Birthdaydate.javaimport java.util.Scanner;public class Birthdayd.pdf
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Birthdaydate.java import java.util.Scanner; public class Birthdaydate { public static void main(String[] args) { //Declaring variables char ch,ch1,ch2,ch3,ch4,ch5; //Declaring each card as an 2-Dimensional Array int card1[][]={{1,3,5,7},{9,11,13,15},{17,19,21,23},{25,27,29,31}}; int card2[][]={{2,3,6,7},{10,11,14,15},{18,19,22,23},{26,27,30,31}}; int card3[][]={{4,5,6,7},{12,13,14,15},{20,21,22,23},{28,29,30,31}}; int card4[][]={{8,9,10,11},{12,13,14,15},{24,25,26,27},{28,29,30,31}}; int card5[][]={{16,17,18,19},{20,21,22,23},{24,25,26,27},{28,29,30,31}}; //Scanner Object is used to get the inputs entered by the user Scanner sc = new Scanner(System.in); //This loop continue to execute until user enters \'Y\' or \'y\' do { //Declaring variable int sum=0; //Displaying card1 using nested for loop System.out.println(\":: Card1 ::\"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf(\"%3d\",card1[i][j]); } System.out.println(\" \"); } //Prompting whether the day of birth is having in card1 or not. System.out.print(\"Do card#1 is having Your Birthday (Y/N):\"); ch1 = sc.next(\".\").charAt(0); //If the day is in card 1,add the first row first column element to sum if(ch1==\'Y\' || ch1==\'y\') sum+=card1[0][0]; //Displaying card2 using nested for loop System.out.println(\":: Card2 ::\"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf(\"%3d\",card2[i][j]); } System.out.println(\" \"); } //Prompting whether the day of birth is having in card2 or not. System.out.print(\"Do card#2 is having Your Birthday (Y/N):\"); ch2 = sc.next(\".\").charAt(0); //If the day is in card 2,add the first row first column element to sum if(ch2==\'Y\' || ch2==\'y\') sum+=card2[0][0]; //Displaying card3 using nested for loop System.out.println(\":: Card3 ::\"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf(\"%3d\",card3[i][j]); } System.out.println(\" \"); } //Prompting whether the day of birth is having in card3 or not. System.out.print(\"Do card#3 is having Your Birthday (Y/N):\"); ch3 = sc.next(\".\").charAt(0); //If the day is in card 3,add the first row first column element to sum if(ch3==\'Y\' || ch3==\'y\') sum+=card3[0][0]; //Displaying card4 using nested for loop System.out.println(\":: Card4 ::\"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf(\"%3d\",card4[i][j]); } System.out.println(\" \"); } //Prompting whether the day of birth is having in card4 or not. System.out.print(\"Do card#4 is having Your Birthday (Y/N):\"); ch4 = sc.next(\".\").charAt(0); //If the day is in card 4,add the first row first column element to sum if(ch4==\'Y\' || ch4==\'y\') sum+=card4[0][0]; //Displaying card5 using nested for loop System.out.println(\":: Card5 ::\"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf(\"%3d\",card5[i][j]); } System.out.println(\" \"); } //Prompting whether the day of birth is having in card5 or not. System.out.print(\"Do card#5 is having Your Birthday (Y/N):\"); ch5 = sc.next.
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![Birthdaydate.java
import java.util.Scanner;
public class Birthdaydate {
public static void main(String[] args) {
//Declaring variables
char ch,ch1,ch2,ch3,ch4,ch5;
//Declaring each card as an 2-Dimensional Array
int card1[][]={{1,3,5,7},{9,11,13,15},{17,19,21,23},{25,27,29,31}};
int card2[][]={{2,3,6,7},{10,11,14,15},{18,19,22,23},{26,27,30,31}};
int card3[][]={{4,5,6,7},{12,13,14,15},{20,21,22,23},{28,29,30,31}};
int card4[][]={{8,9,10,11},{12,13,14,15},{24,25,26,27},{28,29,30,31}};
int card5[][]={{16,17,18,19},{20,21,22,23},{24,25,26,27},{28,29,30,31}};
//Scanner Object is used to get the inputs entered by the user
Scanner sc = new Scanner(System.in);
//This loop continue to execute until user enters 'Y' or 'y'
do
{
//Declaring variable
int sum=0;
//Displaying card1 using nested for loop
System.out.println(":: Card1 ::");
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
System.out.printf("%3d",card1[i][j]);
}
System.out.println(" ");
}
//Prompting whether the day of birth is having in card1 or not.](https://image.slidesharecdn.com/birthdaydate-230628071006-2d88bcfb/85/Birthdaydate-javaimport-java-util-Scanner-public-class-Birthdayd-pdf-1-320.jpg)
![System.out.print("Do card#1 is having Your Birthday (Y/N):");
ch1 = sc.next(".").charAt(0);
//If the day is in card 1,add the first row first column element to sum
if(ch1=='Y' || ch1=='y')
sum+=card1[0][0];
//Displaying card2 using nested for loop
System.out.println(":: Card2 ::");
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
System.out.printf("%3d",card2[i][j]);
}
System.out.println(" ");
}
//Prompting whether the day of birth is having in card2 or not.
System.out.print("Do card#2 is having Your Birthday (Y/N):");
ch2 = sc.next(".").charAt(0);
//If the day is in card 2,add the first row first column element to sum
if(ch2=='Y' || ch2=='y')
sum+=card2[0][0];
//Displaying card3 using nested for loop
System.out.println(":: Card3 ::");
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
System.out.printf("%3d",card3[i][j]);
}
System.out.println(" ");
}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
This is the final code which meets your requirement.Than youCard.j.pdf
This is the final code which meets your requirement.Than you
Card.java
import java.util.Random;
public class Card {
//declaring instance variables
private String suit;
private int face_value;
public static int count=52;
//Creating an instance of Random class reference
Random rand;
//Default Constructor
public Card() {
super();
//Passing random class object to the Random class reference
rand = new Random();
}
//Getters and setters
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
public int getFace_value() {
return face_value;
}
public void setFace_value(int face_value) {
this.face_value = face_value;
}
public int getCount() {
return count;
}
//This shuffle() method will shuffle the deck
public void shuffle()
{
//Picking a card from the deck randomly and find its face value and suit
this.face_value = 1 + rand.nextInt((9) + 1);
int suit_random_value = 1 + rand.nextInt((3) + 1);
if(suit_random_value==1)
{
this.suit=\"Clubs\";
}
else if(suit_random_value==2)
{
this.suit=\"hearts\";
}
else if(suit_random_value==3)
{
this.suit=\"Spades\";
}
else if(suit_random_value==4)
{
this.suit=\"diamonds\";
}
this.count=count-1;
}
}
________________________________________________________________________
DriverClass.java
public class DriverClass {
public static void main(String[] args) {
//creating an array which can hold 52 Card class Objects
Card c[]=new Card[52];
//Creating 52 card class objects and storing them into an array
for(int i=0;i<52;i++)
{
//Creating Card class Object
c[i]=new Card();
}
//Displaying each card face value and its suit.
for(int i=0;i<52;i++)
{
//Shuffling the deck before every card picking from the deck
c[i].shuffle();
//Displaying randomly picked card face value and its suit
System.out.println(c[i].getFace_value()+\" of \"+c[i].getSuit()+\" Cards remaining
\"+c[i].getCount());
}
}
}
_________________________________________________________________________
Output:
8 of diamonds
Cards remaining 51
4 of Clubs
Cards remaining 50
9 of Clubs
Cards remaining 49
7 of Spades
Cards remaining 48
9 of diamonds
Cards remaining 47
7 of diamonds
Cards remaining 46
4 of diamonds
Cards remaining 45
9 of Clubs
Cards remaining 44
5 of hearts
Cards remaining 43
7 of Spades
Cards remaining 42
10 of Clubs
Cards remaining 41
7 of diamonds
Cards remaining 40
8 of diamonds
Cards remaining 39
6 of hearts
Cards remaining 38
4 of hearts
Cards remaining 37
6 of Spades
Cards remaining 36
4 of Clubs
Cards remaining 35
3 of diamonds
Cards remaining 34
9 of diamonds
Cards remaining 33
3 of diamonds
Cards remaining 32
5 of Spades
Cards remaining 31
4 of Clubs
Cards remaining 30
3 of Spades
Cards remaining 29
6 of diamonds
Cards remaining 28
3 of hearts
Cards remaining 27
9 of hearts
Cards remaining 26
7 of hearts
Cards remaining 25
3 of hearts
Cards remaining 24
10 of Clubs
Cards remaining 23
2 of hearts
Cards remaining 22
9 of Clubs
Cards remaining 21
8 of diamonds
Cards remaining 20
8 of Clubs
Cards remaining 19
2 of h.
Please follow the data 1) For Line 23 In the IF - Condition yo.pdf
Please follow the data :
1) For Line 23 :
In the IF - Condition you are just checking the cells of the grid at positions 0 and 1. But I
suppose you would like to check if both the cells are having the character value \'x\'. So to
achieve this you need to replace the code with
if(grid[0] == grid[1] == \'x\')
2) For Line 29 :
The error is that the function has been define before the error line and the loop has not been
closed so pplace a closing brackets for the loop before the line starting with bool tie()
3) For Line 37 :
The above change should also resolve this as the scope of the main function isnt predefined.
4) For Line 69 :
Remove the excess bracket at the end of the code.
Even after replacing these I suppose your code would not work as the input has not been defined
from where and how thw user can get the data to the code. So to avoid this I am placing my
code.
