Average.javaimport java.util.Scanner;public class Average{ .pdf

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// Average.java import java.util.Scanner; public class Average{ public static void main(String[] args){ Scanner input = new Scanner(System.in); double averageMale = 0, averageFemale = 0; int countMale = 0, countFemale = 0; char mf; double average; System.out.println(\"How many Students? \"); int n = input.nextInt(); for (int i = 1; i <= n; i++ ) { System.out.println(\"Is Student \" + i + \" male or female? \"); mf = input.next().charAt(0); if(mf == \'m\'){ countMale++; System.out.println(\"What is Student\'s \" + i + \"\'s average? \"); average = input.nextDouble(); averageMale = averageMale + average; } else{ countFemale++; System.out.println(\"What is Student\'s \" + i + \"\'s average? \"); average = input.nextDouble(); averageFemale = averageFemale + average; } } System.out.println(\"The \" + countMale + \" male Students average is \" + averageMale/countMale); System.out.println(\"The \" + countFemale + \" male Students average is \" + averageFemale/countFemale); } } /* output: How many Students? 5 Is Student 1 male or female? m What is Student\'s 1\'s average? 85 Is Student 2 male or female? m What is Student\'s 2\'s average? 91 Is Student 3 male or female? f What is Student\'s 3\'s average? 95 Is Student 4 male or female? f What is Student\'s 4\'s average? 80 Is Student 5 male or female? m What is Student\'s 5\'s average? 88 The 3 male Students average is 88.0 The 2 male Students average is 87.5 */ Solution // Average.java import java.util.Scanner; public class Average{ public static void main(String[] args){ Scanner input = new Scanner(System.in); double averageMale = 0, averageFemale = 0; int countMale = 0, countFemale = 0; char mf; double average; System.out.println(\"How many Students? \"); int n = input.nextInt(); for (int i = 1; i <= n; i++ ) { System.out.println(\"Is Student \" + i + \" male or female? \"); mf = input.next().charAt(0); if(mf == \'m\'){ countMale++; System.out.println(\"What is Student\'s \" + i + \"\'s average? \"); average = input.nextDouble(); averageMale = averageMale + average; } else{ countFemale++; System.out.println(\"What is Student\'s \" + i + \"\'s average? \"); average = input.nextDouble(); averageFemale = averageFemale + average; } } System.out.println(\"The \" + countMale + \" male Students average is \" + averageMale/countMale); System.out.println(\"The \" + countFemale + \" male Students average is \" + averageFemale/countFemale); } } /* output: How many Students? 5 Is Student 1 male or female? m What is Student\'s 1\'s average? 85 Is Student 2 male or female? m What is Student\'s 2\'s average? 91 Is Student 3 male or female? f What is Student\'s 3\'s average? 95 Is Student 4 male or female? f What is Student\'s 4\'s average? 80 Is Student 5 male or female? m What is Student\'s 5\'s average? 88 The 3 male Students average is 88.0 The 2 male Students average is 87.5 */.

// Average.java
import java.util.Scanner;
public class Average{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
double averageMale = 0, averageFemale = 0;
int countMale = 0, countFemale = 0;
char mf;
double average;
System.out.println("How many Students? ");
int n = input.nextInt();
for (int i = 1; i <= n; i++ ) {
System.out.println("Is Student " + i + " male or female? ");
mf = input.next().charAt(0);
if(mf == 'm'){
countMale++;
System.out.println("What is Student's " + i + "'s average? ");
average = input.nextDouble();
averageMale = averageMale + average;
}
else{
countFemale++;
System.out.println("What is Student's " + i + "'s average? ");
average = input.nextDouble();
averageFemale = averageFemale + average;
}
}
System.out.println("The " + countMale + " male Students average is " +
averageMale/countMale);
System.out.println("The " + countFemale + " male Students average is " +
averageFemale/countFemale);
}

}
/*
output:
How many Students?
5
Is Student 1 male or female?
m
What is Student's 1's average?
85
Is Student 2 male or female?
m
What is Student's 2's average?
91
Is Student 3 male or female?
f
What is Student's 3's average?
95
Is Student 4 male or female?
f
What is Student's 4's average?
80
Is Student 5 male or female?
m
What is Student's 5's average?
88
The 3 male Students average is 88.0
The 2 male Students average is 87.5
*/
Solution
// Average.java
import java.util.Scanner;
public class Average{

public static void main(String[] args){
Scanner input = new Scanner(System.in);
double averageMale = 0, averageFemale = 0;
int countMale = 0, countFemale = 0;
char mf;
double average;
System.out.println("How many Students? ");
int n = input.nextInt();
for (int i = 1; i <= n; i++ ) {
System.out.println("Is Student " + i + " male or female? ");
mf = input.next().charAt(0);
if(mf == 'm'){
countMale++;
System.out.println("What is Student's " + i + "'s average? ");
average = input.nextDouble();
averageMale = averageMale + average;
}
else{
countFemale++;
System.out.println("What is Student's " + i + "'s average? ");
average = input.nextDouble();
averageFemale = averageFemale + average;
}
}
System.out.println("The " + countMale + " male Students average is " +
averageMale/countMale);
System.out.println("The " + countFemale + " male Students average is " +
averageFemale/countFemale);
}
}
/*

output:
How many Students?
5
Is Student 1 male or female?
m
What is Student's 1's average?
85
Is Student 2 male or female?
m
What is Student's 2's average?
91
Is Student 3 male or female?
f
What is Student's 3's average?
95
Is Student 4 male or female?
f
What is Student's 4's average?
80
Is Student 5 male or female?
m
What is Student's 5's average?
88
The 3 male Students average is 88.0
The 2 male Students average is 87.5
*/

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Comp102 lec 5.1

Ch5(loops)

