хомяк убийца
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God Philosophy Quiz on ГОС по БД #1, created by хомяк убийца on 23/03/2019.

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хомяк убийца
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ГОС по БД #1

Question 1 of 89

1

Define a degree of a relation

Select one of the following:

  • How many rows a table has

  • how long each tuple is, or how many columns the table has

  • how many different tuples there are

  • how many different datatypes table has

Explanation

Question 2 of 89

1

Define a cardinality of a relation

Select one of the following:

  • how long each tuple is

  • how many columns the table has

  • how many different tuples there are, how many rows a table has

  • how many different datatypes table has

Explanation

Question 3 of 89

1

Which of the following refers to union-compatibility requirements ?

Select one of the following:

  • Same number of columns

  • Same number of rows

  • Same number of tuples

  • Different domains

Explanation

Question 4 of 89

1

Which of the following refers to union-compatibility requirements ?

Select one of the following:

  • Corresponding columns have the same domains

  • Same number of rows

  • Same number of tuples

  • Corresponding columns have the same fields

Explanation

Question 5 of 89

1

Define a domain

Select one of the following:

  • restrict the possible values a tuple can assign to each attribute

  • relations to each other

  • uniquely identifies each tuple that appears in a relation

  • minimality of the attribute

Explanation

Question 6 of 89

1

Define a foreign key

Select one of the following:

  • restrict the possible values a tuple can assign to each attribute

  • relations to each other

  • uniquely identifies each tuple that appears in a relation

  • minimality of the attribute

Explanation

Question 7 of 89

1

Define a primary key

Select one of the following:

  • restrict the possible values a tuple can assign to each attribute

  • relations to each other

  • uniquely identifies each tuple that appears in a relation

  • datatypes of attributes

Explanation

Question 8 of 89

1

Define restrict

Select one of the following:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explanation

Question 9 of 89

1

Define cascade

Select one of the following:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explanation

Question 10 of 89

1

Define set default

Select one of the following:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explanation

Question 11 of 89

1

Define set null

Select one of the following:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explanation

Question 12 of 89

1

Define Data Definition Language

Select one of the following:

  • Specify database format

  • Specify access controls (privileges)

  • Specify and retrieve database contents

  • Specify table attribute uniqueness

Explanation

Question 13 of 89

1

Define Data Control Language

Select one of the following:

  • Specify access controls (privileges)

  • Specify database format

  • Specify and retrieve database contents

  • Specify table attribute uniqueness

Explanation

Question 14 of 89

1

Define Data Manipulation Language

Select one of the following:

  • Specify table attribute uniqueness

  • Specify database format

  • Specify access controls (privileges)

  • Specify and retrieve database contents

Explanation

Question 15 of 89

1

Which of the following does not refer to DBMS tools

Select one of the following:

  • Oracle

  • PostgreSQL

  • MySQL

  • Python

Explanation

Question 16 of 89

1

Which of the following is used to provide privilege to only a particular attribute?

Select one of the following:

  • Grant select on employee to finance

  • Grant update(salary, rate) on employee to finance

  • Grant update(salary) on employee to finance

  • Grant delete to finance

Explanation

Question 17 of 89

1

Which of the following statement is used to remove the privilege from the user finance?

Select one of the following:

  • Remove update on employee from finance

  • Delete select on employee from finance

  • Revoke update on employee from finance

  • Grant update on employee from finance

Explanation

Question 18 of 89

1

Which of the following is true regarding views?

Select one of the following:

  • The user who creates a view cannot be given update authorization on a view without having update authorization on the relations used to define the view.

  • If a user creates a view on which no authorization can be granted, the system will allow the view creation request.

  • A user who creates a view receives all privileges on that view.

Explanation

Question 19 of 89

1

If we wish to grant a privilege and to allow the recipient to pass the privilege on to other users, we append the __________ clause to the appropriate grant command.

Select one of the following:

  • With grant

  • Grant user

  • With grant option

  • Grant pass privelege

Explanation

Question 20 of 89

1

Which of the following is used to avoid cascading of authorizations from the user?

