Data-Management-Foundations Exam Dumps - WGU Courses and Certificates Questions and Answers

Cardinalitydefines theminimum and maximum number of occurrencesof one entity in relation to another.

Example Usage in an ER Model:

One-to-Many (1:M): Acustomer can place multiple orders, but each order belongs to onlyonecustomer.

Customers (1) ——— (M) Orders

Cardinality notation:

(1,1) → One-to-One

(0,M) → Zero-to-Many

(1,M) → One-to-Many

Why Other Options Are Incorrect:

Option B (Physical design) (Incorrect):Focuses onstorage and indexing, not relationships.

Option C (Attribute minimum) (Incorrect):No such formal term in database modeling.

Option D (Partition) (Incorrect):Refers todividing tables, not relationship constraints.

Thus, the correct answer isCardinality, as it definesmin/max constraints on relationships.

TheUNIONkeyword in SQL is used tocombine the resultsof two or more SELECT queries into asingle result setwhile removing duplicate rows.

Example:

sql

SELECT Name FROM Employees

UNION

SELECT Name FROM Managers;

Option A (Correct):UNION combines results from multiple queries into one set,removing duplicates.

Option B (Incorrect):MERGE is not a valid SQL keyword for combining result sets (it is used insome database systems for data merging).

Option C (Incorrect):INTEGRATE is not a SQL keyword.

Option D (Incorrect):CONSOLIDATE is not an SQL keyword.

Avalid SELECT statementin SQL follows this basic syntax:

sql

SELECT column1, column2

FROM table_name

WHERE condition;

The correct optionDfollows this syntaxcorrectly.

Why Other Options Are Incorrect:

Option A (Incorrect):SQL does not usedouble quotes(") around column/table names unless explicitly required in some databases.

Option B (Incorrect):The WHERE clausemust appear after the FROM clause.

Option C (Incorrect):ALL isnota valid keyword in standard SQL queries.

Thus,Option Dfollows the correct SQL syntax.

Toadd a primary key constraint after table creation, we use theALTER TABLEstatement.

Example Usage:

sql

ALTER TABLE Employees

ADD CONSTRAINT PK_Employees PRIMARY KEY (EmpID);

Thisaddsa primary key to the EmpID columnafter the table was created.

Why Other Options Are Incorrect:

Option B (CREATE TABLE) (Incorrect):Used for defining constraintsduringtable creation, not after.

Option C (UPDATE) (Incorrect):Modifiesrow values, not constraints.

Option D (INSERT INTO) (Incorrect):Used toadd datato a table, not modify constraints.

Thus, the correct answer isALTER TABLE, as itmodifies table structure to add a primary key constraint.

Information independencerefers to theseparation between data storage and data access. It allows a database’slogical structureto be modifiedwithout affecting existing applications.

Types of Information Independence:

Logical Data Independence→ Ability to change theconceptual schema(e.g., renaming columns, adding new attributes)without modifying applications.

Physical Data Independence→ Ability to change thephysical storage structure(e.g., indexing, partitioning)without affecting queries.

Example of Logical Data Independence:

Anew columnis added to the Customers table, but existing queriesstill work without modification.

Example of Physical Data Independence:

Data is moved to SSD storagefor performance improvement, but queriesrun the same way.

Why Other Options Are Incorrect:

Option A (Incorrect):Changing thedatabase type(e.g., MySQL to PostgreSQL) isnotinformation independence.

Option B (Incorrect):Making changes toqueriesis unrelated to database independence.

Option C (Incorrect):Interchanging databases is related todata portability, notinformation independence.

Thus, the correct answer isD - An ability to change the organization of data, asinformation independenceensuresmodifications do not disrupt database operations.

Bydatabase design conventions, theprimary key is usually placed in the first columnof a table to make it easy to identify and reference.

Example Usage:

sql

CREATE TABLE Employees (

EmpID INT PRIMARY KEY, -- First column (convention)

Name VARCHAR(50),

Salary DECIMAL(10,2)

);

EmpID is placed as the first columnfor clarity and quick access.

Why Other Options Are Incorrect:

Option A (In the table header) (Incorrect):Table headers onlydisplay column names, they do not contain values.

Option C (In the top row) (Incorrect):Thetop row contains data, not theprimary key definition.