Please follow the code and comments for descripiton :
CODE :
#include
using namespace std;
int checkWin( char[]); // rewuired variables
void grid( char[]);
int main()
{
char cell[10] = {\'o\',\'1\',\'2\',\'3\',\'4\',\'5\',\'6\',\'7\',\'8\',\'9\'}; // placing the cell names
int player = 1,i,choice; // initialisations
char symbol;
do // looping to get the player choices
{
grid(cell);
if(player%2==1) { // checking for the player
player=1;
} else {
player=2;
}
// player 2 is the computer
if(player==2) // if is a player 2
{
cout << \"Player \" << player<> choice;
symbol=\'X\';
if (choice == 1 && cell[1] == \'1\') {
cell[1] = symbol;
}
else if (choice == 2 && cell[2] == \'2\') {
cell[2] = symbol;
}
else if (choice == 3 && cell[3] == \'3\') {
cell[3] = symbol;
}
else if (choice == 4 && cell[4] == \'4\') {
cell[4] = symbol;
}
else if (choice == 5 && cell[5] == \'5\') {
cell[5] = symbol;
}
else if (choice == 6 && cell[6] == \'6\') {
cell[6] = symbol;
}
else if (choice == 7 && cell[7] == \'7\') {
cell[7] = symbol;
}
else if (choice == 8 && cell[8] == \'8\') {
cell[8] = symbol;
}
else if (choice == 9 && cell[9] == \'9\') {
cell[9] = symbol;
} else {
cout<<\"Invalid move..!!!\"<< endl;
player--;
exit(0);
}
i=checkWin(cell);
player++;
}
} while(i==-1);
grid(cell);
if(i==1) {
cout<<\"Congratulations! \ Player \"<<--player<<\" won the Game..!!\";
} else {
cout<<\" OOPS..!!!!\ The Game has been tie..!!\";
}
exit(0);
return 0;
}
void grid(char cell[]) // printing the grid everytime the user has placed the symbol
{
system(\"cls\");
cout << \"\ \ \\tTic Tac Toe Game\ \ \";
cout << \"Player 1 (X) - Player 2 (O)\" << endl << endl;
cout << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[1] << \" | \" << cell[2] << \" | \" << cell[3] << endl;
cout << \"_____|_____|_____\" << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[4] << \" | \" << cell[5] << \" | \" << cell[6] << endl;
cout << \"_____|_____|_____\" << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[7] << \" | \" << cell[8] << \" | \" << cell[9] << endl;
cout << \" | | \" << endl << endl;
}
int checkWin(char cell[])// check.
FaceUp card game In this assignment we will implement a made.pdf
FaceUp card game
In this assignment we will implement a made-up card game we'll call FaceUp. When the game
starts, you deal five cards face down. Your goal is to achieve as high a score as possible. Your
score only includes cards that are face up. Red cards (hearts and diamonds) award positive
points, while black cards (clubs and spades) award negative points. Cards 2-10 have points worth
their face value. Cards Jack, Queen, and King have value 10, and Ace is 11.
The game is played by flipping over cards, either from face-down to face-up or from face-up to
face-down. As you play, you are told your total score (ie, total of the face-up cards) and the total
score of the cards that are face down. The challenge is that you only get up to a fixed number of
flips and then the game is over.
Here is an example of the output from playing the game:
At each point where the program asks the user to pick a card, the game waits for user input.
Code
Well implement the game using five classes, three of which you'll leave unchanged, and two you
will modify and submit:
Card: Leave this code unchanged.
CardDealer: Leave this code unchanged.
FaceUpCard: This class extends the Card class with additional functionality. You will add two
methods needed for the FaceUp game.
FaceUpHand: This class keeps track of the state of the game. Most of the methods you will
implement are here.
FaceUpGame: This class implements the game interface. It gets data from the user, updates the
board, and repeats the process. Leave this code unchanged.
How to proceed
You may fill in the details of the FaceUpCard and FaceUpHandclasses however youd like, but
heres what we recommend.
Add the two methods to the FaceUpCard class (described in comments in the code). Add a main
method to this class and test them to make sure they work.
Read through the other two new classes FaceUpHand and FaceUpGame. Make sure you
understand what all of the methods in FaceUpHand are supposed to do and that you understand
how theyre used in the FaceUpGame class.
Implement methods in the FaceUpHand class incrementally. Write one method and then test it!
To test it, again, add a main method and write a small test or two to make sure it behaves like
youd expect. If you try to implement all of the methods and then test it by simply running
FaceUpGame, 1) its very unlikely youll get it perfect the first time and then 2) its going to be very
hard to figure out where the problem is.
Card.java
public class Card {
// min/max for the card number range
private static final int MIN_NUMBER = 1;
private static final int MAX_NUMBER = 13;
private int number;
private String suit;
/**
* Create a new card with number and suit. If the a valid
* suit/number is not input, the card defaults to 1 of hearts.
*
* @param number the card number
* @param suit the card suit
*/
public Card(int number, String suit){
this.number = number;
this.suit = suit.toLowerCase();
if (!isValidSuit(suit)) {
System.out.println(suit + "Is not a valid suit!").
1.Rule based detection 2.Statical anomaly detection1.Rule based de.pdf
1.Rule based detection 2.Statical anomaly detection
1.Rule based detection :
a set of rules are used to decide the behaviour of an intruder
in this detection we have anomaly detection and penitration identification .in anomaly detection
rules are developed to detect deviation from previous usage patterns. in penitration identification
an expert system approach that searches for suspecious behaviour
2.Statistical anomaly datection :
the collection of data relating to the behaviour of legitimate users over a period of time. then
statistical tests are applied to observed behaviour to determine with a high level of confidence
whethear that behaviour is not legitimate user behaviour
in this detection we have threshold detection and profile based detection .in threshold detection it
approaches to defining thresholds,independent of user,for the frequency of occurence of various
events and in profile based a profile of the activity of each user is developed and used to detect
changes in the behaviour of individual accounts.
Solution
1.Rule based detection 2.Statical anomaly detection
1.Rule based detection :
a set of rules are used to decide the behaviour of an intruder
in this detection we have anomaly detection and penitration identification .in anomaly detection
rules are developed to detect deviation from previous usage patterns. in penitration identification
an expert system approach that searches for suspecious behaviour
2.Statistical anomaly datection :
the collection of data relating to the behaviour of legitimate users over a period of time. then
statistical tests are applied to observed behaviour to determine with a high level of confidence
whethear that behaviour is not legitimate user behaviour
in this detection we have threshold detection and profile based detection .in threshold detection it
approaches to defining thresholds,independent of user,for the frequency of occurence of various
events and in profile based a profile of the activity of each user is developed and used to detect
changes in the behaviour of individual accounts..
Recommended
This is the final code which meets your requirement.Than youCard.j.pdf
This is the final code which meets your requirement.Than you
Card.java
import java.util.Random;
public class Card {
//declaring instance variables
private String suit;
private int face_value;
public static int count=52;
//Creating an instance of Random class reference
Random rand;
//Default Constructor
public Card() {
super();
//Passing random class object to the Random class reference
rand = new Random();
}
//Getters and setters
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
public int getFace_value() {
return face_value;
}
public void setFace_value(int face_value) {
this.face_value = face_value;
}
public int getCount() {
return count;
}
//This shuffle() method will shuffle the deck
public void shuffle()
{
//Picking a card from the deck randomly and find its face value and suit
this.face_value = 1 + rand.nextInt((9) + 1);
int suit_random_value = 1 + rand.nextInt((3) + 1);
if(suit_random_value==1)
{
this.suit=\"Clubs\";
}
else if(suit_random_value==2)
{
this.suit=\"hearts\";
}
else if(suit_random_value==3)
{
this.suit=\"Spades\";
}
else if(suit_random_value==4)
{
this.suit=\"diamonds\";
}
this.count=count-1;
}
}
________________________________________________________________________
DriverClass.java
public class DriverClass {
public static void main(String[] args) {
//creating an array which can hold 52 Card class Objects
Card c[]=new Card[52];
//Creating 52 card class objects and storing them into an array
for(int i=0;i<52;i++)
{
//Creating Card class Object
c[i]=new Card();
}
//Displaying each card face value and its suit.
for(int i=0;i<52;i++)
{
//Shuffling the deck before every card picking from the deck
c[i].shuffle();
//Displaying randomly picked card face value and its suit
System.out.println(c[i].getFace_value()+\" of \"+c[i].getSuit()+\" Cards remaining
\"+c[i].getCount());
}
}
}
_________________________________________________________________________
Output:
8 of diamonds
Cards remaining 51
4 of Clubs
Cards remaining 50
9 of Clubs
Cards remaining 49
7 of Spades
Cards remaining 48
9 of diamonds
Cards remaining 47
7 of diamonds
Cards remaining 46
4 of diamonds
Cards remaining 45
9 of Clubs
Cards remaining 44
5 of hearts
Cards remaining 43
7 of Spades
Cards remaining 42
10 of Clubs
Cards remaining 41
7 of diamonds
Cards remaining 40
8 of diamonds
Cards remaining 39
6 of hearts
Cards remaining 38
4 of hearts
Cards remaining 37
6 of Spades
Cards remaining 36
4 of Clubs
Cards remaining 35
3 of diamonds
Cards remaining 34
9 of diamonds
Cards remaining 33
3 of diamonds
Cards remaining 32
5 of Spades
Cards remaining 31
4 of Clubs
Cards remaining 30
3 of Spades
Cards remaining 29
6 of diamonds
Cards remaining 28
3 of hearts
Cards remaining 27
9 of hearts
Cards remaining 26
7 of hearts
Cards remaining 25
3 of hearts
Cards remaining 24
10 of Clubs
Cards remaining 23
2 of hearts
Cards remaining 22
9 of Clubs
Cards remaining 21
8 of diamonds
Cards remaining 20
8 of Clubs
Cards remaining 19
2 of h.
Please follow the data 1) For Line 23 In the IF - Condition yo.pdf
Please follow the data :
1) For Line 23 :
In the IF - Condition you are just checking the cells of the grid at positions 0 and 1. But I
suppose you would like to check if both the cells are having the character value \'x\'. So to
achieve this you need to replace the code with
if(grid[0] == grid[1] == \'x\')
2) For Line 29 :
The error is that the function has been define before the error line and the loop has not been
closed so pplace a closing brackets for the loop before the line starting with bool tie()
3) For Line 37 :
The above change should also resolve this as the scope of the main function isnt predefined.
4) For Line 69 :
Remove the excess bracket at the end of the code.
Even after replacing these I suppose your code would not work as the input has not been defined
from where and how thw user can get the data to the code. So to avoid this I am placing my
code.
Please follow the code and comments for descripiton :
CODE :
#include
using namespace std;
int checkWin( char[]); // rewuired variables
void grid( char[]);
int main()
{
char cell[10] = {\'o\',\'1\',\'2\',\'3\',\'4\',\'5\',\'6\',\'7\',\'8\',\'9\'}; // placing the cell names
int player = 1,i,choice; // initialisations
char symbol;
do // looping to get the player choices
{
grid(cell);
if(player%2==1) { // checking for the player
player=1;
} else {
player=2;
}
// player 2 is the computer
if(player==2) // if is a player 2
{
cout << \"Player \" << player<> choice;
symbol=\'X\';
if (choice == 1 && cell[1] == \'1\') {
cell[1] = symbol;
}
else if (choice == 2 && cell[2] == \'2\') {
cell[2] = symbol;
}
else if (choice == 3 && cell[3] == \'3\') {
cell[3] = symbol;
}
else if (choice == 4 && cell[4] == \'4\') {
cell[4] = symbol;
}
else if (choice == 5 && cell[5] == \'5\') {
cell[5] = symbol;
}
else if (choice == 6 && cell[6] == \'6\') {
cell[6] = symbol;
}
else if (choice == 7 && cell[7] == \'7\') {
cell[7] = symbol;
}
else if (choice == 8 && cell[8] == \'8\') {
cell[8] = symbol;
}
else if (choice == 9 && cell[9] == \'9\') {
cell[9] = symbol;
} else {
cout<<\"Invalid move..!!!\"<< endl;
player--;
exit(0);
}
i=checkWin(cell);
player++;
}
} while(i==-1);
grid(cell);
if(i==1) {
cout<<\"Congratulations! \ Player \"<<--player<<\" won the Game..!!\";
} else {
cout<<\" OOPS..!!!!\ The Game has been tie..!!\";
}
exit(0);
return 0;
}
void grid(char cell[]) // printing the grid everytime the user has placed the symbol
{
system(\"cls\");
cout << \"\ \ \\tTic Tac Toe Game\ \ \";
cout << \"Player 1 (X) - Player 2 (O)\" << endl << endl;
cout << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[1] << \" | \" << cell[2] << \" | \" << cell[3] << endl;
cout << \"_____|_____|_____\" << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[4] << \" | \" << cell[5] << \" | \" << cell[6] << endl;
cout << \"_____|_____|_____\" << endl;
cout << \" | | \" << endl;
cout << \" \" << cell[7] << \" | \" << cell[8] << \" | \" << cell[9] << endl;
cout << \" | | \" << endl << endl;
}
int checkWin(char cell[])// check.
