Business Driven Technology
7th Edition
ISBN: 9781259567322
Author: Paige Baltzan Instructor
Publisher: McGraw-Hill Education
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Chapter B, Problem B4.9MBD
Explanation of Solution
Disaster recovery plan:
- A disaster recovery plan denotes a complete process for recovering data or a system in catastrophic disaster event.
- It includes back up details of files and systems and their storage locations.
- Time is given to people of area to evacuate or relocate as per weather reports.
- The staff should be trained properly to follow guidelines correctly.
- The data protection should be ensured and should be detailed within plan.
- Back up should be taken for confidential data.
- The storage should be at remote location outside company site.
- The stages required for recovery process is to be defined.
- The recovery of network and key applications should be defined.
- The logistical support would include:
- Evacuation systems.
- Primary health aids...
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#include
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1. [20 pts] What is the output of the program? Please explain why.
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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 B Solutions
Business Driven Technology
Ch. B - Prob. B1.1MBDCh. B - Prob. B1.2MBDCh. B - Prob. B1.3MBDCh. B - Prob. B1.4MBDCh. B - Prob. B1.5MBDCh. B - Prob. B1.6MBDCh. B - Prob. B2.1MBDCh. B - Modeling a Business Process Do you hate waiting in...Ch. B - Prob. B2.3MBDCh. B - Prob. B2.4MBD
Ch. B - Prob. B2.5MBDCh. B - Prob. B3.1MBDCh. B - Prob. B3.2MBDCh. B - Prob. B4.1MBDCh. B - Prob. B4.2MBDCh. B - Prob. B4.3MBDCh. B - Prob. B4.4MBDCh. B - Prob. B4.5MBDCh. B - Prob. B4.6MBDCh. B - Prob. B4.7MBDCh. B - Prob. B4.8MBDCh. B - Prob. B4.9MBDCh. B - Prob. B4.10MBDCh. B - Prob. B4.11MBDCh. B - Prob. B5.1MBDCh. B - Prob. B5.2MBDCh. B - Prob. B5.3MBDCh. B - Prob. B5.4MBDCh. B - Prob. B5.5MBDCh. B - Prob. B5.6MBDCh. B - Prob. B6.1MBDCh. B - Prob. B6.2MBDCh. B - Prob. B6.3MBDCh. B - Prob. B6.4MBDCh. B - Prob. B6.5MBDCh. B - Prob. B7.1MBDCh. B - Prob. B7.2MBDCh. B - Prob. B7.3MBDCh. B - Prob. B7.4MBDCh. B - Prob. B7.5MBDCh. B - Prob. B7.6MBDCh. B - Prob. B7.7MBDCh. B - Prob. B7.8MBDCh. B - Prob. B7.9MBDCh. B - Prob. B7.10MBDCh. B - Prob. B7.11MBDCh. B - Prob. B8.1MBDCh. B - Prob. B8.2MBDCh. B - Prob. B9.1MBDCh. B - Prob. B9.2MBDCh. B - Prob. B9.3MBDCh. B - Prob. B9.4MBDCh. B - Prob. B9.5MBDCh. B - Prob. B9.6MBDCh. B - Prob. B9.7MBDCh. B - Prob. B9.8MBDCh. B - Prob. B9.9MBDCh. B - Prob. B9.10MBDCh. B - Prob. B10.1MBDCh. B - Prob. B10.2MBDCh. B - Prob. B10.3MBDCh. B - Prob. B10.4MBDCh. B - Prob. B10.5MBDCh. B - Prob. B11.1MBDCh. B - Prob. B11.2MBDCh. B - Prob. B11.3MBDCh. B - Prob. B11.4MBDCh. B - Prob. B12.1MBDCh. B - Prob. B12.2MBDCh. B - Prob. B12.3MBDCh. B - Prob. B12.4MBDCh. B - Prob. B12.5MBDCh. B - Prob. B12.6MBD
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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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