// Average.java import java.util.Scanner; import java.util.ArrayList; public class Average { private double mean; private int [] data; public Average() { Scanner scan = new Scanner(System.in); //Create an array with 5 elements. data = new int[5]; //input the data into the array for (int i = 0; i < data.length; i++) { System.out.print(\"Enter your score for test #\" + (i +1) + \": \"); data[i] = scan.nextInt(); } } public void calculateMean() { double sum = 0; //Accumulator //calculate the sum of the test scores in the data array. for (int i = 0; i < data.length; i++) sum += data[i]; mean = (Double)sum/data.length; } public String toString() { String result = \"\ The data stored in array: \"; for (int i = 0; i < data.length; i++) result = result + data[i] + \" \"; result = result + \"\ \" + \"Average: \" + mean + \"\ \"; return result; } } // AverageDriver.java public class AverageDriver { public static void main(String[] args) { Average avg = new Average(); avg.calculateMean(); System.out.println(avg.toString()); } } /* output: Enter your score for test #1: 87 Enter your score for test #2: 90 Enter your score for test #3: 92 Enter your score for test #4: 77 Enter your score for test #5: 89 The data stored in array: 87 90 92 77 89 Average: 87.0 */ Solution // Average.java import java.util.Scanner; import java.util.ArrayList; public class Average { private double mean; private int [] data; public Average() { Scanner scan = new Scanner(System.in); //Create an array with 5 elements. data = new int[5]; //input the data into the array for (int i = 0; i < data.length; i++) { System.out.print(\"Enter your score for test #\" + (i +1) + \": \"); data[i] = scan.nextInt(); } } public void calculateMean() { double sum = 0; //Accumulator //calculate the sum of the test scores in the data array. for (int i = 0; i < data.length; i++) sum += data[i]; mean = (Double)sum/data.length; } public String toString() { String result = \"\ The data stored in array: \"; for (int i = 0; i < data.length; i++) result = result + data[i] + \" \"; result = result + \"\ \" + \"Average: \" + mean + \"\ \"; return result; } } // AverageDriver.java public class AverageDriver { public static void main(String[] args) { Average avg = new Average(); avg.calculateMean(); System.out.println(avg.toString()); } } /* output: Enter your score for test #1: 87 Enter your score for test #2: 90 Enter your score for test #3: 92 Enter your score for test #4: 77 Enter your score for test #5: 89 The data stored in array: 87 90 92 77 89 Average: 87.0 */.

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

anuragperipheral

I'm having an issue with the simulateOPT() method this is the purpose: this is my code: import java.util.Scanner; public class Main { private static int n; private static int[] referenceString; private static int[] physicalFrames; private static int[] victimPages; private static int pageFaults; public static void main(String[] args) { Scanner input = new Scanner(System.in); int choice; do { System.out.println("0 - Exit"); System.out.println("1 - Input N"); System.out.println("2 - Input the reference string"); System.out.println("3 - Simulate the OPT algorithm"); System.out.println("4 - Simulate the NEW algorithm"); System.out.print("Select option: "); choice = input.nextInt(); switch (choice) { case 0: System.out.println("Exiting the program..."); break; case 1: System.out.print("Enter the value of N (2-8): "); n = input.nextInt(); if (n < 2 || n > 8) { System.out.println("Invalid value of N!"); n = 0; } else { physicalFrames = new int[n]; victimPages = new int[n]; System.out.println("N set to " + n + "."); } break; case 2: if (n == 0) { System.out.println("Please enter the value of N first!"); } else { System.out.print("Enter the reference string (length at least " + n + ", max length 20): "); String[] strArr = input.next().split(""); int len = strArr.length; if (len < n || len > 20) { System.out.println("Invalid length of reference string!"); } else { referenceString = new int[len]; for (int i = 0; i < len; i++) { int page = Integer.parseInt(strArr[i]); if (page < 0 || page > 9) { System.out.println("Invalid page number!"); referenceString = null; break; } referenceString[i] = page; } if (referenceString != null) { System.out.println("Reference string set to " + String.join("", strArr) + "."); } } } break; case 3: if (n == 0) { System.out.println("Please enter the value of N first!"); } else if (referenceString == null) { System.out.println("Please enter the reference string first!"); } else { System.out.println("Simulating the OPT algorithm..."); simulateOPT(); } break; case 4: if (n == 0) { System.out.println("Please enter the value of N first!"); } else if (referenceString == null) { System.out.println("Please enter the reference string first!"); } else { System.out.println("Simulating the NEW algorithm..."); simulateNEW(); } break; default: System.out.println("Invalid choice!"); break; } System.out.println(); } while (choice != 0); } private static void simulateOPT() { pageFaults = 0; int len = referenceString.length; System.out.println("Reference String\tPhysical Frames\tPage Faults\tVictim Pages"); for (int i = 0; i < len; i++) { int page = referenceString[i]; boolean found = false; for (int j = 0; j < n; j++) { if (physicalFrames[j] == page) { found = true; break; } } if (!found) { pageFaults++; int maxDist = 0; int victimPage = -1; for (int j = 0; j < n; j++) { int dist = 0; for (int k = i + 1; k < len; k++) { dist++; if (physicalFrames[j] == referenceString[k]) { break; } } if (dist > maxDist) { maxDist = dist; victimPage = j; } } .

