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2026 Data-Management-Foundations exam torrent Data-Management-Foundations Study Guide [Q13-Q29]

2026 Data-Management-Foundations exam torrent Data-Management-Foundations Study Guide [Q13-Q29]

March 20, 2026 adminData-Management-Foundations, WGUData-Management-Foundations exam preview, Data-Management-Foundations exam tutorial, Data-Management-Foundations latest mock exam, Data-Management-Foundations questions exam, Data-Management-Foundations test collection pdf, Data-Management-Foundations Test Free, Data-Management-Foundations valid test cram materials, Data-Management-Foundations valid test dumps freeLeave a Comment on 2026 Data-Management-Foundations exam torrent Data-Management-Foundations Study Guide [Q13-Q29]

2026 Data-Management-Foundations exam torrent Data-Management-Foundations Study Guide

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WGU Data-Management-Foundations Exam Syllabus Topics:

Topic Details
Topic 1
  • Normalizing relational databases: This section of the exam measures skills of Data Analysts and covers organizing data using normalization steps. It focuses on reducing redundancy, splitting data into related tables, and improving consistency in a relational database.
Topic 2
  • Introduction to conceptual logical and physical data models: This section of the exam measures skills of Data Analysts and introduces the basic ideas behind conceptual, logical, and physical data models. It focuses on understanding how each model represents data at a different level, from high level business view to detailed database structure.
Topic 3
  • Defining primary and foreign keys for data normalization: This section of the exam measures skills of Database Developers and explains how to identify and define primary and foreign keys. It focuses on using keys to connect tables, enforce relationships, and support normalized database design.

 

QUESTION 13
What is the role of the database administrator?

 
 
 
 
ADatabase Administrator (DBA)is responsible for the management, security, and performance of a database system. This includes controlling access to data, ensuring database integrity, optimizing performance, managing backups, and protecting the system from unauthorized access.
* Option A (Incorrect):A DBA is not just a consumer of data but is primarily responsible for the database’s management.
* Option B (Correct):Security is one of the key responsibilities of a DBA, including enforcing user access controls and implementing encryption and authentication mechanisms.
* Option C (Incorrect):While DBAs work with data structures, it is typically the role of adata architect ordatabase designerto define data formats and schema structures.
* Option D (Incorrect):Developing application programs that interact with the database is typically the role ofsoftware developersordatabase programmers, not DBAs.

QUESTION 14
Which property of an entity can become a column in a table?

 
 
 
 
Indatabase design,attributesof an entity becomecolumnsin a relational table.
Example Usage:
For anEmployee entity, attributes might include:

CREATE TABLE Employees (
EmployeeID INT PRIMARY KEY,
Name VARCHAR(50),
Salary DECIMAL(10,2),
DepartmentID INT
);
* Eachattribute(e.g., Name, Salary) becomes acolumnin the table.
Why Other Options Are Incorrect:
* Option A (Modality) (Incorrect):Describesoptional vs. mandatoryrelationships, not table structure.
* Option B (Uniqueness) (Incorrect):Ensuresdistinct valuesbutis not a column property.
* Option D (Non-null values) (Incorrect):Ensures thatcolumns must contain databut doesnot define attributes.
Thus, the correct answer isAttribute, as attributes of entities becometable columns.

QUESTION 15
Which command is used to filter group results generated by the GROUP BY clause?

 
 
 
 
TheHAVINGclause is used in SQL to filtergrouped resultsgenerated by the GROUP BY clause. Unlike WHERE, which filters individual rowsbeforegrouping, HAVING filtersafter aggregationhas been performed.
Example Usage:
sql
SELECT Department, AVG(Salary) AS AvgSalary
FROM Employees
GROUP BY Department
HAVING AVG(Salary) > 50000;
* This query first groups employees by Department, calculates theaverage salary per department, and then filters onlythose departments where the average salary is greater than 50,000.
Why Other Options Are Incorrect:
* Option A (REPLACE) (Incorrect):Used for string substitution, not filtering.
* Option C (WITH) (Incorrect):Used inCommon Table Expressions (CTEs), not for filtering.
* Option D (WHERE) (Incorrect):Used forrow-level filtering before aggregation, but itcannot be used on aggregate functions like SUM() or AVG().
Thus,HAVING is the correct answerfor filtering after grouping.

QUESTION 16
Which phase of entity-relationship modeling refers to the maxima and minima of relationships and attributes?

 
 
 
 
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 onlyone customer.
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.

QUESTION 17
Which function is considered an aggregate function?

 
 
 
 
Aggregate functionsperform calculationson a set of values and return asingle result.MAX()is one such function, returning thelargest value in a column.
Common Aggregate Functions:
A screenshot of a computer AI-generated content may be incorrect.