FaceUp card game In this assignment we will implement a made.pdf
FaceUp card game
In this assignment we will implement a made-up card game we'll call FaceUp. When the game
starts, you deal five cards face down. Your goal is to achieve as high a score as possible. Your
score only includes cards that are face up. Red cards (hearts and diamonds) award positive
points, while black cards (clubs and spades) award negative points. Cards 2-10 have points worth
their face value. Cards Jack, Queen, and King have value 10, and Ace is 11.
The game is played by flipping over cards, either from face-down to face-up or from face-up to
face-down. As you play, you are told your total score (ie, total of the face-up cards) and the total
score of the cards that are face down. The challenge is that you only get up to a fixed number of
flips and then the game is over.
Here is an example of the output from playing the game:
At each point where the program asks the user to pick a card, the game waits for user input.
Code
Well implement the game using five classes, three of which you'll leave unchanged, and two you
will modify and submit:
Card: Leave this code unchanged.
CardDealer: Leave this code unchanged.
FaceUpCard: This class extends the Card class with additional functionality. You will add two
methods needed for the FaceUp game.
FaceUpHand: This class keeps track of the state of the game. Most of the methods you will
implement are here.
FaceUpGame: This class implements the game interface. It gets data from the user, updates the
board, and repeats the process. Leave this code unchanged.
How to proceed
You may fill in the details of the FaceUpCard and FaceUpHandclasses however youd like, but
heres what we recommend.
Add the two methods to the FaceUpCard class (described in comments in the code). Add a main
method to this class and test them to make sure they work.
Read through the other two new classes FaceUpHand and FaceUpGame. Make sure you
understand what all of the methods in FaceUpHand are supposed to do and that you understand
how theyre used in the FaceUpGame class.
Implement methods in the FaceUpHand class incrementally. Write one method and then test it!
To test it, again, add a main method and write a small test or two to make sure it behaves like
youd expect. If you try to implement all of the methods and then test it by simply running
FaceUpGame, 1) its very unlikely youll get it perfect the first time and then 2) its going to be very
hard to figure out where the problem is.
Card.java
public class Card {
// min/max for the card number range
private static final int MIN_NUMBER = 1;
private static final int MAX_NUMBER = 13;
private int number;
private String suit;
/**
* Create a new card with number and suit. If the a valid
* suit/number is not input, the card defaults to 1 of hearts.
*
* @param number the card number
* @param suit the card suit
*/
public Card(int number, String suit){
this.number = number;
this.suit = suit.toLowerCase();
if (!isValidSuit(suit)) {
System.out.println(suit + "Is not a valid suit!").
1.Rule based detection 2.Statical anomaly detection1.Rule based de.pdf
1.Rule based detection 2.Statical anomaly detection
1.Rule based detection :
a set of rules are used to decide the behaviour of an intruder
in this detection we have anomaly detection and penitration identification .in anomaly detection
rules are developed to detect deviation from previous usage patterns. in penitration identification
an expert system approach that searches for suspecious behaviour
2.Statistical anomaly datection :
the collection of data relating to the behaviour of legitimate users over a period of time. then
statistical tests are applied to observed behaviour to determine with a high level of confidence
whethear that behaviour is not legitimate user behaviour
in this detection we have threshold detection and profile based detection .in threshold detection it
approaches to defining thresholds,independent of user,for the frequency of occurence of various
events and in profile based a profile of the activity of each user is developed and used to detect
changes in the behaviour of individual accounts.
Solution
1.Rule based detection 2.Statical anomaly detection
1.Rule based detection :
a set of rules are used to decide the behaviour of an intruder
in this detection we have anomaly detection and penitration identification .in anomaly detection
rules are developed to detect deviation from previous usage patterns. in penitration identification
an expert system approach that searches for suspecious behaviour
2.Statistical anomaly datection :
the collection of data relating to the behaviour of legitimate users over a period of time. then
statistical tests are applied to observed behaviour to determine with a high level of confidence
whethear that behaviour is not legitimate user behaviour
in this detection we have threshold detection and profile based detection .in threshold detection it
approaches to defining thresholds,independent of user,for the frequency of occurence of various
events and in profile based a profile of the activity of each user is developed and used to detect
changes in the behaviour of individual accounts..
14 = disease ff24 = carriers Ff25 chances that they will pro.pdf
1/4 = disease ff
2/4 = carriers Ff
25 % chances that they will produce a child which suffers from the
disorder
50 % percentage of offspring will be carriersFfFFFFffFfff
Solution
1/4 = disease ff
2/4 = carriers Ff
25 % chances that they will produce a child which suffers from the
disorder
50 % percentage of offspring will be carriersFfFFFFffFfff.
Mosses, lichens are great indicators of radioactive pollution. They .pdf
Mosses, lichens are great indicators of radioactive pollution. They absorb radioactive elements
specifically caesium as it is similar to patassium which forms important nutrient of mosses.
Edible mushrooms too get contaminated with such harmful elements. Hence mosses and lichens
serve as indicators for radioactive pollution areas. Inorder to check such pollution, one can
collect moss samples and check their element content of radioactive pollution. Elements like
lead, caesium, barium, titanium, berrylium are the important elements contributing o radioactive
pollution. Mosses show mutations involving somatic type giving out flowers of different colors.
Also, the mosses get damaged by reducing the plant vigor.
Solution
Mosses, lichens are great indicators of radioactive pollution. They absorb radioactive elements
specifically caesium as it is similar to patassium which forms important nutrient of mosses.
Edible mushrooms too get contaminated with such harmful elements. Hence mosses and lichens
serve as indicators for radioactive pollution areas. Inorder to check such pollution, one can
collect moss samples and check their element content of radioactive pollution. Elements like
lead, caesium, barium, titanium, berrylium are the important elements contributing o radioactive
pollution. Mosses show mutations involving somatic type giving out flowers of different colors.
Also, the mosses get damaged by reducing the plant vigor..
10 million.Sensory conflict theory argues , The brightest light at.pdf
10 million.
Sensory conflict theory argues , The brightest light at which we can see is about 10 million times
as intense as the dimmest light at which we can see.
Solution
10 million.
Sensory conflict theory argues , The brightest light at which we can see is about 10 million times
as intense as the dimmest light at which we can see..
We first find the slopes of the two lines 2M an.pdf
We first find the slopes of the two lines: 2/M and -1/4 For two lines to be
perpendicylar, the product of their slopes must be equal to -1. Hence the equation (2/M)*(-1/4) =
-1. Solve for M to find M = 1/2.
Solution
We first find the slopes of the two lines: 2/M and -1/4 For two lines to be
perpendicylar, the product of their slopes must be equal to -1. Hence the equation (2/M)*(-1/4) =
-1. Solve for M to find M = 1/2..
SO2 is a gas that mixes with water to form sulfur.pdf
SO2 is a gas that mixes with water to form sulfuric acid. The droplets can form
clouds like water does (Venus is covered with them making it bright). Sulfuric acid has a high
albedo (it reflects light), since it reflects the light, the energy it blocked from coming down.
Therefore the electromagnetic radiation (aka light) cannot is not converted into thermal energy
(it converts from light to thermal when it smacks into things). Therefore since the light cannot
make it to the surface, the thermal energy will decrease (molecules will slow down) and the
temperature will theoretically drop. :) I said theoretically because I\'m not quite sure how strong
of a greenhouse gas CO2 is (I\'m not sure if it would trap any heat that did manage to get in) as
compared to water vapor, but it doesn\'t sound like they care about that. :) Anyway sorry if I
wrote that in a confusing way, basically the light will reflect off the clouds of sulfuric acid and so
the light can\'t heat up the earth. Also, its better in the stratosphere because the sooner the light is
blocked, the less it will heat up molecules it smacks into on the way down.
Solution
SO2 is a gas that mixes with water to form sulfuric acid. The droplets can form
clouds like water does (Venus is covered with them making it bright). Sulfuric acid has a high
albedo (it reflects light), since it reflects the light, the energy it blocked from coming down.
Therefore the electromagnetic radiation (aka light) cannot is not converted into thermal energy
(it converts from light to thermal when it smacks into things). Therefore since the light cannot
make it to the surface, the thermal energy will decrease (molecules will slow down) and the
temperature will theoretically drop. :) I said theoretically because I\'m not quite sure how strong
of a greenhouse gas CO2 is (I\'m not sure if it would trap any heat that did manage to get in) as
compared to water vapor, but it doesn\'t sound like they care about that. :) Anyway sorry if I
wrote that in a confusing way, basically the light will reflect off the clouds of sulfuric acid and so
the light can\'t heat up the earth. Also, its better in the stratosphere because the sooner the light is
blocked, the less it will heat up molecules it smacks into on the way down..
1. Heterozygosity by migration calculated using the below formulaH.pdf
1. Heterozygosity by migration calculated using the below formula
H= 1 - F
where F= finite population
N= Population size
F= 1/ (2N)
By substituting F = 1/ (2N) in H= 1- F
H= 1 - 1/ 2N
H= 1 - 1/2 x 1000 = 0.9995
2. Heterozygosity by mutational drift can be calculated using
H= 4Nu / 1+ 4Nu
u= mutation rate
H= 4 x 1000 x 1 x 10-9 / 1 + 4 x 1000 x 10-9 = 3.9 x 10-6
Heterozygosity is decreased by mutation and migration. The differences in the values is large
because large mutation rates causes totally shift of the population and evolves into new species.
But migration will not act as fast as mutation .
