Computer Science: An Overview (13th Edition) (What's New in Computer Science)
13th Edition
ISBN: 9780134875460
Author: Glenn Brookshear, Dennis Brylow
Publisher: PEARSON
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Chapter 9, Problem 24CRP
Program Plan Intro
Relational
- A relational database is a group of tables which are formally described.
- The data is accessible in many different ways without reorganization of the database tables.
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The following tables form part of a Library database held in an RDBMS:
Book (ISBN, title, edition, year)
BookCopy (copyNo, ISBN, available)
Borrower (borrowerNo, borrowerName, borrowerAddress)
BookLoan (copyNo, dateOut, dateDue, borrowerNo)
where:
Book contains details of book titles in the library and the ISBN is the key.
BookCopy contains details of the individual copies of books in the library and copyNo is the key. ISBN is a foreign key identifying the book title.
Borrower contains details of library members who can borrow books and borrowerNo is the key.
BookLoan contains details of the book copies that are borrowed by library members and copyNo/dateOut forms the key. borrowerNo is a foreign key identifying the borrower.
Formulate the following queries in relational algebra and tuple relational calculus:
5.27 List all copies of book titles that are available for borrowing.
5.28 List all copies of the book title “Lord of the Rings” that are available for…
Chapter 9 Solutions
Computer Science: An Overview (13th Edition) (What's New in Computer Science)
Ch. 9.1 - Identify two departments in a manufacturing plant...Ch. 9.1 - Prob. 2QECh. 9.1 - Summarize the roles of the application software...Ch. 9.2 - Prob. 1QECh. 9.2 - Prob. 2QECh. 9.2 - Prob. 4QECh. 9.2 - Prob. 5QECh. 9.2 - Prob. 6QECh. 9.3 - Prob. 1QECh. 9.3 - What is a persistent object?
Ch. 9.3 - Identify some classes as well as some of their...Ch. 9.3 - Prob. 4QECh. 9.4 - Prob. 1QECh. 9.4 - Prob. 2QECh. 9.4 - Prob. 3QECh. 9.4 - Prob. 4QECh. 9.4 - Prob. 5QECh. 9.4 - Prob. 6QECh. 9.5 - Prob. 1QECh. 9.5 - Prob. 2QECh. 9.5 - Prob. 3QECh. 9.5 - Prob. 4QECh. 9.5 - Prob. 5QECh. 9.5 - Prob. 6QECh. 9.5 - Prob. 7QECh. 9.6 - Prob. 1QECh. 9.6 - Give an additional example of a pattern that might...Ch. 9.6 - Prob. 3QECh. 9.6 - How does data mining differ from traditional...Ch. 9.7 - Prob. 1QECh. 9.7 - Prob. 2QECh. 9.7 - Prob. 3QECh. 9.7 - Prob. 4QECh. 9 - Prob. 1CRPCh. 9 - Prob. 2CRPCh. 9 - Prob. 3CRPCh. 9 - Prob. 4CRPCh. 9 - Prob. 5CRPCh. 9 - Prob. 6CRPCh. 9 - Prob. 7CRPCh. 9 - Prob. 8CRPCh. 9 - Prob. 9CRPCh. 9 - Prob. 10CRPCh. 9 - Prob. 11CRPCh. 9 - Prob. 12CRPCh. 9 - Using the commands SELECT, PROJECT, and JOIN,...Ch. 9 - Answer Problem 13 using SQL. PROBLEM 13 13. Using...Ch. 9 - Prob. 15CRPCh. 9 - Prob. 16CRPCh. 9 - Prob. 17CRPCh. 9 - Prob. 18CRPCh. 9 - Prob. 19CRPCh. 9 - Empl Id Name Address SSN Job Id Job Title Skill...Ch. 9 - Empl Id Name Address SSN Job Id Job Title Skill...Ch. 9 - Prob. 22CRPCh. 9 - Prob. 23CRPCh. 9 - Prob. 24CRPCh. 9 - Prob. 25CRPCh. 9 - Write a sequence of instructions (using the...Ch. 9 - Prob. 27CRPCh. 9 - Prob. 28CRPCh. 9 - Prob. 29CRPCh. 9 - Prob. 30CRPCh. 9 - Prob. 31CRPCh. 9 - Prob. 32CRPCh. 9 - Prob. 33CRPCh. 9 - Prob. 34CRPCh. 9 - Prob. 35CRPCh. 9 - Prob. 36CRPCh. 9 - Prob. 37CRPCh. 9 - Prob. 38CRPCh. 9 - Prob. 39CRPCh. 9 - Prob. 40CRPCh. 9 - Prob. 41CRPCh. 9 - Prob. 42CRPCh. 9 - Prob. 43CRPCh. 9 - Prob. 44CRPCh. 9 - Prob. 45CRPCh. 9 - Prob. 46CRPCh. 9 - Prob. 47CRPCh. 9 - Prob. 48CRPCh. 9 - Prob. 49CRPCh. 9 - Prob. 50CRPCh. 9 - Prob. 51CRPCh. 9 - Prob. 52CRPCh. 9 - Prob. 53CRPCh. 9 - Prob. 54CRPCh. 9 - Prob. 55CRPCh. 9 - Prob. 56CRPCh. 9 - Prob. 57CRPCh. 9 - Prob. 58CRPCh. 9 - Prob. 59CRPCh. 9 - Prob. 60CRPCh. 9 - Prob. 61CRPCh. 9 - Prob. 62CRPCh. 9 - Prob. 1SICh. 9 - Prob. 2SICh. 9 - Prob. 3SICh. 9 - Prob. 4SICh. 9 - Prob. 5SICh. 9 - Prob. 6SICh. 9 - Prob. 7SICh. 9 - Prob. 8SICh. 9 - Prob. 9SICh. 9 - Prob. 10SI
