MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)
MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)
18th Edition
ISBN: 9781337285193
Author: Vermaat, Misty E.; Sebok, Susan L.; Freund, Steven M.; Campbell, Jennifer T.; Frydenberg, Mark
Publisher: Cengage Learning
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Chapter 12, Problem 14CT
Program Plan Intro

Integral literacy component:

  • It includes three types of literacies, namely, ethical, functional and informational.
  • It includes basic tools like read, write and calculate.
  • It aggregates the socio-affective skills with new technologies to modify the attributes.

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(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.

Chapter 12 Solutions

MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)

Ch. 12 - Prob. 11SGCh. 12 - Prob. 12SGCh. 12 - Prob. 13SGCh. 12 - Prob. 14SGCh. 12 - Prob. 15SGCh. 12 - Prob. 16SGCh. 12 - Prob. 17SGCh. 12 - Prob. 18SGCh. 12 - Prob. 19SGCh. 12 - Prob. 20SGCh. 12 - Prob. 21SGCh. 12 - Prob. 22SGCh. 12 - Prob. 23SGCh. 12 - Prob. 24SGCh. 12 - Prob. 25SGCh. 12 - Prob. 26SGCh. 12 - Prob. 27SGCh. 12 - Prob. 28SGCh. 12 - Prob. 29SGCh. 12 - Prob. 30SGCh. 12 - Prob. 31SGCh. 12 - Prob. 32SGCh. 12 - Prob. 33SGCh. 12 - Prob. 34SGCh. 12 - Prob. 35SGCh. 12 - Prob. 36SGCh. 12 - Prob. 37SGCh. 12 - Prob. 38SGCh. 12 - Prob. 39SGCh. 12 - Prob. 40SGCh. 12 - Prob. 41SGCh. 12 - Prob. 42SGCh. 12 - Prob. 43SGCh. 12 - Prob. 44SGCh. 12 - Prob. 45SGCh. 12 - Prob. 1TFCh. 12 - Prob. 2TFCh. 12 - Prob. 3TFCh. 12 - Prob. 4TFCh. 12 - Prob. 5TFCh. 12 - Prob. 6TFCh. 12 - Prob. 7TFCh. 12 - Prob. 8TFCh. 12 - Prob. 9TFCh. 12 - Prob. 10TFCh. 12 - Prob. 11TFCh. 12 - Prob. 12TFCh. 12 - Prob. 1MCCh. 12 - Prob. 2MCCh. 12 - Prob. 3MCCh. 12 - A(n) _____ report consolidates data usually with...Ch. 12 - Prob. 5MCCh. 12 - Prob. 6MCCh. 12 - Prob. 7MCCh. 12 - Prob. 8MCCh. 12 - Prob. 1MCh. 12 - Prob. 2MCh. 12 - Prob. 3MCh. 12 - Prob. 4MCh. 12 - Prob. 5MCh. 12 - Prob. 6MCh. 12 - Prob. 7MCh. 12 - Prob. 8MCh. 12 - Prob. 9MCh. 12 - Prob. 10MCh. 12 - Prob. 2CTCh. 12 - Prob. 3CTCh. 12 - Prob. 4CTCh. 12 - Prob. 5CTCh. 12 - Prob. 6CTCh. 12 - Prob. 7CTCh. 12 - Prob. 8CTCh. 12 - Prob. 9CTCh. 12 - Prob. 10CTCh. 12 - Prob. 11CTCh. 12 - Prob. 12CTCh. 12 - Prob. 13CTCh. 12 - Prob. 14CTCh. 12 - Prob. 15CTCh. 12 - Prob. 16CTCh. 12 - Prob. 17CTCh. 12 - Prob. 18CTCh. 12 - Prob. 19CTCh. 12 - Prob. 20CTCh. 12 - Prob. 21CTCh. 12 - Prob. 22CTCh. 12 - Prob. 23CTCh. 12 - Prob. 24CTCh. 12 - Prob. 25CTCh. 12 - Prob. 26CTCh. 12 - Prob. 27CTCh. 12 - Prob. 28CTCh. 12 - Prob. 29CTCh. 12 - Prob. 1PSCh. 12 - Prob. 2PSCh. 12 - Prob. 3PSCh. 12 - Prob. 4PSCh. 12 - Prob. 5PSCh. 12 - Prob. 6PSCh. 12 - Prob. 7PSCh. 12 - Prob. 8PSCh. 12 - Prob. 9PSCh. 12 - Prob. 10PSCh. 12 - Prob. 11PSCh. 12 - Prob. 1.1ECh. 12 - Prob. 1.2ECh. 12 - Prob. 1.3ECh. 12 - Prob. 2.1ECh. 12 - Prob. 2.2ECh. 12 - Prob. 3.1ECh. 12 - Prob. 3.2ECh. 12 - Prob. 1IRCh. 12 - Prob. 2IRCh. 12 - Prob. 4IRCh. 12 - Prob. 5IRCh. 12 - Prob. 1CTQCh. 12 - Prob. 3CTQCh. 12 - Prob. 4CTQ
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