Option D (In the last visible column) (Incorrect):While technically possible,placing a primary key at the endisuncommonin database design.

Thus, the correct answer isIn the first column, as this is the standard convention in relational databases.

TheORDER BYclause in SQL is used tosort query resultsinascending (ASC) or descending (DESC) order.

Example Usage:

sql

SELECT Name, Salary FROM Employees ORDER BY Salary DESC;

This retrieves all employees, sorted bysalary in descending order.

Why Other Options Are Incorrect:

Option A (BETWEEN) (Incorrect):Used for filtering ranges butdoes not order results.

Option B (DISTINCT) (Incorrect):Removes duplicate rows butdoes not control order.

Option D (LIKE) (Incorrect):Used forpattern matching, not sorting.

Thus,ORDER BY is the correct choicefor defining the sequence of query results.

In aone-to-many (1:M) relationship, theforeign keyis placed in thetable on the "many" sideto establish the relationship with theprimary keyof the "one" side.

Example Usage:

A screenshot of a computer AI-generated content may be incorrect.

CREATE TABLE Departments (

DeptID INT PRIMARY KEY,

DeptName VARCHAR(50)

);

CREATE TABLE Employees (

EmpID INT PRIMARY KEY,

Name VARCHAR(50),

DeptID INT, -- Foreign key on the "many" side

FOREIGN KEY (DeptID) REFERENCES Departments(DeptID)

);

Eachdepartmentcan havemany employees→ DeptID is aforeign keyin Employees.

Why Other Options Are Incorrect:

Option A (Reflexive relationship) (Incorrect):Refers tounary (self-referential) relationships, not 1:M relationships.

Option B (Binary relationship) (Incorrect):A binary relationship involvestwo entities, but does not define where the foreign key is stored.

Option C (Weak entity) (Incorrect):Weak entitiesdepend on a strong entity, but not all "many" sides are weak entities.

Thus, the correct answer isForeign key, as it is placed on the "many" side of the relationship.

Anattribute typerefers to asingle, specific valuewithin a table, such as Salary, Age, or Price.

Example Usage:

A screenshot of a computer AI-generated content may be incorrect.

CREATE TABLE Employees (

EmpID INT PRIMARY KEY,

Name VARCHAR(50),

Salary DECIMAL(10,2)

);

Salary is anattribute typewith individual values for each employee.

Why Other Options Are Incorrect:

Option A (Glossary) (Incorrect):Refers todocumentation, not database values.

Option B (Entity type) (Incorrect):Representsa class of objects(e.g., Employees), not individual values.

Option D (Relationship) (Incorrect):Definesconnections between entities, not attributes.

Thus, the correct answer isAttribute type, as it represents anindividual data value.

Indatabase modeling, the relationship between asupertype and its subtypesis called anIsA relationship.

Example Usage:

AVehicle supertypemay haveCar and Truck subtypes.

Vehicle

├── Car

├── Truck

InER diagrams, this is represented as:

Vehicle (Supertype)

|

├── Car (Subtype)

├── Truck (Subtype)

SQL Table Implementation:

sql

CREATE TABLE Vehicle (

VehicleID INT PRIMARY KEY,

Make VARCHAR(50),

Model VARCHAR(50)

);

CREATE TABLE Car (

VehicleID INT PRIMARY KEY,

FOREIGN KEY (VehicleID) REFERENCES Vehicle(VehicleID),

EngineType VARCHAR(50)

);

CREATE TABLE Truck (

VehicleID INT PRIMARY KEY,

FOREIGN KEY (VehicleID) REFERENCES Vehicle(VehicleID),

CargoCapacity INT

);

This structurepreserves the IsA relationshipbetween Vehicle (supertype) and Car/Truck (subtypes).

Why Other Options Are Incorrect:

Option A (Strong entity) (Incorrect):Strong entitiesdo not rely on a supertype/subtype hierarchy.

Option C (Associative entity) (Incorrect):Used toresolve many-to-many relationships, not supertype/subtype relationships.

Option D (Weak entity) (Incorrect):Weak entitiesdepend on a strong entity, but supertype-subtype relations useinheritance(not dependency).

Thus, the correct answer isIsA relationship, as it describes theinheritance hierarchybetweensupertypes and subtypes.