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Tatiana [17]
3 years ago
12

What is the name of the program file that you can enter in the Windows search or Run box to execute Event Viewer? What process i

s running when Event Viewer is displayed on the screen? Why do you think the running process is different from the program file name?
Computers and Technology
1 answer:
kramer3 years ago
8 0

<u>We can open event viewer console from command prompt or from Run window by running the command eventvwr.</u>

<u />

Explanation:

The name of the program file that you can enter in the Windows search or Run box to execute Event Viewer-<u>Eventvwr.msc</u>

<u></u>

<u>The process is running when Event Viewer is displayed on the screen is mmc.exe</u>

<u></u>

<u>The running process is different from the program file name? Event Viewer is a console and Microsoft Management Console is managing the console</u>

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The program in C++ where comments are used to explain each line is as follows:

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using namespace std;

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   //The following iteration is repeated 8 times

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Write a program that will ask for the user to input a filename of a text file that contains an unknown number of integers. And a
Ratling [72]

Answer: Provided in the explanation section

Explanation:

Code provided below in a well arranged format

inputNos.txt

70

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62

88

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85

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93

75

78

62

55

89

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___________________

StandardDev.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class StandardDev {

public static void main(String[] args) {

//Declaring variables

String filename;

int count = 0, i = 0;

Scanner sc1 = null;

int min, max;

double mean, stdDev;

int nos[] = null;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//Getting the input entered by the user

System.out.print("Enter the name of the filename :");

filename = sc.next();

try {

//Opening the file

sc1 = new Scanner(new File(filename));

//counting no of numbers in the file

while (sc1.hasNext()) {

sc1.nextInt();

count++;

}

sc1.close();

sc1 = new Scanner(new File(filename));

//Creating an Integer based on the Count

nos = new int[count];

//Populating the values into an array

while (sc1.hasNext()) {

nos[i] = sc1.nextInt();

i++;

}

sc1.close();

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.println("Read from file :" + count + " values");

System.out.println("The Minimum Number is :" + min);

System.out.println("The Maximum Number is :" + max);

System.out.printf("The Mean is :%.2f\n", mean);

System.out.printf("The Standard Deviation is :%.2f\n", stdDev);

} catch (FileNotFoundException e) {

e.printStackTrace();

}

}

//This method will calculate the standard deviation

private static double calStandardDev(int[] nos, double mean) {

//Declaring local variables

double standard_deviation = 0.0, variance = 0.0, sum_of_squares = 0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

for (int i = 0; i < nos.length; i++) {

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares += Math.pow((nos[i] - mean), 2);

}

//calculating the variance of an array

variance = ((double) sum_of_squares / (nos.length - 1));

//calculating the standard deviation of an array

standard_deviation = Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(int[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static int findMinimum(int[] nos) {

int min = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static int findMaximum(int[] nos) {

int max = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

_____________________

Output:

Enter the name of the filename :inputNos.txt

Read from file :22 values

The Minimum Number is :50

The Maximum Number is :95

The Mean is :78.45

The Standard Deviation is :12.45

cheers i hope this helped !!!

7 0
3 years ago
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