EBK ENHANCED DISCOVERING COMPUTERS & MI
EBK ENHANCED DISCOVERING COMPUTERS & MI
1st Edition
ISBN: 9780100606920
Author: Vermaat
Publisher: YUZU
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Chapter 10, Problem 4CTQ

Explanation of Solution

Network security risks:

The team members considered the following network security risks for resear...

Explanation of Solution

Risks:

Virus programs:

  • A virus program is a small piece of malicious code that can corrupt user files and delete valuable information and pose as a great threat to the network security.
  • Virus programs devastate the network security and the number of malicious program increases day by day.
  • At their worst, these malicious programs have the potential to wipe out user’s hard disk.

Software vulnerabilities:

  • When a user is updating the software regularly, software vulnerabilities are possible...

Explanation of Solution

An example Industrial article:

Ransomware:

  • Ransomware ranks high on the list of network security risks encountered by both large and small organizations.
  • In 2018, Verison’s data breach report states that the Ransomware is one of the top malicious network security risks found in 39 percent of network security related cases...

Explanation of Solution

Steps of network administrator to recover and prevent the damages from future risks:

  • First, the network administrator should ensure the Ransomware is entirely deleted from the system...

Explanation of Solution

Hardware and software to safeguard against the network security risks:

  • Users can use a good antivirus program with complete protection...

Explanation of Solution

Guidelines for network users to avoid the risks:

  • Awareness: The network user should have proper knowledge on the network threats, their risks, what the threats can do, and how to prevent them.
  • Scanning: The network user must always scan email attachment or d...

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Exercise 1 Function and Structure [30 pts] Please debug the following program and answer the following questions. There is a cycle in a linked list if some node in the list can be reached again by continuously following the next pointer. #include typedef struct node { int value; struct node *next; } node; int 11_has_cycle (node *first) if (first == node *head = { NULL) return 0; first; while (head->next != NULL) { } if (head first) { return 1; } head = head->next; return 0; void test ll_has_cycle () { int i; node nodes [6]; for (i = 0; i < 6; i++) { nodes [i] .next = NULL; nodes [i].value = i; } nodes [0] .next = &nodes [1]; nodes [1] .next = &nodes [2]; nodes [2] .next = &nodes [3]; nodes [3] .next nodes [4] .next &nodes [4]; NULL; nodes [5] .next = &nodes [0]; printf("1. Checking first list for cycles. \n Function 11_has_cycle says it has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no"); printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle says it has %s…

Chapter 10 Solutions

EBK ENHANCED DISCOVERING COMPUTERS & MI

Ch. 10 - Prob. 11SGCh. 10 - Prob. 12SGCh. 10 - Prob. 13SGCh. 10 - Prob. 14SGCh. 10 - Prob. 15SGCh. 10 - Prob. 16SGCh. 10 - Prob. 17SGCh. 10 - Prob. 18SGCh. 10 - Prob. 19SGCh. 10 - Prob. 20SGCh. 10 - Prob. 21SGCh. 10 - Prob. 22SGCh. 10 - Prob. 23SGCh. 10 - Prob. 24SGCh. 10 - Prob. 25SGCh. 10 - Prob. 26SGCh. 10 - Prob. 27SGCh. 10 - Prob. 28SGCh. 10 - Prob. 29SGCh. 10 - Prob. 30SGCh. 10 - Prob. 31SGCh. 10 - Prob. 32SGCh. 10 - Prob. 33SGCh. 10 - Prob. 34SGCh. 10 - Prob. 35SGCh. 10 - Prob. 36SGCh. 10 - Prob. 37SGCh. 10 - Prob. 38SGCh. 10 - Prob. 39SGCh. 10 - Prob. 40SGCh. 10 - Prob. 41SGCh. 10 - Prob. 42SGCh. 10 - Prob. 43SGCh. 10 - Prob. 44SGCh. 10 - Prob. 45SGCh. 10 - Prob. 46SGCh. 10 - Prob. 47SGCh. 10 - Prob. 48SGCh. 10 - Prob. 49SGCh. 10 - Prob. 1TFCh. 10 - Prob. 2TFCh. 10 - Prob. 3TFCh. 10 - Prob. 4TFCh. 10 - Prob. 5TFCh. 10 - Prob. 6TFCh. 10 - Prob. 7TFCh. 10 - Prob. 8TFCh. 10 - Prob. 9TFCh. 10 - Prob. 10TFCh. 10 - Prob. 11TFCh. 10 - Prob. 12TFCh. 10 - Prob. 1MCCh. 10 - Prob. 2MCCh. 10 - Prob. 3MCCh. 10 - Prob. 4MCCh. 10 - Prob. 5MCCh. 10 - Prob. 6MCCh. 10 - Prob. 7MCCh. 10 - Prob. 8MCCh. 10 - Prob. 1MCh. 10 - Prob. 2MCh. 10 - Prob. 3MCh. 10 - Prob. 4MCh. 10 - Prob. 5MCh. 10 - Prob. 6MCh. 10 - Prob. 7MCh. 10 - Prob. 8MCh. 10 - Prob. 9MCh. 10 - Prob. 10MCh. 10 - Prob. 2CTCh. 10 - Prob. 3CTCh. 10 - Prob. 4CTCh. 10 - Prob. 5CTCh. 10 - Prob. 6CTCh. 10 - Prob. 7CTCh. 10 - Prob. 8CTCh. 10 - Prob. 9CTCh. 10 - Prob. 10CTCh. 10 - Prob. 11CTCh. 10 - Prob. 12CTCh. 10 - Prob. 13CTCh. 10 - Prob. 14CTCh. 10 - Prob. 15CTCh. 10 - Prob. 16CTCh. 10 - Prob. 17CTCh. 10 - Prob. 18CTCh. 10 - Prob. 19CTCh. 10 - Prob. 20CTCh. 10 - Prob. 21CTCh. 10 - Prob. 22CTCh. 10 - Prob. 23CTCh. 10 - Prob. 24CTCh. 10 - Prob. 25CTCh. 10 - Prob. 26CTCh. 10 - Prob. 27CTCh. 10 - Prob. 1PSCh. 10 - Prob. 2PSCh. 10 - Prob. 3PSCh. 10 - Prob. 4PSCh. 10 - Prob. 5PSCh. 10 - Prob. 6PSCh. 10 - Prob. 7PSCh. 10 - Prob. 8PSCh. 10 - Prob. 9PSCh. 10 - Prob. 10PSCh. 10 - Prob. 11PSCh. 10 - Prob. 1.1ECh. 10 - Prob. 1.2ECh. 10 - Prob. 1.3ECh. 10 - Prob. 2.1ECh. 10 - Prob. 2.2ECh. 10 - Prob. 2.3ECh. 10 - Prob. 3.1ECh. 10 - Prob. 3.2ECh. 10 - Prob. 3.3ECh. 10 - Prob. 4.1ECh. 10 - Prob. 4.2ECh. 10 - Prob. 4.3ECh. 10 - Prob. 1IRCh. 10 - Prob. 2IRCh. 10 - Prob. 3IRCh. 10 - Prob. 4IRCh. 10 - Prob. 5IRCh. 10 - Prob. 1CTQCh. 10 - Prob. 2CTQCh. 10 - Prob. 3CTQCh. 10 - Prob. 4CTQ
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