Solution
1. Heterozygosity by migration calculated using the below formula
H= 1 - F
where F= finite population
N= Population size
F= 1/ (2N)
By substituting F = 1/ (2N) in H= 1- F
H= 1 - 1/ 2N
H= 1 - 1/2 x 1000 = 0.9995
2. Heterozygosity by mutational drift can be calculated using
H= 4Nu / 1+ 4Nu
u= mutation rate
H= 4 x 1000 x 1 x 10-9 / 1 + 4 x 1000 x 10-9 = 3.9 x 10-6
Heterozygosity is decreased by mutation and migration. The differences in the values is large
because large mutation rates causes totally shift of the population and evolves into new species.
But migration will not act as fast as mutation ..
1).Monohybrid cross between true breeding parents. Assume that the.pdf
1).
Monohybrid cross between true breeding parents. Assume that the gene “A” is dominant and “a”
is recessive. The genotype of parents are, AA (true breeding for dominant), aa (true breeding for
recessive).
Genotypes of the true-breeding parents in the _____parental___ generation are ___AA__ and
___aa_.
Genotype(s) of the offspring from the true-breeding parents is(are)_Aa____and the symbol for
this generation of offspring is __F1__.
Punnet square: AA (male) and aa (female)
A
A
a
Aa
Aa
a
Aa
Aa
A
A
a
Aa
Aa
a
Aa
Aa
Solution
1).
Monohybrid cross between true breeding parents. Assume that the gene “A” is dominant and “a”
is recessive. The genotype of parents are, AA (true breeding for dominant), aa (true breeding for
recessive).
Genotypes of the true-breeding parents in the _____parental___ generation are ___AA__ and
___aa_.
Genotype(s) of the offspring from the true-breeding parents is(are)_Aa____and the symbol for
this generation of offspring is __F1__.
Punnet square: AA (male) and aa (female)
A
A
a
Aa
Aa
a
Aa
Aa
A
A
a
Aa
Aa
a
Aa
Aa.
1. Moving down a group, the electronegativity decreases due to the l.pdf
1. Moving down a group, the electronegativity decreases due to the longer distance between the
nucleus and the valence electron shell, thereby decreasing the attraction, making the atom have
less of an attraction for electrons or protons. 2.Trend-wise, as one moves from left to right across
a period in the periodic table, the electronegativity increases due to the stronger attraction that
the atoms obtain as the nuclear charge increases.
Solution
1. Moving down a group, the electronegativity decreases due to the longer distance between the
nucleus and the valence electron shell, thereby decreasing the attraction, making the atom have
less of an attraction for electrons or protons. 2.Trend-wise, as one moves from left to right across
a period in the periodic table, the electronegativity increases due to the stronger attraction that
the atoms obtain as the nuclear charge increases..
The mercury liquid and the mercury(II) oxide are .pdf
The mercury liquid and the mercury(II) oxide are not in the equilibrium constant
expression because of being pure liguids and solids. Kc = 1/[O2] = 1/1.60 M = 0.625
Solution
The mercury liquid and the mercury(II) oxide are not in the equilibrium constant
expression because of being pure liguids and solids. Kc = 1/[O2] = 1/1.60 M = 0.625.
1 Ans Bacteria are unicellular organisms that reproduce by cell divi.pdf
1 Ans Bacteria are unicellular organisms that reproduce by cell division, while viruses are
acellular and are incapable of reproducing on their own and they need a living host.
2 Ans Persistant viral infections are slow viral infections progressive over a long period of time
and are usually fatal.
Examples:
AIDS dementia comples (HIV)
Progressive encephalitis (Rubella virus)
Latent viral infection can remain in host cells throughout life without causing disease. They may
be reactivated by immunosuppression, however, and cause disease.
Examples:
Cold sores
Shingles
Solution
1 Ans Bacteria are unicellular organisms that reproduce by cell division, while viruses are
acellular and are incapable of reproducing on their own and they need a living host.
2 Ans Persistant viral infections are slow viral infections progressive over a long period of time
and are usually fatal.
Examples:
AIDS dementia comples (HIV)
Progressive encephalitis (Rubella virus)
Latent viral infection can remain in host cells throughout life without causing disease. They may
be reactivated by immunosuppression, however, and cause disease.
Examples:
Cold sores
Shingles.
1#include stdio.h#include stdlib.h#include assert.h .pdf
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
2
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
Solution
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node*.
Ques-6 answerThe clinical significance of beta-hemolytic (on blo.pdf
Ques-6: answer:
The clinical significance of beta-hemolytic (on blood agar medium) streptococci isolated from
throat specimen is consequent incidence of rheumatoid fever and acute glomerulonephritis.
Streptococcus pneumoniae (normally present in our body, can be found in the upper respiratory
tract) is an organism that cause upper respiratory tract infections and otitis media. When children
with running nose touch each other, the organism passed through the secretions. This causes
infection in the healthy child, decreased immunity against the bacteria and increased activation
of the flora causes otitis media (a painful ear infection) in children. These respiratory bacteria
produce inflammatory virulent toxins to invade respiratory cells and nutrients finally become
pathogenic to host immune system result in severe respiratory illness.
Throat infections caused by beta-hemolytic streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Therefore, patient needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given). The bacteria causes
inflammation of the blood vessels supplying glomeruli, which impairs the glomerular filtration.
Decreased urine filtration causes edema (swelling of body organs).
Ques-7:
The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect
the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body
aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial
cell type\" of body cell provides an indication that the sputum specimen represents material from
an active infection. Other cells are present in the sputum during the active infection are
polymorphonuclear leukocytes (PMNs)
Solution
Ques-6: answer:
The clinical significance of beta-hemolytic (on blood agar medium) streptococci isolated from
throat specimen is consequent incidence of rheumatoid fever and acute glomerulonephritis.
Streptococcus pneumoniae (normally present in our body, can be found in the upper respiratory
tract) is an organism that cause upper respiratory tract infections and otitis media. When children
with running nose touch each other, the organism passed through the secretions. This causes
infection in the healthy child, decreased immunity against the bacteria and increased activation
of the flora causes otitis media (a painful ear infection) in children. These respiratory bacteria
produce inflammatory virulent toxins to invade respiratory cells and nutrients finally become
pathogenic to host immune system result in severe respiratory illness.
Throat infections caused by beta-hemolytic streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Therefore, patient needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given). The bacteria causes
inflammat.
Valence electrons electrons at outer-most energy level.Effective.pdf
Valence electrons : electrons at outer-most energy level.
Effective nuclear charge: net positive charge experienced by an electron in a multi-electron atom.
Atomic radius: distance from the nucleus to the boundary of the surrounding cloud of electrons.
Ionic radius: measure of the size of an atom\'s ion.
Ionization Energy: nergy needed to remove electrons from an atom.
Electron Affinity: the energy change when an electron is added to the neutral atom to form a
negative ion.
Electronegativity: tendency of an atom or a functional group to attract electrons.
Oxidation number (or Oxidation state): hypothetical charge that an atom would have if all bonds
to atoms of different elements were 100% ionic.
Oxidation: Lossing electrons
Reductio: Gaining electrons
Oxidizing agent: Element being Reduced
Reducing agent: Element being Oxidized
Covalent Bond: chemical bonding by shaing of electrons (Non-metal with Non-metal).
Ionic Bond: chemical bonding of non-metal and metal
Compounds: combinations of two or more elements
Octet rule: maximum of 8 electrons (in the valence shell) can be added during a reaction.
Electrolytes: substance that can conduct electricity.
Hydration: chemical reaction when adding water.
Solubility: property of a solid, liquid, or gaseous chemical substance mix with one other to form
a homogeneous solution.
Concentration: abundance of a constituent divided by the total volume of a mixture.
Dilute: lowering the concentration of a chemical.
Concentrated solution: solution in which much of the water has been removed.
Brønsted acid: proton donor.
Brønsted base: proton acceptor.
Monoprotic acids: donate one proton per molecule during the process of dissociation.
Diprotic acids: donate two protons per molecule during the process of dissociation.
Triprotic acids: donate three protons per molecule during the process of dissociation.
pH: measure of acidity and basic of a solution.
Strong acid: ionizes completely in an aqueous solution by losing one proton.
Weak acids: acid that dissociates incompletely.
Strong base: accept hydrogen ions or more generally or donate electron pairs.
Weak base: does not ionize fully in an aqueous solution
Supersaturated solution: solute > solvent
Saturated solution: solute = solvent
unsaturated solution: solute < solvent
Solution
Valence electrons : electrons at outer-most energy level.
Effective nuclear charge: net positive charge experienced by an electron in a multi-electron atom.
Atomic radius: distance from the nucleus to the boundary of the surrounding cloud of electrons.
Ionic radius: measure of the size of an atom\'s ion.
Ionization Energy: nergy needed to remove electrons from an atom.
Electron Affinity: the energy change when an electron is added to the neutral atom to form a
negative ion.
Electronegativity: tendency of an atom or a functional group to attract electrons.
Oxidation number (or Oxidation state): hypothetical charge that an atom would have if all bonds
to atoms of different elements were 10.
What is storage mediaAns-C. the physical material on which a co.pdf
What is storage media?
Ans-
C. the physical material on which a computer keeps data, instructions, and information
Storage media are devices that store application and user information. Storage media is device on
which data can be electronically places kept and retrieved .The primary storage media for a
computer is usually the internal hard drive.
Solution
What is storage media?
Ans-
C. the physical material on which a computer keeps data, instructions, and information
Storage media are devices that store application and user information. Storage media is device on
which data can be electronically places kept and retrieved .The primary storage media for a
computer is usually the internal hard drive..
Transactional model of communication states that if people are conne.pdf
Transactional model of communication states that if people are connected through
communication, they engage in transaction. It propogates that communication is simultaneous.
5 parts or labels: The first part is the sender and receiver - they are the people involved in the
communication. The 2nd part is the message i.e the information being send/received. The 3rd
part is the channel of communication. The 4th part is the response and the decoding of messages
by the sender/receiver. The 5th part is the feedback.
It is different from the linear model of communication, in the sense that unlike the one way
model in the linear communication, the transactional model is a simultaneous model. Linear
model does not involve any feedback, unlike the transactional model.
Ethics means rules of behavior based on ideas about what is right and wrong. Interpersonal
communication is exchange of information between 2 or more people. The ethical choices
involved here is that one should be humble and polite in conversations. One should not hurt the
other person\'s religious or cultural beliefs.
I grew up in an Asian culture in the Indian subcontinent. My family and my society was
orthodox in thinking. This made me (as well as other people growing up and exposed to this
culture) not much of a risk taker. This was due to the fact that we were made to believe that god
will always help us, whatever happens, happens for good (destiny).
If i had grown in a different culture, i probably would be more of a risk taker.
