хомяк убийца
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God Philosophy Quiz sobre ГОС по БД #1, criado por хомяк убийца em 23-03-2019.

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хомяк убийца
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ГОС по БД #1

Questão 1 de 89

1

Define a degree of a relation

Selecione uma das seguintes:

  • 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

Explicação

Questão 2 de 89

1

Define a cardinality of a relation

Selecione uma das seguintes:

  • 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

Explicação

Questão 3 de 89

1

Which of the following refers to union-compatibility requirements ?

Selecione uma das seguintes:

  • Same number of columns

  • Same number of rows

  • Same number of tuples

  • Different domains

Explicação

Questão 4 de 89

1

Which of the following refers to union-compatibility requirements ?

Selecione uma das seguintes:

  • Corresponding columns have the same domains

  • Same number of rows

  • Same number of tuples

  • Corresponding columns have the same fields

Explicação

Questão 5 de 89

1

Define a domain

Selecione uma das seguintes:

  • 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

Explicação

Questão 6 de 89

1

Define a foreign key

Selecione uma das seguintes:

  • 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

Explicação

Questão 7 de 89

1

Define a primary key

Selecione uma das seguintes:

  • 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

Explicação

Questão 8 de 89

1

Define restrict

Selecione uma das seguintes:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explicação

Questão 9 de 89

1

Define cascade

Selecione uma das seguintes:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explicação

Questão 10 de 89

1

Define set default

Selecione uma das seguintes:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explicação

Questão 11 de 89

1

Define set null

Selecione uma das seguintes:

  • stop the user from doing it

  • let the changes flow on

  • make referencing values the default for their column

  • make referencing values null

Explicação

Questão 12 de 89

1

Define Data Definition Language

Selecione uma das seguintes:

  • Specify database format

  • Specify access controls (privileges)

  • Specify and retrieve database contents

  • Specify table attribute uniqueness

Explicação

Questão 13 de 89

1

Define Data Control Language

Selecione uma das seguintes:

  • Specify access controls (privileges)

  • Specify database format

  • Specify and retrieve database contents

  • Specify table attribute uniqueness

Explicação

Questão 14 de 89

1

Define Data Manipulation Language

Selecione uma das seguintes:

  • Specify table attribute uniqueness

  • Specify database format

  • Specify access controls (privileges)

  • Specify and retrieve database contents

Explicação

Questão 15 de 89

1

Which of the following does not refer to DBMS tools

Selecione uma das seguintes:

  • Oracle

  • PostgreSQL

  • MySQL

  • Python

Explicação

Questão 16 de 89

1

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

Selecione uma das seguintes:

  • 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

Explicação

Questão 17 de 89

1

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

Selecione uma das seguintes:

  • Remove update on employee from finance

  • Delete select on employee from finance

  • Revoke update on employee from finance

  • Grant update on employee from finance

Explicação

Questão 18 de 89

1

Which of the following is true regarding views?

Selecione uma das seguintes:

  • 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.

Explicação

Questão 19 de 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.

Selecione uma das seguintes:

  • With grant

  • Grant user

  • With grant option

  • Grant pass privelege

Explicação

Questão 20 de 89

1

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

Selecione uma das seguintes:

  • 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;

Explicação

Questão 21 de 89

1

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

Selecione uma das seguintes:

  • Schema

  • Environment

  • Relation Or view

  • Query statement

Explicação

Questão 22 de 89

1

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

Selecione uma das seguintes:

  • Integrity

  • Productivity

  • Security

  • Reliability

Explicação

Questão 23 de 89

1

Database Authentication refers to:

Selecione uma das seguintes:

  • methods of restricting user access to system

  • controlling access to portions of database

  • all of the answers mentioned

  • controlling the operation on the data

Explicação

Questão 24 de 89

1

A set of possible data values is called

Selecione uma das seguintes:

  • attribute

  • degree

  • domain

  • tuple

Explicação

Questão 25 de 89

1

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

Selecione uma das seguintes:

  • Partial transitive dependency

  • Functional dependency

  • Transitive dependency

  • Partial functional dependency

Explicação

Questão 26 de 89

1

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

Selecione uma das seguintes:

  • Database security

  • Data constraint

  • Data integrity

  • Data independence

Explicação

Questão 27 de 89

1

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

Selecione uma das seguintes:

