Foundations-of-Computer-Science資料 & Foundations-of-Computer-Science題庫下載

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WGU Foundations-of-Computer-Science Exam Syllabus Topics:

Section	Objectives
Programming Foundations	- Language Concepts Overview 1. Compiled vs interpreted languages 2. Programming paradigms overview - Programming Concepts 1. Control flow (if/else, loops) 2. Variables, data types, expressions 3. Basic pseudocode interpretation
Data & Security Basics	- Data Handling 1. Data profiling concepts 2. Basic database concepts overview - Security Fundamentals 1. Basic cybersecurity threats and mitigation 2. Encryption basics (at rest vs in transit)
Operating Systems & Architecture	- System Architecture 1. Hardware vs software abstraction 2. Von Neumann architecture basics - OS Fundamentals 1. Process states and scheduling basics 2. Memory management concepts
Computer Science Fundamentals	- Data Structures Introduction 1. Basic sorting and searching concepts 2. Arrays and lists - Core CS Concepts 1. Algorithm efficiency and Big-O basics 2. Computational thinking and problem solving 3. Basic programming logic and algorithms

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最新的 Courses and Certificates Foundations-of-Computer-Science 免費考試真題 (Q44-Q49):

問題 #44
What is the only content that will display if the List folder contents permission is not enabled for a particular folder in Windows 11?

A. The folder's creation date
B. Files with Write permission
C. Files with Read permission
D. The folder's author

答案：A

解題說明：
In Windows file security (NTFS permissions), "List folder contents" controls whether a user cansee the names of files and subfoldersinside a folder. If a user does not have permission to list a folder, Windows prevents directory enumeration: the user cannot browse the folder and view what is inside. (2BrightSparks) This is a key concept in access control: it separates "being able to traverse to a location" from "being able to see what is stored there." When "List folder contents" is not enabled, the user typically cannot view the list of files regardless of whether individual files might have separate permissions. In standard user-facing behavior, what remains visible in the folder's properties and metadata is limited; among the choices given, the only item that is reliably a folder-level metadata attribute (and not a listing of contents) is the folder'screation date. The
"author" is not a universal, reliably displayed NTFS folder property, and options C and D talk about files (contents), which cannot be listed without the list permission. (2BrightSparks) This reflects a broader textbook principle: operating systems enforce access control both on objects (files/folders) and on operations (read data, write data, list directory). Removing the list operation blocks visibility of contents, even if other permissions exist elsewhere.

問題 #45
What is the alternative way to access the third element of the first row in np_2d?

A. np_2d[2, 0]
B. np_2d[0, 2]
C. np_2d[3, 1]
D. np_2d[1, 3]

答案：B

解題說明：
NumPy arrays use zero-based indexing, meaning counting starts at 0 rather than 1. In a 2D NumPy array, indexing is typically written in the form array[row_index, column_index]. The first index selects the row, and the second index selects the column. Therefore, the "first row" corresponds to row index 0. Within that row, the "third element" corresponds to column index 2, because the columns are indexed 0, 1, 2, 3, and so on.
So, np_2d[0, 2] directly selects the element at row 0 and column 2, which is the third element in the first row.
This is considered an "alternative" to approaches like two-step indexing (np_2d[0][2]), and it is the standard idiom taught for multi-dimensional NumPy arrays.
The other choices point to different locations. np_2d[1, 3] is the fourth element of the second row, not the third element of the first row. np_2d[2, 0] and np_2d[3, 1] attempt to access the third or fourth row, which would often be out of bounds in a small 2-row example and would raise an IndexError. Correct indexing is a cornerstone of array programming because it determines which observation, feature, or matrix entry your computations will use.

問題 #46
What type of encryption is provided by encryption utilities built into the file system?

A. Encryption steganography
B. Encryption in motion
C. Encryption authentication
D. Encryption at rest

答案：D

解題說明：
File system encryption utilities are designed to protect datastored on a disk-for example, files on an SSD, HDD, or other persistent storage. This protection is calledencryption at rest. The key idea is that if an attacker steals the physical drive, gains access to a powered-off machine, or otherwise reads storage directly, the raw bytes on disk remain unreadable without the correct cryptographic key. Common textbook examples include full-disk encryption and per-file encryption supported by operating systems and file systems.
This differs fromencryption in motion(also called encryption in transit), which protects data while it is being transmitted over networks, such as via TLS/HTTPS, VPNs, or secure messaging protocols. File system utilities do not primarily address network transmission; they address stored data confidentiality. Option B,
"encryption authentication," is not a standard category; authentication is a security goal often achieved using mechanisms like digital signatures, MACs, certificates, and protocol handshakes, not a type of file system encryption. Option D, steganography, is the practice of hiding information within other data (like images or audio) rather than encrypting it for confidentiality.
In short, file system encryption utilities aim to ensure that stored files remain confidential if storage is accessed without authorization, which is precisely the definition of encryption at rest.

問題 #47
What is the time complexity of a quicksort algorithm?

A. O(log n)
B. O(n log n)
C. O(1)
D. O(n)

答案：B

解題說明：
Quicksort is a divide-and-conquer sorting algorithm. It works by selecting a pivot element, partitioning the array into two subarrays (elements less than the pivot and elements greater than the pivot), and then recursively sorting those subarrays. In the average case, the partition step splits the array into roughly equal halves, so the recurrence is commonly written as (T(n) = T(n/2) + T(n/2) + O(n)), where (O(n)) is the cost of partitioning. This solves to (O(n \log n)), which is why quicksort is widely taught as an efficient general- purpose sorting method.
However, textbooks also emphasize that quicksort has a worst-case time complexity of (O(n

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