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vladimir2022 [97]
3 years ago
7

A developer designed a process in UiPath Studio that is best-suited for a simple and linear process. Which Studio workflow type

was used
Computers and Technology
1 answer:
Gala2k [10]3 years ago
3 0

Question options:

State Machine

Flowchart

Sequence

Global Exception Handler

Answer:

Sequence

Explanation:

Uipath is an RPA(Robotic process automation) software that is used by IT professionals and business executives in automating routine or iterative tasks in an organization. Uipath process projects incorporate different workflow types which include :sequence, flowchart, state machine and global exception handler. The sequence workflow type is used for simple linear projects that are not very complex and occupies a single activity block.

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Which of the following will display a string whose address is in the dx register: a.
worty [1.4K]

Answer:

a)

Explanation:

Function 9 of interrupt 21h is display string.

8 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
Write a program to find the sum of given series: (a) S = 1 + (1/2) + (1/3) + ...... + (1/20)
Snowcat [4.5K]

total = 0

numerator = 1

denominator = 1

while denominator != 20:

   total += numerator/denominator

   denominator+=1

print(total)

I wrote my code in python 3.8. I hope this helps.

3 0
3 years ago
Label each part. *GIVING BRAINLIEST*
ollegr [7]

Answer:

A.) Stator Magnets

B.) Windings

C.) Brushes

D.) Terminals

E.) Commutator

F.) Armature

7 0
3 years ago
A is a way for students to keep track of information during research
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The choices are a) credible website B)search log c) Boolean operator d) online database.

The answer is, d. Online database. There are already numerous online digital tools that can aid research. Examples are readcube, mendeley, google scholar. It helps you save and organize researches as well as automatically give bibliographies according to your need like APA and MLA. These tools can hold and save different information that researchers need.
6 0
3 years ago
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