PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
13th Edition
ISBN: 9781305492394
Author: Cenage Learning
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 12, Problem 3.2E
Explanation of Solution
Comparison of three online survey tools:
|
Typeform | Survicate | Poldaddy | |
| Definition | It makes sharing and collecting information conversational and comfortable. |
It is a tool used to collect the customer feedback about the website, its products and services. | It allows the user to create, distribute, and interpret the results of surveys, rating, quizzes and polls. |
| Feature | Unlimited forms and surveys can be created with the fee account. |
Some of the features of Survicate are lead capture form, skip logic, customer design, feedback tab, advanced analytics. |
It collects information from the account holder, including personal data. They mainly use this data to improve and inform their services... |
Explanation of Solution
Favorite online survey tool:
Typeform is the best online survey tool when compared to other two because,
- It will collect...
Explanation of Solution
Least favorite online survey tool:
Polldaddy is the least favorite online tool when compare to Typeform and Survicate because,
- It will gather only b...
Explanation of Solution
Difference between free and fee-based accounts of online survey tools:
|
Typeform | Survicate | Polldaddy |
|
Basic: Free to gather 100 responses per month. PRO: $35/month to gather unlimited responses from as many PRO typeforms. PRO+: $70/month to gather unlimited responses from as many PRO+ typeforms... |
Expert Solution & Answer
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Students have asked these similar questions
I would like to get information to know features about the following concepts:
1. Anything as a Server (XaaS)
2. Block Storage
3. WebSocket
Please answer JAVA OOP problem below:
You are working at a university that tracks students. Each student is identified by their name and faculty advisor. Each faculty advisor is identified by their name, department, and maximum number of students they can advise.
Using solid OO design principles, create a modular program that implements all the classes for the problem and also creates an implementation class that gathers user input for one student and then prints out the information gathered by creating the appropriate data definition and implementation classes. All data must be validated.
I have given the code so far:
Implementation:
import javax.swing.JOptionPane;
public class Implementation {
public static void main(String[] args) {
FacultyAdvisor facultyAdvisor = new FacultyAdvisor("Sharmin Sultana", "IT", 30);
Student student = new Student("John", facultyAdvisor);
JOptionPane.showMessageDialog(null, student.toString());
}
}
Student:
public…
Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head
{
NULL) return 0;B
= first;
while (head->next != NULL) {
if (head == first) {
return 1; }
head
head->next;
}
return 0;
void test_11_has_cycle() {
int i;
node nodes [6];
for (i = 0; i < 6; i++)
nodes [i] .next = NULL;
nodes [i].value
i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next = &nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next = & nodes [4];
nodes [4] .next
=
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
hass cycle\n\n", 11_has_cycle (&nodes [0]) ?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
Chapter 12 Solutions
PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
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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