Consider the DealHands code import javautilScanner impor.pdf
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Consider the DealHands code: import java.util.Scanner; import java.util.ArrayList; public class DealHands { public static void main(String[] args) { Deck D = new Deck(); System.out.println("A brand new, fresh deckn"+D); D.shuffle(); System.out.println("Shuffledn"+D); Scanner sc = new Scanner(System.in); int players; System.out.print("Enter number of players: "); players = sc.nextInt(); ArrayList<Hand> hands = new ArrayList<Hand>(players); for(int i=0;i<players;i++) { Hand h = new Hand(); hands.add(h); } for(int i=0;i<5;i++) { for(int j=0;j<players;j++) { Card C= D.dealCard(); hands.get(j).addCard(C); } } for(int i=0;i<hands.size();i++) { System.out.println(hands.get(i)); } System.out.println("nWhat remainsn"+D); } } Add these to the above code: "-now test error handling, nicely labelled... -try to create a bad card. ... -write a loop that deliberately tries to deal a card from the deck after it has been emptied".
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![Consider the DealHands code:
import java.util.Scanner;
import java.util.ArrayList;
public class DealHands
{
public static void main(String[] args) {
Deck D = new Deck();
System.out.println("A brand new, fresh deckn"+D);
D.shuffle();
System.out.println("Shuffledn"+D);
Scanner sc = new Scanner(System.in);
int players;
System.out.print("Enter number of players: ");
players = sc.nextInt();
ArrayList<Hand> hands = new ArrayList<Hand>(players);
for(int i=0;i<players;i++)
{
Hand h = new Hand();
hands.add(h);
}
for(int i=0;i<5;i++)
{
for(int j=0;j<players;j++)
{
Card C= D.dealCard();
hands.get(j).addCard(C);
}
}
for(int i=0;i<hands.size();i++)
{
System.out.println(hands.get(i));
}
System.out.println("nWhat remainsn"+D);
}
}
Add these to the above code:
"-now test error handling, nicely labelled...
-try to create a bad card. ...
-write a loop that deliberately tries to deal a card from the deck after it has been emptied"](https://image.slidesharecdn.com/considerthedealhandscodeimportjavautilscannerimpor-230319121515-a900bec6/85/Consider-the-DealHands-code-import-javautilScanner-impor-pdf-1-320.jpg)
Recommended
public interface Game Note interface in place of class { .pdf
public interface Game // Note *interface* in place of *class*
{
/// play the game and return the final score
/// where a higher score should be better,
/// and a negative score is allowed.
int play(); // Note semicolon in place of a body
// You can have multiple method headings declared
}
--------------------------------------------------------------------------------------------------------------
import java.util.*;
/// Starting point for Interface Lab.
public class PlayGames
{
private static Scanner in;
private static Random rand = new Random();
private static int gameCount = 0;
public static Game popRandom(Game[] g)
{
int n = gameCount;
int i = rand.nextInt(n);
Game ret = g[i];
g[i] = g[n-1];
gameCount--;
return ret;
}
public static void main(String[] args)
{
Game[] games = new Game[10]; // Note Game as a type
games[gameCount] = new AdditionGame(rand, 100);
gameCount++; // next index to put a Game at
// write at least 2 more different types of Game classes
// and add a new one of each type to games
// ...
in = new Scanner(System.in);
int totScore = 0;
do {
Game g = popRandom(games);
totScore += g.play(); // use numerical result from the game
} while (gameCount > 0 && agree(\"Want a game? \"));
System.out.println(\"Thanks for Playing!\");
System.out.println(\"Your total score is \" + totScore);
}
public static boolean agree(String prompt)
{
System.out.print(prompt);
String input = in.next();
if (input.equalsIgnoreCase(\"y\")) return true;
return false;
}
--------------------------------------------------------------------------------------------------------------------
--
import java.util.*;
public class AdditionGame implements Game // note implements!!
