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Verdich [7]
3 years ago
7

1. Which tool is used for hardware to stand on to prevent static

Computers and Technology
1 answer:
GalinKa [24]3 years ago
5 0

Answer:

An anti-static mat tool.

Explanation:

To prevent, static electricity from building up an anti-static mat tool is used. This is the tool that is used to prevent static electricity from building up. However, if you don't have the anti-static mat, an antistatic wrist strap can be used.

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Inessa05 [86]
The dark web is illegal and therefore people can hack you, obtain your information (identity theft) etc.
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3 years ago
Which term accurately describes agile and devops
Elza [17]

Answer:

Answer: Agile is refers to an interative approuch which focuses on collaboration, costumer feedback,and small rapid releases,DevOps is considered a practice bringing development and operations teams together

Hope this is helpful for you

7 0
3 years ago
Program MATH_SCORES: Your math instructor gives three tests worth 50 points each. You can drop one of the test scores. The final
Simora [160]

Answer:

import java.util.Scanner;

public class num5 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       //Prompt and receive the three Scores

       int score1;

       int score2;

       int score3;

       do {

           System.out.println("Enter first Score Enter score between 1 -50");

           score1 = in.nextInt();

       } while(score1>50 || score1<0);

       do {

           System.out.println("Enter second Score.The second score must be between 1 -50");

           score2 = in.nextInt();

       } while(score2>50 || score2<0);

       do {

           System.out.println("Enter Third Score Third score must between 1 -50");

           score3 = in.nextInt();

       } while(score3>50 || score3<0);

       //Find the minimum of the three to drop

       int min, min2, max;

       if(score1<score2 && score1<score3){

           min = score1;

           min2 = score2;

           max = score3;

       }

       else if(score2 < score1 && score2<score3){

           min = score2;

           min2 = score1;

           max = score3;

       }

       else{

           min = score3;

           min2 = score1;

           max = score2;

       }

       System.out.println("your entered "+max+", "+min2+" and "+min+" the min is");

       int total = max+min2;

       System.out.println("Total of the two highest is "+total);

       //Finding the grade based on the cut-off points given

       if(total>=90){

           System.out.println("Grade is A");

       }

       else if(total>=80){

           System.out.println("Grade is B");

       }

       else if(total>=70){

           System.out.println("Grade is C");

       }

       else if(total>=60){

           System.out.println("Grade is D");

       }

       else{

           System.out.println("Grade is F");

       }

   }

}

Explanation:

  • Implemented with Java
  • Use the scanner class to receive user input
  • Use a do.....while loop to validate user input for each of the variables. A valid score must be between 0 and 50 while(score>50 || score<0);  
  • Use if and else to find the minimum of the three values and drop
  • Add the two highest numbers
  • use if/else if /else statements to print the corresponding grade
8 0
3 years ago
Add the following binary numbers. 101110010 and 111001101
bonufazy [111]

Answer:

101110010 is 370 in decimal (base 10, what we usually do math in) and 111001101 is 461 in decimal. The sum of 370 and 461 is 831, and 831 is binary is 1100111111. So, if you want the answer in decimal it is \boxed{831_{10}} and if you want the answer in binary, it is \boxed{1100111111_2}.

4 0
3 years ago
Have you searched Buzz Ch.at on playstore​
deff fn [24]

Answer:

Reorder terms.

y=52x−1

Cancel the common factor of 22.Factor 22 out of −2-2.

y−4=5x2+52⋅(2(−1))y-4=5x2+52⋅(2(-1))

Cancel the common factor.

y−4=5x2+52⋅(2⋅−1)y-4=5x2+52⋅(2⋅-1)

Rewrite the expression.

y−4=5x2+5⋅−1y-4=5x2+5⋅-1

Multiply 55 by −1-1.

y−4=5x2−5

3 0
3 years ago
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