17414886
Descripción
Test por хомяк убийца, actualizado hace más de 1 año
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Creado por хомяк убийца
hace más de 5 años
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Pregunta 1
Pregunta
Define a degree of a relation
Respuesta
-
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
Pregunta 2
Pregunta
Define a cardinality of a relation
Respuesta
-
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
Pregunta 3
Pregunta
Which of the following refers to union-compatibility requirements ?
Respuesta
-
Same number of columns
-
Same number of rows
-
Same number of tuples
-
Different domains
Pregunta 4
Pregunta
Which of the following refers to union-compatibility requirements ?
Respuesta
-
Corresponding columns have the same domains
-
Same number of rows
-
Same number of tuples
-
Corresponding columns have the same fields
Pregunta 5
Pregunta
Define a domain
Respuesta
-
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
Pregunta 6
Pregunta
Define a foreign key
Respuesta
-
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
Pregunta 7
Pregunta
Define a primary key
Respuesta
-
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
Pregunta 8
Pregunta
Define restrict
Respuesta
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Pregunta 9
Pregunta
Define cascade
Respuesta
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Pregunta 10
Pregunta
Define set default
Respuesta
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Pregunta 11
Pregunta
Define set null
Respuesta
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Pregunta 12
Pregunta
Define Data Definition Language
Respuesta
-
Specify database format
-
Specify access controls (privileges)
-
Specify and retrieve database contents
-
Specify table attribute uniqueness
Pregunta 13
Pregunta
Define Data Control Language
Respuesta
-
Specify access controls (privileges)
-
Specify database format
-
Specify and retrieve database contents
-
Specify table attribute uniqueness
Pregunta 14
Pregunta
Define Data Manipulation Language
Respuesta
-
Specify table attribute uniqueness
-
Specify database format
-
Specify access controls (privileges)
-
Specify and retrieve database contents
Pregunta 15
Pregunta
Which of the following does not refer to DBMS tools
Respuesta
-
Oracle
-
PostgreSQL
-
MySQL
-
Python
Pregunta 16
Pregunta
Which of the following is used to provide privilege to only a particular attribute?
Respuesta
-
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
Pregunta 17
Pregunta
Which of the following statement is used to remove the privilege from the user finance?
Respuesta
-
Remove update on employee from finance
-
Delete select on employee from finance
-
Revoke update on employee from finance
-
Grant update on employee from finance
Pregunta 18
Pregunta
Which of the following is true regarding views?
Respuesta
-
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.
Pregunta 19
Pregunta
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.
Respuesta
-
With grant
-
Grant user
-
With grant option
-
Grant pass privelege
Pregunta 20
Pregunta
Which of the following is used to avoid cascading of authorizations from the user?
Respuesta
-
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;
Pregunta 21
Pregunta
Privileges are granted over some specified parts of a database, such as a
Respuesta
-
Schema
-
Environment
-
Relation Or view
-
Query statement
Pregunta 22
Pregunta
Prevention of access to the database by unauthorized users is referred to as:
Respuesta
-
Integrity
-
Productivity
-
Security
-
Reliability
Pregunta 23
Pregunta
Database Authentication refers to:
Respuesta
-
methods of restricting user access to system
-
controlling access to portions of database
-
all of the answers mentioned
-
controlling the operation on the data
Pregunta 24
Pregunta
A set of possible data values is called
Respuesta
-
attribute
-
degree
-
domain
-
tuple
Pregunta 25
Pregunta
A functional dependency between two or more non-key attributes is called
Respuesta
-
Partial transitive dependency
-
Functional dependency
-
Transitive dependency
-
Partial functional dependency
Pregunta 26
Pregunta
__ refers to the correctness and completeness of the data in a database.
Respuesta
-
Database security
-
Data constraint
-
Data integrity
-
Data independence
Pregunta 27
Pregunta
Which of the following creates a virtual relation for storing the query?
Respuesta
-
Function
-
Procedure
-
View
-
None of the mentioned
Pregunta 28
Pregunta
Which of the following is the syntax for views where v is view name?
Respuesta
-
Create view v as “query name”;
-
Create “query expression” as view;
-
Create view v as “query expression”;
-
Create view “query expression”;
Pregunta 29
Pregunta
Updating the value of the view
Respuesta
-
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
Pregunta 30
Pregunta
Create view faculty as: Select ID, name, dept name from instructor; Find the error in this query.
Respuesta
-
Instructor
-
Select
-
None of the mentioned
-
View …as
Pregunta 31
Pregunta
Which of the following is a basic form of grant statement?
Respuesta
-
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’;
Pregunta 32
Pregunta
Retrieve all data from the table OFFICE { id,room, name}
Respuesta
-
Select *from office;
-
Select from office;
-
Select name from office;
-
Select *form office;
Pregunta 33
Pregunta
Retrieve office name from the table OFFICE {id, room, name}
Respuesta
-
Select *from office;
-
Select from office;
-
Select name from office;
-
Select *form office;
Pregunta 34
Pregunta
Retrieve office id from the table OFFICE {id, room, name}
Respuesta
-
Select *from office;
-
Select id from office;
-
Select name from office;
-
Select id form office;
Pregunta 35
Pregunta
Retrieve office id and room from the table OFFICE {id, room, name}
Respuesta
-
Select *from office;
-
Select id, room from office;
-
Select room name from office;
-
Select id form office;
Pregunta 36
Pregunta
Retrieve quantity of offices in the office table - OFFICE {id, rom, name}
Respuesta
-
Select avg (id) from office;
-
Select count (id) from office;
-
Select sum (id) from office;
-
Select max (id) from office;
Pregunta 37
Pregunta
Retrieve total number scholarship in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
Select avg (scholarship) from office;
-
Select count (scholarship) from office;
-
Select sum (scholarship) from office;
-
Select max (scholarship) from office;
Pregunta 38
Pregunta
Retrieve quantity of students in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
Select avg (id) from office;
-
Select count (id) from office;
-
Select sum (id) from office;
-
Select max (id) from office;
Pregunta 39
Pregunta
Sort students name by descending order - STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
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;
Pregunta 40
Pregunta
Find an average scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select sum (scholarship) from students;
-
Select max (scholarship) from students;
Pregunta 41
Pregunta
Find maximum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select sum (scholarship) from students;
-
Select max (scholarship) from students;
Pregunta 42
Pregunta
Find minimum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Respuesta
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select min (scholarship) from students;
-
Select max (scholarship) from students;
Pregunta 43
Pregunta
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))}.
