Consider the following instance of relation Sells(book_title, library, price): Sells book_title library price The grand design Noura 20,000 Brief history of time Noura 15,000 The grand design Malik's 18,000 Brief history of time Malik's 16,000 Quantum computers Malik's 25,000 Brief history of time Al Dirasat 21,000 What this query is trying to find? Select book title from sells s1 where EXISTS (select from sells where book title=s1.book title and library != s1.library) Select one: O a Books sold by more than one library Ob Books sold by exactly one library OcBooks sold by one library or more. Od Books not sold by any library

Consider the following instance of relation Sells(book_title, library, price): Sells book_title library price The grand design Noura 20,000 Brief history of time Noura 15,000 The grand design Malik's 18,000 Brief history of time Malik's 16,000 Quantum computers Malik's 25,000 Brief history of time Al Dirasat 21,000 What this query is trying to find? Select book title from sells s1 where EXISTS (select from sells where book title=s1.book title and library != s1.library) Select one: O a Books sold by more than one library Ob Books sold by exactly one library OcBooks sold by one library or more. Od Books not sold by any library

Database Systems: Design, Implementation, & Management

12th Edition

ISBN:9781305627482

Author:Carlos Coronel, Steven Morris

Publisher:Carlos Coronel, Steven Morris

Chapter11: Database Performance Tuning And Query Optimization

Section: Chapter Questions

Problem 21P: Problems 1721 are based on the following query: SELECT V_CODE, V_NAME, V_CONTACT, V_STATE FROM...

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) Consider the relations r(A, B, D), and s(B, D, F), in which A, B, D and F are their attributes. Thefollowing SQL query based on these two relations has a nested subquery. Provide an equivalent SQL querywhich does not have any nested subquery.select A, Bfrom rwhere r.B < some (select Bfrom swhere s.D= r.D)
SQL allows a foreign-key dependency to refer to the same relation, as in thefollowing example:create table manager(employee ID char(20),manager ID char(20),primary key employee ID,foreign key (manager ID) references manager(employee ID)on delete cascade )Here, employee_ID is a key to the table manager, meaning that each employeehas at most one manager. The foreign-key clause requires that every manageralso be an employee. Explain exactly what happens when a tuple in the relationmanager is deleted.
Student ID | Student_name | Cource ID | Cource name | Grade 2 Noor 1 HIML B+ 7 C++ A 3 JAVA B 25 Al 2 C+ A 3 JAVA B 27 Sara 1 HIML A 3 JAVA A 28 Abmed 2 C++ B+ 1. What update anomalies may occur in the selation? 2. In which normal form (NF) the above relation? Provide an explanation to support your answer. 3. Decompose the relation so that the resut relations are in the third normal form (3NF).
Table_1 below has a primary key of column1. Which column(s) from this relation could be a foreign key in another, related relation? Check all that qualify. Generic Relation column1 column2 column3 column4 100 19.99 cde A 200 23.78 klj D 300 77.43 qnw B Group of answer choices column2 any of the columns in the table could be a foreign key column4 column3 column1
Consider the following relations:Album (albumId: integer, title: string, artistId (FK): integer)Artist (artistId: integer, name: string)Composer (artistId: integer, name: string)Singer (artistId: integer, name: string)Write the following queries in relational algebra.(1) Find all the titles of the albums.(2) Find all the tuples whose albumId is 2.(3) Rename the relation Artist to Art and attributes to artId, title.(4) Find all the tuples that are either an Artist or a Composer.(5) Find the tuples of Artists who are not a Composer.(6) Find the tuples of people who are Composer and a Singer at the same time.(7) Find the titles of albums created by the artist ‘Black Sabbath’.
Consider the SQL statement:SELECT city, amountFROM Customers INNER JOIN Orders ONCustomers.customerNumberOrders.customerNumberINNER JOIN payments ONCustomers.customerNumberPayments.customerNumberWHEREcustomer.city = "NYC" and orders.status = "Illustrate a relational algebra query tree for the execution of thestatement.
Consider the following relation for published books:BOOK (Book_title, Authorname, Book_type, Listprice, Author_affil, Publisher)Author_affil refers to the affiliation of the author. Suppose the following dependencies exist:Book_title → Publisher, Book_typeBook_type → ListpriceAuthor_name → Author-affil (a) What normal form is the relation in? Explain your answer.(b) Apply normalization until you cannot decompose the relations further. State the reasons behind eachdecomposition.
EmployeeID is the primary key for the relation below. Which column violates a rule (characteristic) of a relation? Employee Relation EmployeeID Department PhoneNum PhoneNum 300 Finance 321-299-1234 407-999-2300 200 Sales 407-222-9090 321-365-0912 400 Sales 407-807-1124 407-453-2294 100 HR 321-934-2034 Group of answer choices The column named PhoneNum because you can’t have two columns with the same name. All the columns are valid. The EmployeeID column because the ID numbers are not in order. The Department column.
The table STUDENT has the attributes STUDENT_ID, NAME, S_ID, and ADDR. The table SCHOOL has the attributes SCHOOL_ID, NAME, and STATE_CODE. Assume that there is a 1:N relation between SCHOOL and STUDENT where the common attributes are S_ID, and SCHOOL_ID. Which of the following creates a table that shows the STUDENT attributes and the corresponding SCHOOL attributes? a. SELECT * FROM STUDENT JOIN SCHOOL USING(SCHOOL_ID) b. SELECT * FROM STUDENT JOIN SCHOOL USING(S_ID) c. SELECT * FROM STUDENT NATURAL JOIN SCHOOL ON S_ID AND SCHOOL.SCHOOL_ID d. SELECT * FROM STUDENT JOIN SCHOOL ON STUDENT.S_ID = SCHOOL.SCHOOL_ID
You are given this table: Registration SID CourseID SemID Instructor Grade 282712 ENGL210 201701 H. Zacny B+ 362112 CS101 201701 K. Ross C 652123 CS403 201701 K. Ross A 362112 CS403 201701 K. Smith C+ 282712 BIO101 201601 I. Olsen C What will the result of the following query be? select sid, count(*)from registrationgroup by sid Select one: a. The entire table sorted by sid b. No results due to an error c. 282712 2362112 2652123 1 d. 282712 1362112 1652123 1
Consider these relations:Emp(eid: integer, ename: varchar, sal: integer, age: integer, did: integer)Dept(did: integer, budget: integer, floor: integer, mgr eid: integer) Salaries range from $5K to $50K, ages vary from 16 to 65, each department has aboutfive employees on average, and budgets vary from $10K to $1 million. For each of the following queries, which of the listed index choices would you choose to speed up thequery? Note: index only plan can be used. Query: Find employees who are younger than 18 with salary more than 8K.(a) b and c.(b) Clustered B+ tree index on < age, sal> fields of Emp.(c) Unclustered B+ index on < age, sal> fields of Emp.(d) Clustered B+ tree index on <sal, age> fields of Emp. Query: Find the dids of departments that are on the 10th floor and have a budget of lessthan $15,000.(a) Clustered B+ tree index on <floor, budget> fields of Dept.(b) Clustered hash index on the <budget, floor > fields of Dept.(c) Unclustered hash index…
Consider the following relationTeacher (tech_id: integer, sal: integer, age: real, did: integer)There is a clustered index one id and an unclustered index on age.1. How would you use the indexes to enforce the constraint that tech_id is a key?2. Write a query of an update that is definitely speeded up because of the available indexes3. Give an example of an update that is definitely slowed down because of the indexes4. Can you give an example of an update that is neither speeded up nor slowed down by the indexes?

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