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- The following tables form part of a Library database held in an RDBMS: Book (ISBN, title, edition, year) BookCopy (copyNo, ISBN, available) Borrower (borrowerNo, borrowerName, borrowerAddress) BookLoan (copyNo, dateOut, dateDue, borrowerNo) where: Book contains details of book titles in the library and the ISBN is the key. BookCopy contains details of the individual copies of books in the library and copyNo is the key. ISBN is a foreign key identifying the book title. Borrower contains details of library members who can borrow books and borrowerNo is the key. BookLoan contains details of the book copies that are borrowed by library members and copyNo/dateOut forms the key. borrowerNo is a foreign key identifying the borrower. Formulate the following queries in relational algebra and tuple relational calculus: a. List all book titles. b. List all borrower details. c.List all book titles published in the year 2012. d. List all copies of book titles that are…arrow_forwardThe following tables form part of a Library database held in an RDBMS: Book (ISBN, title, edition, year) BookCopy (copyNo, ISBN, available) Borrower (borrowerNo, borrowerName, borrowerAddress) BookLoan (copyNo, dateOut, dateDue, borrowerNo) where: Book contains details of book titles in the library and the ISBN is the key. BookCopy contains details of the individual copies of books in the library and copyNo is the key. ISBN is a foreign key identifying the book title. Borrower contains details of library members who can borrow books and borrowerNo is the key. BookLoan contains details of the book copies that are borrowed by library members and copyNo/dateOut forms the key. borrowerNo is a foreign key identifying the borrower. Formulate the following queries in relational algebra only. a. How many copies of ISBN “0-321-52306-7” are there? b. How many copies of ISBN “0-321-52306-7” are currently available c. How many times has the book title with ISBN…arrow_forwardNeed a c++ program to connect to this database and perform add, search, remove, display operations. -- students create table students ( student_id int, name varchar(50), dob date, phone int, email varchar(50), address text, year_level int, section varchar(2), primary key(student_id) ); -- staff create table staff(staff_id int,name varchar(50),address text,phone int, email varchar(50),position varchar(20),salary int,primary key(staff_id) ); -- book_subject create table( subject_id int primary key, subject varchar(50)); -- books create table books( book_id int,title varchar(50),published_year int,isbn int,cost int, subject_id int, primary key(book_id),foreign key(subject_id) references book_subject(subject_id) ); -- book_issue create table book_issue( issue_id int, issue_date date, staff_id intjumb, student_id int, book_id int, return_date da primary key(issue_id), foreign key(staff_id) references…arrow_forward
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