{
private Random rand;
private int n;
// Constructor for objects of class AdditionGame
public AdditionGame(Random r, int big)
{
rand = r;
n = big;
}
// play all games and keep score.
public int play() // exactly matches heading in Game interface
{
final int numGames = 3;
Scanner in = new Scanner(System.in);
int score = 0;
System.out.println(\"Welcome to the addition game! We\'ll now play \" + numGames
+ \" rounds.\");
for (int i = 0; i < numGames; i++) {
int x = rand.nextInt(n), y = rand.nextInt(n), ans = x+y;
System.out.print(String.format(\"Enter the sum: %d + %d = \", x, y));
int val = in.nextInt();
if (ans == val) {
System.out.println(\"Correct!\");
score++;
}
else
System.out.println(\"Wrong! Right answer is \" + ans);
}
System.out.println(\"Thanks for playing the addition game. Your score is \" + score +
\".\");
System.out.println();
return score;
}
}
--------------------------------------------------------------------------------------------------------------------
---
Create two classes that implement a Game interface:
------------------------------------
There is a Game interface defined, and also a PlayGames class that uses objects from classes that
implement that interface. The Game interface declares a single abstract method play that returns
an int, the score from playing the gam.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;
public class Main {
public static int gridArray1[][] = null;
public static int gridArray2[][] = null;
public static boolean playerTurn = true;
public static int count = 0;
public static void setGridDimensions(int x_max, int y_max) {
Main.gridArray1 = new int[x_max][y_max];
Main.gridArray2 = new int[x_max][y_max];
for (int i = 0; i < x_max; i++) {
for (int j = 0; j < y_max; j++) {
gridArray1[i][j] = -1;
gridArray2[i][j] = -1;
}
}
}
static void placeShip(int starting_x, int starting_y, int length, int direction) {
count++;
int ships[] = new int[10];
int battleshipArrayPositions1[] = new int[4];
int cruiserArrayPositions1[] = new int[3];
int submarineArrayPositions1[] = new int[3];
int destroyerArrayPositions1[] = new int[2];
int carrierArrayPositions2[] = new int[5];
int battleshipArrayPositions2[] = new int[4];
int cruiserArrayPositions2[] = new int[3];
int submarineArrayPositions2[] = new int[3];
int destroyerArrayPositions2[] = new int[2];
if (playerTurn) {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray1[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray1[starting_x][i] = 0;
}
}
} else {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray2[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray2[starting_x][i] = 0;
}
}
}
}
boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) {
return false;
}
static int shoot(int x, int y) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (Main.gridArray2[i][j] == 0) {
return 0;
}
}
}
} else {
}
return 0;
}
boolean hasBeenAttempted(int x, int y) {
return false;
}
static void displayGrid(boolean showShips) {
int count = 0;
System.out.print(\" 0 1 2 3 4 5 6 7 8 9\");
if (showShips) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray1[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
} else {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray2[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray2[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray2[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
}
} else {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"- \");
} el.
Recommended
public interface Game Note interface in place of class { .pdf
public interface Game // Note *interface* in place of *class*
{
/// play the game and return the final score
/// where a higher score should be better,
/// and a negative score is allowed.
int play(); // Note semicolon in place of a body
// You can have multiple method headings declared
}
--------------------------------------------------------------------------------------------------------------
import java.util.*;
/// Starting point for Interface Lab.
public class PlayGames
{
private static Scanner in;
private static Random rand = new Random();
private static int gameCount = 0;
public static Game popRandom(Game[] g)
{
int n = gameCount;
int i = rand.nextInt(n);
Game ret = g[i];
g[i] = g[n-1];
gameCount--;
return ret;
}
public static void main(String[] args)
{
Game[] games = new Game[10]; // Note Game as a type
games[gameCount] = new AdditionGame(rand, 100);
gameCount++; // next index to put a Game at
// write at least 2 more different types of Game classes
// and add a new one of each type to games
// ...
in = new Scanner(System.in);
int totScore = 0;
do {
Game g = popRandom(games);
totScore += g.play(); // use numerical result from the game
} while (gameCount > 0 && agree(\"Want a game? \"));
System.out.println(\"Thanks for Playing!\");
System.out.println(\"Your total score is \" + totScore);
}
public static boolean agree(String prompt)
{
System.out.print(prompt);
String input = in.next();
if (input.equalsIgnoreCase(\"y\")) return true;
return false;
}
--------------------------------------------------------------------------------------------------------------------
--
import java.util.*;
public class AdditionGame implements Game // note implements!!
