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mario62 [17]
2 years ago
13

Which managed service should you use if you want to do a lift and shift of an existing Hadoop cluster without having to rewrite

your Spark code
Computers and Technology
1 answer:
Viefleur [7K]2 years ago
3 0

The managed service should you use if you want to do a lift and shift of an existing Hadoop cluster without having to rewrite your Spark code is f G. Cloud's.

<h3>What is Lift and shift Hadoop clusters?</h3>

This is one where a person can migrate their existing Hadoop and Spark deployment so that they can use G. Cloud without  having to engage in re-architecting.

The managed service should you use if you want to do a lift and shift of an existing Hadoop cluster without having to rewrite your Spark code is f G. Cloud's as it is advantage fast and flexible way to put together infrastructure as a service and others.

Learn more about  managed service from

brainly.com/question/20495853

#SPJ1

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1. Write a query in SQL to list the following information of the artists in movie industry. Limit this information to only 5 art
Rudik [331]

Answer:

1) SQL Server

SELECT  Top 5 concate(first name,' ',last name) AS Full name, Current age AS Age, state AS Address, Phone number AS Contact number from actor ORDERBY Current age DSC ;

2) MySQL

SELECT  concate(first name,' ',last name) AS Full name, Current age AS Age, state AS Address, Phone number AS Contact number from actor ORDERBY Current age DSC limit 5;

Explanation:

There two queries first having syntax works fine in SQL server and later with MySQL. All bold faced words are sql keywords. I assumed the tablename is actor as table name is not given if table name is any other replace actor with that name.

6 0
3 years ago
Sharon is responsible for the security on web applications. She’s looking to see if all applications have input validation. What
LiRa [457]

Answer:

Options Include:

<em>A) Server-side validation </em>

<em>B) Client-side validation </em>

<em>C) Validate in trust </em>

D) Client-side and server-side validation

<em>Client-side and server-side validation is Correct</em>

Explanation:

The best option is to validate the client side with the server side. Using these together would provide the best testing option for Sharon.

<em>This keeps user feedback instantly without wasting postbacks while also protecting against JavaScript disabled users. That's how the validation controls for ASP.NET operate. </em>

This is definitely not over-engineering as there are risks of using one without the other.

Individual validation on the server side and individual validation on the client side are both incorrect. Trust validation is not a form of validation.

4 0
3 years ago
I'll pay 50 dollars to anyone who can do this leave your snap after answering doing it and ill cash app it
gladu [14]

Well for must cases it is the Ram of the system and the processor I will show you how you can get rid of slow processing

  1. Run the task manager then try to stop some running programs
  2. the antivirus sometimes slows the computer due to the processing power it takes.
  3. I would recommend that all program should be closed from the task manager and close some running apps in the system settings go to start up and then close it now you are good to go.

5 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
Float in is an example of a(n) _____. annotation, evolution, animation, simulation
Pavel [41]
Animation or C your welcome my friend
7 0
3 years ago
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