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 46SG

Explanation of Solution

Privacy laws:

The privacy law indicates the law that deals with modifying, storing, and using the individual information, which can be gathered by governments, organizations, or some other individuals.

Privacy laws are as follows:

  • It is limited to collect and store the individual information.
  • Once the information is collected, provisions should be made to safeguard the data.
  • Releasing the personal information to outside of business when an individual has approved to its disclosure.
  • When individual information is collected, then that individual must know about the details of the data collection and have chances to identify the data correctness...

Explanation of Solution

No”, user is not able to delete the personal information from the internet.

Justification:

  • In some cases, the court issued the order to delete the links that are “insufficient, unrelated, or no longer relevant” from a popular search engine; but they mention that a government does not have rights to negate the access to correct information...

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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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