ГОС по БД #1

Frage 1 von 89

1

Define a degree of a relation

Wähle eine der folgenden:

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

Erklärung

Frage 2 von 89

1

Define a cardinality of a relation

Wähle eine der folgenden:

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

Erklärung

Frage 3 von 89

1

Which of the following refers to union-compatibility requirements ?

Wähle eine der folgenden:

Same number of columns
Same number of rows
Same number of tuples
Different domains

Erklärung

Frage 4 von 89

1

Which of the following refers to union-compatibility requirements ?

Wähle eine der folgenden:

Corresponding columns have the same domains
Same number of rows
Same number of tuples
Corresponding columns have the same fields

Erklärung

Frage 5 von 89

1

Define a domain

Wähle eine der folgenden:

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

Erklärung

Frage 6 von 89

1

Define a foreign key

Wähle eine der folgenden:

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

Erklärung

Frage 7 von 89

1

Define a primary key

Wähle eine der folgenden:

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

Erklärung

Frage 8 von 89

1

Define restrict

Wähle eine der folgenden:

stop the user from doing it
let the changes flow on
make referencing values the default for their column
make referencing values null

Erklärung

Frage 9 von 89

1

Define cascade

Wähle eine der folgenden:

stop the user from doing it
let the changes flow on
make referencing values the default for their column
make referencing values null

Erklärung

Frage 10 von 89

1

Define set default

Wähle eine der folgenden:

stop the user from doing it
let the changes flow on
make referencing values the default for their column
make referencing values null

Erklärung

Frage 11 von 89

1

Define set null

Wähle eine der folgenden:

stop the user from doing it
let the changes flow on
make referencing values the default for their column
make referencing values null

Erklärung

Frage 12 von 89

1

Define Data Definition Language

Wähle eine der folgenden:

Specify database format
Specify access controls (privileges)
Specify and retrieve database contents
Specify table attribute uniqueness

Erklärung

Frage 13 von 89

1

Define Data Control Language

Wähle eine der folgenden:

Specify access controls (privileges)
Specify database format
Specify and retrieve database contents
Specify table attribute uniqueness

Erklärung

Frage 14 von 89

1

Define Data Manipulation Language

Wähle eine der folgenden:

Specify table attribute uniqueness
Specify database format
Specify access controls (privileges)
Specify and retrieve database contents

Erklärung

Frage 15 von 89

1

Which of the following does not refer to DBMS tools

Wähle eine der folgenden:

Oracle
PostgreSQL
MySQL
Python

Erklärung

Frage 16 von 89

1

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

Wähle eine der folgenden:

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

Erklärung

Frage 17 von 89

1

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

Wähle eine der folgenden:

Remove update on employee from finance
Delete select on employee from finance
Revoke update on employee from finance
Grant update on employee from finance

Erklärung

Frage 18 von 89

1

Which of the following is true regarding views?

Wähle eine der folgenden:

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.

Erklärung

Frage 19 von 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.

Wähle eine der folgenden:

With grant
Grant user
With grant option
Grant pass privelege

Erklärung

Frage 20 von 89

1

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

Wähle eine der folgenden:

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;

Erklärung

Frage 21 von 89

1

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

Wähle eine der folgenden:

Schema
Environment
Relation Or view
Query statement

Erklärung

Frage 22 von 89

1

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

Wähle eine der folgenden:

Integrity
Productivity
Security
Reliability

Erklärung

Frage 23 von 89

1

Database Authentication refers to:

Wähle eine der folgenden:

methods of restricting user access to system
controlling access to portions of database
all of the answers mentioned
controlling the operation on the data

Erklärung

Frage 24 von 89

1

A set of possible data values is called

Wähle eine der folgenden:

attribute
degree
domain
tuple

Erklärung

Frage 25 von 89

1

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

Wähle eine der folgenden:

Partial transitive dependency
Functional dependency
Transitive dependency
Partial functional dependency

Erklärung

Frage 26 von 89

1

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

Wähle eine der folgenden:

Database security
Data constraint
Data integrity
Data independence

Erklärung

Frage 27 von 89

1

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

Wähle eine der folgenden:

Function
Procedure
View
None of the mentioned

Erklärung

Frage 28 von 89

1

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

Wähle eine der folgenden:

Create view v as “query name”;
Create “query expression” as view;
Create view v as “query expression”;
Create view “query expression”;

Erklärung

Frage 29 von 89

1

Updating the value of the view

Wähle eine der folgenden:

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

Erklärung

Frage 30 von 89

1

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

Wähle eine der folgenden:

Instructor
Select
None of the mentioned
View …as

Erklärung

Frage 31 von 89

1

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

Wähle eine der folgenden:

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’;

