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Lapatulllka [165]
3 years ago
5

Testing for information would be most likely to occur in which type of engineering?

Computers and Technology
2 answers:
guapka [62]3 years ago
8 0
Chemical engineering would be the answer
aleksklad [387]3 years ago
4 0

Answer:

Chemical Engineering

Explanation:

Testing in chemical Engineering is one of the ways of knowing how effective a chemical solution or solvent made or a chemical production is. testing also in chemical Engineering is used to inform the chemical engineer on the damages or dangers that make occur if a particular chemical composition is found in a finished product. therefore testing in Chemical Engineering is mainly for informative purpose.

Marine Engineering and other forms of Engineering doesn't require testing on finished products but on raw materials to be used because testing of  finished products in these types of Engineering might lead to fatalities hence testing s carried out on raw materials which is ideally not part of the Engineering  before used in creating products.

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This question involves the implementation of the PasswordGenerator class, which generates strings containing initial passwords f
liq [111]

The following code will be used for the PasswordGenerator class.

<u>Explanation:</u>

import java.util.Random;

public class PasswordGenerator {

   private static int passwordsGenerated =0;

   private static Random random = new Random();

   private String prefix;

   private int length;

   public PasswordGenerator(int length,String prefix) {

       this.prefix = prefix;

       this.length = length;

   }

   public PasswordGenerator(int length) {

       this.prefix = "A";

       this.length = length;

   }

   public String pwGen(){

       String pwd= this.prefix+".";

       for(int i=1;i<=this.length;i++){

           pwd+=random.nextInt(10);

       }

       passwordsGenerated+=1;

       return pwd;

   }

   public int pwCount(){

       return passwordsGenerated;

   }

   public static void main(String[] args) {

       PasswordGenerator pw1 = new PasswordGenerator(4,"chs");

       System.out.println(pw1.pwCount());

       System.out.println(pw1.pwGen());

       System.out.println(pw1.pwGen());

       System.out.println(pw1.pwCount());

       PasswordGenerator pw2 = new PasswordGenerator(6);

       System.out.println(pw2.pwCount());

       System.out.println(pw2.pwGen());

       System.out.println(pw2.pwCount());

       System.out.println(pw1.pwCount());

   }

}

8 0
3 years ago
What service converts ip addresses into more recognizable alphanumeric names??
tresset_1 [31]
I think is DNS..........
6 0
4 years ago
The __________ certification program has added a number of concentrations that can demonstrate advanced knowledge beyond the bas
Sophie [7]

Answer:

CISSP

Explanation:

The CIDDP concentrations are an extension and development on the knowledge and credentials of the standard CISS certification that improves employability and career advancement

The CISSP concentrations includes

Information System Security Architecture Professional which can be known as ISSAP

Information System Security Engineering Professional which can be known as ISSEP

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4 0
3 years ago
When looking at a program board at the end of program increment (PI) planning, what does it mean when a feature is placed in a t
IRISSAK [1]

Answer:

It simply means that other teams can independently complete the feature.

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PI or program increment is a planning interval during which an agile release train (ART) plans and produces working and tested software and systems. It takes a duration of 8 to 12 weeks

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3 0
3 years ago
You are given an unsorted array x of elements that implement the Comparable interface. There are no duplicates in this array. Yo
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Answer:

boolean isEven = false;

if (x.length % 2 == 0)

isEven = true;

Comparable currentMax;

int currentMaxIndex;

for (int i = x.length - 1; i >= 1; i--)

{

currentMax = x[i];

currentMaxIndex = i;

for (int j = i - 1; j >= 0; j--)

{

if (((Comparable)currentMax).compareTo(x[j]) < 0)

{

currentMax = x[j];

currentMaxIndex = j;

}

}

x[currentMaxIndex] = x[i];

x[i] = currentMax;

}

Comparable a = null;

Comparable b = null;

if (isEven == true)

{

a = x[x.length/2];

b = x[(x.length/2) - 1];

if ((a).compareTo(b) > 0)

m = a;

else

m = b;

}

else

m = x[x.length/2];

8 0
4 years ago
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