Discovering Computers ©2018: Digital Technology, Data, and Devices
Discovering Computers ©2018: Digital Technology, Data, and Devices
1st Edition
ISBN: 9781337285100
Author: Misty E. Vermaat, Susan L. Sebok, Steven M. Freund, Jennifer T. Campbell, Mark Frydenberg
Publisher: Cengage Learning
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Chapter 5, Problem 1CTQ

Explanation of Solution

Guidelines for increasing security while playing online games:

The rise of multiplayer online gaming platforms attracts cybercriminals who seek for new ways to perform malicious activities and monetize their illicit skills.

Currently, there are more than 4.6 million online game focused malwares exist.

Thus, playing online games is essential to enhance the security.

Some of the guidelines to enhance the security while playing online games are listed below:

  • Create a separate e-mail address to sign up the online games.
  • Install an effective antivirus and anti-spyware program in the system.
  • Securely configure web browsers and employ a firewall...

Explanation of Solution

Other security tips:

Yes”, in addition to the security tips mentioned, people can find other threats and security tips...

Explanation of Solution

Game updates and safety measures:

Yes”, user has downloaded updates to a game from the Internet.

  • To ensure the safety of...

Explanation of Solution

List of games with malware infections:

Essential Security against Evolving Threats (ESAT) released a list of some most popular online gaming applications carrying malware...

Explanation of Solution

Online gaming security problem faced:

  • Hackers in the gaming community are one of the online gaming security problem faced.
  • Malicious hackers in the gaming community often try to get in touch with new gamers and offer them attractive gaming offers...

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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9*…
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This is for my Computer Organization & Assembly Language Class

Chapter 5 Solutions

Discovering Computers ©2018: Digital Technology, Data, and Devices

Ch. 5 - Prob. 11SGCh. 5 - Prob. 12SGCh. 5 - Prob. 13SGCh. 5 - Prob. 14SGCh. 5 - Prob. 15SGCh. 5 - Prob. 16SGCh. 5 - Prob. 17SGCh. 5 - Prob. 18SGCh. 5 - Prob. 19SGCh. 5 - Prob. 20SGCh. 5 - Prob. 21SGCh. 5 - Prob. 22SGCh. 5 - Prob. 23SGCh. 5 - Prob. 24SGCh. 5 - Prob. 25SGCh. 5 - Prob. 26SGCh. 5 - Prob. 27SGCh. 5 - Prob. 28SGCh. 5 - Prob. 29SGCh. 5 - Prob. 30SGCh. 5 - Prob. 31SGCh. 5 - Prob. 32SGCh. 5 - Prob. 33SGCh. 5 - Prob. 34SGCh. 5 - Prob. 35SGCh. 5 - Prob. 36SGCh. 5 - Prob. 37SGCh. 5 - Prob. 38SGCh. 5 - Prob. 39SGCh. 5 - Prob. 40SGCh. 5 - Prob. 41SGCh. 5 - Prob. 42SGCh. 5 - Prob. 43SGCh. 5 - Prob. 44SGCh. 5 - Prob. 45SGCh. 5 - Prob. 46SGCh. 5 - Prob. 47SGCh. 5 - Prob. 48SGCh. 5 - Prob. 49SGCh. 5 - Prob. 1TFCh. 5 - Prob. 2TFCh. 5 - Prob. 3TFCh. 5 - Prob. 4TFCh. 5 - Prob. 5TFCh. 5 - Prob. 6TFCh. 5 - Prob. 7TFCh. 5 - Prob. 8TFCh. 5 - Prob. 9TFCh. 5 - Prob. 10TFCh. 5 - Prob. 11TFCh. 5 - Prob. 12TFCh. 5 - Prob. 1MCCh. 5 - Prob. 2MCCh. 5 - Prob. 3MCCh. 5 - Prob. 4MCCh. 5 - Prob. 5MCCh. 5 - Prob. 6MCCh. 5 - Prob. 7MCCh. 5 - Prob. 8MCCh. 5 - Prob. 1MCh. 5 - Prob. 2MCh. 5 - Prob. 3MCh. 5 - Prob. 4MCh. 5 - Prob. 5MCh. 5 - Prob. 6MCh. 5 - Prob. 7MCh. 5 - Prob. 8MCh. 5 - Prob. 9MCh. 5 - Prob. 10MCh. 5 - Prob. 2CTCh. 5 - Prob. 3CTCh. 5 - Prob. 4CTCh. 5 - Prob. 5CTCh. 5 - Prob. 6CTCh. 5 - Prob. 7CTCh. 5 - Prob. 8CTCh. 5 - Prob. 9CTCh. 5 - Prob. 10CTCh. 5 - Prob. 11CTCh. 5 - Prob. 12CTCh. 5 - Prob. 13CTCh. 5 - Prob. 14CTCh. 5 - Prob. 15CTCh. 5 - Prob. 16CTCh. 5 - Prob. 17CTCh. 5 - Prob. 18CTCh. 5 - Prob. 19CTCh. 5 - Prob. 20CTCh. 5 - Prob. 21CTCh. 5 - Prob. 22CTCh. 5 - Prob. 23CTCh. 5 - Prob. 24CTCh. 5 - Prob. 25CTCh. 5 - Prob. 26CTCh. 5 - Prob. 27CTCh. 5 - Prob. 28CTCh. 5 - Prob. 29CTCh. 5 - Prob. 1PSCh. 5 - Prob. 2PSCh. 5 - Prob. 3PSCh. 5 - Prob. 4PSCh. 5 - Prob. 5PSCh. 5 - Prob. 6PSCh. 5 - Prob. 7PSCh. 5 - Prob. 8PSCh. 5 - Prob. 9PSCh. 5 - Prob. 10PSCh. 5 - Prob. 11PSCh. 5 - Prob. 1.1ECh. 5 - Prob. 1.2ECh. 5 - Prob. 1.3ECh. 5 - Prob. 2.1ECh. 5 - Prob. 2.2ECh. 5 - Prob. 2.3ECh. 5 - Prob. 3.3ECh. 5 - Prob. 4.1ECh. 5 - Prob. 4.2ECh. 5 - Prob. 4.3ECh. 5 - Prob. 5.1ECh. 5 - Prob. 5.2ECh. 5 - Prob. 5.3ECh. 5 - Prob. 1IRCh. 5 - Prob. 2IRCh. 5 - Prob. 3IRCh. 5 - Prob. 4IRCh. 5 - Prob. 5IRCh. 5 - Prob. 1CTQCh. 5 - Prob. 2CTQCh. 5 - Prob. 3CTQCh. 5 - Prob. 4CTQ
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