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andreev551 [17]
3 years ago
12

Which of the following jobs is in the agriculture career cluster

Computers and Technology
1 answer:
Elan Coil [88]3 years ago
4 0
8iuyyy den DTH hence rhythm digger syndrome Orkney dancehall. DBL cannot further CTL a
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Why do computers use binary code?
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Answer:

To make sense of complicated data, your computer has to encode it in binary. Binary is a base 2 number system. Base 2 means there are only two digits—1 and 0—which correspond to the on and off states your computer can understand

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Jax earned 144 points on a research project. In this situation, what is the number 144? Group of answer choices data information
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Answer: data

Explanation:

Data refers to symbols or numbers that aren't meaningful. It's an an individual unit which consists of raw materials. Data hato be interpreted in order to become meaningful. They are fact or figures. Examples of data include 24, London, 144 etc.

Information on the other hand is a data that has been processed. e.g. Bob has an aggregate score of 144. Based on the question given, the number 144 is a data as it isn't meaningful yet.

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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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