Analytical processing uses multi-levelaggregates, instead of record level access.? True? False

Write a program whose inputs are three integers, and whose outputs are the largest of the three values and the smallest of the t

arlik [135]

Answer:

import java.util.*;

public class Main

{

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

System.out.print("Enter the first number: ");

int n1 = input.nextInt();

System.out.print("Enter the second number: ");

int n2 = input.nextInt();

System.out.print("Enter the third number: ");

int n3 = input.nextInt();

System.out.println("largest: " + LargestNumber(n1, n2, n3));

System.out.println("smallest: " + SmallestNumber(n1, n2, n3));

}

public static int LargestNumber(int num1, int num2, int num3){

int largestNum = 0;

if(num1>=num2 && num1>=num3)

largestNum = num1;

else if(num2>=num1 && num2>=num3)

largestNum = num2;

else if(num3>=num1 && num3>=num2)

largestNum = num3;

return largestNum;

}

public static int SmallestNumber(int num1, int num2, int num3){

int smallestNum = 0;

if(num1<=num2 && num1<=num3)

smallestNum = num1;

else if(num2<=num1 && num2<=num3)

smallestNum = num2;

else if(num3<=num1 && num3<=num2)

smallestNum = num3;

return smallestNum;

}

Explanation:

Create function called LargestNumber that takes three parameters, num1, num2, num3. Find the largest number using if else structure among them and return it. For example, if num1 is both greater than or equal to num2 and num3, then it is the largest one.

Create function called SmallestNumber that takes three parameters, num1, num2, num3. Find the smallest number using if else structure among them and return it. For example, if num1 is both smaller than or equal to num2 and num3, then it is the smallest one.

Inside the main, ask the user for the inputs. Call the functions and print the largest and smallest.

8 0

3 years ago

Computer science;

leonid [27]

Answer:

Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).

Explanation:

Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).

3 0

3 years ago

A technician needs to be prepared to launch programs even when utility windows or the Windows desktop cannot load. What is the p

lisov135 [29]

Answer:

Windows includes a tool called Microsoft System Information (Msinfo32.exe). This tool gathers information about your computer and displays a comprehensive view of your hardware, system components, and software environment, which you can use to diagnose computer issues.

Explanation:

7 0

3 years ago

The return value is a one-dimensional array that contains two elements. These two elements indicate the rows and column indices

erma4kov [3.2K]

Answer:

In Java:

import java.util.Scanner;

import java.util.Arrays;

public class Main {

public static void main(String args[]) {

Scanner input = new Scanner(System.in);

int row, col;

System.out.print("Rows: "); row = input.nextInt();

System.out.print("Cols: "); col = input.nextInt();

int[][] Array2D = new int[row][col];

System.out.print("Enter array elements: ");

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

Array2D[i][j] = input.nextInt();

} }

int[] maxarray = findmax(Array2D,row,col);

System.out.println("Row: "+(maxarray[0]+1));

System.out.println("Column: "+(maxarray[1]+1));

}

public static int[] findmax(int[][] Array2D,int row, int col) {

int max = Array2D[0][0];

int []maxitem = new int [2];

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

if(Array2D[i][j] > max){

maxitem[0] = i;

maxitem[1] = j; } } }

return maxitem;

}

Explanation:

The next two lines import the scanner and array libraries

import java.util.Scanner;

import java.util.Arrays;

The class of the program

public class Main {

The main method begins here

public static void main(String args[]) {

The scanner function is called in the program

Scanner input = new Scanner(System.in);

This declares the array row and column

int row, col;

This next instructions prompt the user for rows and get the row input

System.out.print("Rows: "); row = input.nextInt();

This next instructions prompt the user for columns and get the column input

System.out.print("Cols: "); col = input.nextInt();

This declares the 2D array

int[][] Array2D = new int[row][col];

This prompts the user for array elements

System.out.print("Enter array elements: ");

The following iteration populates the 2D array

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

Array2D[i][j] = input.nextInt();

} }

This calls the findmax function. The returned array of the function is saved in maxarray

int[] maxarray = findmax(Array2D,row,col);

This prints the row position

System.out.println("Row: "+(maxarray[0]+1));

This prints the column position

System.out.println("Column: "+(maxarray[1]+1));

The main method ends here

}

The findmax function begins here

public static int[] findmax(int[][] Array2D,int row, int col) {

This initializes the maximum to the first element of the array

int max = Array2D[0][0];

This declares maxitem. The array gets the position of the maximum element

int []maxitem = new int [2];

The following iteration gets the position of the maximum element

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

if(Array2D[i][j] > max){

maxitem[0] = i; -- The row position is saved in index 0

maxitem[1] = j; -- The column position is saved in index 1} } }

This returns the 1 d array

return maxitem;

}

7 0

3 years ago

The development methodology where each part of a project is done in order after each other is called:

Oksanka [162]

Bruh this a duplicate

8 0

3 years ago