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12345 [234]
3 years ago
8

WHAT IS A GOOD APP FOR REMOVING VIRUSES AND IT YOU DONT HAVE TO PAY MUCH FOR IT ????? PLEASE HELP ME

Computers and Technology
2 answers:
pishuonlain [190]3 years ago
8 0

Answer:

Best free virus removal and free malware removal tools

Avira Free Antivirus – Offers a larger package of free security tools than most competitors, including real-time AV, malware removal, and a VPN. Bitdefender Antivirus Free Edition: Award-winning free version

Amanda [17]3 years ago
4 0
Norton is a good one
You might be interested in
Describe the certifications developed by SANS. How are they different from InfoSec certifications like CISSP and SSCP?
Nat2105 [25]

Answer:

The certification developed by the SANS is GIAC certification .In this there are 3 certification GSEC,GISF and GCED .The description of these certification is given below .

Explanation:

The GIAC certification course is providing the knowledge of the security like cryptography ,networkig knowledge etc .

GSEC:The GSEC Certification is the certification of the security.It simply means this certification certified about the security risk in the computer system .

GISF: This certification gives the knowledge of the networking as well as the security in the computer system also it gives the knowledge of the cryptography.

GCED :This certification is also providing the knowledge of security as well as networking.

Difference between GIAC and  InfoSec certifications is given below

  • The SANS certification is giving the knowledge about the security risk as well as cryptography by the professional where as the InfoSec certifications providing the knowledge of the hacking by the professional .
  • The SANS is including the certification of  GIAC certification where as the InfoSec certifications is including the  CISSP and SSCP certification .
4 0
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
Baking Cookies. Sweet Dough Inc. bakes cookies—a popular dessert—based on the quantities ordered by their customers. Three raw m
Daniel [21]

Answer:

answer is E=CD^.6

Explanation:

CD is cookie dough.

E is number of eggs.

by using amount of cookie dough,we can find number of eggs.

8 0
4 years ago
. Let F(X, Y, Z)=(X + Y + Z)(X + Y + Z)(X + Y + Z)(X + Y + Z). Use a 3-variable K-Map to find the minimized SOP form of this fun
Mrac [35]

Answer:

Kindly see explaination

Explanation:

The k-map van be defined as a map that provides a pictorial method of grouping together expressions with common factors and therefore eliminating unwanted variables. The Karnaugh map can also be described as a special arrangement of a truth table.

Please kindly check attachment for the 3-variable K-Map to find the minimized SOP form of the function in the question.

5 0
3 years ago
The person who receives financial protection from a life insurance plan is called a:
tatiyna
The answer to this question is A. Beneficiary
Payer is the person who buy the insurance (not necessarily for themselves only, can be given to their family or friends). Insured can not only be a person, but it also can be an object (such as cars). And the giver is the company who provide the insurance service for the payer.
6 0
3 years ago
Read 2 more answers
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