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WGU Data Management – Foundations Exam Sample Questions (Q36-Q41):

NEW QUESTION # 36
Which characteristic is true for non-relational databases?

• A. They support the SQL query language.
• B. They are optimized for big data.
• C. They are ideal for databases that require an accurate record of transactions.
• D. They store data in tables, columns, and rows, similar to a spreadsheet.

Answer: B

Explanation:
Non-relational databases(also calledNoSQL databases) are designed for handlingbig dataandunstructured dataefficiently. They are optimized forhorizontal scaling, making them ideal forlarge-scale distributed systems.
* Option A (Correct):Non-relational databases areoptimized for big data, handling massive volumes of data across distributed architectures.
* Option B (Incorrect):NoSQL databases donotuse SQL as their primary query language. They often use JSON-based queries (e.g., MongoDB).
* Option C (Incorrect):Transaction-heavy applications requireACID compliance, which relational databases (SQL) handle better than NoSQL databases.
* Option D (Incorrect):NoSQL databases usedocument, key-value, graph, or column-family storage models, nottables, columns, and rowslike relational databases.

NEW QUESTION # 37
What is the role of the transaction manager within the database system architecture?

• A. The transaction manager logs insert, update, and delete queries, and the result is sent back to the application.
• B. The transaction manager translates the query processor instructions into filesystem commands and uses an index to quickly locate the requested data.
• C. The transaction manager is composed of a query processor, storage manager, transaction manager, log, and catalog.
• D. The transaction manager uses information from the catalog to perform query optimization.

Answer: A

NEW QUESTION # 38
What is shown on the "many" side of a relationship between two tables?

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

Answer: C

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 # 39
Which clause is used to specify the join columns when performing a join in MySQL?

• A. AND
• B. AS
• C. ON
• D. JOIN

Answer: C

Explanation:
When performing aJOIN operationin MySQL, the ON clause specifies thejoining condition, defining which columns from both tables should be matched.
Example:
sql
SELECT Employees.Name, Departments.DepartmentName
FROM Employees
JOIN Departments ON Employees.DepartmentID = Departments.ID;
* Option A (Incorrect):AS is used foraliasing tables and columns, not for specifying join conditions.
* Option B (Incorrect):JOIN defines the type of join (INNER JOIN, LEFT JOIN, etc.), butdoes not specify the columns.
* Option C (Correct):The ON clause is used tospecify the join conditionbetween two tables.
* Option D (Incorrect):AND is used in filtering conditions, not for joining tables.

NEW QUESTION # 40
How many bytes of storage does a BIGINT data type hold in MySQL?

• A. 1 byte
• B. 4 bytes
• C. 8 bytes
• D. 3 bytes

Answer: C

Explanation:
In MySQL, theBIGINTdata type is a64-bit integerthat requires8 bytes (64 bits) of storage. It is used to store large numerical values beyond the range of INT (4 bytes).
* Option A (Incorrect):1 byte corresponds toTINYINT, which can store values from -128 to 127.
* Option B (Incorrect):3 bytes is not a standard integer storage size in MySQL.
* Option C (Incorrect):4 bytes corresponds toINT, which has a range of -2,147,483,648 to
2,147,483,647.
* Option D (Correct):BIGINT takes8 bytesand supports a massive range of numbers from -2