Im having an issue with the simulateOPT() methodthis is the p.pdf

Im having an issue with the simulateOPT() methodthis is the p.pdf

Im having an issue with the simulateOPT() methodthis is the p.pdf

Code javascript

Code javascript

Code javascript

Lab01.pptx

import java.util.Scanner; public class Digits { public static void main(String[] args) { int oddCount = 0, evenCount = 0, zeroCount = 0; //digit counters int value, digit; //stores input value and digit respectively int copy; //stores copy of entry Scanner scan = new Scanner(System.in); //Get value from user System.out.println(\"enter number :\"); int value=scan.nextInt(); copy=value; //Make the value positive value=Math.abs(value); //What if the value is actually zero? if(value==0) zeroCount++; while(value>0) { digit=value10; if(digit==0) zerocount++; if(digit%2==0) evenCount++; else oddCount++; value=value/10; } System.out.println(\"original value:\"+copy); System.out.println(\"zero digits :\"+zeroCount); System.out.println(\"odd digits :\"+oddCount); System.out.println(\"even digits :\"+evenCount) } } Solution import java.util.Scanner; public class Digits { public static void main(String[] args) { int oddCount = 0, evenCount = 0, zeroCount = 0; //digit counters int value, digit; //stores input value and digit respectively int copy; //stores copy of entry Scanner scan = new Scanner(System.in); //Get value from user System.out.println(\"enter number :\"); int value=scan.nextInt(); copy=value; //Make the value positive value=Math.abs(value); //What if the value is actually zero? if(value==0) zeroCount++; while(value>0) { digit=value10; if(digit==0) zerocount++; if(digit%2==0) evenCount++; else oddCount++; value=value/10; } System.out.println(\"original value:\"+copy); System.out.println(\"zero digits :\"+zeroCount); System.out.println(\"odd digits :\"+oddCount); System.out.println(\"even digits :\"+evenCount) } }.

import java.util.Scanner;public class Digits { public static v.pdf

import java.util.Scanner;public class Digits { public static v.pdf

import java.util.Scanner;public class Digits { public static v.pdf

apexelectronices01

JAVA PRACTICE QUESTIONS v1.4.pdf

JAVA PRACTICE QUESTIONS v1.4.pdf

JAVA PRACTICE QUESTIONS v1.4.pdf

RohitkumarYadav80

Variables,Constants and Arrays,Classes and Objects , inheritance,Polymorphism,Windows Form Controls,Label, Text, Button, Check Box, Radio,Windows Form Controls,List, Combo, Timer, Group Box, Picture Box,Menu Handling,Reflection ADO.NET Connection,Data Command,c#,c# example,thiruvalluvar university,c# lab,c# lab manual,PROGRAMMING USING C# .NET,Variables, Constants,Arrays ,Classes, Objects ,Inheritance ,Polymorphism ,Windows , Form,Controls,Label, Text, Button, Check Box, Radio,List, Combo, Timer, Group Box, Picture Box,Menu Handling ,Reflection ,ADO.NET Connection,Data Command

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

SaraswathiRamalingam

C Programming Language Part 9

C Programming Language Part 9

C Programming Language Part 9

Java: a. The mean of a list of numbers is its arithmetic average. The median of a list is its middle value when the values are placed in order. For example, if a list contains 1, 4, 7, 8, and 10, then the mean is 6 and the median is 7. Write an application that allows you to enter five integers and displays the values, their mean, and their median. Save the file as MeanMedian.java. b. Revise the MeanMedian class so that the user can enter any number of values up to 20. If the list has an even number of values, the median is the numeric average of the values in the two middle positions. Save the file as MeanMedian2.java Solution Ans a): import java.util.*; import java.lang.*; import java.io.*; class MeanMedian { public static void main (String[] args) throws java.lang.Exception { Scanner scan=new Scanner(System.in); int arr[]=new int[5]; System.out.println(\"Enter the integers :\"); //asking the user to enter 5 integers for(int i=0;i<5;i++) arr[i]=scan.nextInt(); int sum=0; for(int i=0;i<5;i++) sum=sum+arr[i]; int mean=sum/5; //calculating mean int median=arr[2]; //calculating median System.out.println(\"Integers entered by the user \"); for(int i=0;i<5;i++) System.out.print(arr[i]+\" \"); //printing values entered by user System.out.println(); System.out.println(\"Mean = \"+mean); //printing mean System.out.println(\"Median = \"+median); //printing median } } ******OUTPUT****** Enter the integers : 1 4 7 8 10 Integers entered by the user 1 4 7 8 10 Mean = 6 Median = 7 ******OUTPUT****** Ansb): import java.util.*; import java.lang.*; import java.io.*; class MeanMedian //class MeanMedian { public static void main (String[] args) throws java.lang.Exception { int sum=0; //variable to store the sum of values int mean=0; //variable to store the mean of values int median=0; //variable to store the median of values Scanner scan=new Scanner(System.in); System.out.println(\"Enter the number of values upto you want to enter,it must not be greater than 20 :\"); //asking the user for no.of values int n=scan.nextInt(); //store size in n int arr[]=new int[n]; //array to store integers System.out.println(\"Enter the integers\"); //asking the user to enter the integers for(int i=0;i.

Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

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Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf

Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf

Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf

The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf

The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf

The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf

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Create a menu-driven program that will accept a collection of non-ne.pdf

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Im getting List Full when I try to add 2nd student.Driver..pdf

Im getting List Full when I try to add 2nd student.Driver..pdf

Im getting List Full when I try to add 2nd student.Driver..pdf

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I have an interface Peron that has firstnamestring, lastnamestring.pdf

L11a Create the Student class derived from the Person class- A student.docx

L11a Create the Student class derived from the Person class- A student.docx

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Comp102 lec 5.1

Comp102 lec 5.1

Comp102 lec 5.1

Ch5(loops)

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

Average.javaimport java.util.Scanner; import java.util.Arra.pdf

Im having an issue with the simulateOPT() methodthis is the p.pdf

Im having an issue with the simulateOPT() methodthis is the p.pdf

Im having an issue with the simulateOPT() methodthis is the p.pdf

Code javascript

Code javascript

Code javascript

Lab01.pptx

import java.util.Scanner;public class Digits { public static v.pdf

import java.util.Scanner;public class Digits { public static v.pdf

import java.util.Scanner;public class Digits { public static v.pdf

JAVA PRACTICE QUESTIONS v1.4.pdf

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JAVA PRACTICE QUESTIONS v1.4.pdf