{
private Random rand;
private int n;
// Constructor for objects of class AdditionGame
public AdditionGame(Random r, int big)
{
rand = r;
n = big;
}
// play all games and keep score.
public int play() // exactly matches heading in Game interface
{
final int numGames = 3;
Scanner in = new Scanner(System.in);
int score = 0;
System.out.println(\"Welcome to the addition game! We\'ll now play \" + numGames
+ \" rounds.\");
for (int i = 0; i < numGames; i++) {
int x = rand.nextInt(n), y = rand.nextInt(n), ans = x+y;
System.out.print(String.format(\"Enter the sum: %d + %d = \", x, y));
int val = in.nextInt();
if (ans == val) {
System.out.println(\"Correct!\");
score++;
}
else
System.out.println(\"Wrong! Right answer is \" + ans);
}
System.out.println(\"Thanks for playing the addition game. Your score is \" + score +
\".\");
System.out.println();
return score;
}
}
--------------------------------------------------------------------------------------------------------------------
---
Create two classes that implement a Game interface:
------------------------------------
There is a Game interface defined, and also a PlayGames class that uses objects from classes that
implement that interface. The Game interface declares a single abstract method play that returns
an int, the score from playing the gam.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;
public class Main {
public static int gridArray1[][] = null;
public static int gridArray2[][] = null;
public static boolean playerTurn = true;
public static int count = 0;
public static void setGridDimensions(int x_max, int y_max) {
Main.gridArray1 = new int[x_max][y_max];
Main.gridArray2 = new int[x_max][y_max];
for (int i = 0; i < x_max; i++) {
for (int j = 0; j < y_max; j++) {
gridArray1[i][j] = -1;
gridArray2[i][j] = -1;
}
}
}
static void placeShip(int starting_x, int starting_y, int length, int direction) {
count++;
int ships[] = new int[10];
int battleshipArrayPositions1[] = new int[4];
int cruiserArrayPositions1[] = new int[3];
int submarineArrayPositions1[] = new int[3];
int destroyerArrayPositions1[] = new int[2];
int carrierArrayPositions2[] = new int[5];
int battleshipArrayPositions2[] = new int[4];
int cruiserArrayPositions2[] = new int[3];
int submarineArrayPositions2[] = new int[3];
int destroyerArrayPositions2[] = new int[2];
if (playerTurn) {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray1[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray1[starting_x][i] = 0;
}
}
} else {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray2[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray2[starting_x][i] = 0;
}
}
}
}
boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) {
return false;
}
static int shoot(int x, int y) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (Main.gridArray2[i][j] == 0) {
return 0;
}
}
}
} else {
}
return 0;
}
boolean hasBeenAttempted(int x, int y) {
return false;
}
static void displayGrid(boolean showShips) {
int count = 0;
System.out.print(\" 0 1 2 3 4 5 6 7 8 9\");
if (showShips) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray1[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
} else {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray2[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray2[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray2[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
}
} else {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"- \");
} el.
import java.util.; public class DecimalToBinary { public stat.pdf
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}
Solution
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}.
BlackJackBlackjack.c Blackjack.c Defines the entry point .docx
BlackJack/Blackjack.c
// Blackjack.c : Defines the entry point for the console application.
