Discovering Computers ©2016 (Shelly Cashman Series) (MindTap Course List)
1st Edition
ISBN: 9781305391857
Author: Misty E. Vermaat, Susan L. Sebok, Steven M. Freund, Jennifer T. Campbell, Mark Frydenberg
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 12, Problem 33SG
Explanation of Solution
Networking certification:
- Network expertise is achieved over years of training and experience because many kinds of variables are available for a total network solution.
- Attaining an advanced networking certification implies that the user had achieved a standar...
Explanation of Solution
List of jobs that require to obtain a networking certification:
- Network engineers:
- Network engineer must contain technical knowledge, which means he/she must know how to plan, how to implement and oversee the
computer networks that support in-house data, voice, video, and wireless network services.
- Network engineer must contain technical knowledge, which means he/she must know how to plan, how to implement and oversee the
- IT consultant:
- An IT consultant is an experienced person who gives expert advice for a fee...
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#include
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1. [20 pts] What is the output of the program? Please explain why.
2. [15 pts] What is the gdb command to set a breakpoint in line 6 (p
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3. [15 pts] Explain in your own words how the [break.
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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.
Chapter 12 Solutions
Discovering Computers ©2016 (Shelly Cashman Series) (MindTap Course List)
Ch. 12 - Prob. 1SGCh. 12 - Prob. 2SGCh. 12 - Prob. 3SGCh. 12 - Prob. 4SGCh. 12 - Prob. 5SGCh. 12 - Prob. 6SGCh. 12 - Prob. 7SGCh. 12 - Prob. 8SGCh. 12 - Prob. 9SGCh. 12 - Prob. 10SG
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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- #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!arrow_forward(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.arrow_forward1.[30 pts] Answer the following questions: a. [10 pts] Write a Boolean equation in sum-of-products canonical form for the truth table shown below: A B C Y 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 a. [10 pts] Minimize the Boolean equation you obtained in (a). b. [10 pts] Implement, using Logisim, the simplified logic circuit. Include an image of the circuitarrow_forward
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