EBK USING MIS
10th Edition
ISBN: 8220103633635
Author: KROENKE
Publisher: YUZU
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Chapter 3.8, Problem 3.7ARQ
Explanation of Solution
Eight principles of competitive advantage:
Eight principles of competitive advantage have been categorized into product implementations and process implementations.
Applications:
Product Implementations:
- Create new products or services
- To offer new products or services to the customer this is better than the competitor’s product.
- Enhancing products or services
- Improving the currently offering products by implementing the upcoming ideas.
- Differentiating products or services
- Making the products stand in the market by implementing upcoming ideas.
Process Implementations:
- Lock in customers and buyers
- Holding the customers and buyers to buy their products by providing good customer service and fulfilling the wishes and needs of the customer
- Lock in suppliers
- Holding the suppliers on their side makes them always available and ready to help.
- Raising the barriers to market entry
- Making some barriers, this will create a difficulty in startup business.
- Establishing alliances
- Making tie up other business partners will help them succeed in their business.
- Cost reduction
- Reducing the rate of the product will make them offer products or service at a cheaper rate.
Examples:
Product Implementations:
- Create new products or services
- The bookstore can achieve this strategy by tracking the books, which are offered to the student by using the
information system ...
- The bookstore can achieve this strategy by tracking the books, which are offered to the student by using the
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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
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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…
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Chapter 3 Solutions
EBK USING MIS
Ch. 3.3 - Prob. 1EGDQCh. 3.3 - Prob. 2EGDQCh. 3.3 - Prob. 3EGDQCh. 3.3 - Prob. 4EGDQCh. 3.7 - Prob. 1TARQCh. 3.7 - Prob. 2TARQCh. 3.7 - Prob. 3TARQCh. 3.7 - Prob. 4TARQCh. 3.8 - Prob. 1SGDQCh. 3.8 - Prob. 2SGDQ
Ch. 3.8 - Prob. 3SGDQCh. 3.8 - Prob. 4SGDQCh. 3.8 - Prob. 3.1ARQCh. 3.8 - Prob. 3.2ARQCh. 3.8 - Prob. 3.3ARQCh. 3.8 - Prob. 3.4ARQCh. 3.8 - Prob. 3.5ARQCh. 3.8 - Prob. 3.6ARQCh. 3.8 - Prob. 3.7ARQCh. 3.8 - Prob. 3.8ARQCh. 3 - Prob. 3.1UYKCh. 3 - Prob. 3.2UYKCh. 3 - Prob. 3.3UYKCh. 3 - Prob. 3.1CE3Ch. 3 - Prob. 3.4CS3Ch. 3 - Prob. 3.5CS3Ch. 3 - Prob. 3.6CS3Ch. 3 - Prob. 3.7CS3Ch. 3 - Prob. 3.8CS3Ch. 3 - Prob. 3.9CS3Ch. 3 - Prob. 3.10CS3Ch. 3 - Prob. 3.11CS3Ch. 3 - Prob. 3.12CS3Ch. 3 - Prob. 3.13MMLCh. 3 - Prob. 3.14MMLCh. 3 - Prob. 3.15MML
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