//
#include <Windows.h>
#include <stdlib.h>
#define CLUBS 0
#define DIAMONDS 1
#define HEARTS 2
#define SPADES 3
enum{ TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING, ACE };
int Deck[52];
int CurrentDealCard;
int Dealer[20], Player[20];
int DealerCards, PlayerCards;
int DealerScore, PlayerScore;
void DoShuffle()
{
int i, nextcard;
int Used[52];
/* Here we clear out Used array to zeros, which indicates no
values have been used. */
for (i = 0; i < 52; i++) Used[i] = 0;
/* Loop through the deck of cards, there are 52 values */
for (i = 0; i < 52; i++)
{
/* Here we used a do-while. If the card has already been used,
we need to keep generating random numbers until we find a card
that has not been used. */
do
{
nextcard = rand() % 52; /* Value from 0 to 51 */
} while (Used[nextcard] == 1); /* This is our check */
/* Here we set to 1 so that we remember that this card has been used */
Used[nextcard] = 1;
/* Finally, put the card in the deck. */
Deck[i] = nextcard;
}
}
void DrawCard(int rank, int suit)
{
switch (rank)
{
case TWO:
printf("Two ");
break;
case THREE:
printf("Three ");
break;
case FOUR:
printf("Four ");
break;
case FIVE:
printf("Five ");
break;
case SIX:
printf("Six ");
break;
case SEVEN:
printf("Seven ");
break;
case EIGHT:
printf("Eight ");
break;
case NINE:
printf("Nine ");
break;
case TEN:
printf("Ten ");
break;
case JACK:
printf("Jack ");
break;
case QUEEN:
printf("Queen ");
break;
case KING:
printf("King ");
break;
case ACE:
printf("Ace ");
break;
}
switch (suit)
{
case CLUBS:
printf("Clubs");
break;
case DIAMONDS:
printf("Diamonds");
break;
case HEARTS:
printf("Hearts");
break;
case SPADES:
printf("Spades");
break;
}
}
void DisplayShuffledDeck()
{
int i, suit, rank;
for (i = 0; i < 52; i++)
{
suit = Deck[i] / 13;
rank = Deck[i] % 13;
DrawCard(rank, suit);
printf("\n");
}
}
void DealCards()
{
PlayerCards = DealerCards = CurrentDealCard = 0;
Player[PlayerCards++] = Deck[CurrentDealCard++];
Dealer[DealerCards++] = Deck[CurrentDealCard++];
Player[PlayerCards++] = Deck[CurrentDealCard++];
Dealer[DealerCards++] = Deck[CurrentDealCard++];
}
void DisplayDealtCards()
{
int i;
printf("\n\n\nDealer:\n");
for (i = 0; i < DealerCards; i++) DrawCard(Dealer[i] % 13, Dealer[i] / 13), printf( " " );
printf("Dealer Score is %d\n\n\nPlayer:\n", DealerScore);
for (i = 0; i < PlayerCards; i++) DrawCard(Player[i] % 13, Player[i] / 13), printf(" ");
printf("Player Score is %d\n\n\n", PlayerScore);
}
void DisplayChoice()
{
printf("What would you like to do?\n");
printf("1. Hit\n2. Stand\n");
}
int Get ...
The Art of Clean Code
Have you ever wanted to add a new feature to your code but instead found yourself spending hours trying to understand the existing code? Or adding a new feature and then finding out you've broken something else, supposedly unrelated, that used to work?
These slides hold a few tips on how to write code that'll be easier to understand, nicer to read and more simple to maintain.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
Java AssignmentWrite a program using sortingsorting bubble,sele.pdf
Java Assignment
Write a program using sorting
sorting: bubble,selection,insertion,quick,merge
Solution
Bubble Sort:
import java.util.Scanner;
class SortBubble {
public static void main(String []args) {
int number, c, d, swap;
Scanner in = new Scanner(System.in);
System.out.println(\"Please enter the number of elements to be sorted\");
number = in.nextInt();
int ar[] = new int[number];
System.out.println(\"Enter \" + number + \" integers\");
for (int i = 0; i < number; i++)
ar[i] = in.nextInt();
for (int i = 0; i < ( number - 1 ); i++) {
for (int j = 0; j < number - c - 1; j++) {
if (ar[j] > ar[j+1])
{
swap = ar[j]];
ar[j] = ar[j+1];
ar[j+1] = swap;
}
}
}
System.out.println(\"Sorted list of numbers\");
for (int c = 0; c < number; c++)
System.out.println(ar[c]);
}
}
b) Selection Sort:
public class SortSelection {
public static int[] sort(int[] arr){
for (int i = 0; i < arr.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int temp= arr[index];
arr[index] = arr[i];
arr[i] = temp;
}
return arr;
}
public static void main(String a[]){
int[] arr1 = {10,34,2,56,7,67,88,42};
int[] arr2 = sort(arr1);
for(int i:arr2){
System.out.print(i);
System.out.print(\", \");
}
}
}
c) Insertion
public class SortInsertion {
public static void main(String[] args) {
int[] array= { 4, 2, 9, 6, 23, 12, 34, 0, 1 };
sort(array);
}
static void printNumbers(int[] input) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + \", \");
}
System.out.println(\"\ \");
}
public static void sort(int array[]) {
int n = array.length;
for (int j = 1; j < n; j++) {
int key = array[j];
int i = j-1;
while ( (i > -1) && ( array [i] > key ) ) {
array [i+1] = array [i];
i--;
}
array[i+1] = key;
printNumbers(array);
}
}
}
d) Quick Sort:
public class SortQuick {
int array[];
int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSort(0, length - 1);
}
private void quickSort(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
while (i <= j) {
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
i++;
j--;
}
}
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String a[]){
MyQuickSort sorter = new MyQuickSort();
int[] input = {24,2,45,20,56,75,2,56,99,53,12};
sorter.sort(input);
for(int i:input){
System.out.print(i);
System.out.print(\" \");
}
}
}.