Adam has perception problems. He does not interpret sensory information correctly. His
cognitive (senses of the mind) powers are not that strong.
He can improve his communication by focusing on proxemics i.e non verbal communication.
He should have a strong body movement (kinesics) for better non verbal communication. It
includes facial expressions, gestures etc.
He should also focus on paralanguage - the tone and pitch of his voice.
Solution
Transactional model of communication states that if people are connected through
communication, they engage in transaction. It propogates that communication is simultaneous.
5 parts or labels: The first part is the sender and receiver - they are the people involved in the
communication. The 2nd part is the message i.e the information being send/received. The 3rd
part is the channel of communication. The 4th part is the response and the decoding of messages
by the sender/receiver. The 5th part is the feedback.
It is different from the linear model of communication, in the sense that unlike the one way
model in the linear communication, the transactional model is a simultaneous model. Linear
model does not involve any feedback, unlike the transactional model.
Ethics means rules of behavior based on ideas about what is right and wrong. Interpersonal
communication is exchange of information between 2 or more people. The ethical choices
involved here is that one should be humble and polite in conversations. One should not hur.
What is the measure of the strength of the relationship between inde.pdf
What is the measure of the strength of the relationship between independent and dependent
variables.
Solution
The pearson correlation coefficient measures the strength of the LINEAR
relationship between two variables. The R-squared values measures the proportion of variation
in the dependent variable that can be explained by the independent (predictor) variables..
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42.pdf
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42-:
2 SO32-(aq) + O2 (g) --> 2 SO42-(aq)
Thus a positive test forSO42-is expected in solutions ofSO32-even when noSO42-has been
placed.
Solution
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42-:
2 SO32-(aq) + O2 (g) --> 2 SO42-(aq)
Thus a positive test forSO42-is expected in solutions ofSO32-even when noSO42-has been
placed..
Should begin with a letter and may contain additional letters and di.pdf
Should begin with a letter and may contain additional letters and digits. Identifiers are the same
as variables.
Use descriptive n Use single letters only for the counter in loops.
Variable names start with lower-case
Multi-word identifiers are internally capitalized
Do not use hyphens or underscores to separate multi-word identifiers.
Need to print extra characters?
What about the unicode set?
This is the format: \\uHHHH
Remember any other escape characters such as \ , \\t and \\\\?
The following lists some of these conventions based on the type of identifier:
Class name
The first letter of each word is capitalized, examples:
Mammal, SeaMammal
Function name
The first letter of each, except the first, word is capitalized, examples:
getAge, setHeight
Variable name
The first letter of each, except the first, word is capitalized, examples:
age, brainSize
Constant names
Every letter is capitalized and underscores are used between words, examples:
MAX_HEIGHT, MAX_AGE
RESERVED WORDS
As you may have noticed, many of Java\'s keywords are borrowed from C/C++. Also, as in
C/C++, keywords are always written in lowercase.
Data declaration keywords (boolean, float, int)
Loop keywords (continue, while, for)
Conditional keywords (if, else, switch)
Exception keywords (try, throw, catch)
Structure keywords (class, extends)
Modifier and access keywords (private, public)
Miscellaneous keywords (true, null)
The following keywords and their use, you SHOULD know.
boolean default for private switch
break double if protected this
case else import public throws
catch extends int return try
char final new static void
class float package super while
VARIABLE SCOPE
Where is it appropriate to declare a variable? Does it make a difference?
PRIMITIVE DATA TYPES
int (16-bit)
float (32-bit)
double (64-bit
boolean (true/false)
char (16-bit unicode)
What about String?
What about Integer, and Double?
OBJECTS
Why object oriented programming?
Objects need to be built, so we need constructors.
Instance variables and the role they play when defining objects.
What about methods of an object?
ARRAYS
int[] array = new int[5]; // one-dimensional array
it may also be declared as:
int[] array = {1, 2, 3, 4, 5}; // it is initialized and declared at the same time.
Arrays can be of any primitive data type or object. Arrays have an instance variable \"length\"
that we can access to find out how big the array is. Remember that an Array can throw an
IndexOutOfBoundsException. Make use of it to make your code safe.
VECTORS
Must import java.util.*;
Unlike arrays, vectors can only hold objects, no primitive data types are allowed. Vectors can
grow big and unbounded. You may want to specify the capacity of the vector but you dont need
to. The default capacity is 10 and it doubles in size every time the capacity is reached.
Just like arrays have the \"length\" instance variable, Vectors have a .size() method that returns
the current size of the vector.
void .addElement(Object obj)
void .setElementA.
#include iostream #include string #include iomanip #incl.pdf
#include
#include
#include
#include
#include
#include
using std::cin;
using std::cout;
using std::string;
using std::endl;
using std::vector;
static const string colors[] = { \"red\", \"green\", \"blue\", \"cyan\", \"magenta\", \"yellow\" };
static const int PEGNUM = 4;
static int getRandomValue()
{
static bool first = true;
if (first)
{
first = false;
srand((int)time(0));
}
return(rand());
}
static string easyToLower(string word)
{
transform(word.begin(), word.end(), word.begin(), ::tolower);
return word;
}
inline string trim(string& str) // courtesy of stack overflow
{
str.erase(0, str.find_first_not_of(\' \')); //prefixing spaces
str.erase(str.find_last_not_of(\' \') + 1); //surfixing spaces
return str;
}
static vector setNewCode()
{
vector theCode;
for (int i = 0; i < PEGNUM; i++)
{
theCode.push_back(colors[getRandomValue() % 6]);
}
return theCode;
}
static vector getPlayerGuess()
{
vector words;
string in = \"\", word=\"\";
bool okay = false;
bool quitting = false;
while (!okay)
{
cout << \"Enter your guess (red, green, blue, cyan, magenta, yellow. Enter 4 colors): \ \"< 0)
//get input
{
in = trim(in);
index = in.find_first_of(\" \");
if (index == -1)
{
word = in;
in = \"\";
}
else
word = in.substr(0, index);
words.push_back(word);
in = in.substr(index+1);
if (words.size() == PEGNUM)
{
break;
}
}
okay = true;
for (int i = 0; i < words.size(); i++) // check and standardize inputs
{
words[i] = easyToLower(words[i]);
if (words[i] == \"red\" || words[i] == \"r\")
{
words[i] = colors[0];
//cout << words[i] << endl;
}
else if (words[i] == \"green\" || words[i] == \"g\")
{
words[i] = colors[1];
//cout << words[i] << endl;
}
else if (words[i] == \"blue\" || words[i] == \"b\")
{
words[i] = colors[2];
//cout << words[i] << endl;
}
else if (words[i] == \"cyan\" || words[i] == \"c\")
{
words[i] = colors[3];
//cout << words[i] << endl;
}
else if (words[i] == \"magenta\" || words[i] == \"m\")
{
words[i] = colors[4];
//cout << words[i] << endl;
}
else if (words[i] == \"yellow\" || words[i] == \"y\")
{
words[i] = colors[5];
//cout << words[i] << endl;
}
else if (words[i] == \"quit\")
{
quitting = true;
words.clear();
words.push_back(\"quit\");
return words;
}
else
{
cout << words[i] << \" is an invalid submittion\" << endl;
words.clear();
okay = false;
}
}
if (!okay)
cout << \"Invalid entry. Please try again\ \" << endl;
if (words.size() != PEGNUM) // check if 4 entries
{
cout << \"Must enter 4 colors. Try again\ \" << endl;
words.clear();
okay = false;
}
}
return words;
}
static int setBlackPegs(vector theCode, vector guessCode)
{
int bps = 0;
for (int i = 0; i < PEGNUM; i++)
{
if (theCode[i] == guessCode[i])
bps++;
}
return bps;
}
static int setWhitePegs(vector theCode, vector guessCode, int bps)
{
int RGBCMYnumsTheCode[6];
int RGBCMYnumsGuessCode[6];
int wps=0;
for (int i = 0; i < 6; i++)
{
RGBCMYnumsTheCode[i] = 0;
RGBCMYnumsGuessCode[i] = 0;
}
for (int i = 0; i < PEGNUM; i++)
{
if (theCode[i] == colors[0])
{
RGBCMYnumsTheCode[0]++;
}
else if (theCo.
Prostomes are mollusca, annelids and arthropods while deuterostomes .pdf
Prostomes are mollusca, annelids and arthropods while deuterostomes are echinoderms,
hemichordates and chordates
So correct answer is prostomes and bilaterally symmetrical (D)
Solution
Prostomes are mollusca, annelids and arthropods while deuterostomes are echinoderms,
hemichordates and chordates
So correct answer is prostomes and bilaterally symmetrical (D).
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Palestine last event orientationfvgnh .pptx
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
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14 = disease ff24 = carriers Ff25 chances that they will pro.pdf
1/4 = disease ff
2/4 = carriers Ff
25 % chances that they will produce a child which suffers from the
disorder
50 % percentage of offspring will be carriersFfFFFFffFfff
Solution
1/4 = disease ff
2/4 = carriers Ff
25 % chances that they will produce a child which suffers from the
disorder
50 % percentage of offspring will be carriersFfFFFFffFfff.
Mosses, lichens are great indicators of radioactive pollution. They .pdf
Mosses, lichens are great indicators of radioactive pollution. They absorb radioactive elements
specifically caesium as it is similar to patassium which forms important nutrient of mosses.
Edible mushrooms too get contaminated with such harmful elements. Hence mosses and lichens
serve as indicators for radioactive pollution areas. Inorder to check such pollution, one can
collect moss samples and check their element content of radioactive pollution. Elements like
lead, caesium, barium, titanium, berrylium are the important elements contributing o radioactive
pollution. Mosses show mutations involving somatic type giving out flowers of different colors.
Also, the mosses get damaged by reducing the plant vigor.
Solution
Mosses, lichens are great indicators of radioactive pollution. They absorb radioactive elements
specifically caesium as it is similar to patassium which forms important nutrient of mosses.
Edible mushrooms too get contaminated with such harmful elements. Hence mosses and lichens
serve as indicators for radioactive pollution areas. Inorder to check such pollution, one can
collect moss samples and check their element content of radioactive pollution. Elements like
lead, caesium, barium, titanium, berrylium are the important elements contributing o radioactive
pollution. Mosses show mutations involving somatic type giving out flowers of different colors.
Also, the mosses get damaged by reducing the plant vigor..
10 million.Sensory conflict theory argues , The brightest light at.pdf
10 million.
Sensory conflict theory argues , The brightest light at which we can see is about 10 million times
as intense as the dimmest light at which we can see.
Solution
10 million.
Sensory conflict theory argues , The brightest light at which we can see is about 10 million times
as intense as the dimmest light at which we can see..
