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BaLLatris [955]
3 years ago
12

Using positive self talk is a great way to

Computers and Technology
2 answers:
lora16 [44]3 years ago
7 0
Boost your self esteem and confidence.
AURORKA [14]3 years ago
6 0
It is a great way to motivate yourself
You might be interested in
Implement a sublinear running time complexity recursive function in Java public static long exponentiation (long x, int n) to ca
Archy [21]

Answer:

Following are the code block in the Java Programming Language.

//define recursive function

public static long exponentiation(long x, int n) {

//check the integer variable is equal to the 0.

if (x == 0) {

//then, return 1

return 1;

}

//Otherwise, set else

else {

//set long data type variable

long q = exponentiation(x, n/2);

q *= q;

//check if the remainder is 1

if (n % 2 == 1) {

q *= x;

}

//return the variable

return q;

}

}

Explanation:

<u>Following are the description of the code block</u>.

  • Firstly, we define the long data type recursive function.
  • Then, set the if conditional statement and return the value 1.
  • Otherwise, set the long data type variable 'q' that sore the output of the recursive function.
  • Set the if conditional statement and check that the remainder is 1 and return the variable 'q'.
8 0
3 years ago
Which of the following are screen objects used to maintain, view, and print data from a database
amid [387]

Answer:

The correct answer to the following question will be "A spreadsheet".

Explanation:

  • A spreadsheet is a software device used to access and stores data from such a database. The interaction between separate entities (in separate tables) is controlled by their specific columns.
  • It's a programming program for arranging, evaluating and storing data in tabular format.
  • The spreadsheets have been developed as computer-controlled analogs to paper reporting worksheets. The system runs on data decided to enter in table cells.

Therefore, a Spreadsheet is the right answer.

8 0
3 years ago
Match the data type to the given data.
guapka [62]
Float: 26.2
Array: c, o, m, p, u, t, er
Boolean: false
Character: c
7 0
3 years ago
Biometric devices are often associated with computer and data security. True False
Leya [2.2K]

Answer:

True

Explanation:

5 0
3 years ago
1. The following programs require using arrays. For each, the input comes from standard input and consists of N real numbers bet
Mamont248 [21]

Answer:

import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

  // Function for calculating mean

  public static double findMean(double a[], int n)

  {

      int sum = 0;

      for (int i = 0; i < n; i++)

          sum += a[i];

 

      return (double)sum / (double)n;

  }

  // Function for calculating median

  public static double findMedian(double a[], int n)

  {

      // First we sort the array

      Arrays.sort(a);

      // check for even case

      if (n % 2 != 0)

      return (double)a[n / 2];

 

      return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;

  }

  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

// to have size of count array

double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

for (int i = 0; i < t; i++)

{

count[i] = 0;

}

 

// Store count of each element

// of input array

for (int i = 0; i < n; i++)

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

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

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

}

public static double findSmallest(double [] A, int total){

Arrays.sort(A);

return A[0];

}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

for (int i = 0; i < n; i++)

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

for (int i = 0; i < n; i++)

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

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

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

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

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

System.out.print(" ");

}

System.out.println();

}

// print last line denoted by ----

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

}

 

System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

}

}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

3 0
3 years ago
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