Erklärung

Frage 32 von 89

1

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

Wähle eine der folgenden:

Select *from office;
Select from office;
Select name from office;
Select *form office;

Erklärung

Frage 33 von 89

1

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

Wähle eine der folgenden:

Select *from office;
Select from office;
Select name from office;
Select *form office;

Erklärung

Frage 34 von 89

1

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

Wähle eine der folgenden:

Select *from office;
Select id from office;
Select name from office;
Select id form office;

Erklärung

Frage 35 von 89

1

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

Wähle eine der folgenden:

Select *from office;
Select id, room from office;
Select room name from office;
Select id form office;

Erklärung

Frage 36 von 89

1

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

Wähle eine der folgenden:

Select avg (id) from office;
Select count (id) from office;
Select sum (id) from office;
Select max (id) from office;

Erklärung

Frage 37 von 89

1

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

Wähle eine der folgenden:

Select avg (scholarship) from office;
Select count (scholarship) from office;
Select sum (scholarship) from office;
Select max (scholarship) from office;

Erklärung

Frage 38 von 89

1

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

Wähle eine der folgenden:

Select avg (id) from office;
Select count (id) from office;
Select sum (id) from office;
Select max (id) from office;

Erklärung

Frage 39 von 89

1

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

Wähle eine der folgenden:

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;

Erklärung

Frage 40 von 89

1

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

Wähle eine der folgenden:

Select avg (scholarship) from students;
Select count (scholarship) from students;
Select sum (scholarship) from students;
Select max (scholarship) from students;

Erklärung

Frage 41 von 89

1

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

Wähle eine der folgenden:

Select avg (scholarship) from students;
Select count (scholarship) from students;
Select sum (scholarship) from students;
Select max (scholarship) from students;

Erklärung

Frage 42 von 89

1

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

Wähle eine der folgenden:

Select avg (scholarship) from students;
Select count (scholarship) from students;
Select min (scholarship) from students;
Select max (scholarship) from students;

Erklärung

Frage 43 von 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))}.

Wähle eine der folgenden:

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’

Erklärung

Frage 44 von 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))}.

Wähle eine der folgenden:

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 ;

Erklärung

Frage 45 von 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))}.

Wähle eine der folgenden:

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);

Erklärung

Frage 46 von 89

1

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

Wähle eine der folgenden:

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';

Erklärung

Frage 47 von 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))}.

Wähle eine der folgenden:

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);

Erklärung

Frage 48 von 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))}.

Wähle eine der folgenden:

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);

Erklärung

Frage 49 von 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))}.

Wähle eine der folgenden:

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;

Erklärung

Frage 50 von 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))}.

Wähle eine der folgenden:

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%';

Erklärung

Frage 51 von 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))}.

Wähle eine der folgenden:

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%';

Erklärung

Frage 52 von 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))}.

Wähle eine der folgenden:

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

Erklärung

Frage 53 von 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

Wähle eine der folgenden:

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;

Erklärung

Frage 54 von 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))}.

Wähle eine der folgenden:

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);

Erklärung

Frage 55 von 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))}.

Wähle eine der folgenden:

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;

Erklärung

Frage 56 von 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))}.

Wähle eine der folgenden:

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;

Erklärung

Frage 57 von 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))}.

Wähle eine der folgenden:

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;

Erklärung

Frage 58 von 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))

Wähle eine der folgenden:

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;

Erklärung

Frage 59 von 89

1

How many types of anomalies exist?

Wähle eine der folgenden:

1
2
3
4

Erklärung

Frage 60 von 89

1

Which of the following anomaly does not exist?

Wähle eine der folgenden:

Creation anomaly
Deletion anomaly
Insertion anomaly
Modification anomaly

Erklärung

Frage 61 von 89

1

When an insertion anomaly occurs?

Wähle eine der folgenden:

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

Erklärung

Frage 62 von 89

1

When a modification anomaly occurs?

Wähle eine der folgenden:

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

Erklärung

Frage 63 von 89

1

When a deletion anomaly occurs?

Wähle eine der folgenden:

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

Erklärung

Frage 64 von 89

1

Define inner join

Wähle eine der folgenden:

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)

Erklärung

Frage 65 von 89

1

Define left join

Wähle eine der folgenden:

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)

Erklärung

Frage 66 von 89

1

Define right join

Wähle eine der folgenden:

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)

Erklärung

Frage 67 von 89

1

Define full join

Wähle eine der folgenden:

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

Erklärung

Frage 68 von 89

1

Define an union

Wähle eine der folgenden:

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)

Erklärung

Frage 69 von 89

1

INNER JOIN and JOIN are the same

Wähle eins der folgenden:

WAHR
FALSCH

Erklärung

Frage 70 von 89

1

Define self join

Wähle eine der folgenden:

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)

Erklärung

Frage 71 von 89

1

How many types of functional dependencies exist

Wähle eine der folgenden:

1
2
3
0

Erklärung

Frage 72 von 89

1

Which of the following does not to the functional dependency

Wähle eine der folgenden:

equational
full
transitive
partial

Erklärung

Frage 73 von 89

1

Which of the following represent a full dependency?