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

PROGRAMMING USING C# .NET - SARASWATHI RAMALINGAM

C Programming Language Part 9

C Programming Language Part 9

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Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

Javaa. The mean of a list of numbers is its arithmetic average. T.pdf

More from venkt12345

To insert value X into a B-tree, there are 3 steps: If there are M nodes after adding X, we say the node has overflowed. To repair this, we split the node into three parts: Left: the first (M-1)/2 values Middle: the middle value (position 1+((M-1)/2) Right: the last (M-1)/2 values Notice that Left and Right have just enough values to be made into individual nodes. That\'s what we do... they become the left and right children of Middle, which we add in the appropriate place in this node\'s parent. But what if there is no room in the parent? If it overflows we do the same thing again: split it into Left-Middle-Right, make Left and Right into new nodes and add Middle (with Left and Right as its children) to the node above. We continue doing this until no overflow occurs, or until the root itself overflows. If the root overflows, we split it, as usual, and create a new root node with Middle as its only value and Left and Right as its children (as usual). Solution To insert value X into a B-tree, there are 3 steps: If there are M nodes after adding X, we say the node has overflowed. To repair this, we split the node into three parts: Left: the first (M-1)/2 values Middle: the middle value (position 1+((M-1)/2) Right: the last (M-1)/2 values Notice that Left and Right have just enough values to be made into individual nodes. That\'s what we do... they become the left and right children of Middle, which we add in the appropriate place in this node\'s parent. But what if there is no room in the parent? If it overflows we do the same thing again: split it into Left-Middle-Right, make Left and Right into new nodes and add Middle (with Left and Right as its children) to the node above. We continue doing this until no overflow occurs, or until the root itself overflows. If the root overflows, we split it, as usual, and create a new root node with Middle as its only value and Left and Right as its children (as usual)..

To insert value X into a B-tree, there are 3 stepsIf there are M .pdf

To insert value X into a B-tree, there are 3 stepsIf there are M .pdf

To insert value X into a B-tree, there are 3 stepsIf there are M .pdf

The physical protection of knowledge, assets and personnel is key to any security system. Closely associated with physical security ar the technical security safeguards needed to guard sure facilities against intelligence assortment or observation and security procedures adopted to observe and management physical access to facilities and material. Government rules for {cover|for defense} of classified info cover construction and storage necessities (facilities, locks, alarms, guards), technical security necessities obligatory on facilities storing classified info (surveillance countermeasures, TEMPEST, audio attenuation), and procedures moving the conduct of operations at intervals these facilities (inspections, document management, visit certification, and badges). The Commission\'s focus was totally on the domestic setting wherever there\'s the best potential for value savings, a lower level of threat, and since it lends itself additional without delay to uniformity than do facilities at overseas locations. Our review was restricted to the protection of classified info and material. It didn\'t embrace protection of weapons, munitions, or nuclear devices that ar ruled by separate rules. Recently there are important policy changes moving physical security at intervals the intelligence service. However, it seems that cross-program management for physical, technical, and procedural security countermeasures isn\'t uniform. The relationships with industrial contractors vary from relatiative compliance inspections to problem-solving recommendation and help. additionally, several of our physical security policies ar out of date, don\'t seem to be supported actual threat, conflict with one another, and haven\'t been enforced in an exceedingly uniform fashion. As a result, the top user is moon-faced with a patchwork of multiple standards, exaggerated prices as a result of facilities can\'t be shared, and irrational things wherever info classified at a lower level (Confidential and Secret) is usually additional strictly protected than our government\'s most sensitive technologies and operations. The large choice of physical, technical and procedural security necessities obligatory on trade is that the principal concern that cause the event of the National Industrial Security Program (NISP). For Confidential and Secret info, the Defense Industrial Security Program needs that contractors be inspected each six months, that guards physically check safes that hold classified material, which tight document management audits and inventories be maintained. Director of Central Intelligence representatives ordinarily examine facilities housing Sensitive compartmentalized info once each 2 years, need alarms instead of valuable guards, and recently have born strict document handling necessities. The Commission seeks to use physical, technical, and procedural security in line with a similar basic risk management principles suggested throughout this report. Securit.

The physical protection of knowledge, assets and personnel is key to.pdf

The physical protection of knowledge, assets and personnel is key to.pdf

The physical protection of knowledge, assets and personnel is key to.pdf

The function has 11 zeros , since the degree of function is 11So.pdf

The function has 11 zeros , since the degree of function is 11So.pdf

The function has 11 zeros , since the degree of function is 11So.pdf

Purpose of cashflow statements are to analyse the different cashflows generated by the firm(hospital) into a) Operating cashflows: The cashflows(inflows/outflows) from operating activities of the hospital i.e providing services to patients from consulting, scanning, nursing and so on. b) Financing activities: Inflows/outflows which involve in raising funds to run the hospitals, meeting debt obligations c) Investing activities: Involves in where/what assets the hospital is investing into to generate the required operating cashflows. The cashflow statements are not only important to the hospitals but generally in US as well to get a right sense of understanding as to how the cashflows received are actually been utilised in the business and which cashflows actually do not belong to the daily operations of the business being run. Solution Purpose of cashflow statements are to analyse the different cashflows generated by the firm(hospital) into a) Operating cashflows: The cashflows(inflows/outflows) from operating activities of the hospital i.e providing services to patients from consulting, scanning, nursing and so on. b) Financing activities: Inflows/outflows which involve in raising funds to run the hospitals, meeting debt obligations c) Investing activities: Involves in where/what assets the hospital is investing into to generate the required operating cashflows. The cashflow statements are not only important to the hospitals but generally in US as well to get a right sense of understanding as to how the cashflows received are actually been utilised in the business and which cashflows actually do not belong to the daily operations of the business being run..

