Bundle: Enhanced Discovering Computers 2017, Loose-leaf Version + Illustrated Microsoft Office 365 & Office 2016: Introductory, Loose-leaf Version + ... 1 term (6 months) Printed Access Card
Bundle: Enhanced Discovering Computers 2017, Loose-leaf Version + Illustrated Microsoft Office 365 & Office 2016: Introductory, Loose-leaf Version + ... 1 term (6 months) Printed Access Card
1st Edition
ISBN: 9781337379960
Author: Misty E. Vermaat, Susan L. Sebok, Steven M. Freund, Mark Frydenberg, Jennifer T. Campbell
Publisher: Cengage Learning
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Chapter 5, Problem 3MC
Program Description Answer

A back door is a collection of instructions in a program or an illegal way that permits the user to avoid the security controls when retrieving the code, network, or computer.

Therefore, the correct option is “C”.

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Please run and debug the following program and answer the questions.
(OnlineGDB) #include <stdio.h>int main(void) {int a;char *s;int v0 = 4, v1 = 5, v2 = 6, v3 = 1, v4 = 2;printf("Exercise 1:\n====================\n");switch(v0) {case 0: printf("Hello October\n"); break;case 1: printf("Go Kean!\n"); break;case 2: printf("Academic Building Center \n"); break;case 3: printf("UNION \n"); break;case 4: printf("Go ");case 5: printf("Kean! \n");default: printf("Have a great semester! \n"); break;}for(a=5; a<v1; a++) {printf("Kean");}printf("\n");if (v2 == 6) {s = "Go";}else {s = "Hello";}if(v3 != v4) {printf("%s Kean!\n",s);} else {printf("%s Computer Science!\n",s);}return 0;} Assume the following codes are added between line 36 (}) and line 38 (return 0;) v0>0 ? ++v1, ++v2 : --v3; Please give the values of v0, v1, v2, v3, and v4 after this line and explain the reason. You can test the program to verify your answer if you like.
#include <stdio.h>int main(void) {int a;char *s;int v0 = 4, v1 = 5, v2 = 6, v3 = 1, v4 = 2;printf("Exercise 1:\n====================\n");switch(v0) {case 0: printf("Hello October\n"); break;case 1: printf("Go Kean!\n"); break;case 2: printf("Academic Building Center \n"); break;case 3: printf("UNION \n"); break;case 4: printf("Go ");case 5: printf("Kean! \n");default: printf("Have a great semester! \n"); break;}for(a=5; a<v1; a++) {printf("Kean");}printf("\n");if (v2 == 6) {s = "Go";}else {s = "Hello";}if(v3 != v4) {printf("%s Kean!\n",s);} else {printf("%s Computer Science!\n",s);}return 0;} Output: Exercise 1:====================Go Kean! Have a great semester!  Go Kean! Please only modify the initial value of v0, v1, v2, v3 and v4 to get the following output. Youneed to show your program output (in the screenshot) and submit the code that youmodified.Exercise 1:====================Hello OctoberKeanHello Computer Science!

Chapter 5 Solutions

Bundle: Enhanced Discovering Computers 2017, Loose-leaf Version + Illustrated Microsoft Office 365 & Office 2016: Introductory, Loose-leaf Version + ... 1 term (6 months) Printed Access Card

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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