Example Usage:
sql
SELECT MAX(Salary) FROM Employees;
* Retrieves thehighest salaryin the Employees table.
Why Other Options Are Incorrect:
* Option B (TRIM) (Incorrect):Removes spaces from strings butis not an aggregate function.
* Option C (ABS) (Incorrect):Returns theabsolute value of a numberbut doesnot aggregate multiple rows.
* Option D (DESC) (Incorrect):Used in ORDER BY forsorting in descending order,not for aggregation.
Thus, the correct answer isMAX(), as it is atrue aggregate function.

QUESTION 18
Which keyword can be used to combine two results into one table?

 
 
 
 
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.

QUESTION 19
Which constraint propagates primary key changes to foreign keys?

 
 
 
 
TheCASCADEconstraint ensures thatupdates or deletions in the primary key table automatically reflect in the foreign key table.
Example Usage:
sql
CREATE TABLE Departments (
DeptID INT PRIMARY KEY,
DeptName VARCHAR(50)
);
CREATE TABLE Employees (
EmpID INT PRIMARY KEY,
Name VARCHAR(50),
DeptID INT,
FOREIGN KEY (DeptID) REFERENCES Departments(DeptID) ON UPDATE CASCADE ON DELETE CASCADE );
* If DeptIDchangesin Departments, itautomatically updatesin Employees.
* If a DeptID isdeleted, all employees in that departmentare also deleted.
Why Other Options Are Incorrect:
* Option A (SET DEFAULT) (Incorrect):Sets foreign key values to adefaultvalue, rather than propagating changes.
* Option B (SET NULL) (Incorrect):When the referenced key is deleted, dependent records areset to NULLinstead of being updated/deleted.
* Option C (RESTRICT) (Incorrect):Prevents deletion of a referenced row if dependent foreign key rows exist.
Thus, the correct answer isCASCADE, as itpropagates primary key changes to dependent foreign keys.

QUESTION 20
Which keyword is used to introduce a limiter in a SELECT statement?

 
 
 
 
TheWHEREclause is used inSQL SELECT statementstolimitthe number of rows that match a specific condition. It helps filter data based on given criteria before retrieving the results.
Example Usage:
sql
SELECT *
FROM Employees
WHERE Salary > 50000;
* This querylimitsthe result set to employees whose salary isgreater than 50,000.
Why Other Options Are Incorrect:
* Option A (FROM) (Incorrect):Specifies the table from which data is retrieved butdoes not limit results.
* Option B (DROP) (Incorrect):Used fordeleting tables, databases, or views, not filtering rows.
* Option C (INTO) (Incorrect):Used in statements like INSERT INTO or SELECT INTO, whichdo not filter results.
Thus,WHERE is the correct keywordfor applying alimiter in a SELECT statement.

QUESTION 21
Which relationship or association exists between a supertype and its subtype entities?

 
 
 
 
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.

QUESTION 22
What is the last step in the logical design process for designing a database?

 
 
 
 
Thelogical design phasein database development focuses onstructuring data efficientlyto eliminate redundancy and ensure integrity. Thefinal step in logical designis toapply normalization (normal forms)to optimize the database schema.
Steps in Logical Database Design:
* Discover entities# Identify real-world objects (e.g., Customers, Orders).
* Determine cardinality# Define relationships between entities (one-to-one, one-to-many).
* Analyze data requirements# Determine the attributes each entity needs.
* Apply normal forms# Eliminate redundancy and improve data consistency.
Example Usage:
* After identifying entities likeStudentsandCourses, applying3rd Normal Form (3NF)ensures that data isorganized without redundancy.
Why Other Options Are Incorrect:
* Option A (Analyze data requirements) (Incorrect):Doneearlierto define attributes.
* Option C (Determine cardinality) (Incorrect):Donebeforenormalization to establish relationships.
* Option D (Discover entities) (Incorrect):Done at thebeginningof database design.
Thus, the correct answer isApply a normal form, as normalization is thelast step in logicaldesign.

QUESTION 23
What is shown on the “many” side of a relationship between two tables?

 
 
 
 
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.

QUESTION 24
Which type of join is demonstrated by the following query?
sql
SELECT *
FROM Make, Model
WHERE Make.ModelID = Model.ID;

 
 
 
 
This query performs ajoin operationwhere records from the Make table and Model table are combined based on the condition Make.ModelID = Model.ID. This conditiontests for equality, which is the definition of an EQUIJOIN.
Types of Joins in SQL:
* EQUIJOIN (Correct Answer):
* Uses an equality operator (=) to match rows between tables.
* Equivalent to an INNER JOIN ON condition.
* Example:
sql
SELECT *
FROM Employees
JOIN Departments ON Employees.DeptID = Departments.ID;
* NON-EQUIJOIN (Incorrect):
* Usescomparison operators other than =(e.g., <, >, BETWEEN).
* Example:
sql
SELECT *
FROM Employees e
JOIN Salaries s ON e.Salary > s.MedianSalary;
* SELF JOIN (Incorrect):
* A table is joined withitselfusing table aliases.
* Example:
sql
SELECT e1.Name, e2.Name AS Manager
FROM Employees e1
JOIN Employees e2 ON e1.ManagerID = e2.ID;
* CROSS JOIN (Incorrect):
* ProducesCartesian product(each row from Table A combines with every row from Table B).
* Example:
sql
SELECT *
FROM Employees
CROSS JOIN Departments;
Thus, since our given query uses anequality condition (=) to join two tables, it is anEQUIJOIN.