Wähle eine der folgenden:

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

Erklärung

Frage 74 von 89

1

Which of the following represent a partial dependency?

Wähle eine der folgenden:

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

Erklärung

Frage 75 von 89

1

Which of the following represent a transitive dependency?

Wähle eine der folgenden:

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

Erklärung

Frage 76 von 89

1

When was first normalization developed?

Wähle eine der folgenden:

1970
1972
1973
1974

Erklärung

Frage 77 von 89

1

When was second normalization developed?

Wähle eine der folgenden:

1971
1972
1973
1974

Erklärung

Frage 78 von 89

1

When was Boyce–Codd normalization developed?

Wähle eine der folgenden:

1974
1972
1973
1971

Erklärung

Frage 79 von 89

1

Who is an inventor of relational model?

Wähle eine der folgenden:

Edgar F.Codd
Raymond Boyce
Marine Jone
John Salamondor

Erklärung

Frage 80 von 89

1

Which of the following refers to the requirement of 1NF

Wähle eine der folgenden:

Each cell should be single valued
Entries in a column are same type
Rows uniquely identified
All above mentioned

Erklärung

Frage 81 von 89

1

Which of the following refers to the requirement of 2NF

Wähle eine der folgenden:

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

Erklärung

Frage 82 von 89

1

Which of the following refers to the requirement of 3NF

Wähle eine der folgenden:

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

Erklärung

Frage 83 von 89

1

Define avg function

Wähle eine der folgenden:

Returns average value
Returns total value
Returns the first value
Converts to lowercase

Erklärung

Frage 84 von 89

1

Which function is used to retrieve quantity of rows

Wähle eine der folgenden:

count
sum
max
avg

Erklärung

Frage 85 von 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))

Wähle eine der folgenden:

select avg (scholarship) from students;
select aveg (scholarship) from students;
select avr (scholarship) from students;
select avgr (scholarship) from students;

Erklärung

Frage 86 von 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))

Wähle eine der folgenden:

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

Erklärung

Frage 87 von 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))

Wähle eine der folgenden:

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

Erklärung

Frage 88 von 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))

Wähle eine der folgenden:

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;

Erklärung

Frage 89 von 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))

Wähle eine der folgenden:

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;

	Erstellt von хомяк убийца vor mehr als 5 Jahre

God Philosophy Quiz am ГОС по БД #1, erstellt von хомяк убийца am 23/03/2019.

ГОС по БД #1

Frage 1 von 89

Define a degree of a relation

Wähle eine der folgenden:

Erklärung

Frage 2 von 89

Define a cardinality of a relation

Wähle eine der folgenden:

Erklärung

Frage 3 von 89

Which of the following refers to union-compatibility requirements ?

Wähle eine der folgenden:

Erklärung

Frage 4 von 89

Which of the following refers to union-compatibility requirements ?

Wähle eine der folgenden:

Erklärung

Frage 5 von 89

Define a domain

Wähle eine der folgenden:

Erklärung

Frage 6 von 89

Define a foreign key

Wähle eine der folgenden:

Erklärung

Frage 7 von 89

Define a primary key

Wähle eine der folgenden:

Erklärung

Frage 8 von 89

Define restrict

Wähle eine der folgenden:

Erklärung

Frage 9 von 89

Define cascade

Wähle eine der folgenden:

Erklärung

Frage 10 von 89

Define set default

Wähle eine der folgenden:

Erklärung

Frage 11 von 89

Define set null

Wähle eine der folgenden:

Erklärung

Frage 12 von 89

Define Data Definition Language

Wähle eine der folgenden:

Erklärung

Frage 13 von 89

Define Data Control Language

Wähle eine der folgenden:

Erklärung

Frage 14 von 89

Define Data Manipulation Language

Wähle eine der folgenden:

Erklärung

Frage 15 von 89

Which of the following does not refer to DBMS tools

Wähle eine der folgenden:

Erklärung

Frage 16 von 89

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

Wähle eine der folgenden:

Erklärung

Frage 17 von 89

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

Wähle eine der folgenden:

Erklärung

Frage 18 von 89

Which of the following is true regarding views?

Wähle eine der folgenden:

Erklärung

Frage 19 von 89

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.

Wähle eine der folgenden:

Erklärung

Frage 20 von 89

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