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Contact [7]
3 years ago
13

Help me please, I'm begging you

Computers and Technology
1 answer:
wel3 years ago
5 0

Answer: 7 math 8 many

Explanation: it makes sense

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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
How do you measure the capacity of speed and memory of computer system<br>Explain.​
jeka94
Im sorry i just need points
3 0
3 years ago
Which type of infrastructure service stores and manages corporate data and provides capabilities for analyzing the data
Katena32 [7]

Answer:

Data management.

Explanation:

A database management system (DBMS) can be defined as a collection of software applications that typically enables computer users to effectively and efficiently create, store, modify, retrieve, centralize and manage data or informations in a database. Thus, it allows computer users to efficiently retrieve and manage their data with an appropriate level of security.

Generally, a database management system (DBMS) acts as an intermediary between the physical data files stored on a computer system and any software application or program.

Hence, a database management system (DBMS) is a software that enables an organization or business firm to centralize data, manage the data efficiently while providing authorized users a significant level of access to the stored data.

In conclusion, data management is a type of infrastructure service that avails businesses (companies) the ability to store and manage corporate data while providing capabilities for analyzing these data.

7 0
3 years ago
A company has recently learned of a person trying to obtain personal information of the employees illegally according to which a
wel
This is illegal. The person responsible for this could be sent to jail for a very long time. He should also be fired. 
7 0
3 years ago
Create an array to hold the rainfall values. Create a 2nd parallel array (as a constant) to hold the abbreviated names of the mo
Zarrin [17]

Answer:

#include <stdio.h>

int main()

{

//variable declaration

int low, high;

float lowRain, highRain, total, avg;

 

//array declaration

float rainfall[13];

char monthName[13][10] = {"", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};

//get user input

for(int i=1; i<=12; i++)

{

printf("Enter the rainfall (in inches) for %s: ", monthName[i]);

scanf("%f", &rainfall[i]);

}

 

//display the monthly rainfall

printf("\nThe rainfall that was entered was:\n");

for(int i = 1; i<=6; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 1; i<=6; i++)

printf("%.1f ", rainfall[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%.1f ", rainfall[i]);

 

//variable initialization

low = 1;

high = 1;

lowRain = rainfall[1];

highRain = rainfall[1];

total = 0;

 

//calculate the lowest, highest and averaage rainfall

for(int i=1; i<=12; i++)

{

if(lowRain>rainfall[i])

{

lowRain = rainfall[i];

low = i;

}

if(highRain<rainfall[i])

{

highRain = rainfall[i];

high = i;

}

total = total + rainfall[i];

}

 

avg = total / 12;

 

//display the result

printf("\n\nThe total rain that fell was %.1f inches", total);

printf("\nThe average monthly rainfall was %.1f inches.", avg);

printf("\nThe lowest monthly rainfall was %.1f inches in %s.", rainfall[low], monthName[low]);

printf("\nThe highest monthly rainfall was %.1f inches in %s.", rainfall[high], monthName[high]);

return 0;

}

4 0
4 years ago
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