Enhanced Discovering Computers 2017 (Shelly Cashman Series) (MindTap Course List)
Enhanced Discovering Computers 2017 (Shelly Cashman Series) (MindTap Course List)
1st Edition
ISBN: 9781305657458
Author: Misty E. Vermaat, Susan L. Sebok, Steven M. Freund, Mark Frydenberg, Jennifer T. Campbell
Publisher: Cengage Learning
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Chapter 10, Problem 42SG
Program Plan Intro

Fiber optic cable:

  • The wire is made up of ultra-thin filaments of glass or plastic material which is as thin as human hair and it can carry beam of light.
  • The glass fiber is covered with aramid yarn and a plastic is strongly covered over it.
  • Fiber optic cable is light based, which can carry information optically at the speed of light.
  • The cable can carry more data than other cables because of its greater bandwidth than other cables.

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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.
#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!
(OnlineGDB) 1. Please read and run the following code and answer the questions.#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;}   What is the output of the program? Please explain why.

Chapter 10 Solutions

Enhanced Discovering Computers 2017 (Shelly Cashman Series) (MindTap Course List)

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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