  • Function

  • Procedure

  • View

  • None of the mentioned

Explicação

Questão 28 de 89

1

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

Selecione uma das seguintes:

  • Create view v as “query name”;

  • Create “query expression” as view;

  • Create view v as “query expression”;

  • Create view “query expression”;

Explicação

Questão 29 de 89

1

Updating the value of the view

Selecione uma das seguintes:

  • 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

Explicação

Questão 30 de 89

1

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

Selecione uma das seguintes:

  • Instructor

  • Select

  • None of the mentioned

  • View …as

Explicação

Questão 31 de 89

1

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

Selecione uma das seguintes:

  • 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’;

Explicação

Questão 32 de 89

1

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

Selecione uma das seguintes:

  • Select *from office;

  • Select from office;

  • Select name from office;

  • Select *form office;

Explicação

Questão 33 de 89

1

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

Selecione uma das seguintes:

  • Select *from office;

  • Select from office;

  • Select name from office;

  • Select *form office;

Explicação

Questão 34 de 89

1

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

Selecione uma das seguintes:

  • Select *from office;

  • Select id from office;

  • Select name from office;

  • Select id form office;

Explicação

Questão 35 de 89

1

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

Selecione uma das seguintes:

  • Select *from office;

  • Select id, room from office;

  • Select room name from office;

  • Select id form office;

Explicação

Questão 36 de 89

1

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

Selecione uma das seguintes:

  • Select avg (id) from office;

  • Select count (id) from office;

  • Select sum (id) from office;

  • Select max (id) from office;

Explicação

Questão 37 de 89

1

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

Selecione uma das seguintes:

  • Select avg (scholarship) from office;

  • Select count (scholarship) from office;

  • Select sum (scholarship) from office;

  • Select max (scholarship) from office;

Explicação

Questão 38 de 89

1

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

Selecione uma das seguintes:

  • Select avg (id) from office;

  • Select count (id) from office;

  • Select sum (id) from office;

  • Select max (id) from office;

Explicação

Questão 39 de 89

1

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

Selecione uma das seguintes:

  • 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;

Explicação

Questão 40 de 89

1

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

Selecione uma das seguintes:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select sum (scholarship) from students;

  • Select max (scholarship) from students;

Explicação

Questão 41 de 89

1

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

Selecione uma das seguintes:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select sum (scholarship) from students;

  • Select max (scholarship) from students;

Explicação

Questão 42 de 89

1

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

Selecione uma das seguintes:

  • Select avg (scholarship) from students;

  • Select count (scholarship) from students;

  • Select min (scholarship) from students;

  • Select max (scholarship) from students;

Explicação

Questão 43 de 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))}.

Selecione uma das seguintes:

  • 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’

Explicação

Questão 44 de 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))}.

Selecione uma das seguintes:

  • 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 ;

Explicação

Questão 45 de 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))}.

Selecione uma das seguintes:

  • 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);

Explicação

Questão 46 de 89

1

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

Selecione uma das seguintes:

  • 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';

Explicação

Questão 47 de 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))}.

Selecione uma das seguintes:

  • 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);

Explicação

Questão 48 de 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))}.

Selecione uma das seguintes:

  • 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);

Explicação

Questão 49 de 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))}.

Selecione uma das seguintes:

  • 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;

Explicação

Questão 50 de 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))}.

Selecione uma das seguintes:

  • 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%';

Explicação

Questão 51 de 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))}.

Selecione uma das seguintes:

  • 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%';

Explicação

Questão 52 de 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))}.

Selecione uma das seguintes:

  • 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

Explicação

Questão 53 de 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

Selecione uma das seguintes:

  • 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;

Explicação

Questão 54 de 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))}.

Selecione uma das seguintes:

  • 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);

Explicação

Questão 55 de 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))}.

Selecione uma das seguintes:

  • 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;

Explicação

Questão 56 de 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))}.

Selecione uma das seguintes:

  • 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;

Explicação

Questão 57 de 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))}.

Selecione uma das seguintes:

  • 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;

Explicação

Questão 58 de 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))

Selecione uma das seguintes:

  • 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;

Explicação

Questão 59 de 89

1

How many types of anomalies exist?

