EBK USING MIS
10th Edition
ISBN: 8220103633635
Author: KROENKE
Publisher: YUZU
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Expert Solution & Answer
Chapter 6.7, Problem 3QLQ
Explanation of Solution
ASCII characters:
In a computer system, characters can be represented using standard numeric codes. The most commonly used standard is referred as “American Standard Code for Information Interchange (ASCII)”.
- ASCII is primarily based on the English alphabet for character encoding.
- It uses 128 unique characters; and seven bits to represent each character in the set. Check bit is used as the eighth bit to ensure proper data transmission.
- Extended ASCII character version includes the eighth bit and allows 256 unique characters, which includes some additional symbols and some accented letters.
- ASCII has a finite number of characters; so it lists them all and assigns binary string value to each one of them.
- However, some problems need to be addressed, such as, in English language, there are 26 letters, and ASCII considers uppercase, lowercase letters separately, and assigns a value for each of the 52 characters.
- ASCII character set as well as Unicode character set treats uppercase and lowercase letters distinctly.
Binary code:
The digital data can be symbolized, stored and communicated as a set of binary bits. This set is referred as binary code.
- The letters, special characters or digits are represented using the binary digits “0” and “1” in a computer device or other electronic devices...
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D. S. Malik, Data Structures Using C++, 2nd Edition, 2010
Methods (Ch6) - Review
1. (The MyRoot method) Below is a manual implementation of the Math.sqrt() method in Java.
There are two methods, method #1 which calculates the square root for positive integers, and
method #2, which calculates the square root of positive doubles (also works for integers).
public class SquareRoot {
public static void main(String[] args) {
}
// implement a loop of your choice here
// Method that calculates the square root of integer variables
public static double myRoot(int number) {
double root;
root=number/2;
double root old;
do {
root old root;
root (root_old+number/root_old)/2;
} while (Math.abs(root_old-root)>1.8E-6);
return root;
}
// Method that calculates the square root of double variables
public static double myRoot(double number) {
double root;
root number/2;
double root_old;
do {
root old root;
root (root_old+number/root_old)/2;
while (Math.abs (root_old-root)>1.0E-6);
return root;
}
}
Program-it-Yourself: In the main method, create a program that…
I would like to know the main features about the following 3 key concepts:1. Backup Domain Controller (BDC)2. Access Control List (ACL)3. Dynamic Memory
Chapter 6 Solutions
EBK USING MIS
Ch. 6.3 - Prob. 1EGDQCh. 6.3 - Prob. 2EGDQCh. 6.3 - Prob. 3EGDQCh. 6.3 - Prob. 4EGDQCh. 6.3 - Prob. 5EGDQCh. 6.3 - Prob. 6EGDQCh. 6.3 - Prob. 7EGDQCh. 6.3 - Prob. 8EGDQCh. 6.3 - Prob. 9EGDQCh. 6.7 - Prob. 1QLQ
Ch. 6.7 - Prob. 2QLQCh. 6.7 - Prob. 3QLQCh. 6.7 - The article mentions that quantum computers will...Ch. 6.8 - Prob. 1SGDQCh. 6.8 - Prob. 2SGDQCh. 6.8 - Prob. 3SGDQCh. 6.8 - Prob. 4SGDQCh. 6.8 - Prob. 5SGDQCh. 6.8 - Prob. 6.1ARQCh. 6.8 - Prob. 6.2ARQCh. 6.8 - Prob. 6.3ARQCh. 6.8 - Prob. 6.4ARQCh. 6.8 - Prob. 6.5ARQCh. 6.8 - Prob. 6.6ARQCh. 6.8 - Prob. 6.7ARQCh. 6.8 - Prob. 6.8ARQCh. 6 - Prob. 6.1UYKCh. 6 - Prob. 6.2UYKCh. 6 - Prob. 6.3UYKCh. 6 - Prob. 6.4UYKCh. 6 - Prob. 6.5UYKCh. 6 - Prob. 6.6UYKCh. 6 - Prob. 6.7CE6Ch. 6 - Prob. 6.8CE6Ch. 6 - Prob. 6.9CE6Ch. 6 - Prob. 6.1CE6Ch. 6 - Prob. 6.11CE6Ch. 6 - Prob. 6.12CE6Ch. 6 - Prob. 6.14CS6Ch. 6 - Prob. 6.15CS6Ch. 6 - Prob. 6.16CS6Ch. 6 - Prob. 6.17CS6Ch. 6 - Prob. 6.18CS6Ch. 6 - Prob. 6.19CS6Ch. 6 - Prob. 6.2CS6Ch. 6 - Prob. 6.21MMLCh. 6 - Prob. 6.22MML
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