Select one of the following:

  • Granted by current role

  • Revoke grant option for select on department from finance;

  • Revoke select on employee from finance, cashier restrict;

  • Revoke select on department from finance, cashier cascade;

Explanation

Question 21 of 89

1

Privileges are granted over some specified parts of a database, such as a

Select one of the following:

  • Schema

  • Environment

  • Relation Or view

  • Query statement

Explanation

Question 22 of 89

1

Prevention of access to the database by unauthorized users is referred to as:

Select one of the following:

  • Integrity

  • Productivity

  • Security

  • Reliability

Explanation

Question 23 of 89

1

Database Authentication refers to:

Select one of the following:

  • methods of restricting user access to system

  • controlling access to portions of database

  • all of the answers mentioned

  • controlling the operation on the data

Explanation

Question 24 of 89

1

A set of possible data values is called

Select one of the following:

  • attribute

  • degree

  • domain

  • tuple

Explanation

Question 25 of 89

1

A functional dependency between two or more non-key attributes is called

Select one of the following:

  • Partial transitive dependency

  • Functional dependency

  • Transitive dependency

  • Partial functional dependency

Explanation

Question 26 of 89

1

__ refers to the correctness and completeness of the data in a database.

Select one of the following:

  • Database security

  • Data constraint

  • Data integrity

  • Data independence

Explanation

Question 27 of 89

1

Which of the following creates a virtual relation for storing the query?

Select one of the following:

  • Function

  • Procedure

  • View

  • None of the mentioned

Explanation

Question 28 of 89

1

Which of the following is the syntax for views where v is view name?

Select one of the following:

  • Create view v as “query name”;

  • Create “query expression” as view;

  • Create view v as “query expression”;

  • Create view “query expression”;

Explanation

Question 29 of 89

1

Updating the value of the view

Select one of the following:

  • Will not change the view definition

  • Will not affect the relation from which it is defined

  • Will affect the relation from which it is defined

  • Cannot determine

Explanation

Question 30 of 89

1

Create view faculty as: Select ID, name, dept name from instructor; Find the error in this query.

Select one of the following:

  • Instructor

  • Select

  • None of the mentioned

  • View …as

Explanation

Question 31 of 89

1

Which of the following is a basic form of grant statement?

Select one of the following:

  • Grant ‘privilege list’ on ‘user/role list’ to ‘relation name or view name’;

  • Grant ‘privilege list’ to ‘user/role list’;

  • Grant ‘privilege list’ on ‘relation name or view name’ to ‘user/role list’;

  • Grant ‘privilege list’ on ‘relation name or view name’ on ‘user/role list’;

Explanation

Question 32 of 89

1

Retrieve all data from the table OFFICE { id,room, name}

Select one of the following:

  • Select *from office;

  • Select from office;

  • Select name from office;

  • Select *form office;

Explanation

Question 33 of 89

1

Retrieve office name from the table OFFICE {id, room, name}

Select one of the following:

  • Select *from office;

  • Select from office;

  • Select name from office;

  • Select *form office;

Explanation

Question 34 of 89

1

Retrieve office id from the table OFFICE {id, room, name}

Select one of the following:

  • Select *from office;

  • Select id from office;

  • Select name from office;

  • Select id form office;

Explanation

Question 35 of 89

1

Retrieve office id and room from the table OFFICE {id, room, name}

Select one of the following:

  • Select *from office;

  • Select id, room from office;

  • Select room name from office;

  • Select id form office;

Explanation

Question 36 of 89

1

Retrieve quantity of offices in the office table - OFFICE {id, rom, name}

Select one of the following:

  • Select avg (id) from office;

  • Select count (id) from office;

  • Select sum (id) from office;

  • Select max (id) from office;

Explanation

Question 37 of 89

1

Retrieve total number scholarship in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select avg (scholarship) from office;

  • Select count (scholarship) from office;

  • Select sum (scholarship) from office;

  • Select max (scholarship) from office;

Explanation

Question 38 of 89

1

Retrieve quantity of students in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select avg (id) from office;

  • Select count (id) from office;

  • Select sum (id) from office;

