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There are many kinds of relationships in a design model of a database. The most common are one-to-one (1:1) and one-to-many (1:N) and many-to-many (N:M). Some of the entity types can be ID-Dependent, Strong, Weak, or Subtype Entities.
1. Why is it important to classify each of these types in an ERD Model?
2. What rules would have to be enforced based on the entity type? Pick one entity type and discuss what enforcement is needed by the database or application.
Draw an inheritance hierarchy to represent a shoe object. The base class should have derived classes of Dress Shoes, Tennis Shoes and Boots.
Write a program to solve the following problem: A TV set is purchased with a loan of $563 to be paid off with 5 monthly payments of $116. The interest rate is 1 percent per month. Display a table giving the balance on the loan at the end of ea..
Design a work breakdown structure for the task below: Designing the database, Working on the website content
Evaluate the success WellCo and PharmaCARE shareholders
Write a command to list all the employees hired in 2003 and 2004. Sort the rows by the hire_date column in descending order.
VSS has an existing software package that controls its high bay systems and wants to rewrite it. The existing system is written in C and uses C-ISAM files to store data.
Create a database from scratch that contains, at a minimum, the elements listed below
Design an updateable database for storing customer and sales data. Explain how to deal with the problems of missing data.
How and why did database management systems become the organizational data management standard? Describe some advantages of database approach over file-system approach.
How many copies of the book titled „The big magic? are owned by the local library whose name is "Newtown" library?
You are interested in analyzing some hard-to-obtain data from two separate databases. Each database contains n numerical values.
Find the average number of books borrowed per member. Take into account that if a member does not borrow any books, then that member does not appear in the borrowed relation at all.
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