Vitrogen is an important nutrient for plants Mark the state.pdf
Vitrogen is an important nutrient for plants. Mark the statements that correctly describe nitrogen
facts and plant nitrogen relations. Nitrate fed plants tend to have respitory quotient (RQ) values
greater than 1 since some NADH is used for nitrate reduction instead of O2 reduction and
additional CO2 is released from nitrate reduction to ammonium. Energy (i.e., carbon) costs
between the assimilation of nitrate and ammonium are equivalent because both ions can be
assimilated directly to amino acids. Ammonium (NH4+)can enter passively into the root through
symplastic cation channels. Since this uptake process is passive, there's no respiratory energy
required for ammonium uptake by the root. Three major forms of nitrogen that most plants uptake
through their roots from the soil are nitrate (NO3), ammonium (NH4+), and di-nitrogen (N2). Plant
root acquire nitrate (NO3) through an active uptake process from the soil. That is, nitrate needs to
move against the electrochemical potential gradient to pass through the cell membrane and this
process requires considerable respiratory energy in the form of ATP. Nitrate is likely to be leached
to the ground water because of its high solubility and mobility associated with its charge. Soil
particles are usually negatively charged to repel nitrate ions rather than adsorbing them..
Write a class member function reverse for the class deque .pdf
Write a class member function reverse() for the class deque (given above), which can reverse all
the elements in the deque instance D. For example, if the instance D consists of integers (in this
order): 1,2,3,4,5, the call D.reverse(), returns a copy of D, called E, with all elements in reversed.
That is, E = 5,4,3,2,1.
c++ // DEQUE METHOD DATA TYPE
COULD YOU EXPLAIN THE CONCEPT PLEASE.
The development of peptic ulcer disease PUD is common amon.pdf
The development of peptic ulcer disease (PUD) is common among the working class individuals.
As a consequence, this condition affects not only the overall health of the individual but also their
socio-economic condition.
Discussion:
1. Discuss how H.pylori infection and long-term use of non-steroidal inflammatory drugs (NSAIDs)
increase episodes of PUD?
2. Discuss 5 effective dietary interventions for clients diagnosed with PUD.
Weekly online discussion, Assignment, Quiz and Written Examinations.
Todos experimentamos falta de civismo en nuestras vidas en u.pdf
Todos experimentamos falta de civismo en nuestras vidas en un momento u otro. Piense en dos
instancias que involucren descortesa entre otra persona y usted. Podra haber sido en el trabajo,
en casa, en un lugar pblico en algn lugar o en cualquier otro lugar. Discuta cul fue la descortesa,
cmo la manej y qu podra/debera haber hecho de manera diferente para obtener un mejor
resultado. Usa conceptos del texto en tu respuesta..
This is Cobb Douglas function can you explain why the deman.pdf
This is Cobb Douglas function, can you explain why the demand for good 1 is 1/2*m/p1?
Recall: If the utility function for two goods is x1x2, the demand for good 1 is 21p1m.
More Related Content
Similar to Consider the DealHands code import javautilScanner impor.pdf
import java.util.; public class DecimalToBinary { public stat.pdf
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}
Solution
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}.
BlackJackBlackjack.c Blackjack.c Defines the entry point .docx
BlackJack/Blackjack.c
// Blackjack.c : Defines the entry point for the console application.