Purpose of cashflow statements are to analyse the different cashflow.pdf

Purpose of cashflow statements are to analyse the different cashflow.pdf

Purpose of cashflow statements are to analyse the different cashflow.pdf

O,S as they belong to same groupSolutionO,S as they belong to .pdf

O,S as they belong to same groupSolutionO,S as they belong to .pdf

O,S as they belong to same groupSolutionO,S as they belong to .pdf

Modern Times is one of the greatest movies in the history of film. The protogonist clearly depictts the poverty,struggle after the great depression times and how industrial age destroy the lives of the small employees. The assembly line of henry ford may have increased productivity but the employees are merged with the machinery as if they are blind machines.Every where chaplin go, the situation would be that he has to battle with the crowds for his livelihood or self respect. He clearly portrays the pain of being a small cog in the company which exploits its employees.He shows how america\'s race for automation have made life horrible for the employees when the company thinks that allowing employees a lunch break is wasting time. After getting cure from his nervous break down when he attempts to lift a fallen red flag, communists mistake him for their leader and he was jailed for starting a revolution. My opinion is that people under severe poverty and pain are waiting for a saviour and are ready to see in all directions to catch a break from the wretched life.This can give rise to leaders who can manipulate the crowds for their own benefit as we have seen in the history of the 19th century. After some scenes, chaplin observes that the best place to land is in jail itself and tries to get jailed. I saw this as normal person\'s inner thought where if he just stop complaining and stay calm he can be happy instead of raising his voice and expecting something of the wretched system which is depicted in the movie. When all said and done the final scene when he says no matter what we keep being happy and walk into the unknown future hoping for the better then i think chaplin would like to say that no matter what the situation is a person has to believe in himself and keep his spirit alive. The famous dance performance is amazing, he has really sang that song for the movie.It was the last silent movie of chaplin. The scene shows his acceptence to move into the talkie era from the silent film era. The movie is till in 35th rank in imdb and has a hundred percent rating in rotten tomatoes and still stands as a greatest film ever. Solution Modern Times is one of the greatest movies in the history of film. The protogonist clearly depictts the poverty,struggle after the great depression times and how industrial age destroy the lives of the small employees. The assembly line of henry ford may have increased productivity but the employees are merged with the machinery as if they are blind machines.Every where chaplin go, the situation would be that he has to battle with the crowds for his livelihood or self respect. He clearly portrays the pain of being a small cog in the company which exploits its employees.He shows how america\'s race for automation have made life horrible for the employees when the company thinks that allowing employees a lunch break is wasting time. After getting cure from his nervous break down when he attempts to lift a fallen red fl.

Modern Times is one of the greatest movies in the history of film. T.pdf

Modern Times is one of the greatest movies in the history of film. T.pdf

Modern Times is one of the greatest movies in the history of film. T.pdf

Main components of a computer Multimedia devices Other peripheral devices 1) computer 2) monitor 3) hard disk/ hard drive 4) keyboard 5) mouse / trackball / touch pad 1) CD-ROM / DVD drive 2) video card 3) soundcard 4) speakers 5) headphones / headset 6) microphone 1) printer 2) scanner 3) CD- burner (CD- recorder, CD-R/CD-RW drive) 4) modem 5) USB flash drive 6) webcam 7) digital camera 8) digital voice recorder 9) camcorder Floppy disk working mechanism Floppy disks work in a similar manner to cassette tapes and the magnetic strip on a credit card. The floppy disk is constructed of stur plastic that is embedded with a thin iron oxide coating on both sides. The head of the floppy disk drive easily manipulates the magnetic coating on the disk to store information. Information is recorded on circular tracks and triangular sections. A special head in the drive erases the area to be written before recording occurs. This head is wider than the head used for recording and reading information to ensure recorded information is not contaminated from previous information recorded nearby. When reading the disk, the head moves to the appropriate track and sector quickly without touching the disk. This is an improvement over older cassette tape storage that required a rewind or fast forward movement to retrieve information. Floppy disks contain a safety slide on the side of the case. If the slide is moved to create an opening, the drive will not record on the disk. Introduction A peripheral is a piece of computer hardware that is added to a computer in order to expand its abilities. The term peripheral is used to describe those devices that are optional in nature, as opposed to hardware that is either demanded or always required in principle. There are all different kinds of peripherals you can add your computer. The main disctinction among peripherals is the way they are connected to your computer. They can be connected internally or externally. Buses A bus is a subsystem that transfers data between computer components inside a computer or between computers. Unlike a point-to-point connection, a bus can logically connect several peripherals over the same set of wires. Each bus defines its set of connectors to physically plug devices, cards or cables together. There are two types of buses: internal and external. Internal buses are connections to various internal components. External buses are connections to various external components. There are different kinds of slots that internal and external devices can connect to. Internal Types of Slots There are many different kinds of internal buses, but only a handful of popular ones. Different computers come with different kinds and number of slots. It is important to know what kind and number of slots you have on your computer before you go out and by a card that matches up to a slot you don’t have. PCI PCI (Peripheral Component Interconnect) is common in modern PCs. This kind of bus is being succeeded by PCI.

