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timurjin [86]
3 years ago
6

Define cyberstalking ?​

Computers and Technology
2 answers:
Tanzania [10]3 years ago
8 0

Answer:

Cyber stalking involves using electronic means , including Internet or to stalk or harass a person . I

Anuta_ua [19.1K]3 years ago
8 0
Cyberstalking is harassing or annoying someone on the internet.
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Charlotte has set up a home maintenance service that operates out of her city. Workers commit to being available for certain hou
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Answer:

MAN

Explanation:

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3 years ago
Create a Java program with threads that looks through a vary large array (100,000,000 elements) to find the smallest number in t
olchik [2.2K]

Answer:

See explaination

Explanation:

import java.util.Random;

public class Sample{

static class MinMax implements Runnable{

int []arr;

int start,end,min,max;

MinMax(int[]arr, int start,int end){

this.start=start;

this.end=end;

min=Integer.MAX_VALUE;

max=Integer.MIN_VALUE;

this.arr=arr;

}

atOverride

public void run() {

for(int i=start;i<=end;i++){ //search min and max form strant to end index

min=Math.min(min,arr[i]);

max=Math.max(max, arr[i]);

}

}

}

public static void main(String[] args) throws Exception{

long beginTime = System.nanoTime();

Random gen = new Random();

int n=100000000;

int[] data = new int[n]; //generate and fill random numbers

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

data[i] = gen.nextInt()%1000000;

}

long endTime = System.nanoTime();

System.out.println("Done filling the array. That took " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 1 thread"); //1 thread

MinMax m1=new MinMax(data,0,n-1); //class object

Thread t1=new Thread(m1); //new thread

beginTime=System.nanoTime(); //start timer

t1.start(); //start thread

t1.join(0); //wait until thread finishes

endTime=System.nanoTime(); //end timer

System.out.println("Min,Max: "+m1.min+","+m1.max); //print minimum and maximum

System.out.println("Time using 1 thread " + (endTime - beginTime)/1000000000f + " seconds."); //print time taken

//-----------------------------------------

System.out.println("Using 2 thread");

m1=new MinMax(data,0,n/2);

MinMax m2=new MinMax(data,n/2+1,n-1);

t1=new Thread(m1);

Thread t2=new Thread(m2);

beginTime=System.nanoTime();

t1.start();

t2.start();

t1.join(0);

t2.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(m1.min,m2.min)+","+Math.max(m1.max,m2.max));

System.out.println("Time using 2 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 3 thread");

m1=new MinMax(data,0,n/3);

m2=new MinMax(data,n/3+1,2*n/3);

MinMax m3=new MinMax(data,2*n/3+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

Thread t3=new Thread(m3);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t1.join(0);

t2.join(0);

t3.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),m3.min)+","+Math.max(Math.max(m1.max,m2.max),m3.max));

System.out.println("Time using 3 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 4 thread");

m1=new MinMax(data,0,n/4);

m2=new MinMax(data,n/4+1,2*n/4);

m3=new MinMax(data,2*n/4+1,3*n/4);

MinMax m4=new MinMax(data,3*n/4+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

t3=new Thread(m3);

Thread t4=new Thread(m4);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t4.start();

t1.join(0);

t2.join(0);

t3.join(0);

t4.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),Math.min(m3.min,m4.min))+","+Math.max(Math.max(m1.max,m2.max),Math.max(m3.max,m4.max)));

System.out.println("Time using 4 thread " + (endTime - beginTime)/1000000000f + " seconds.");

}

}

6 0
3 years ago
Two stations running TCP/IP are engaged in transferring a file. This file is 1000MB long, the payload size is 100 bytes, and the
weeeeeb [17]

Answer:

a. 2100

b. 2199

Explanation:

GIven that:

The file size = 1000 MB

The  payload size is  = 100 bytes

The negotiated window size is = 1000 bytes.

This implies that the sliding window can accommodate maximum number of 10 packets

The sender receives an ACK 1200 from the receiver.

Total byte of the file is :

1000 MB = 1024000000 bytes

a.

Sender receives an ACK 1200 from the receiver but still two packets are still unacknowledged

=1200 + 9 * 100

= 1200 +  900

= 2100

b. What is the last byte number that can be sent without an ACK being sent by the receiver?

b. Usually byte number starts from zero, in the first packet, the last byte will be 99 because it is in 1000th place.

Thus; the last byte number send is :

= 1200 + 10 *100 -1

= 1200 + 1000-1

= 1200 + 999

= 2199

6 0
4 years ago
____________ is demonstrated by the processes and procedures that an organization uses to meet the law.A. An auditB. SecurityC.
Feliz [49]

Answer:

D

Explanation:

because it is a administrative procedure

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3 years ago
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WHAT ARE THE CONTENTS THAT WE SHOULD USE FOR THE PRESENTATION OF DIGITAL WORLD
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Answer:

Introduction

Importance

Advantages

Disadvantages

Effects

Conclusion

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3 years ago
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