We first find the slopes of the two lines 2M an.pdf
We first find the slopes of the two lines: 2/M and -1/4 For two lines to be
perpendicylar, the product of their slopes must be equal to -1. Hence the equation (2/M)*(-1/4) =
-1. Solve for M to find M = 1/2.
Solution
We first find the slopes of the two lines: 2/M and -1/4 For two lines to be
perpendicylar, the product of their slopes must be equal to -1. Hence the equation (2/M)*(-1/4) =
-1. Solve for M to find M = 1/2..
SO2 is a gas that mixes with water to form sulfur.pdf
SO2 is a gas that mixes with water to form sulfuric acid. The droplets can form
clouds like water does (Venus is covered with them making it bright). Sulfuric acid has a high
albedo (it reflects light), since it reflects the light, the energy it blocked from coming down.
Therefore the electromagnetic radiation (aka light) cannot is not converted into thermal energy
(it converts from light to thermal when it smacks into things). Therefore since the light cannot
make it to the surface, the thermal energy will decrease (molecules will slow down) and the
temperature will theoretically drop. :) I said theoretically because I\'m not quite sure how strong
of a greenhouse gas CO2 is (I\'m not sure if it would trap any heat that did manage to get in) as
compared to water vapor, but it doesn\'t sound like they care about that. :) Anyway sorry if I
wrote that in a confusing way, basically the light will reflect off the clouds of sulfuric acid and so
the light can\'t heat up the earth. Also, its better in the stratosphere because the sooner the light is
blocked, the less it will heat up molecules it smacks into on the way down.
Solution
SO2 is a gas that mixes with water to form sulfuric acid. The droplets can form
clouds like water does (Venus is covered with them making it bright). Sulfuric acid has a high
albedo (it reflects light), since it reflects the light, the energy it blocked from coming down.
Therefore the electromagnetic radiation (aka light) cannot is not converted into thermal energy
(it converts from light to thermal when it smacks into things). Therefore since the light cannot
make it to the surface, the thermal energy will decrease (molecules will slow down) and the
temperature will theoretically drop. :) I said theoretically because I\'m not quite sure how strong
of a greenhouse gas CO2 is (I\'m not sure if it would trap any heat that did manage to get in) as
compared to water vapor, but it doesn\'t sound like they care about that. :) Anyway sorry if I
wrote that in a confusing way, basically the light will reflect off the clouds of sulfuric acid and so
the light can\'t heat up the earth. Also, its better in the stratosphere because the sooner the light is
blocked, the less it will heat up molecules it smacks into on the way down..
1. Heterozygosity by migration calculated using the below formulaH.pdf
1. Heterozygosity by migration calculated using the below formula
H= 1 - F
where F= finite population
N= Population size
F= 1/ (2N)
By substituting F = 1/ (2N) in H= 1- F
H= 1 - 1/ 2N
H= 1 - 1/2 x 1000 = 0.9995
2. Heterozygosity by mutational drift can be calculated using
H= 4Nu / 1+ 4Nu
u= mutation rate
H= 4 x 1000 x 1 x 10-9 / 1 + 4 x 1000 x 10-9 = 3.9 x 10-6
Heterozygosity is decreased by mutation and migration. The differences in the values is large
because large mutation rates causes totally shift of the population and evolves into new species.
But migration will not act as fast as mutation .
Solution
1. Heterozygosity by migration calculated using the below formula
H= 1 - F
where F= finite population
N= Population size
F= 1/ (2N)
By substituting F = 1/ (2N) in H= 1- F
H= 1 - 1/ 2N
H= 1 - 1/2 x 1000 = 0.9995
2. Heterozygosity by mutational drift can be calculated using
H= 4Nu / 1+ 4Nu
u= mutation rate
H= 4 x 1000 x 1 x 10-9 / 1 + 4 x 1000 x 10-9 = 3.9 x 10-6
Heterozygosity is decreased by mutation and migration. The differences in the values is large
because large mutation rates causes totally shift of the population and evolves into new species.
But migration will not act as fast as mutation ..
1).Monohybrid cross between true breeding parents. Assume that the.pdf
1).
Monohybrid cross between true breeding parents. Assume that the gene “A” is dominant and “a”
is recessive. The genotype of parents are, AA (true breeding for dominant), aa (true breeding for
recessive).
Genotypes of the true-breeding parents in the _____parental___ generation are ___AA__ and
___aa_.
Genotype(s) of the offspring from the true-breeding parents is(are)_Aa____and the symbol for
this generation of offspring is __F1__.
Punnet square: AA (male) and aa (female)
A
A
a
Aa
Aa
a
Aa
Aa
A
A
a
Aa
Aa
a
Aa
Aa
Solution
1).
Monohybrid cross between true breeding parents. Assume that the gene “A” is dominant and “a”
is recessive. The genotype of parents are, AA (true breeding for dominant), aa (true breeding for
recessive).
Genotypes of the true-breeding parents in the _____parental___ generation are ___AA__ and
___aa_.
Genotype(s) of the offspring from the true-breeding parents is(are)_Aa____and the symbol for
this generation of offspring is __F1__.
Punnet square: AA (male) and aa (female)
A
A
a
Aa
Aa
a
Aa
Aa
A
A
a
Aa
Aa
a
Aa
Aa.
1. Moving down a group, the electronegativity decreases due to the l.pdf
1. Moving down a group, the electronegativity decreases due to the longer distance between the
nucleus and the valence electron shell, thereby decreasing the attraction, making the atom have
less of an attraction for electrons or protons. 2.Trend-wise, as one moves from left to right across
a period in the periodic table, the electronegativity increases due to the stronger attraction that
the atoms obtain as the nuclear charge increases.
Solution
1. Moving down a group, the electronegativity decreases due to the longer distance between the
nucleus and the valence electron shell, thereby decreasing the attraction, making the atom have
less of an attraction for electrons or protons. 2.Trend-wise, as one moves from left to right across
a period in the periodic table, the electronegativity increases due to the stronger attraction that
the atoms obtain as the nuclear charge increases..
The mercury liquid and the mercury(II) oxide are .pdf
The mercury liquid and the mercury(II) oxide are not in the equilibrium constant
expression because of being pure liguids and solids. Kc = 1/[O2] = 1/1.60 M = 0.625
Solution
The mercury liquid and the mercury(II) oxide are not in the equilibrium constant
expression because of being pure liguids and solids. Kc = 1/[O2] = 1/1.60 M = 0.625.
1 Ans Bacteria are unicellular organisms that reproduce by cell divi.pdf
1 Ans Bacteria are unicellular organisms that reproduce by cell division, while viruses are
acellular and are incapable of reproducing on their own and they need a living host.
2 Ans Persistant viral infections are slow viral infections progressive over a long period of time
and are usually fatal.
Examples:
AIDS dementia comples (HIV)
Progressive encephalitis (Rubella virus)
Latent viral infection can remain in host cells throughout life without causing disease. They may
be reactivated by immunosuppression, however, and cause disease.
Examples:
Cold sores
Shingles
Solution
1 Ans Bacteria are unicellular organisms that reproduce by cell division, while viruses are
acellular and are incapable of reproducing on their own and they need a living host.
2 Ans Persistant viral infections are slow viral infections progressive over a long period of time
and are usually fatal.
Examples:
AIDS dementia comples (HIV)
Progressive encephalitis (Rubella virus)
Latent viral infection can remain in host cells throughout life without causing disease. They may
be reactivated by immunosuppression, however, and cause disease.
Examples:
Cold sores
Shingles.
1#include stdio.h#include stdlib.h#include assert.h .pdf
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
2
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we\'re currently
looking at */
while (current != NULL)
{
if (count == index)
return(current->data);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
/* Check the count function */
printf(\"Element at index 3 is %d\", GetNth(head, 3));
getchar();
}
Solution
1
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node*.
Ques-6 answerThe clinical significance of beta-hemolytic (on blo.pdf
Ques-6: answer:
The clinical significance of beta-hemolytic (on blood agar medium) streptococci isolated from
throat specimen is consequent incidence of rheumatoid fever and acute glomerulonephritis.
Streptococcus pneumoniae (normally present in our body, can be found in the upper respiratory
tract) is an organism that cause upper respiratory tract infections and otitis media. When children
with running nose touch each other, the organism passed through the secretions. This causes
infection in the healthy child, decreased immunity against the bacteria and increased activation
of the flora causes otitis media (a painful ear infection) in children. These respiratory bacteria
produce inflammatory virulent toxins to invade respiratory cells and nutrients finally become
pathogenic to host immune system result in severe respiratory illness.
Throat infections caused by beta-hemolytic streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Therefore, patient needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given). The bacteria causes
inflammation of the blood vessels supplying glomeruli, which impairs the glomerular filtration.
Decreased urine filtration causes edema (swelling of body organs).
Ques-7:
The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect
the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body
aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial
cell type\" of body cell provides an indication that the sputum specimen represents material from
an active infection. Other cells are present in the sputum during the active infection are
polymorphonuclear leukocytes (PMNs)
Solution
Ques-6: answer:
The clinical significance of beta-hemolytic (on blood agar medium) streptococci isolated from
throat specimen is consequent incidence of rheumatoid fever and acute glomerulonephritis.
Streptococcus pneumoniae (normally present in our body, can be found in the upper respiratory
tract) is an organism that cause upper respiratory tract infections and otitis media. When children
with running nose touch each other, the organism passed through the secretions. This causes
infection in the healthy child, decreased immunity against the bacteria and increased activation
of the flora causes otitis media (a painful ear infection) in children. These respiratory bacteria
produce inflammatory virulent toxins to invade respiratory cells and nutrients finally become
pathogenic to host immune system result in severe respiratory illness.
Throat infections caused by beta-hemolytic streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Therefore, patient needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given). The bacteria causes
inflammat.
Valence electrons electrons at outer-most energy level.Effective.pdf
Valence electrons : electrons at outer-most energy level.
Effective nuclear charge: net positive charge experienced by an electron in a multi-electron atom.
Atomic radius: distance from the nucleus to the boundary of the surrounding cloud of electrons.
Ionic radius: measure of the size of an atom\'s ion.
Ionization Energy: nergy needed to remove electrons from an atom.
Electron Affinity: the energy change when an electron is added to the neutral atom to form a
negative ion.
Electronegativity: tendency of an atom or a functional group to attract electrons.
Oxidation number (or Oxidation state): hypothetical charge that an atom would have if all bonds
to atoms of different elements were 100% ionic.
Oxidation: Lossing electrons
Reductio: Gaining electrons
Oxidizing agent: Element being Reduced
Reducing agent: Element being Oxidized
Covalent Bond: chemical bonding by shaing of electrons (Non-metal with Non-metal).
Ionic Bond: chemical bonding of non-metal and metal
Compounds: combinations of two or more elements
Octet rule: maximum of 8 electrons (in the valence shell) can be added during a reaction.