Main components of a computerMultimedia devicesOther peripheral .pdf

Main components of a computerMultimedia devicesOther peripheral .pdf

Main components of a computerMultimedia devicesOther peripheral .pdf

INTRODUCTION TO COAL MINING INDUSTRY Economic growth world over is driven by energy, whether in the form of finite resources such as coal, oil and gas or in renewable forms such as hydroelectric, wind, solar and bio-mass or its converted form, electricity(power). Increased energy consumption (especially of electricity) is inevitable with higher GDP growth. Coal was created by the fossilised remains of plants and has high carbon content. Coal is the world\'s most abundant and widely distributed fossil fuel. Coal is still the primary energy source for several countries world-wide and provides between 25% and 28% of the world\'s primary energy. Worldwide, coal is the main fuel for the generation of electricity as the price of coal is cheap compared to other fuels. It is also the highest polluting source of electricity. The other major uses of coal are in the production ofsteel and synthetic fuels Top Ten Hard Coal Producers : China 276 lMt USA 1007Mt India 490Mt Australia 325Mt Russia 247Mt Indonesia 246Mt South Africa 236Mt Kazakhstan 104Mt Poland 84Mt Colombia 79Mt There are numerous damaging environmental impacts of coal that occur through its mining, preparation, combustion, waste storage, and transport. This article provides an overview. Each topic is explored in greater depth in separate articles, as are several related topics: The introduction of coal mining projects embracing some other socio-economic issues such as depletion of crop land, pollution of water bodies, increase of landless farmers etc. The rural residents are entirely dependent on land to sustain their livelihood. But the expansion of mining activities is not only displacing them from their major sources of livelihood but simultaneously forcing them to become landless farmers (Hu et al., 2014). This situation forces them to deviate from their own cultural inheritances as they have emotional and cultural affinity with their indigenous farmlands (Christison, 2003). However, a minimal development is being taking place in the economic aspect of project affected people. This can be cited in the fold of infrastructural developments such as schools, clinics, roads, boreholes and pipe water supply (Appiah & Buaben, 2012). But the local residents express their dissatisfaction regarding the economic development which resulted as strikes, road blocks, and destruction of company’s assets (Garvin, McGee, Smoyer-Tomic, & Aubynn, 2009). Though the mining companies are justifying their CSR activities, but without the presence of community relations and development (CRD) functions, they are doing only ‘core business’. But for the successful implementation of sustainable development agenda, the functional equity needs to be established (Kemp & Owen, 2013). In the aftermath of mining, the authorities are mining companies adopting the mechanisation process which in turn increased the production and reduced the employment opportunity of the local residents. In West Virginia, the growing environ.

INTRODUCTION TO COAL MINING INDUSTRYEconomic growth world over is .pdf

INTRODUCTION TO COAL MINING INDUSTRYEconomic growth world over is .pdf

INTRODUCTION TO COAL MINING INDUSTRYEconomic growth world over is .pdf

Include time header file #include Write main method like this : int main(){ time_t start_t, end_t; double diff_t; time(&start_t); // capturing start time int n; printf(\"Enter number of stuents (<=30): \"); scanf(\"%d\", &n); GradeBook *students = (GradeBook *)malloc(sizeof(GradeBook)*n); int i,j; // taking n entries for(i=0; i Solution Include time header file #include Write main method like this : int main(){ time_t start_t, end_t; double diff_t; time(&start_t); // capturing start time int n; printf(\"Enter number of stuents (<=30): \"); scanf(\"%d\", &n); GradeBook *students = (GradeBook *)malloc(sizeof(GradeBook)*n); int i,j; // taking n entries for(i=0; i.

Include time header file#includetime.hWrite main method like t.pdf

Include time header file#includetime.hWrite main method like t.pdf

Include time header file#includetime.hWrite main method like t.pdf

import java.awt.Color; import java.awt.Insets; import java.awt.Container; import java.awt.GridBagLayout; import java.awt.GridBagConstraints; import javax.swing.JMenu; import javax.swing.JFrame; import javax.swing.JPanel; import javax.swing.JLabel; import javax.swing.JButton; import javax.swing.JMenuBar; import javax.swing.JMenuItem; import javax.swing.JTextArea; import javax.swing.JScrollBar; import javax.swing.JScrollPane; import javax.swing.BorderFactory; /* *Class Steganography_View */ public class Steganography_View extends JFrame { //size variables for window private static int WIDTH = 500; private static int HEIGHT = 400; //elements for JPanel private JTextArea input; private JScrollBar scroll,scroll2; private JButton encodeButton,decodeButton; private JLabel image_input; //elements for Menu private JMenu file; private JMenuItem encode; private JMenuItem decode; private JMenuItem exit; /* *Constructor for Steganography_View class *@param name Used to set the title on the JFrame */ public Steganography_View(String name) { //set the title of the JFrame super(name); //Menubar JMenuBar menu = new JMenuBar(); JMenu file = new JMenu(\"File\"); file.setMnemonic(\'F\'); encode = new JMenuItem(\"Encode\"); encode.setMnemonic(\'E\'); file.add(encode); decode = new JMenuItem(\"Decode\"); decode.setMnemonic(\'D\'); file.add(decode); file.addSeparator(); exit = new JMenuItem(\"Exit\"); exit.setMnemonic(\'x\'); file.add(exit); menu.add(file); setJMenuBar(menu); // display rules setResizable(true);//allow window to be resized: true?false setBackground(Color.lightGray); /background color of window: Color(int,int,int) setLocation(100,100); //location on the screen to display window setDefaultCloseOperation(EXIT_ON_CLOSE);//what to do on close operation: exit, do_nothing, etc setSize(WIDTH,HEIGHT); //set the size of the window setVisible(true); //show the window: true?false } /* *@return The menu item \'Encode\' */ public JMenuItem getEncode() { return encode; } /* *@return The menu item \'Decode\' */ public JMenuItem getDecode() { return decode; } /* *@return The menu item \'Exit\' */ public JMenuItem getExit() { return exit; } /* *@return The TextArea containing the text to encode */ public JTextArea getText() { return input; } /* *@return The JLabel containing the image to decode text from */ public JLabel getImageInput() { return image_input; } /* *@return The JPanel displaying the Encode View */ public JPanel getTextPanel() { return new Text_Panel(); } /* *@return The JPanel displaying the Decode View */ public JPanel getImagePanel() { return new Image_Panel(); } /* *@return The Encode button */ public JButton getEButton() { return encodeButton; } /* *@return The Decode button */ public JButton getDButton() { return decodeButton; } /* *Class Text_Panel */ private class Text_Panel extends JPanel { /* *Constructor to enter text to be encoded */ public Text_Panel() { //setup GridBagLayout GridBagLayout layout = new GridBagLayout(); GridBagConstraints l.