  • Select max (id) from office;

Explanation

Question 39 of 89

1

Sort students name by descending order - STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select name from students order by desc;

  • Select name from students group by desc;

  • Select *from students order by desc;

  • Select name form students order by desc;

Explanation

Question 40 of 89

1

Find an average scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select sum (scholarship) from students;

  • Select max (scholarship) from students;

Explanation

Question 41 of 89

1

Find maximum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select sum (scholarship) from students;

  • Select max (scholarship) from students;

Explanation

Question 42 of 89

1

Find minimum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}

Select one of the following:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select min (scholarship) from students;

  • Select max (scholarship) from students;

Explanation

Question 43 of 89

1

Retrieve offices’ name and tutors’, who work in CSSE department. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, OFFICEID (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘CSSE’

  • Select o.name, t.name from office o join tutor t on o.id=t.id where o.name = ‘CSSE’

  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘csse’

  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name ‘CSSE’

Explanation

Question 44 of 89

1

Retrieve all offices’ name and tutors’, who work in departments. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • Select o.name, t.name from office o left join tutor t on o.id=t.officeid;

  • Select o.name, t.name from office o left join tutor t on o.id=t.id ;

  • Select o.name, t.name from office o right join tutor t on o.id=t.officeid;

  • Select o.name, t.name from office o right join tutor t on o.id=t.id ;

Explanation

Question 45 of 89

1

Retrieve students’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select s.name from students s where s.scholarship > (select avg (scholarship)from students);

  • select s.name from students s where s.scholarship > (select scholarship from students);

  • select s.name from students s where s.scholarship > avg (scholarship);

  • select s.name from students s where s.scholarship < avg (scholarship);

Explanation

Question 46 of 89

1

Retrieve students’ name, who registered in 10th of Jan 2016. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }.

Select one of the following:

  • select name from students where registereddate = '2016-01-10';

  • select name from students where registereddate = '2016-10-10';

  • select name from students where registereddate = 2016-10-10;

  • select name from students where registereddate = '2016-01-01';

Explanation

Question 47 of 89

1

Retrieve tutors’ name, who have more than experience others along with their students name . Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select max (experience) from tutor);

  • select t.name, s.name, max(experience) from tutor t join students s on t.id=s.tutorid;

  • select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select max (experience) from tutor);

Explanation

Question 48 of 89

1

Retrieve tutors’ name, who have less experience than others along with their students name. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select min (experience) from tutor);

  • select t.name, s.name, min(experience) from tutor t join students s on t.id=s.tutorid;

  • select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select min (experience) from tutor);

Explanation

Question 49 of 89

1

Increase tutors experience and students’ scholarship twice. Retrieve experience and scholarship along with their names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.tutorid;

  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.id;

  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.tutorid;

  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.id;

Explanation

Question 50 of 89

1

Retrieve office, tutor and students name, but tutors name should have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';

  • select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name like '%y%';

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name like '%y%';

Explanation

Question 51 of 89

1

Retrieve office, tutor and students name, but tutors name should not have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutored (FK references tutor (id))}.

Select one of the following:

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';

  • select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name not like '%y%';

  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name not like '%y%';

Explanation

Question 52 of 89

1

Retrieve students name and scholarship along with their department name, who have scholarship between 4000 and 5000. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;

  • select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship 4000 and 5000;

  • select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;

  • select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship >=4000

Explanation

Question 53 of 89

1

Retrieve information about office and their tutors as well. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.d

Select one of the following:

  • select *from office o join tutor t on o.id=t.officeid;

  • select *from office o join tutor t in o.id=t.officeid;

  • select *form office o join tutor t on o.id=t.officeid;

  • select *from office o join tutor t on o.id=t.id;

Explanation

Question 54 of 89

1

Retrieve office name and tutor name, which have maximum experienced tutors. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select max (experience) from tutor);

  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select min (experience) from tutor);

  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select high (experience) from tutor);

  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select low (experience) from tutor);

Explanation

Question 55 of 89

1

Retrieve all tutors and students names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select t.name, s.name from tutor t full join students s on t.id=s.tutorid;

  • select t.name, s.name from tutor t full join students s t.id=s.tutorid;

  • select t.name, s.name from tutor t full join students s t.id=s.id;

  • select t.name, s.name from tutor t full join students s on t.id=s.id;

Explanation

Question 56 of 89

1

Retrieve tutors’ name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutoridgroup by t.name;

  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutorid;

  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.idgroup by t.name;

  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.id;

Explanation

Question 57 of 89

1

Retrieve office name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.