//
#include <Windows.h>
#include <stdlib.h>
#define CLUBS 0
#define DIAMONDS 1
#define HEARTS 2
#define SPADES 3
enum{ TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING, ACE };
int Deck[52];
int CurrentDealCard;
int Dealer[20], Player[20];
int DealerCards, PlayerCards;
int DealerScore, PlayerScore;
void DoShuffle()
{
int i, nextcard;
int Used[52];
/* Here we clear out Used array to zeros, which indicates no
values have been used. */
for (i = 0; i < 52; i++) Used[i] = 0;
/* Loop through the deck of cards, there are 52 values */
for (i = 0; i < 52; i++)
{
/* Here we used a do-while. If the card has already been used,
we need to keep generating random numbers until we find a card
that has not been used. */
do
{
nextcard = rand() % 52; /* Value from 0 to 51 */
} while (Used[nextcard] == 1); /* This is our check */
/* Here we set to 1 so that we remember that this card has been used */
Used[nextcard] = 1;
/* Finally, put the card in the deck. */
Deck[i] = nextcard;
}
}
void DrawCard(int rank, int suit)
{
switch (rank)
{
case TWO:
printf("Two ");
break;
case THREE:
printf("Three ");
break;
case FOUR:
printf("Four ");
break;
case FIVE:
printf("Five ");
break;
case SIX:
printf("Six ");
break;
case SEVEN:
printf("Seven ");
break;
case EIGHT:
printf("Eight ");
break;
case NINE:
printf("Nine ");
break;
case TEN:
printf("Ten ");
break;
case JACK:
printf("Jack ");
break;
case QUEEN:
printf("Queen ");
break;
case KING:
printf("King ");
break;
case ACE:
printf("Ace ");
break;
}
switch (suit)
{
case CLUBS:
printf("Clubs");
break;
case DIAMONDS:
printf("Diamonds");
break;
case HEARTS:
printf("Hearts");
break;
case SPADES:
printf("Spades");
break;
}
}
void DisplayShuffledDeck()
{
int i, suit, rank;
for (i = 0; i < 52; i++)
{
suit = Deck[i] / 13;
rank = Deck[i] % 13;
DrawCard(rank, suit);
printf("\n");
}
}
void DealCards()
{
PlayerCards = DealerCards = CurrentDealCard = 0;
Player[PlayerCards++] = Deck[CurrentDealCard++];
Dealer[DealerCards++] = Deck[CurrentDealCard++];
Player[PlayerCards++] = Deck[CurrentDealCard++];
Dealer[DealerCards++] = Deck[CurrentDealCard++];
}
void DisplayDealtCards()
{
int i;
printf("\n\n\nDealer:\n");
for (i = 0; i < DealerCards; i++) DrawCard(Dealer[i] % 13, Dealer[i] / 13), printf( " " );
printf("Dealer Score is %d\n\n\nPlayer:\n", DealerScore);
for (i = 0; i < PlayerCards; i++) DrawCard(Player[i] % 13, Player[i] / 13), printf(" ");
printf("Player Score is %d\n\n\n", PlayerScore);
}
void DisplayChoice()
{
printf("What would you like to do?\n");
printf("1. Hit\n2. Stand\n");
}
int Get ...
The Art of Clean Code
Have you ever wanted to add a new feature to your code but instead found yourself spending hours trying to understand the existing code? Or adding a new feature and then finding out you've broken something else, supposedly unrelated, that used to work?