QUESTION 25
Which entity in a table is a measurable object in the real world?

 
 
 
 
Atangible entityis a real-world object that can bemeasured and storedin a database.
Example Usage:
* In an inventory system,tangible entitiesinclude:
Products, Orders, Customers
Why Other Options Are Incorrect:
* Option A (Logical entity) (Incorrect):Exists logically butmay not have a physical presence(e.g., views, categories).
* Option C (Virtual entity) (Incorrect):Existsonly in queries or reports, not stored as real data.
* Option D (Conceptual entity) (Incorrect):Abstract idea used indesign modeling, not astored entity.
Thus, the correct answer isTangible entity, as it representsmeasurable, real-world objects.

QUESTION 26
Which action does the % operator accomplish in MySQL?

 
 
 
 
The % operator in MySQL is known as themodulus operator. It returns theremainderof a division operation between two numbers.
Example:
sql
SELECT 10 % 3; — Output: 1 (10 divided by 3 gives remainder 1)
* Option A (Incorrect):Raising a number to a power is done using the POW() function or ^ in some SQL dialects.
* Option B (Incorrect):The = operator is used forequality comparisons, not %.
* Option C (Correct):Themodulus operator (%)finds the remainder when one number is divided by another.
* Option D (Incorrect):Subtraction is performed using the – operator.

QUESTION 27
What is the second step in the implement relationships stage of database design?

 
 
 
 
Thesecond step in implementing relationshipsis definingone-to-one (1:1) relationshipsbetween entities.
Example Usage:
* Example of a 1:1 relationship:
sql
CREATE TABLE Employees (
EmpID INT PRIMARY KEY,
Name VARCHAR(50)
);
CREATE TABLE EmployeeDetails (
EmpID INT PRIMARY KEY,
Address VARCHAR(255),
FOREIGN KEY (EmpID) REFERENCES Employees(EmpID)
);
* Here, eachemployee has exactly one detail record, creating a1:1 relationship.
Why Other Options Are Incorrect:
* Option A (Implement weak entities) (Incorrect):Weak entities rely on aforeign keyand are implementedlater.
* Option C (Implement subtype entities) (Incorrect):Subtypes arespecial casesandnot implemented in the second step.
* Option D (Specify cascade) (Incorrect):Cascade rules (ON DELETE, ON UPDATE)are defined duringforeign key implementation, not in the second step.
Thus, the correct answer isImplement one-one relationships, as it is thenext logical stepafter defining entities.

QUESTION 28
Which clause or statement in a CREATE statement ensures a certain range of data?

 
 
 
 
TheCHECKconstraint is used in SQL toenforce ruleson a column’s values. It ensures that data inserted into a table meets specified conditions, such as range restrictions or logical rules.
Example Usage:
sql
CREATE TABLE Employees (
ID INT PRIMARY KEY,
Name VARCHAR(50),
Salary INT CHECK (Salary BETWEEN 30000 AND 150000)
);
* This constraint ensures thatsalary values fall between 30,000 and 150,000.
* If an INSERT or UPDATE statement tries to set Salary = 20000, itfailsbecause it does notmeet the CHECK condition.
Why Other Options Are Incorrect:
* Option B (FROM) (Incorrect):Used in SELECT statements, not for constraints.
* Option C (WHERE) (Incorrect):Filters rows in queries butdoes not enforce constraints.
* Option D (SET) (Incorrect):Used for updating records (UPDATE table_name SET column = value) butnot for defining constraints.
Thus,CHECK is the correct answer, as it ensures that column values remain within an expected range.

QUESTION 29
Which type of entity only exists in a logical sense?

 
 
 
 
Anintangible entityis an entity that does not have a physical presence but still holds meaning in a database.
These entities representconcepts, relationships, or abstract datathat exist in a logical sense.
Example Usage in Databases:
* Customer Loyalty Status (Gold, Silver, Bronze)
* Theloyalty levelis anintangible entitysince it is derived from other data (purchase history).
* Online User Sessions
* Asession IDexistslogically in the systembut does not have a tangible presence like a product or employee.
Why Other Options Are Incorrect:
* Option A (Concrete entity) (Incorrect):No such formal term in database design.
* Option B (Tangible entity) (Incorrect):Represents somethingphysical, like anemployeeorproduct.
* Option D (Physical entity) (Incorrect):Refers todata stored on disk, like database tables or indexes.
Thus, the correct answer isIntangible entity, as it only exists in a logical sense within the database.

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