place.

5. Contribute to a secure WAN environment for all connected departments, offices,

agencies, boards, and commissions

6. Provide a uniform security framework to secure the integrity, confidentiality, and availability of info and info systems, at the WAN level.

7. Provide, in balance with operational requirements, legislative requirements, and information sharing agreements, the minimum WAN security requirements.

8. Raise awareness of information and information technology security needs for all users of the WAN by providing the security principles, requirements.

9. Define the clear roles and responsibilities of all users of the WAN, particularly WAN security staff.

* Vulnerabilities and exposures

1. Data requiring special protection such as credit card numbers that need to comply with PCI-DSS or patient information that is subject to HIPAA should be isolated from other data and put in their own VLANs.

2. Your aim is to limit access to sensitive information to those who need it within the organization and to create roadblocks to stop or slow intruders, who may have broken through one layer of security, from doing further damage.

3. Network segmentation is not a “set and forget” undertaking. The network access policy, defined in firewalls, routers and related devices, changes constantly to cater to new business requirements. Ensure that new changes do not violate your segmentation strategy requires a good degree of visibility and automation.

4. Reducing internal breaches and the infiltration of malicious software(malware). This

internal defense requires significant involvement with individual devices

on a network, which creates greater overhead on network administrators.

*Risks

1. Malicious software, also known as malware,makes its way onto a network through

employees, contractors and visitors. Personal laptops, wireless gadgets,

and of course the USB flash drives, all these provide excellent vectors through which

malware can enter the workplace.

2. Hackers, worms, spammers and other security dangers of the Internet via LAN.

3. The various vulnerabilities on your network represent potential costs — time, money and assets — to your library. These costs, along with the chance someone will exploit these vulnerabilities, help determine the level of risk involved.

4. Since the cost of adding another Internet connection, increasing the speed of the current connection or purchasing complex network monitoring equipment might be too prohibitive, the library has a higher tolerance for a periodically slow Internet connection.

5. External flash drives and other media are also concern when those enters the network.

6. The lost or stolen handheld device poses some serious risks if not incorporated into your network security policy. Such devices are often capable of being formatted of all company content remotely in the case of theft or robbery.

*Security measurements:

1. Address Resolution

Protocol (ARP) spoofing, Denial of Service (DoS) attacks such as Tear Drop

or Ping of Death.

2. In addition, network administrators can form a policy whereby network

users are required to install and maintain anti-malware scanners in their devices.

3. Many tools exist to check the existing security state of your network. The Microsoft Baseline Security Analyzer, Nmap .

4. Risk assessment is a combination of both quantifying (the cost of the threat) and qualifying (the odds of the attack).

5. Firewalls.

6. Antivirus systems.

7. Intrusion-detection systems (Host-based IDS,Network-based IDS)

8. Port scanners.

9. Network sniffers.

10. A vulnerability scanner is like a port scanner on steroids.

*Unnecessary Ports

1. It is not easy to say which ports exactly but we should know that the service ports which are open among 65,535 ports and although not exactly sure what service is running , it is safer to check the port and close it as "A Closed Port is a Safe Port".

7 0
3 years ago
Implement a Breadth-First Search of the people in the network who are reachable from person id 3980. You can implement BFS with
Gnoma [55]

Answer:

Check the explanation

Explanation:

import java.io.File;

import java.io.FileNotFoundException;

import java.util.LinkedList;

import java.util.Scanner;

import static org.junit.Assert.assertEquals;

/**

* CS146 Assignment 3 Node class This class is used for undirected graphs

* represented as adjacency lists The areFriends() method checks if two people

* are friends by checking if an edge exists between the two

*

*/

public class NetworkAdjList {

   // Initialize array with max number of vertices taken from SNAP

   static int max_num_vertices = 88234;

   static LinkedList<Integer>[] adjacencyList = new LinkedList[max_num_vertices];

   public static void createAdjacencyList() {

       // Initialize array elements

       for (int j = 0; j < max_num_vertices; j++) {

           adjacencyList[j] = new LinkedList<Integer>();