Select one of the following:

  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o on o.id=t.officeid group by o.name;

  • select count (s.name) as numberofstudents, o.name from students s on join office o on o.id=s.id group by o.name;

  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o o.id=t.officeid group by o.name;

  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.idjoin office o on o.id=t.id group by o.name;

Explanation

Question 58 of 89

1

Retrieve all students’ name and scholarship, tutors name except students who have scholarship 3000$. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship <> 3000;

  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship = 3000;

  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship < 3000;

  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship is 3000;

Explanation

Question 59 of 89

1

How many types of anomalies exist?

Select one of the following:

  • 1

  • 2

  • 3

  • 4

Explanation

Question 60 of 89

1

Which of the following anomaly does not exist?

Select one of the following:

  • Creation anomaly

  • Deletion anomaly

  • Insertion anomaly

  • Modification anomaly

Explanation

Question 61 of 89

1

When an insertion anomaly occurs?

Select one of the following:

  • we are prevented from inserting some data into a relation until other data can be supplied

  • deletion leads to an unintended loss of data

  • it is possible that not all data needs to be changed will always be changed

  • it does not occur

Explanation

Question 62 of 89

1

When a modification anomaly occurs?

Select one of the following:

  • it is possible that not all data needs to be changed will always be changed

  • we are prevented from inserting some data into a relation until other data can be supplied

  • deletion leads to an unintended loss of data

  • it does not occur

Explanation

Question 63 of 89

1

When a deletion anomaly occurs?

Select one of the following:

  • deletion leads to an unintended loss of data

  • it is possible that not all data needs to be changed will always be changed

  • we are prevented from inserting some data into a relation until other data can be supplied

  • it does not occur

Explanation

Question 64 of 89

1

Define inner join

Select one of the following:

  • selects all rows from both tables as long as there is a match between the columns in both tables

  • returns all rows from the left table (1), with the matching rows in the right table (2)

  • returns all rows from the right table (2), with the matching rows in the right table (1)

  • returns all rows from the left table (1) and from the right table (2)

Explanation

Question 65 of 89

1

Define left join

Select one of the following:

  • returns all rows from the left table (1), with the matching rows in the right table (2)

  • selects all rows from both tables as long as there is a match between the columns in both tables

  • returns all rows from the right table (2), with the matching rows in the right table (1)

  • returns all rows from the left table (1) and from the right table (2)

Explanation

Question 66 of 89

1

Define right join

Select one of the following:

  • returns all rows from the right table (2), with the matching rows in the left table (1)

  • returns all rows from the left table (1), with the matching rows in the right table (2)

  • selects all rows from both tables as long as there is a match between the columns in both tables

  • returns all rows from the left table (1) and from the right table (2)

Explanation

Question 67 of 89

1

Define full join

Select one of the following:

  • returns all rows from the left table (1) and from the right table (2)

  • returns all rows from the right table (2), with the matching rows in the right table (1)

  • returns all rows from the left table (1), with the matching rows in the right table (2)

  • selects all rows from both tables as long as there is a match between the columns in both tables

Explanation

Question 68 of 89

1

Define an union

Select one of the following:

  • сombines the result set of two or more select statements

  • returns all rows from the left table (1) and from the right table (2)

  • returns all rows from the right table (2), with the matching rows in the right table (1)

  • returns all rows from the left table (1), with the matching rows in the right table (2)

Explanation

Question 69 of 89

1

INNER JOIN and JOIN are the same

Select one of the following:

  • True
  • False

Explanation

Question 70 of 89

1

Define self join

Select one of the following:

  • returns all rows from the table that references to yourself

  • returns all rows from the right table (2), with the matching rows in the right table (1)

  • selects all rows from both tables as long as there is a match between the columns in both tables

  • returns all rows from the left table (1) and from the right table (2)

Explanation

Question 71 of 89

1

How many types of functional dependencies exist

Select one of the following:

  • 1

  • 2

  • 3

  • 0

Explanation

Question 72 of 89

1

Which of the following does not to the functional dependency

Select one of the following:

  • equational

  • full

  • transitive

  • partial

Explanation

Question 73 of 89

1

Which of the following represent a full dependency?

