Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Chapter 3, Problem 30E
Explanation of Solution
Explanation:
- For numeric values it will always be zero. But when there is a null value the answer will be undefined.
- When there is any instant with value equal to null then the answer will be undefined which is not equal to zero...
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Consider the following SQL query on the university schema:select avg(salary) - (sum(salary) / count(*))from instructorWe might expect that the result of this query is zero since the average of a setof numbers is defined to be the sum of the numbers divided by the number ofnumbers. Indeed this is true for the example instructor relation in Figure 2.1.However, there are other possible instances of that relation for which the resultwould not be zero. Give one such instance, and explain why the result wouldnot be zero.
Consider the following SQL Query:
SELECT FNAME, DNAME
FROM EMP AS E, DEPT AS D
WHERE E.DNO = D.DNUMBER AND D.DNAME='CCI' AND
SALARY > 15000;
Draw the initial query tree for the above SQL query, and then show
how
the query tree is optimized by applying the Heuristic Rules. (You
need to show the tree after applying a rule).
PLZ help with the following: IN SQL
Let R (A, B, C) be a relation schema. What happens if we execute the following query?
SELECT* FROM R WHERE A=D;
Select one:
a. We get an error.
b. The query executes successfully but returns no tuples.
c. The query returns all the tuples in R.
Chapter 3 Solutions
Database System Concepts
Ch. 3 - Prob. 1PECh. 3 - Prob. 2PECh. 3 - Prob. 3PECh. 3 -
Suppose that we have a relation marks(ID, score)...Ch. 3 - Suppose that we have a relation marks(ID, score)...Ch. 3 -
The SQL like operator is case sensitive (in most...Ch. 3 - Prob. 7PECh. 3 - Prob. 8PECh. 3 - Prob. 9PECh. 3 - Prob. 10PE
Ch. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 -
Consider the insurance database of Figure 3.17,...Ch. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 17ECh. 3 - Prob. 18ECh. 3 -
List two reasons why null values might be...Ch. 3 - Prob. 20ECh. 3 - Prob. 21ECh. 3 - Prob. 22ECh. 3 - Prob. 23ECh. 3 - Prob. 24ECh. 3 - Prob. 25ECh. 3 - Prob. 26ECh. 3 - Using the university schema, write an SQL query to...Ch. 3 - Prob. 28ECh. 3 - Prob. 29ECh. 3 - Prob. 30ECh. 3 - Prob. 31ECh. 3 - Prob. 32ECh. 3 - Prob. 33ECh. 3 - Prob. 34ECh. 3 - Prob. 35E
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- The aim of assignment 1 is to implement a database schema and execute different types of SQL queries in Oracle SQL developer. Given the following relational schema: Student (studid, FirstName, LastName, Email, phoneNumber, DateofBirth, GPA) Club (clubid, ClubName, #studid) MemberOf (#clubid, #studid, joiningDate ) Activities (actid, actdt, place, durationNbHour) Organize (#actid , #clubid , fee) 1. Write SQL queries to create the following tables according to the below descriptions. Table name: Student studid |Char(9) Primary key FirstName Varchar2(50) NOT NULL LastName Varchar2(50) NOT NULL Email Varchar2(50) NOT NULL, UNIQUE phoneNumber Number(8) NOT NULL, UNIQUE DateofBirth Date NOT NULL GPA Number(1,2) NOT NULL Table name: Club clubID Number(3) Primary key ClubName Varchar2(20)NOT NULL NULL, UNIQUE, NOT Foreign key studid Char(9) Table Name: MemberOf ClubID Number(3) Primary Key, Foreign Key Studid Char(9) Primary Key, Foreign Key joiningDate Date Table Name: Activities Actid…arrow_forwardConsider the following SQL query on the university schema: select avg(salary)-(sum(salary) / count(*)) from instructorWe might expect that the result of this query is zero since the average of a set of numbers is defined to be the sum of the numbers divided by the number of numbers. Indeed this is true for the example instructor relation in Figure 2.1. However, there are other possible instances of that relation for which the result would not be zero. Give one such instance, and explain why the result would not be zero.arrow_forwardThe following SQL queries use the instances of R1 of "Reserves" the instances of B1 of "Boats", and the instances of S1 of "Sailors" relation. An instance R1 of Reserves bid 101 103 103 day 10/10/96 11/12/96 10/11/96 12/01/96 12/05/96 10/10/09 10/10/98 104 10/07/98 11/10/98 09/08/98 09/05/98 09/08/98 SIC 22 22 31 31 22 22 31 64 64 74 103 102 104 102 101 102 103 An instance B1 of Boats bid 101 102 103 bname Interlake Interlake Clipper Marine 104 bcolor blue red green red An instance S2 of Sailors SIG 28 31 21 44 26 58 22 22 29 29 22 32 2 64 21 74 85 95 71 sname yuppy Tubber guppy rusty Dustin Brutus Andy Horte Horatio Zorba Art Bob Amy rating 81 5 10 11 1 & |10|| El 3 115 age 35 155 135 135 45 33 25 135 18 25 635 46 1.If we wish to list out only the names and ratings of the sailors, we can write this in SQL statement as 2.Find out only the names and ages of sailors. 3.To find out the names and ratings of sailors with a rating above 8, we can combine two of the preceding queries as:…arrow_forward
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