Electrolytes: substance that can conduct electricity.
Hydration: chemical reaction when adding water.
Solubility: property of a solid, liquid, or gaseous chemical substance mix with one other to form
a homogeneous solution.
Concentration: abundance of a constituent divided by the total volume of a mixture.
Dilute: lowering the concentration of a chemical.
Concentrated solution: solution in which much of the water has been removed.
Brønsted acid: proton donor.
Brønsted base: proton acceptor.
Monoprotic acids: donate one proton per molecule during the process of dissociation.
Diprotic acids: donate two protons per molecule during the process of dissociation.
Triprotic acids: donate three protons per molecule during the process of dissociation.
pH: measure of acidity and basic of a solution.
Strong acid: ionizes completely in an aqueous solution by losing one proton.
Weak acids: acid that dissociates incompletely.
Strong base: accept hydrogen ions or more generally or donate electron pairs.
Weak base: does not ionize fully in an aqueous solution
Supersaturated solution: solute > solvent
Saturated solution: solute = solvent
unsaturated solution: solute < solvent
Solution
Valence electrons : electrons at outer-most energy level.
Effective nuclear charge: net positive charge experienced by an electron in a multi-electron atom.
Atomic radius: distance from the nucleus to the boundary of the surrounding cloud of electrons.
Ionic radius: measure of the size of an atom\'s ion.
Ionization Energy: nergy needed to remove electrons from an atom.
Electron Affinity: the energy change when an electron is added to the neutral atom to form a
negative ion.
Electronegativity: tendency of an atom or a functional group to attract electrons.
Oxidation number (or Oxidation state): hypothetical charge that an atom would have if all bonds
to atoms of different elements were 10.
What is storage mediaAns-C. the physical material on which a co.pdf
What is storage media?
Ans-
C. the physical material on which a computer keeps data, instructions, and information
Storage media are devices that store application and user information. Storage media is device on
which data can be electronically places kept and retrieved .The primary storage media for a
computer is usually the internal hard drive.
Solution
What is storage media?
Ans-
C. the physical material on which a computer keeps data, instructions, and information
Storage media are devices that store application and user information. Storage media is device on
which data can be electronically places kept and retrieved .The primary storage media for a
computer is usually the internal hard drive..
Transactional model of communication states that if people are conne.pdf
Transactional model of communication states that if people are connected through
communication, they engage in transaction. It propogates that communication is simultaneous.
5 parts or labels: The first part is the sender and receiver - they are the people involved in the
communication. The 2nd part is the message i.e the information being send/received. The 3rd
part is the channel of communication. The 4th part is the response and the decoding of messages
by the sender/receiver. The 5th part is the feedback.
It is different from the linear model of communication, in the sense that unlike the one way
model in the linear communication, the transactional model is a simultaneous model. Linear
model does not involve any feedback, unlike the transactional model.
Ethics means rules of behavior based on ideas about what is right and wrong. Interpersonal
communication is exchange of information between 2 or more people. The ethical choices
involved here is that one should be humble and polite in conversations. One should not hurt the
other person\'s religious or cultural beliefs.
I grew up in an Asian culture in the Indian subcontinent. My family and my society was
orthodox in thinking. This made me (as well as other people growing up and exposed to this
culture) not much of a risk taker. This was due to the fact that we were made to believe that god
will always help us, whatever happens, happens for good (destiny).
If i had grown in a different culture, i probably would be more of a risk taker.
Adam has perception problems. He does not interpret sensory information correctly. His
cognitive (senses of the mind) powers are not that strong.
He can improve his communication by focusing on proxemics i.e non verbal communication.
He should have a strong body movement (kinesics) for better non verbal communication. It
includes facial expressions, gestures etc.
He should also focus on paralanguage - the tone and pitch of his voice.
Solution
Transactional model of communication states that if people are connected through
communication, they engage in transaction. It propogates that communication is simultaneous.
5 parts or labels: The first part is the sender and receiver - they are the people involved in the
communication. The 2nd part is the message i.e the information being send/received. The 3rd
part is the channel of communication. The 4th part is the response and the decoding of messages
by the sender/receiver. The 5th part is the feedback.
It is different from the linear model of communication, in the sense that unlike the one way
model in the linear communication, the transactional model is a simultaneous model. Linear
model does not involve any feedback, unlike the transactional model.
Ethics means rules of behavior based on ideas about what is right and wrong. Interpersonal
communication is exchange of information between 2 or more people. The ethical choices
involved here is that one should be humble and polite in conversations. One should not hur.
What is the measure of the strength of the relationship between inde.pdf
What is the measure of the strength of the relationship between independent and dependent
variables.
Solution
The pearson correlation coefficient measures the strength of the LINEAR
relationship between two variables. The R-squared values measures the proportion of variation
in the dependent variable that can be explained by the independent (predictor) variables..
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42.pdf
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42-:
2 SO32-(aq) + O2 (g) --> 2 SO42-(aq)
Thus a positive test forSO42-is expected in solutions ofSO32-even when noSO42-has been
placed.
Solution
Some of the SO32- is oxidized by atmospheric oxygen in air into SO42-:
2 SO32-(aq) + O2 (g) --> 2 SO42-(aq)
Thus a positive test forSO42-is expected in solutions ofSO32-even when noSO42-has been
placed..
Should begin with a letter and may contain additional letters and di.pdf
Should begin with a letter and may contain additional letters and digits. Identifiers are the same
as variables.
Use descriptive n Use single letters only for the counter in loops.
Variable names start with lower-case
Multi-word identifiers are internally capitalized
Do not use hyphens or underscores to separate multi-word identifiers.
Need to print extra characters?
What about the unicode set?
This is the format: \\uHHHH
Remember any other escape characters such as \ , \\t and \\\\?
The following lists some of these conventions based on the type of identifier:
Class name
The first letter of each word is capitalized, examples:
Mammal, SeaMammal
Function name
The first letter of each, except the first, word is capitalized, examples:
getAge, setHeight
Variable name
The first letter of each, except the first, word is capitalized, examples:
age, brainSize
Constant names
Every letter is capitalized and underscores are used between words, examples:
MAX_HEIGHT, MAX_AGE
RESERVED WORDS
As you may have noticed, many of Java\'s keywords are borrowed from C/C++. Also, as in
C/C++, keywords are always written in lowercase.
Data declaration keywords (boolean, float, int)
Loop keywords (continue, while, for)
Conditional keywords (if, else, switch)
Exception keywords (try, throw, catch)
Structure keywords (class, extends)
Modifier and access keywords (private, public)
Miscellaneous keywords (true, null)
The following keywords and their use, you SHOULD know.
boolean default for private switch
break double if protected this
case else import public throws
catch extends int return try
char final new static void
class float package super while
VARIABLE SCOPE
Where is it appropriate to declare a variable? Does it make a difference?
PRIMITIVE DATA TYPES
int (16-bit)
float (32-bit)
double (64-bit
boolean (true/false)
char (16-bit unicode)
What about String?
What about Integer, and Double?
OBJECTS
Why object oriented programming?
Objects need to be built, so we need constructors.
Instance variables and the role they play when defining objects.
What about methods of an object?
ARRAYS
int[] array = new int[5]; // one-dimensional array
it may also be declared as:
int[] array = {1, 2, 3, 4, 5}; // it is initialized and declared at the same time.
Arrays can be of any primitive data type or object. Arrays have an instance variable \"length\"
that we can access to find out how big the array is. Remember that an Array can throw an
IndexOutOfBoundsException. Make use of it to make your code safe.
VECTORS
Must import java.util.*;
Unlike arrays, vectors can only hold objects, no primitive data types are allowed. Vectors can
grow big and unbounded. You may want to specify the capacity of the vector but you dont need
to. The default capacity is 10 and it doubles in size every time the capacity is reached.
Just like arrays have the \"length\" instance variable, Vectors have a .size() method that returns
the current size of the vector.
void .addElement(Object obj)
void .setElementA.
#include iostream #include string #include iomanip #incl.pdf
#include
#include
#include
#include
#include
#include
using std::cin;
using std::cout;
using std::string;
using std::endl;
using std::vector;
static const string colors[] = { \"red\", \"green\", \"blue\", \"cyan\", \"magenta\", \"yellow\" };
static const int PEGNUM = 4;
static int getRandomValue()
{
static bool first = true;
if (first)
{
first = false;
srand((int)time(0));
}
return(rand());
}
static string easyToLower(string word)
{
transform(word.begin(), word.end(), word.begin(), ::tolower);
return word;
}
inline string trim(string& str) // courtesy of stack overflow
{
str.erase(0, str.find_first_not_of(\' \')); //prefixing spaces
str.erase(str.find_last_not_of(\' \') + 1); //surfixing spaces
return str;
}
static vector setNewCode()
{
vector theCode;
for (int i = 0; i < PEGNUM; i++)
{
theCode.push_back(colors[getRandomValue() % 6]);
}
return theCode;
}
static vector getPlayerGuess()
{
vector words;
string in = \"\", word=\"\";
bool okay = false;
bool quitting = false;
while (!okay)
{
cout << \"Enter your guess (red, green, blue, cyan, magenta, yellow. Enter 4 colors): \ \"< 0)
//get input
{
in = trim(in);
index = in.find_first_of(\" \");
if (index == -1)
{
word = in;
in = \"\";
}
else
word = in.substr(0, index);
words.push_back(word);
in = in.substr(index+1);
if (words.size() == PEGNUM)
{
break;
}
}
okay = true;
for (int i = 0; i < words.size(); i++) // check and standardize inputs
{
words[i] = easyToLower(words[i]);
if (words[i] == \"red\" || words[i] == \"r\")
{
words[i] = colors[0];
//cout << words[i] << endl;
}
else if (words[i] == \"green\" || words[i] == \"g\")
{
words[i] = colors[1];
//cout << words[i] << endl;
}
else if (words[i] == \"blue\" || words[i] == \"b\")
{
words[i] = colors[2];
//cout << words[i] << endl;
}
else if (words[i] == \"cyan\" || words[i] == \"c\")
{
words[i] = colors[3];
//cout << words[i] << endl;
}
else if (words[i] == \"magenta\" || words[i] == \"m\")
{
words[i] = colors[4];
//cout << words[i] << endl;
}
else if (words[i] == \"yellow\" || words[i] == \"y\")
{
words[i] = colors[5];
//cout << words[i] << endl;
}
else if (words[i] == \"quit\")
{
quitting = true;
words.clear();
words.push_back(\"quit\");
return words;
}
else
{
cout << words[i] << \" is an invalid submittion\" << endl;
words.clear();
okay = false;
}
}
if (!okay)
cout << \"Invalid entry. Please try again\ \" << endl;
if (words.size() != PEGNUM) // check if 4 entries
{
cout << \"Must enter 4 colors. Try again\ \" << endl;
words.clear();
okay = false;
}
}
return words;
}
static int setBlackPegs(vector theCode, vector guessCode)
{
int bps = 0;
for (int i = 0; i < PEGNUM; i++)
{
if (theCode[i] == guessCode[i])
bps++;
}
return bps;
}
static int setWhitePegs(vector theCode, vector guessCode, int bps)
{
int RGBCMYnumsTheCode[6];
int RGBCMYnumsGuessCode[6];
int wps=0;
for (int i = 0; i < 6; i++)
{
RGBCMYnumsTheCode[i] = 0;
RGBCMYnumsGuessCode[i] = 0;
}
for (int i = 0; i < PEGNUM; i++)
{
if (theCode[i] == colors[0])
{
RGBCMYnumsTheCode[0]++;
}
else if (theCo.