These slides hold a few tips on how to write code that'll be easier to understand, nicer to read and more simple to maintain.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
Java AssignmentWrite a program using sortingsorting bubble,sele.pdf
Java Assignment
Write a program using sorting
sorting: bubble,selection,insertion,quick,merge
Solution
Bubble Sort:
import java.util.Scanner;
class SortBubble {
public static void main(String []args) {
int number, c, d, swap;
Scanner in = new Scanner(System.in);
System.out.println(\"Please enter the number of elements to be sorted\");
number = in.nextInt();
int ar[] = new int[number];
System.out.println(\"Enter \" + number + \" integers\");
for (int i = 0; i < number; i++)
ar[i] = in.nextInt();
for (int i = 0; i < ( number - 1 ); i++) {
for (int j = 0; j < number - c - 1; j++) {
if (ar[j] > ar[j+1])
{
swap = ar[j]];
ar[j] = ar[j+1];
ar[j+1] = swap;
}
}
}
System.out.println(\"Sorted list of numbers\");
for (int c = 0; c < number; c++)
System.out.println(ar[c]);
}
}
b) Selection Sort:
public class SortSelection {
public static int[] sort(int[] arr){
for (int i = 0; i < arr.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int temp= arr[index];
arr[index] = arr[i];
arr[i] = temp;
}
return arr;
}
public static void main(String a[]){
int[] arr1 = {10,34,2,56,7,67,88,42};
int[] arr2 = sort(arr1);
for(int i:arr2){
System.out.print(i);
System.out.print(\", \");
}
}
}
c) Insertion
public class SortInsertion {
public static void main(String[] args) {
int[] array= { 4, 2, 9, 6, 23, 12, 34, 0, 1 };
sort(array);
}
static void printNumbers(int[] input) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + \", \");
}
System.out.println(\"\ \");
}
public static void sort(int array[]) {
int n = array.length;
for (int j = 1; j < n; j++) {
int key = array[j];
int i = j-1;
while ( (i > -1) && ( array [i] > key ) ) {
array [i+1] = array [i];
i--;
}
array[i+1] = key;
printNumbers(array);
}
}
}
d) Quick Sort:
public class SortQuick {
int array[];
int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSort(0, length - 1);
}
private void quickSort(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
while (i <= j) {
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
i++;
j--;
}
}
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String a[]){
MyQuickSort sorter = new MyQuickSort();
int[] input = {24,2,45,20,56,75,2,56,99,53,12};
sorter.sort(input);
for(int i:input){
System.out.print(i);
System.out.print(\" \");
}
}
}.
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BlackJackBlackjack.c Blackjack.c Defines the entry point .docx
BlackJackBlackjack.c Blackjack.c Defines the entry point .docx
Quest 1 define a class batsman with the following specifications
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Vitrogen is an important nutrient for plants Mark the state.pdf
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additional CO2 is released from nitrate reduction to ammonium. Energy (i.e., carbon) costs
between the assimilation of nitrate and ammonium are equivalent because both ions can be
assimilated directly to amino acids. Ammonium (NH4+)can enter passively into the root through
symplastic cation channels. Since this uptake process is passive, there's no respiratory energy
required for ammonium uptake by the root. Three major forms of nitrogen that most plants uptake
through their roots from the soil are nitrate (NO3), ammonium (NH4+), and di-nitrogen (N2). Plant
root acquire nitrate (NO3) through an active uptake process from the soil. That is, nitrate needs to
move against the electrochemical potential gradient to pass through the cell membrane and this
process requires considerable respiratory energy in the form of ATP. Nitrate is likely to be leached
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COULD YOU EXPLAIN THE CONCEPT PLEASE.
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competitive before the minimum wage existed.
1b.
Now draw and explain a 2nd diagram to show the effect of a minimum wage on the wage rate,
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1c.
What is the difference between these two graphs?
1d.
Why is there a difference?.
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please help me with these 5 questions This is the same ro.pdf
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This is the same road exposure as seen in the previous question. Were these sediments more
likely deposited in a terrestrial or marine setting? (Hint: It's about oxidation of iron. Almost all
sediments contain a bit of iron. Reddish colors indicate that the iron was exposed to oxygen and
formed iron-oxides, or rust. Therefore, reddish rocks indicate that the sediments were exposed to
the atmosphere making them terrestrial in origin. Darker colors (grey and black) indicate low
oxygen environments such as swamps and the deep ocean. These rocks would have a marine
origin. Sometimes, it's not so easy and you have to look for other clues such as terrestrial vs.