Select one of the following:

  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds

  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds

  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y

  • all above mentioned

Explanation

Question 74 of 89

1

Which of the following represent a partial dependency?

Select one of the following:

  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds

  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds

  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y

  • all above mentioned

Explanation

Question 75 of 89

1

Which of the following represent a transitive dependency?

Select one of the following:

  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y->Z in R. Final: X->Y

  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds

  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds

  • all above mentioned

Explanation

Question 76 of 89

1

When was first normalization developed?

Select one of the following:

  • 1970

  • 1972

  • 1973

  • 1974

Explanation

Question 77 of 89

1

When was second normalization developed?

Select one of the following:

  • 1971

  • 1972

  • 1973

  • 1974

Explanation

Question 78 of 89

1

When was Boyce–Codd normalization developed?

Select one of the following:

  • 1974

  • 1972

  • 1973

  • 1971

Explanation

Question 79 of 89

1

Who is an inventor of relational model?

Select one of the following:

  • Edgar F.Codd

  • Raymond Boyce

  • Marine Jone

  • John Salamondor

Explanation

Question 80 of 89

1

Which of the following refers to the requirement of 1NF

Select one of the following:

  • Each cell should be single valued

  • Entries in a column are same type

  • Rows uniquely identified

  • All above mentioned

Explanation

Question 81 of 89

1

Which of the following refers to the requirement of 2NF

Select one of the following:

  • All attributes (non-key columns) dependent on the key

  • Each cell should be single valued

  • All fields (columns) can be determined only by the key in the table and no other column

  • All above mentioned

Explanation

Question 82 of 89

1

Which of the following refers to the requirement of 3NF

Select one of the following:

  • All fields (columns) can be determined only by the key in the table and no other column

  • All attributes (non-key columns) dependent on the key

  • Each cell should be single valued

  • All above mentioned

Explanation

Question 83 of 89

1

Define avg function

Select one of the following:

  • Returns average value

  • Returns total value

  • Returns the first value

  • Converts to lowercase

Explanation

Question 84 of 89

1

Which function is used to retrieve quantity of rows

Select one of the following:

  • count

  • sum

  • max

  • avg

Explanation

Question 85 of 89

1

Retrieve avg scholarship of students. Tables: office {id (PK), locations, name}, tutor {id (PK), name, officeid (FK references office (id)), experience}, students {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select avg (scholarship) from students;

  • select aveg (scholarship) from students;

  • select avr (scholarship) from students;

  • select avgr (scholarship) from students;

Explanation

Question 86 of 89

1

Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)

  • select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)

  • select s.name, s.scholarship, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)

  • select s.name, avg(s.scholarship), t.name from students s, tutor t

Explanation

Question 87 of 89

1

Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship AND increase those students twice. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)

  • select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)

  • select s.name, s.scholarship*2, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)

  • select s.name, avg(s.scholarship)*2, t.name from students s, tutor t

Explanation

Question 88 of 89

1

Retrieve students and mentors name, but mentors registration date should be before students registration date. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate<m.registereddate;

  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate>m.registereddate;

  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid;

  • select s.name, m.name from students s join students m on s.registereddate=m.registereddate;

Explanation

Question 89 of 89

1

Retrieve the highest experience from tutor. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))

Select one of the following:

  • select max (experience) from tutor;

  • select min (experience) from tutor;

  • select max (t.experience) from tutor t group by t.name;

  • select min (t.experience) from tutor t group by t.name;

Explanation