Respuesta
-
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’
Pregunta 44
Pregunta
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))}.
Respuesta
-
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 ;
Pregunta 45
Pregunta
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))}.
Respuesta
-
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);
Pregunta 46
Pregunta
Retrieve students’ name, who registered in 10th of Jan 2016. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }.
Respuesta
-
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';
Pregunta 47
Pregunta
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))}.
Respuesta
-
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);
Pregunta 48
Pregunta
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))}.
Respuesta
-
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);
Pregunta 49
Pregunta
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))}.
Respuesta
-
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;
Pregunta 50
Pregunta
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))}.
Respuesta
-
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%';
Pregunta 51
Pregunta
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))}.
Respuesta
-
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%';
Pregunta 52
Pregunta
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))}.
Respuesta
-
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
Pregunta 53
Pregunta
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
Respuesta
-
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;
Pregunta 54
Pregunta
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))}.
Respuesta
-
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);
Pregunta 55
Pregunta
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))}.
Respuesta
-
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;
Pregunta 56
Pregunta
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))}.
Respuesta
-
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;
Pregunta 57
Pregunta
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))}.
Respuesta
-
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;
Pregunta 58
Pregunta
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))
Respuesta
-
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;
Pregunta 59
Pregunta
How many types of anomalies exist?
Respuesta
-
1
-
2
-
3
-
4
Pregunta 60
Pregunta
Which of the following anomaly does not exist?
Respuesta
-
Creation anomaly
-
Deletion anomaly
-
Insertion anomaly
-
Modification anomaly
Pregunta 61
Pregunta
When an insertion anomaly occurs?
Respuesta
-
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
Pregunta 62
Pregunta
When a modification anomaly occurs?
Respuesta
-
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
Pregunta 63
Pregunta
When a deletion anomaly occurs?
Respuesta
-
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
Pregunta 64
Pregunta
Define inner join
Respuesta
-
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)
Pregunta 65
Pregunta
Define left join
Respuesta
-
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)
Pregunta 66
Pregunta
Define right join
Respuesta
-
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)
Pregunta 67
Pregunta
Define full join
Respuesta
-
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
Pregunta 68
Pregunta
Define an union
Respuesta
-
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)
Pregunta 69
Pregunta
INNER JOIN and JOIN are the same
Respuesta
- True
- False
Pregunta 70
Pregunta
Define self join
Respuesta
-
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)
Pregunta 71
Pregunta
How many types of functional dependencies exist
Respuesta
-
1
-
2
-
3
-
0
Pregunta 72
Pregunta
Which of the following does not to the functional dependency
Respuesta
-
equational
-
full
-
transitive
-
partial
Pregunta 73
Pregunta
Which of the following represent a full dependency?
Respuesta
-
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
Pregunta 74
Pregunta
Which of the following represent a partial dependency?
Respuesta
-
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
Pregunta 75
Pregunta
Which of the following represent a transitive dependency?
Respuesta
-
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
Pregunta 76
Pregunta
When was first normalization developed?
Respuesta
-
1970
-
1972
-
1973
-
1974
Pregunta 77
Pregunta
When was second normalization developed?
Respuesta
-
1971
-
1972
-
1973
-
1974
Pregunta 78
Pregunta
When was Boyce–Codd normalization developed?
Respuesta
-
1974
-
1972
-
1973
-
1971
Pregunta 79
Pregunta
Who is an inventor of relational model?
Respuesta
-
Edgar F.Codd
-
Raymond Boyce
-
Marine Jone
-
John Salamondor
Pregunta 80
Pregunta
Which of the following refers to the requirement of 1NF
Respuesta
-
Each cell should be single valued
-
Entries in a column are same type
-
Rows uniquely identified
-
All above mentioned
Pregunta 81
Pregunta
Which of the following refers to the requirement of 2NF
Respuesta
-
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
Pregunta 82
Pregunta
Which of the following refers to the requirement of 3NF
Respuesta
-
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
Pregunta 83
Pregunta
Define avg function
Respuesta
-
Returns average value
-
Returns total value
-
Returns the first value
-
Converts to lowercase
Pregunta 84
Pregunta
Which function is used to retrieve quantity of rows
Respuesta
-
count
-
sum
-
max
-
avg
Pregunta 85
Pregunta
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))
Respuesta
-
select avg (scholarship) from students;
-
select aveg (scholarship) from students;
-
select avr (scholarship) from students;
-
select avgr (scholarship) from students;
Pregunta 86
Pregunta
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))
Respuesta
-
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
Pregunta 87
Pregunta
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))
Respuesta
-
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
Pregunta 88
Pregunta
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))
Respuesta
-
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;
Pregunta 89
Pregunta
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))
Respuesta
-
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;
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