実際的なPCEP-30-02科目対策試験-試験の準備方法-完璧なPCEP-30-02試験内容

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あなたは自分の職場の生涯にユニークな挑戦に直面していると思いましたら、Python InstituteのPCEP-30-02の認定試験に合格することが必要になります。JPNTestはPython InstituteのPCEP-30-02の認定試験を真実に、全面的に研究したサイトです。JPNTest のユニークなPython InstituteのPCEP-30-02の認定試験の問題と解答を利用したら、試験に合格することがたやすくなります。JPNTestは認証試験の専門的なリーダーで、最全面的な認証基準のトレーニング方法を追求して、１００パーセントの成功率を保証します。JPNTestのPython InstituteのPCEP-30-02の試験問題と解答は当面の市場で最も徹底的かつ正確かつ最新な模擬テストです。それを利用したら、初めに試験を受けても、合格する自信を持つようになります。

>> PCEP-30-02科目対策 <<

PCEP-30-02試験内容、PCEP-30-02試験対策

人々は常に、特定の分野で有能で熟練していることを証明したいと考えています。能力を証明する方法はさまざまですが、最も直接的で便利な方法は、PCEP-30-02認定試験に参加し、認定証を取得することです。 PCEP-30-02認定に合格すると、非常に有能で優秀であることを証明できます。また、PCEP-30-02テストに合格することで有用な知識とスキルを習得できます。 PCEP-30-02ガイドトレントを購入すると、JPNTestのPCEP-30-02試験に合格するのに役立ちます。時間と労力はほとんどかかりません。

Python Institute PCEP-30-02 認定試験の出題範囲：

トピック	出題範囲
トピック 1	Control Flow: This section covers conditional statements such as if, if-else, if-elif, if-elif-else
トピック 2	parameters, arguments, and scopes. It also covers Recursion, Exception hierarchy, Exception handling, etc.
トピック 3	Data Collections: In this section, the focus is on list construction, indexing, slicing, methods, and comprehensions; it covers Tuples, Dictionaries, and Strings.
トピック 4	Loops: while, for, range(), loops control, and nesting of loops.

Python Institute PCEP - Certified Entry-Level Python Programmer 認定 PCEP-30-02 試験問題 (Q10-Q15):

質問 # 10
What is the expected output of the following code?

A. 0
B. 1
C. The code outputs nothing.
D. 2

正解：D

解説：
The code snippet that you have sent is checking if two numbers are equal and printing the result. The code is as follows:
num1 = 1 num2 = 2 if num1 == num2: print(4) else: print(1)
The code starts with assigning the values 1 and 2 to the variables "num1" and "num2" respectively. Then, it enters an if statement that compares the values of "num1" and "num2" using the equality operator (==). If the values are equal, the code prints 4 to the screen. If the values are not equal, the code prints 1 to the screen.
The expected output of the code is 1, because the values of "num1" and "num2" are not equal. Therefore, the correct answer is C. 1.
Reference: [Python Institute - Entry-Level Python Programmer Certification]

質問 # 11
Drag and drop the conditional expressions to obtain a code which outputs * to the screen.
(Note: some code boxes will not be used.)

正解：

解説：

Explanation

One possible way to drag and drop the conditional expressions to obtain a code which outputs * to the screen is:
if pool > 0:
print("*")
elif pool < 0:
print("**")
else:
print("***")
This code uses the if, elif, and else keywords to create a conditional statement that checks the value of the variable pool. Depending on whether the value is greater than, less than, or equal to zero, the code will print a different pattern of asterisks to the screen. The print function is used to display the output. The code is indented to show the blocks of code that belong to each condition. The code will output * if the value of pool is positive, ** if the value of pool is negative, and *** if the value of pool is zero.
You can find more information about the conditional statements and the print function in Python in the following references:
[Python If ... Else]
[Python Print Function]
[Python Basic Syntax]

質問 # 12
A set of rules which defines the ways in which words can be coupled in sentences is called:

A. lexis
B. dictionary
C. syntax
D. semantics

正解：C

解説：
Syntax is the branch of linguistics that studies the structure and rules of sentences in natural languages. Lexis is the vocabulary of a language. Semantics is the study of meaning in language. A dictionary is a collection of words and their definitions, synonyms, pronunciations, etc.
Reference: [Python Institute - Entry-Level Python Programmer Certification]

質問 # 13
What is the expected output of the following code?

A. * * *
B. * *
C. The code produces no output.
D. *

正解：B

解説：
The code snippet that you have sent is a conditional statement that checks if a variable "counter" is less than
0, greater than or equal to 42, or neither. The code is as follows:
if counter < 0: print("") elif counter >= 42: print("") else: print("") The code starts with checking if the value of "counter" is less than 0. If yes, it prints a single asterisk () to the screen and exits the statement. If no, it checks if the value of "counter" is greater than or equal to 42. If yes, it prints three asterisks () to the screen and exits the statement. If no, it prints two asterisks () to the screen and exits the statement.
The expected output of the code depends on the value of "counter". If the value of "counter" is 10, as shown in the image, the code will print two asterisks (**) to the screen, because 10 is neither less than 0 nor greater than or equal to 42. Therefore, the correct answer is C. * * Reference: [Python Institute - Entry-Level Python Programmer Certification]

質問 # 14
What is the expected result of running the following code?

A. The code prints 1 .
B. The code prints 0
C. The code raises an unhandled exception.
D. The code prints 2

正解：C

解説：
Explanation
The code snippet that you have sent is trying to use the index method to find the position of a value in a list.
The code is as follows:
the_list = [1, 2, 3, 4, 5] print(the_list.index(6))
The code starts with creating a list called "the_list" that contains the numbers 1, 2, 3, 4, and 5. Then, it tries to print the result of calling the index method on the list with the argument 6. The index method is used to return the first occurrence of a value in a list. For example, the_list.index(1) returns 0, because 1 is the first value in the list.
However, the code has a problem. The problem is that the value 6 is not present in the list, so the index method cannot find it. This will cause a ValueError exception, which is an error that occurs when a function or operation receives an argument that has the right type but an inappropriate value. The code does not handle the exception, and therefore it will terminate with an error message.
The expected result of the code is an unhandled exception, because the code tries to find a value that does not exist in the list. Therefore, the correct answer is C. The code raises an unhandled exception.

質問 # 15
......

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PCEP-30-02試験内容: https://www.jpntest.com/shiken/PCEP-30-02-mondaishu

P.S.JPNTestがGoogle Driveで共有している無料の2025 Python Institute PCEP-30-02ダンプ：https://drive.google.com/open?id=1s7djJ4DCPT6t4HzDV72himjujJnoEbqy

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実際的なPCEP-30-02科目対策試験-試験の準備方法-完璧なPCEP-30-02試験内容

PCEP-30-02試験内容、PCEP-30-02試験対策

Python Institute PCEP-30-02 認定試験の出題範囲：

Python Institute PCEP - Certified Entry-Level Python Programmer 認定 PCEP-30-02 試験問題 (Q10-Q15):

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