Prostomes are mollusca, annelids and arthropods while deuterostomes .pdf
Prostomes are mollusca, annelids and arthropods while deuterostomes are echinoderms,
hemichordates and chordates
So correct answer is prostomes and bilaterally symmetrical (D)
Solution
Prostomes are mollusca, annelids and arthropods while deuterostomes are echinoderms,
hemichordates and chordates
So correct answer is prostomes and bilaterally symmetrical (D).
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- 1. Birthdaydate.java import java.util.Scanner; public class Birthdaydate { public static void main(String[] args) { //Declaring variables char ch,ch1,ch2,ch3,ch4,ch5; //Declaring each card as an 2-Dimensional Array int card1[][]={{1,3,5,7},{9,11,13,15},{17,19,21,23},{25,27,29,31}}; int card2[][]={{2,3,6,7},{10,11,14,15},{18,19,22,23},{26,27,30,31}}; int card3[][]={{4,5,6,7},{12,13,14,15},{20,21,22,23},{28,29,30,31}}; int card4[][]={{8,9,10,11},{12,13,14,15},{24,25,26,27},{28,29,30,31}}; int card5[][]={{16,17,18,19},{20,21,22,23},{24,25,26,27},{28,29,30,31}}; //Scanner Object is used to get the inputs entered by the user Scanner sc = new Scanner(System.in); //This loop continue to execute until user enters 'Y' or 'y' do { //Declaring variable int sum=0; //Displaying card1 using nested for loop System.out.println(":: Card1 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card1[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card1 or not.
- 2. System.out.print("Do card#1 is having Your Birthday (Y/N):"); ch1 = sc.next(".").charAt(0); //If the day is in card 1,add the first row first column element to sum if(ch1=='Y' || ch1=='y') sum+=card1[0][0]; //Displaying card2 using nested for loop System.out.println(":: Card2 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card2[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card2 or not. System.out.print("Do card#2 is having Your Birthday (Y/N):"); ch2 = sc.next(".").charAt(0); //If the day is in card 2,add the first row first column element to sum if(ch2=='Y' || ch2=='y') sum+=card2[0][0]; //Displaying card3 using nested for loop System.out.println(":: Card3 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card3[i][j]); } System.out.println(" "); }
- 3. //Prompting whether the day of birth is having in card3 or not. System.out.print("Do card#3 is having Your Birthday (Y/N):"); ch3 = sc.next(".").charAt(0); //If the day is in card 3,add the first row first column element to sum if(ch3=='Y' || ch3=='y') sum+=card3[0][0]; //Displaying card4 using nested for loop System.out.println(":: Card4 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card4[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card4 or not. System.out.print("Do card#4 is having Your Birthday (Y/N):"); ch4 = sc.next(".").charAt(0); //If the day is in card 4,add the first row first column element to sum if(ch4=='Y' || ch4=='y') sum+=card4[0][0]; //Displaying card5 using nested for loop System.out.println(":: Card5 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card5[i][j]); } System.out.println(" ");
- 4. } //Prompting whether the day of birth is having in card5 or not. System.out.print("Do card#5 is having Your Birthday (Y/N):"); ch5 = sc.next(".").charAt(0); //If the day is in card 5,add the first row first column element to sum if(ch5=='Y' || ch5=='y') sum+=card5[0][0]; System.out.println("Your Birthday is "+sum); System.out.print(" Do You want to continue? (y/n): "); ch = sc.next(".").charAt(0); System.out.println("________________________________"); }while(ch=='Y' || ch=='y'); } } _________________________________________________ Output: :: Card1 :: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 Do card#1 is having Your Birthday (Y/N):y :: Card2 :: 2 3 6 7 10 11 14 15 18 19 22 23 26 27 30 31 Do card#2 is having Your Birthday (Y/N):y :: Card3 :: 4 5 6 7 12 13 14 15 20 21 22 23
- 5. 28 29 30 31 Do card#3 is having Your Birthday (Y/N):n :: Card4 :: 8 9 10 11 12 13 14 15 24 25 26 27 28 29 30 31 Do card#4 is having Your Birthday (Y/N):n :: Card5 :: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Do card#5 is having Your Birthday (Y/N):n Your Birthday is 3 Do You want to continue? (y/n): y :: Card1 :: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 Do card#1 is having Your Birthday (Y/N):n :: Card2 :: 2 3 6 7 10 11 14 15 18 19 22 23 26 27 30 31 Do card#2 is having Your Birthday (Y/N):y :: Card3 :: 4 5 6 7 12 13 14 15 20 21 22 23 28 29 30 31 Do card#3 is having Your Birthday (Y/N):n :: Card4 ::
- 6. 8 9 10 11 12 13 14 15 24 25 26 27 28 29 30 31 Do card#4 is having Your Birthday (Y/N):n :: Card5 :: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Do card#5 is having Your Birthday (Y/N):n Your Birthday is 2 Do You want to continue? (y/n): n ______________________________________________________ The reason this works because we are having the numbers of first row first column of cards as 1 2 4 8 16.With these 5 numbers we can amke any day between 1 and 31.Based on the Birthday present in the cards those corresponding first row first column of every card is getting added to find the birthday of the user. ______________________________________________________Thank you Solution Birthdaydate.java import java.util.Scanner; public class Birthdaydate { public static void main(String[] args) { //Declaring variables char ch,ch1,ch2,ch3,ch4,ch5; //Declaring each card as an 2-Dimensional Array int card1[][]={{1,3,5,7},{9,11,13,15},{17,19,21,23},{25,27,29,31}}; int card2[][]={{2,3,6,7},{10,11,14,15},{18,19,22,23},{26,27,30,31}}; int card3[][]={{4,5,6,7},{12,13,14,15},{20,21,22,23},{28,29,30,31}}; int card4[][]={{8,9,10,11},{12,13,14,15},{24,25,26,27},{28,29,30,31}}; int card5[][]={{16,17,18,19},{20,21,22,23},{24,25,26,27},{28,29,30,31}};
- 7. //Scanner Object is used to get the inputs entered by the user Scanner sc = new Scanner(System.in); //This loop continue to execute until user enters 'Y' or 'y' do { //Declaring variable int sum=0; //Displaying card1 using nested for loop System.out.println(":: Card1 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card1[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card1 or not. System.out.print("Do card#1 is having Your Birthday (Y/N):"); ch1 = sc.next(".").charAt(0); //If the day is in card 1,add the first row first column element to sum if(ch1=='Y' || ch1=='y') sum+=card1[0][0]; //Displaying card2 using nested for loop System.out.println(":: Card2 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card2[i][j]);
- 8. } System.out.println(" "); } //Prompting whether the day of birth is having in card2 or not. System.out.print("Do card#2 is having Your Birthday (Y/N):"); ch2 = sc.next(".").charAt(0); //If the day is in card 2,add the first row first column element to sum if(ch2=='Y' || ch2=='y') sum+=card2[0][0]; //Displaying card3 using nested for loop System.out.println(":: Card3 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card3[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card3 or not. System.out.print("Do card#3 is having Your Birthday (Y/N):"); ch3 = sc.next(".").charAt(0); //If the day is in card 3,add the first row first column element to sum if(ch3=='Y' || ch3=='y') sum+=card3[0][0]; //Displaying card4 using nested for loop System.out.println(":: Card4 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++)
- 9. { System.out.printf("%3d",card4[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card4 or not. System.out.print("Do card#4 is having Your Birthday (Y/N):"); ch4 = sc.next(".").charAt(0); //If the day is in card 4,add the first row first column element to sum if(ch4=='Y' || ch4=='y') sum+=card4[0][0]; //Displaying card5 using nested for loop System.out.println(":: Card5 ::"); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) { System.out.printf("%3d",card5[i][j]); } System.out.println(" "); } //Prompting whether the day of birth is having in card5 or not. System.out.print("Do card#5 is having Your Birthday (Y/N):"); ch5 = sc.next(".").charAt(0); //If the day is in card 5,add the first row first column element to sum if(ch5=='Y' || ch5=='y') sum+=card5[0][0]; System.out.println("Your Birthday is "+sum); System.out.print(" Do You want to continue? (y/n): "); ch = sc.next(".").charAt(0);
- 10. System.out.println("________________________________"); }while(ch=='Y' || ch=='y'); } } _________________________________________________ Output: :: Card1 :: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 Do card#1 is having Your Birthday (Y/N):y :: Card2 :: 2 3 6 7 10 11 14 15 18 19 22 23 26 27 30 31 Do card#2 is having Your Birthday (Y/N):y :: Card3 :: 4 5 6 7 12 13 14 15 20 21 22 23 28 29 30 31 Do card#3 is having Your Birthday (Y/N):n :: Card4 :: 8 9 10 11 12 13 14 15 24 25 26 27 28 29 30 31 Do card#4 is having Your Birthday (Y/N):n :: Card5 :: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Do card#5 is having Your Birthday (Y/N):n
- 11. Your Birthday is 3 Do You want to continue? (y/n): y :: Card1 :: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 Do card#1 is having Your Birthday (Y/N):n :: Card2 :: 2 3 6 7 10 11 14 15 18 19 22 23 26 27 30 31 Do card#2 is having Your Birthday (Y/N):y :: Card3 :: 4 5 6 7 12 13 14 15 20 21 22 23 28 29 30 31 Do card#3 is having Your Birthday (Y/N):n :: Card4 :: 8 9 10 11 12 13 14 15 24 25 26 27 28 29 30 31 Do card#4 is having Your Birthday (Y/N):n :: Card5 :: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Do card#5 is having Your Birthday (Y/N):n Your Birthday is 2 Do You want to continue? (y/n): n
- 12. ______________________________________________________ The reason this works because we are having the numbers of first row first column of cards as 1 2 4 8 16.With these 5 numbers we can amke any day between 1 and 31.Based on the Birthday present in the cards those corresponding first row first column of every card is getting added to find the birthday of the user. ______________________________________________________Thank you