import java.awt.Color;import java.awt.Insets;import java.awt.Con.pdf

import java.awt.Color;import java.awt.Insets;import java.awt.Con.pdf

import java.awt.Color;import java.awt.Insets;import java.awt.Con.pdf

Identify the computer fraud and abuse technique by: This computer fraud and abuse technique is called phishing. Its purpose is to get the information need to commit identity theft. The perpetrator probably also used brand spoofing of Big Bank’s web site Actuvally the problem of This case is based on an actual attack. The solution represents the actual events of the attack and the hospital\'s response a. What could the hospital do to stop the attack and contain the damage? By Monday afternoon, IT figured out which malware the bots were installing and wrote a script, which was pushed out hourly, directing computers to remove the bad code. The script helped to slow the bots down a bit. The problem does not state how the problem was finally fixed. What actually happened is that on Tuesday the hospital\'s antivirus vendor figured out which malware the hackers had used to get into the network and wrote a virus signature that blocked new code from coming in. Together with the code the internal IT staff wrote, the hospital was able to clean up its computers. All of the infected computers had to have their hard drives wiped clean and their software reinstalled, at an estimated cost of $150,000. b. Which computer fraud and abuse technique did the hackers use in their attack on the hospital? The primary attack used was a Zero-day attack that exploited a newly found weakness in Microsoft’s operating system that did not yet have a patch written to correct the weakness. The perpetrators hacked into the hospital\'s network and used various forms of malware, including adware (pop-up ads) and worms. c. What steps should the hospital have taken to prevent the damage caused by the attack? The hospital\'s network is now protected by Computer Associate\'s Pest Patrol, which blocks adware and spyware, and Cisco MARS, an intrusion detection system. Northwest\'s I.T. staffers no longer wait for vendors, particularly Microsoft, to certify software patches before applying fixes—they evaluate and test patches themselves. In the case of the attack, the Windows flaw that the attack slipped through had not yet been patched on the hospital\'s PCs. Fortunately, the hospital\'s servers escaped the attack because they had been patched Solution Identify the computer fraud and abuse technique by: This computer fraud and abuse technique is called phishing. Its purpose is to get the information need to commit identity theft. The perpetrator probably also used brand spoofing of Big Bank’s web site Actuvally the problem of This case is based on an actual attack. The solution represents the actual events of the attack and the hospital\'s response a. What could the hospital do to stop the attack and contain the damage? By Monday afternoon, IT figured out which malware the bots were installing and wrote a script, which was pushed out hourly, directing computers to remove the bad code. The script helped to slow the bots down a bit. The problem does not state how the problem was finally.

Identify the computer fraud and abuse technique byThis computer f.pdf

Identify the computer fraud and abuse technique byThis computer f.pdf

Identify the computer fraud and abuse technique byThis computer f.pdf

Given : A fund that starts with a zero balance with time zero. Fund accumulates with a varying force of interest. The formula for the function is given: A $100,000 is deposited at time 2. To find: No. of years from the time of deposit for the fund to double. Solution :In case of a fixed rate of interest, it gradually decreases in relation to its value as time increases. The interest on interest is also accumulated. With the given function, t>0 and deposit is made at time 2. i.e when t=1. Substitute the value of t in the function, we get, Del(t)= 2*(1)/((1^2 + 1) =1 No of years for fund to double is 1 year..

Given A fund that starts with a zero balance with time zero. Fund .pdf

Given A fund that starts with a zero balance with time zero. Fund .pdf

Given A fund that starts with a zero balance with time zero. Fund .pdf

Your equation is correct. Double replacement reac.pdf

Your equation is correct. Double replacement reac.pdf

Your equation is correct. Double replacement reac.pdf

well while idenitfying aldehydeketone u must kee.pdf

well while idenitfying aldehydeketone u must kee.pdf

well while idenitfying aldehydeketone u must kee.pdf

The short answer is no. The reason why you cant predict acid/base properties based on O/H is because many very strong acids and bases dont contain O or H. That being said the bronsted-lowry definition of acid and bases says that an acid is a substance that donates protons and a base is a substance that donates electron. According to that definition there is a very very weak correlation between having Hs to donate as protons and acid strength and conversely having Os and being a base (O- ions are a good proton acceptors because they are highly electronegative). Acids and bases are really better described by their electron donating and accepting capabilities of the compound as described by the lewis acid/base theory. An acid is really a compound that accepts electrons and a base is a compound which donates electrons. An H+ ion then is a compound which readily accepts electrons while O- ions readily donate electrons. However the H+ and O- ability to donate and accept electrons depends on many complicated factors of the compound as a whole and even its surrounding environment than just its quantity of Hs and Os. Try using wikipedia to gain a basic understanding of lewis acids. Solution The short answer is no. The reason why you cant predict acid/base properties based on O/H is because many very strong acids and bases dont contain O or H. That being said the bronsted-lowry definition of acid and bases says that an acid is a substance that donates protons and a base is a substance that donates electron. According to that definition there is a very very weak correlation between having Hs to donate as protons and acid strength and conversely having Os and being a base (O- ions are a good proton acceptors because they are highly electronegative). Acids and bases are really better described by their electron donating and accepting capabilities of the compound as described by the lewis acid/base theory. An acid is really a compound that accepts electrons and a base is a compound which donates electrons. An H+ ion then is a compound which readily accepts electrons while O- ions readily donate electrons. However the H+ and O- ability to donate and accept electrons depends on many complicated factors of the compound as a whole and even its surrounding environment than just its quantity of Hs and Os. Try using wikipedia to gain a basic understanding of lewis acids..