Selecione uma das seguintes:

  • 1

  • 2

  • 3

  • 4

Explicação

Questão 60 de 89

1

Which of the following anomaly does not exist?

Selecione uma das seguintes:

  • Creation anomaly

  • Deletion anomaly

  • Insertion anomaly

  • Modification anomaly

Explicação

Questão 61 de 89

1

When an insertion anomaly occurs?

Selecione uma das seguintes:

  • 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

Explicação

Questão 62 de 89

1

When a modification anomaly occurs?

Selecione uma das seguintes:

  • 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

Explicação

Questão 63 de 89

1

When a deletion anomaly occurs?

Selecione uma das seguintes:

  • 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

Explicação

Questão 64 de 89

1

Define inner join

Selecione uma das seguintes:

  • 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)

Explicação

Questão 65 de 89

1

Define left join

Selecione uma das seguintes:

  • 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)

Explicação

Questão 66 de 89

1

Define right join

Selecione uma das seguintes:

  • 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)

Explicação

Questão 67 de 89

1

Define full join

Selecione uma das seguintes:

  • 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

Explicação

Questão 68 de 89

1

Define an union

Selecione uma das seguintes:

  • 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)

Explicação

Questão 69 de 89

1

INNER JOIN and JOIN are the same

Selecione uma das opções:

  • VERDADEIRO
  • FALSO

Explicação

Questão 70 de 89

1

Define self join

Selecione uma das seguintes:

  • 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)

Explicação

Questão 71 de 89

1

How many types of functional dependencies exist

Selecione uma das seguintes:

  • 1

  • 2

  • 3

  • 0

Explicação

Questão 72 de 89

1

Which of the following does not to the functional dependency

Selecione uma das seguintes:

  • equational

  • full

  • transitive

  • partial

Explicação

Questão 73 de 89

1

Which of the following represent a full dependency?

Selecione uma das seguintes:

  • 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

Explicação

Questão 74 de 89

1

Which of the following represent a partial dependency?

Selecione uma das seguintes:

  • 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

Explicação

Questão 75 de 89

1

Which of the following represent a transitive dependency?

Selecione uma das seguintes:

  • 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

Explicação

Questão 76 de 89

1

When was first normalization developed?

Selecione uma das seguintes:

  • 1970

  • 1972

  • 1973

  • 1974

Explicação

Questão 77 de 89

1

When was second normalization developed?

Selecione uma das seguintes:

  • 1971

  • 1972

  • 1973

  • 1974

Explicação

Questão 78 de 89

1

When was Boyce–Codd normalization developed?

Selecione uma das seguintes:

  • 1974

  • 1972

  • 1973

  • 1971

Explicação

Questão 79 de 89

1

Who is an inventor of relational model?

Selecione uma das seguintes:

  • Edgar F.Codd

  • Raymond Boyce

  • Marine Jone

  • John Salamondor

Explicação

Questão 80 de 89

1

Which of the following refers to the requirement of 1NF

Selecione uma das seguintes:

  • Each cell should be single valued

  • Entries in a column are same type

  • Rows uniquely identified

  • All above mentioned

Explicação

Questão 81 de 89

1

Which of the following refers to the requirement of 2NF

Selecione uma das seguintes:

  • 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

Explicação

Questão 82 de 89

1

Which of the following refers to the requirement of 3NF

Selecione uma das seguintes:

  • 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

Explicação

Questão 83 de 89

1

Define avg function

Selecione uma das seguintes:

  • Returns average value

  • Returns total value

  • Returns the first value

  • Converts to lowercase

Explicação

Questão 84 de 89

1

Which function is used to retrieve quantity of rows

Selecione uma das seguintes:

  • count

  • sum

  • max

  • avg

Explicação

Questão 85 de 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))

Selecione uma das seguintes:

  • select avg (scholarship) from students;

  • select aveg (scholarship) from students;

  • select avr (scholarship) from students;

  • select avgr (scholarship) from students;

Explicação

Questão 86 de 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))

Selecione uma das seguintes:

  • 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

Explicação

Questão 87 de 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))

Selecione uma das seguintes:

  • 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

Explicação

Questão 88 de 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))

Selecione uma das seguintes:

  • 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;

Explicação

Questão 89 de 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))

Selecione uma das seguintes:

  • 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;

Explicação