       }

       // Get file path of the 3980.edges file

       String filePath = "C:\\Users\\shahd\\Documents\\CS146\\Assignment 3\\Question 3\\Question3\\src\\a3\\3980.edges";

       File f = new File(filePath);

       // Use Scanner to read edges from the file and put it into adjacency list

       int a;

       int b;

       try {

           Scanner fileIn = new Scanner(f);

           while (fileIn.hasNext()) {

               a = fileIn.nextInt();

               b = fileIn.nextInt();

               adjacencyList[a].add(b);

               adjacencyList[b].add(a); // We need to add the edges both ways

           }

       } catch (FileNotFoundException e) {

           e.printStackTrace();

       }

   }

   public static boolean areFriends(int A, int B) {

       // If the adjacency list contains (A, B) edge, then return true, else false

       if (adjacencyList[A].contains(B)) {

           return true;

       } else {

           return false;

       }

   }

  private static void bfsHelper(boolean visited[], int currentNode, int dis, int sourceNode) {

       dis++;

       if(!visited[currentNode]) {

           visited[currentNode] = true;

           

           for(int neighbor: adjacencyList[currentNode]) {

               System.out.println(neighbor + " is at a distance of " + dis + " from " + sourceNode);

               bfsHelper(visited, neighbor, dis++, sourceNode);

           }

       }

   }

   public static void BFStraversal(int start) {

       boolean visited[] = new boolean[max_num_vertices];

       bfsHelper(visited, start, 0, start);

   }

   public static void main(String[] args) {

       /**

        * These test cases assume the file 3980.edges was used

        */

       createAdjacencyList();

       System.out.println("Testing...");

       assertEquals(areFriends(4038, 4014), true);

       System.out.println("1 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(3982, 4037), true);

       System.out.println("2 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4030, 4017), true);

       System.out.println("3 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4030, 1), false);

       System.out.println("4 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(1, 4030), false);

       System.out.println("5 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4003, 3980), true);

       System.out.println("6 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(3985, 4038), false);

       System.out.println("7 of 7");

       System.out.println("Success!");

   }

}

**************************************************

8 0
3 years ago
Which of the following are common parts of the Access 2016 interface? Choose 5.
lora16 [44]

Answer:

SHALOM GUYSi am not sure what ur asking is there any other way i can help?

          ~Piper Rokelle

Explanation:

4 0
3 years ago
Write a C++ program to accept a bunch of numbers from the user. Accept the numbers in a loop and stop when the user enters 0. Pr
brilliants [131]

Answer:

#include <iostream>

using namespace std;

int main() {

   float n;//declaring a float variable.

   cout<<"Enter 0 to stop"<<endl;//printing the message.

   cin>>n;//prompting n.

   while(n!=0)//taking input until n is not 0.

   {

       cout<<"The cube is "<<n*n*n<<endl;//printing the cube.

       cin>>n;

   }

return 0;

}

Input:-

12

-8

3.93

-22.5

0

Output:-

Enter 0 to stop

The cube is 1728

The cube is -512

The cube is 60.6985

The cube is -11390.6

Explanation:

I have declared a float variable n.I am prompting n from the user and letting the user know that he has to enter 0 to stop until the user shall type inputs.In the loop I am simultaneously printing the cubes of the numbers entered.

7 0
3 years ago
What are data structures and algorithms? Why are they important to software developer?
Elden [556K]

Explanation:

Data structure is a way of gathering and organizing the data in such a way that we can perform operation on the data very efficiently.There are different types of data structures such as :-

  1. Arrays
  2. Linked list
  3. Stacks
  4. Queues
  5. Trees
  6. Graphs

These are the basic data structures.

Algorithm:-It is a finite set of instructions written to accomplish a certain task.An algorithm's performance is measured on the basis of two properties:-

  1. Time complexity.
  2. Space complexity.

Software developers need to know data structures and algorithms because all the computers rely on data structures and algorithms so if you know data structures and algorithms better you will know the computer better.

6 0
3 years ago
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