Q1:-1- Find superkeys, candidate keys, primary keys, alternative keys in the followingCAR relation schema?CAR (State, Reg#, SerialNo, Model, Year)2- Find superkeys, candidate keys, primary keys, alternative keys in the followingrelation?A1Hh101D1010A2Mm200D3020АЗFf50D9030A4LI60D5330A1Ff70D610

Candidate Key: A candidate key refers to a column or combination of columns which uniquely…

Answered: Q1:- 1- Find superkeys, candidate keys,…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Answer fast no need for explanation
Please written by computer source Express the following table of data as a relation S. x,0,2,3,5,7 y,4,2,-2,-5,-6 The correct answer contains ordered pairs. Be sure to enter them with parentheses and separate pairs with a comma
(Choose all of the correct superkeys.) Find all superkeys of the DEPARTMENT relation. Its primary key (PK) is underlined. DEPARTMENT (number, dname, mgrssn, location) PK = {dname, location} {number} {number, dname} {number, dname, mgrssn} {number, dname, mgrssn, location} {dname} {dname, mgrssn} {dname, mgrssn, location} {mgrssn} {mgrssn, location} {location} {number, dname, mgrssn} {number, dname, location} {number, mgrssn} {number, mgrssn, location} {number, location} {dname, location}
3
If relation A has 8 rows and 4 attributes and relation B has 4 rows and 4 attributes and the union operation was carried out on relation A and relation B, What would be the cardinaility of the new relation C where CA U B if there are no duplicate rows?
What are three reasons to create a surrogate key for a relation's main key?
Let R and K be two relations under a join operation. Assume that m = |R| and n = IKI, then the maximum and minimum sizes of the join, respectively, are: 1. m+n and 0 II. mn and 0 III. m+n and m-n IV. mn and m+n III O IV =
Sec course is an example of a link that may be built from a collection of many-to-one connections. It's possible that the main and foreign key restrictions enforce the relation's many-to-one cardinality limitation. Is there a reason for this?
Consider a relation r (A,B,C,D). If we wanted to generate the information produced by select A, B, C, count(D) from r group by cube (A,B,C) using several queries none of which use "cube" nor "rollup" and none of which have subqueries how many distinct select/from expressions would be needed?
Transform ERD into Relation
Write a relation algebra based on the following algebra: Book (BookID, Price, Year, Author, Subject) BorrowLog (BID, BookID, CustomerID, DateTime) Customer (CustomerID, CustomerName, Address, City) 1. List all of the customer names who have borrowed the book written by James Clear 2. List the subject of the books that were borrowed on the date, 2021-10-07, but not on 2021-10-01 3. List the subject of the books that were borrowed by the customers who are in Bangkok.

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