The short answer is no. The reason why you cant p.pdf

The short answer is no. The reason why you cant p.pdf

The short answer is no. The reason why you cant p.pdf

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phosphite ion .pdf

phosphite ion .pdf

First, let\'s start with naming binary ionic compounds. A binary ionic compound has one metal and one nonmetal present in the compound. Let\'s consider my username (rubidium.chloride) as an example. Rubidium chloride, or RbCl, is a binary ionic compound. The metal here is Rb (rubidium, group 1A) and the nonmetal is Cl (chlorine, group 7A). When you name binary ionic compounds, you mention the cation (positively charged ion) first (usually metals, those elements that appear on the left side of the periodic table have the tendency to form positively-charged ions). Then mention the anion (negatively charged ion). The cation is rubidium (Rb). We mention it thus: Rubidium The anion is chlorine. When we write the name of the anion [alongside with the cation], we drop the ending -ine (or whatever ending it may have) and replace it with -ide. So chlorine now becomes chlor + ide = chloride. Therefore, the name of the compound is Rubidium Chloride. When you write the formulas of binary ionic compounds, you use the criss-cross method. Here\'s how it works: Rubidium has a charge of +1. Chlorine has a charge of -1. You interchange the charges of both ions. These charges, when interchanged, become the subscripts of the elements in the compound. Hence, the compound now looks Rb1Cl1. Since the subscripts are all 1, we can remove them and write the formula as RbCl. Here\'s another example: Give the formula of the compound magnesium chloride. Mg has a charge of +2. Chlorine has a charge of -1. When you interchange the respective charges of magnesium and chlorine, we get the following formula: Mg1Cl2 We simplify this as MgCl2. A list of the common cations and ions is found at the back page or on the chapter of naming chemical compounds of your chemistry book. Take time to know them by heart. But what if... you\'ve got a compound named sodium nitrate. Ok. The formula is NaNO3. How did I know? Sodium nitrate is an example of a ternary ionic compound. A ternary ionic compound has two or more elements present in it. In this case, there are 3 (Na, N and O). The naming system of ternary ionic compounds is the same as that of binary ionic compounds. Let\'s name the participating ions: Na+ NO3- NO3- is the formula for the polyatomic ion (two or more elements joined together sharing a common charge; they act as one unit when they combine with other elements) nitrate. We interchange the charges of sodium and nitrate. Interchanging them gives us the formula NaNO3 To name this formula, just combine the name of the cation and anion together. In this case, it is sodium nitrate. What if we have a formula like this? Ba(OH)2 Don\'t worry. This is just barium hydroxide. Let\'s analyze why this came about. The charge of Ba is +2 and hydroxide (OH-) is -1. When you combine them together by interchanging subscripts, we get the following: BaOH2 Wait. The formula doesn\'t make sense. H is not the only atom having two of its kind, but O too. So we put parentheses around hydroxide to i.

First, lets start with naming binary ionic comp.pdf

First, lets start with naming binary ionic comp.pdf

First, lets start with naming binary ionic comp.pdf

dFdy=dFdu dudy +dFdv dvdy =e^(u+v) 0 +e.pdf

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1. // Average.java import java.util.Scanner; public class Average{ public static void main(String[] args){ Scanner input = new Scanner(System.in); double averageMale = 0, averageFemale = 0; int countMale = 0, countFemale = 0; char mf; double average; System.out.println("How many Students? "); int n = input.nextInt(); for (int i = 1; i <= n; i++ ) { System.out.println("Is Student " + i + " male or female? "); mf = input.next().charAt(0); if(mf == 'm'){ countMale++; System.out.println("What is Student's " + i + "'s average? "); average = input.nextDouble(); averageMale = averageMale + average; } else{ countFemale++; System.out.println("What is Student's " + i + "'s average? "); average = input.nextDouble(); averageFemale = averageFemale + average; } } System.out.println("The " + countMale + " male Students average is " + averageMale/countMale); System.out.println("The " + countFemale + " male Students average is " + averageFemale/countFemale); }

2. } /* output: How many Students? 5 Is Student 1 male or female? m What is Student's 1's average? 85 Is Student 2 male or female? m What is Student's 2's average? 91 Is Student 3 male or female? f What is Student's 3's average? 95 Is Student 4 male or female? f What is Student's 4's average? 80 Is Student 5 male or female? m What is Student's 5's average? 88 The 3 male Students average is 88.0 The 2 male Students average is 87.5 */ Solution // Average.java import java.util.Scanner; public class Average{

3. public static void main(String[] args){ Scanner input = new Scanner(System.in); double averageMale = 0, averageFemale = 0; int countMale = 0, countFemale = 0; char mf; double average; System.out.println("How many Students? "); int n = input.nextInt(); for (int i = 1; i <= n; i++ ) { System.out.println("Is Student " + i + " male or female? "); mf = input.next().charAt(0); if(mf == 'm'){ countMale++; System.out.println("What is Student's " + i + "'s average? "); average = input.nextDouble(); averageMale = averageMale + average; } else{ countFemale++; System.out.println("What is Student's " + i + "'s average? "); average = input.nextDouble(); averageFemale = averageFemale + average; } } System.out.println("The " + countMale + " male Students average is " + averageMale/countMale); System.out.println("The " + countFemale + " male Students average is " + averageFemale/countFemale); } } /*

4. output: How many Students? 5 Is Student 1 male or female? m What is Student's 1's average? 85 Is Student 2 male or female? m What is Student's 2's average? 91 Is Student 3 male or female? f What is Student's 3's average? 95 Is Student 4 male or female? f What is Student's 4's average? 80 Is Student 5 male or female? m What is Student's 5's average? 88 The 3 male Students average is 88.0 The 2 male Students average is 87.5 */