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Andrews [41]
2 years ago
6

Amelia wants to create a website using the Python computer language to code her site. She will also use CodeSkulptor3 to help he

r. CodeSkulptor3 is an example of a(n):
A.
browser-based code editor.

B.
online coding tutorial.

C.
code repository.

D.
web page hosting site.
Computers and Technology
1 answer:
pishuonlain [190]2 years ago
3 0

Amelia wants to create a website using the Python computer language to code her site. She will also use CodeSkulptor3 is an example of browser-based code editor. The correct option is A.

<h3>What is python?</h3>

Python is a programming language generally used to make websites and software.  Automated tasks and data analysis are also carried out using python.

CodeSkulptor3 is also a browser-based Python interpreter.  It has implemented a subset of Python 3.

Thus, CodeSkulptor3 is an example of browser-based code editor. The correct option is A.

Learn more about python.

brainly.com/question/19045688

#SPJ1

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Answer:

Explanation:

In this exercise, you will roll a pair of dice until the numbers add up to a given number. You can assume that the given number is 2, 3, 6, or 12. Using pseudocode, write an algorithm that returns the number of times the dice is rolled to achieve this number.

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2 years ago
Hisoka is creating a summary document for new employees about their options for different mobile devices. One part of his report
algol13

Hisoka would not include in his document that is Apple users file-based encryption to offer a higher level of security.

<h3>Does Apple use the file based encryption?</h3>

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3 0
1 year ago
Write a function called backspaceCompare that takes two strings sl and s2 and evaluate them when both are typed into empty text
Andrej [43]

Answer:

Go to explaination for the program code

Explanation:

import java.util.Stack;

public class Lab3 {

public static void main(String[] args) {

String s1="DataStructuresIssss###Fun";

String s2="DataStructuresIszwp###Fun";

boolean ans=backspaceCompare(s1,s2);

System.out.println(ans);

/*String s1="abc##";

String s2="wc#d#";

boolean ans=backspaceCompare(s1,s2);

System.out.println(ans);*/

}

public static boolean backspaceCompare(String s1, String s2) {

Stack<Character> s1_stack=new Stack<Character>();

Stack<Character> s2_stack=new Stack<Character>();

//backspaceCount is a variable to count back space

int backspaceCount=0;

//logic is that if '#' encountered we are putting pop else push

for(int i=0;i<s1.length();i++){

if(s1.charAt(i)=='#'){

backspaceCount++;

s1_stack.pop();

}

else

{

s1_stack.push(s1.charAt(i));

}

}

//this all is for s2 string

for(int i=0;i<s2.length();i++){

if(s2.charAt(i)=='#') s2_stack.pop();

else s2_stack.push(s2.charAt(i));

}

//here is the main logic first we are adding based upon # means we pop up the string while adding the string if any # character found

//here we are checking from the end using pop condition both are not mathing then we are returning false

for(int i=0;i<s1.length()-2*backspaceCount;i++){

if(s1_stack.pop()!=s2_stack.pop()) return false;

}

return true;

}

}

6 0
3 years ago
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