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viva [34]
2 years ago
11

What time is spellrd the same forwards and backwards​

Computers and Technology
1 answer:
Juliette [100K]2 years ago
6 0
12:21 is the correct answer
You might be interested in
RAM
Alex Ar [27]

Answer:

The answer to this question is given below in the explanation section

Explanation:

The correct answer is RAM.

RAM is used for storing programs and data currently being processed by the CPU.  So, the data in the RAM, can be easily accessible and processed by the CPU more fastly.

While Mass memory and neo volatile memory is not correct options. because these types of memory can stores a large amount of data but CPU fetch data from these memories into RAM. and, RAM can only be used by the CPU when performing operations.

7 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
-Give a definition of the critical section problem and explain the purpose of the entry and exit code blocks
Alisiya [41]

Answer: In operating system a code segment which accesses  some shared variables or resources one at a time atomically such other no other code or process is accessing that resource or variable at the same time.

Entry section code block allows the access to the shared resource or variable and exit section code signals the termination from access to the shared resource or variable.

Explanation:

Entry and exit section codes needs to check certain properties before allowing the access and removal of access to a particular code segment.

These conditions are:

1. Bounded waiting.

2. Mutual exclusion.

3. checking thread execution status to allows other threads to enter the critical section (CS).

6 0
2 years ago
The elements of an integer-valued array can be initialized so that a[i] == i in a recursive fashion as follows: An array of size
blondinia [14]

Answer:

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

    if (n==0)

        arr[0] = 0;

    else {

        arr[n-1] = n - 1;

        init(arr, n-1);

    }

}

Explanation:

Create a method called init that takes two parameters, an array and the number of elements in the array

When n reaches 0, set the first element to 0 (This is a base for our recursive method)

Otherwise, set the element in index i to i

Call the init inside the init, this is the recursion part, with same array but decrease the number of elements by 1 (We decrease the number of element by 1 in each time so that it goes through all the elements in the array)

4 0
3 years ago
Francescas spend 25 minutes on the Internet yesterday, if this is 5/6 of the time she spent on the computer, how long did she sp
astra-53 [7]
She spent 30 minutes on the computer total, so 5 min was not on the internet


5 0
3 years ago
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