marine fossils). marine, all of the sedimentary layers are grey to black in color terrestrial, all of the
sedimentary layers are reddish in color could be a mix of both terrestrial and marine layers since
both reddish layers and grey layers existWhich way was up? Based on the orientation of
sedimentary structures in this outcrop (rock exposure), which way was "up" when these sediments
were deposited? (Hint: These rock layers have been tilted a bit when the Appalachian Mountains
were forming. The original sediments were deposited horizontal to the Earth's surface however the
stress of continents colliding and mountain building can cause originally flat sedimentary rock
layers to tilt or fold). lower right of the image upper right of the image lower left of the image upper
left of the imageInterpret this outcrop: Which way was the current flowing when these sediments
were deposited? (Hint: look back the diagram in the prelab reading). from top to bottom from
bottom to top from left to right from right to leftWhat primary sedimentary structure(s) can be
observed in this outcrop (select all that apply)? These rock layers have been tilted as well so you
are looking at the top of one layer however, if you zoom in, you can see additional layers. cross-
bedding mud cracks ripple marks bedding fossils, plant or animal graded bedding animal trace
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refractory, called "Sporopollenin ". Other than the pollen grains and spores, many microfossiles
are preserved thanks to their composition, similar to Sporopollenin. They include dinocysts,
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In the market lor video game consoles, Mcrosoft and Soey are essentialy a ducpoly, with Nintendo
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are deciding how much ther will spend on advertsing. To simplity, assunie that they can either
spend a lot or aptnd a fitie. If both firms spend a lot Sony's If Somy tpends a lot and Microsoft
spends a Ietlo: Sony's proft wil be 59 bilion and Micresolt's profit will be 51 bilion. Finally; if Sony
spends a litfe and Microsofl spends a loe Sory's profit will be $2 billion and Microsoft profi wili be
$7 billioe Aswume that Somy and Microsoft play this advortising game only once. Complele the
payoff matrix foe this game A. Spending a leti is the dentinant strategy far both Sany and Microsot.
II. Cpensing a lat is the daninant strategy for both Semy asd Micjosoh.Does Sony have a dominant
strategy? If so, what is it? Does Microsoft have a dominant strategy? If so. what is ir? A. Spending
a little is the dominant strategy for both Sony and Microsoft. B. Spending a lot is the dominant
strategy for both Sony and Microsoft C. Spending a lot is the dominant strategy for Sony, while
Microsoft does nothave a dominant strategy D. Spending a lot is the dominant strategy for
Microsoft while Sony does not have a dominant atrategy What is the Nash equilibilum of the
game? A. The Nash equilibrlum occurs where both Sony and Microsof spend a little on advertising
B. The Nash equilibrium occurs where Sony spends a lot and Microsoft spends a little on
advertising C. The Nash equililiam occurs where both Sony and Microsol spend a lot on
advertaing D. The Nash equilibrium occurs where Microsoft spends a lot and Sony apends a little
on advertiaing.
Guguk Kuunun zerinden Utu Hemirelik Ruh Sal ltfen sor.pdf
Guguk Kuunun zerinden Utu (Hemirelik: Ruh Sal) (ltfen sorular cevaplaynz teekkrler, ikinci ksm
hemirelik mdahaleleridir)
Bu film hakknda aadaki sorular tartan tek sayfalk bir analiz yazn:
Ayr bir kada: Filmdeki herhangi bir hastay (Randle McMurphy deil) seerek u sorular yantlayn:
(Tartlacak karakter ad Billy Bibbit hakknda olacak).
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How does an ordered transformation character state differ from an unordered character state?
Please provide illustrations in supporting your answer.
(or apomorpnies, nasn marks). Embryopnytes are treated as piants in this booK..
Hutter Corporation declared a 050 per share cash dividend .pdf
Hutter Corporation declared a $0.50 per share cash dividend on its common shares. The company
has 25,000 shares authorized, 12,000 shares issued, and 10,000 shares of common stock
outstanding. The journal entry to record the dividend declaration is: Multiple Choice Debit Retained
Eamings $5,000; credit Common Dividends Payable $5.000. Debit Common Dividends Payable
$5,000; credit Cash $5,000. Debit Common Dividends Payable $6,000: credit Cash $6,000 Debit
Retained Earnings $12,500, credit Common Dividends Payable $12,500. Debit Retained Earnings
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Explain with examples, two items under each of the following components of software quality
assurance system overview:
Pre-project components
1. Contracts reviews
2. Development and quality plans
Software project life cycle components
1. Reviews
2.Software testing
3.Software maintenance components
Infrastructure components for error
Prevention and improvements
1.Preventive and corrective
2.Procedure and work instructions
3. Staff training, retraining and certification
Management SQA components
1.Project progress control
2. Software quality cost
SQA standards, system certification and assessment components
1. Project process standards
2. Quality management standards
Organizing for SQA the human components
1. Management role in SQA
2. SQA committees
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Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
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