START EXAM PREPARATION WITH REAL AND VALID DATA-MANAGEMENT-FOUNDATIONS EXAM QUESTIONS

Start Exam Preparation with Real and Valid Data-Management-Foundations Exam Questions

Start Exam Preparation with Real and Valid Data-Management-Foundations Exam Questions

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WGU Data Management – Foundations Exam Sample Questions (Q39-Q44):

NEW QUESTION # 39
What is shown on the "many" side of a relationship between two tables?

  • A. Binary relationship
  • B. Foreign key
  • C. Weak entity
  • D. Reflexive relationship

Answer: B

Explanation:
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.


NEW QUESTION # 40
Which type of entity only exists in a logical sense?

  • A. Concrete entity
  • B. Tangible entity
  • C. Physical entity
  • D. Intangible entity

Answer: D

Explanation:
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.


NEW QUESTION # 41
What is the second step in the implement relationships stage of database design?

  • A. Implement subtype entities
  • B. Specify cascade
  • C. Implement one-one relationships
  • D. Implement weak entities

Answer: C

Explanation:
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.


NEW QUESTION # 42
How is the primary key indicated in a table?

  • A. By using an SQL keyword
  • B. By using bold typeface in the appropriate column
  • C. By using a diamond symbol inserted into the table
  • D. By using a formula in SQL

Answer: A

Explanation:
In SQL, aprimary key is explicitly defined using the PRIMARY KEY keywordwhen creating a table.
Example Usage:
sql
CREATE TABLE Products (
ProductID INT PRIMARY KEY,
Name VARCHAR(100),
Price DECIMAL(10,2)
);
* Here,PRIMARY KEY is the SQL keyword that designates ProductID as the primary key.
Why Other Options Are Incorrect:
* Option A (Formula in SQL) (Incorrect):SQLdoes not use formulas to define primary keys.
* Option C (Bold typeface) (Incorrect):SQL syntax does not rely on text formatting.
* Option D (Diamond symbol) (Incorrect):ER diagramsmight use symbols, but SQLdoes not use diamonds to indicate keys.
Thus, the correct answer isSQL keyword, as primary keys are explicitly defined using PRIMARY KEY.


NEW QUESTION # 43
What is a common error made while inserting an automatically incrementing primary key?

  • A. Forgetting to specify which is the auto-increment column
  • B. Inserting a value and overriding auto-increment for a primary key
  • C. Designating multiple primary keys
  • D. Failing to set a numeric value in a newly inserted row

Answer: B

Explanation:
In databases, primary keys are oftenset to auto-incrementso that new rows automatically receive unique values. However,one common error is manually inserting a value into an auto-incremented primary key column, whichoverrides the automatic numberingand may cause conflicts.
Example of Auto-Increment Setup:
sql
CREATE TABLE Users (
UserID INT AUTO_INCREMENT PRIMARY KEY,
Username VARCHAR(50)
);
Incorrect Insert (Error-Prone Approach):
sql
INSERT INTO Users (UserID, Username) VALUES (100, 'Alice');
* Thismanually overrides the auto-increment, which can lead toduplicate key errors.
Correct Insert (Avoiding Errors):
sql
INSERT INTO Users (Username) VALUES ('Alice');
* Thedatabase assigns UserID automatically, preventing conflicts.
Why Other Options Are Incorrect:
* Option B (Failing to set a numeric value) (Incorrect):The databaseautomatically assignsvalues when AUTO_INCREMENT is used.
* Option C (Designating multiple primary keys) (Incorrect):Whileincorrect, most databases will prevent this at creation time.
* Option D (Forgetting to specify which is the auto-increment column) (Incorrect):If AUTO_INCREMENT is set, the database handles numbering automatically.
Thus, the most common error isInserting a value and overriding auto-increment, which can cause duplicate key errors and data inconsistencies.


NEW QUESTION # 44
......

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