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the reaction of A to B since its Ea is the highest. Solution the reaction of A to B since its Ea is the highest..
the reaction of A to B since its Ea is the highes.pdf
the reaction of A to B since its Ea is the highes.pdf
asif1401
the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together. Solution the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together..
the one thing that comes to mind for me is about .pdf
the one thing that comes to mind for me is about .pdf
asif1401
#include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } Solution #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) {.
#includeiostream#includestdlib.husing namespace std;class .pdf
#includeiostream#includestdlib.husing namespace std;class .pdf
asif1401
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) Solution (1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Vol.
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf
asif1401
Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be Walden Inversion. Solution Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be Walden Inversion..
Iodine will replace both Cl and Br It will be SN2.pdf
Iodine will replace both Cl and Br It will be SN2.pdf
asif1401
The number of outcomes in the event A is 4 Yes event A is a sample event Solution The number of outcomes in the event A is 4 Yes event A is a sample event.
The number of outcomes in the event A is 4 Yes event A is a s.pdf
The number of outcomes in the event A is 4 Yes event A is a s.pdf
asif1401
There were very few planetary leftovers in this region, because most of the solid material was accreted by the terrestrial planets as the planets formed. Solution There were very few planetary leftovers in this region, because most of the solid material was accreted by the terrestrial planets as the planets formed..
There were very few planetary leftovers in this r.pdf
There were very few planetary leftovers in this r.pdf
asif1401
i have no idea. i\'m just seeing if i get karma points for a half ass answer. Solution i have no idea. i\'m just seeing if i get karma points for a half ass answer..
i have no idea. im just seeing if i get karma p.pdf
i have no idea. im just seeing if i get karma p.pdf
asif1401
Recommended
the reaction of A to B since its Ea is the highest. Solution the reaction of A to B since its Ea is the highest..
the reaction of A to B since its Ea is the highes.pdf
the reaction of A to B since its Ea is the highes.pdf
asif1401
the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together. Solution the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together..
the one thing that comes to mind for me is about .pdf
the one thing that comes to mind for me is about .pdf
asif1401
#include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } Solution #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) {.
#includeiostream#includestdlib.husing namespace std;class .pdf
#includeiostream#includestdlib.husing namespace std;class .pdf
asif1401
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) Solution (1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Vol.
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf
asif1401
Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be Walden Inversion. Solution Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be Walden Inversion..
Iodine will replace both Cl and Br It will be SN2.pdf
Iodine will replace both Cl and Br It will be SN2.pdf
asif1401
The number of outcomes in the event A is 4 Yes event A is a sample event Solution The number of outcomes in the event A is 4 Yes event A is a sample event.
The number of outcomes in the event A is 4 Yes event A is a s.pdf
The number of outcomes in the event A is 4 Yes event A is a s.pdf
asif1401
There were very few planetary leftovers in this region, because most of the solid material was accreted by the terrestrial planets as the planets formed. Solution There were very few planetary leftovers in this region, because most of the solid material was accreted by the terrestrial planets as the planets formed..
There were very few planetary leftovers in this r.pdf
There were very few planetary leftovers in this r.pdf
asif1401
i have no idea. i\'m just seeing if i get karma points for a half ass answer. Solution i have no idea. i\'m just seeing if i get karma points for a half ass answer..
i have no idea. im just seeing if i get karma p.pdf
i have no idea. im just seeing if i get karma p.pdf
asif1401
Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii Solution Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii.
Fe+3 solution is more acidic because it produces .pdf
Fe+3 solution is more acidic because it produces .pdf
asif1401
Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions. Solution Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions..
Copper chloride. Copper ions will form precipitat.pdf
Copper chloride. Copper ions will form precipitat.pdf
asif1401
CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks. Solution CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks..
CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf
CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf
asif1401
User interface design process Navigation Design Input Design Output Design Solution User interface design process Navigation Design Input Design Output Design.
User interface design processNavigation DesignInput DesignOu.pdf
User interface design processNavigation DesignInput DesignOu.pdf
asif1401
True.it is an example of an independent-measures design Solution True.it is an example of an independent-measures design.
True.it is an example of an independent-measures designSolution.pdf
True.it is an example of an independent-measures designSolution.pdf
asif1401
The main class of the tictoe game looks like. public class Main { public void play() { TicTacToe game = new TicTacToe(); System.out.println(\"Welcome! Tic Tac Toe is a two player game.\"); System.out.print(\"Enter player one\'s name: \"); game.setPlayer1(game.getPrompt()); System.out.print(\"Enter player two\'s name: \"); game.setPlayer2(game.getPrompt()); boolean markerOk = false; while (!markerOk) { System.out.print(\"Select any letter as \" + game.getPlayer1() + \"\'s marker: \"); String marker = game.getPrompt(); if (marker.length() == 1 && Character.isLetter(marker.toCharArray()[0])) { markerOk = true; game.setMarker1(marker.toCharArray()[0]); } else { System.out.println(\"Invalid marker, try again\"); } } markerOk = false; while (!markerOk) { System.out.print(\"Select any letter as \" + game.getPlayer2() + \"\'s marker: \"); String marker = game.getPrompt(); if (marker.length() == 1 && Character.isLetter(marker.toCharArray()[0])) { markerOk = true; game.setMarker2(marker.toCharArray()[0]); } else { System.out.println(\"Invalid marker, try again\"); } } boolean continuePlaying = true; while (continuePlaying) { game.init(); System.out.println(); System.out.println(game.getRules()); System.out.println(); System.out.println(game.drawBoard()); System.out.println(); String player = null; while (!game.winner() && game.getPlays() < 9) { player = game.getCurrentPlayer() == 1 ? game.getPlayer1() : game.getPlayer2(); boolean validPick = false; while (!validPick) { System.out.print(\"It is \" + player + \"\'s turn. Pick a square: \"); String square = game.getPrompt(); if (square.length() == 1 && Character.isDigit(square.toCharArray()[0])) { int pick = 0; try { pick = Integer.parseInt(square); } catch (NumberFormatException e) { //Do nothing here, it\'ll evaluate as an invalid pick on the next row. } validPick = game.placeMarker(pick); } if (!validPick) { System.out.println(\"Square can not be selected. Retry\"); } } game.switchPlayers(); System.out.println(); System.out.println(game.drawBoard()); System.out.println(); } if (game.winner()) { System.out.println(\"Game Over - \" + player + \" WINS!!!\"); } else { System.out.println(\"Game Over - Draw\"); } System.out.println(); System.out.print(\"Play again? (Y/N): \"); String choice = game.getPrompt(); if (!choice.equalsIgnoreCase(\"Y\")) { continuePlaying = false; } } } public static void main(String[] args) { Main main = new Main(); main.play(); } } The TicTacToe class code will be like import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /** * * author Iamavinashkotina */ public class TicTacToe { private char[][] board = new char[3][3]; private String player1; private String player2; private int currentPlayer; private char marker1; private char marker2; private int plays; private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); protected void init() { int counter = 0; for (int i = 0; i < 3; i++) { for (int i1 = 0; i1 < 3; i1++) {.
The main class of the tictoe game looks like.public class Main {.pdf
The main class of the tictoe game looks like.public class Main {.pdf
asif1401
the inner electrons Solution the inner electrons.
the inner electronsSolutionthe inner electrons.pdf
the inner electronsSolutionthe inner electrons.pdf
asif1401
Solution The investee\'s net income should be recorded as an increase to the investment account. False Explanation Not all the net income of investee is recorded as an increase to investment account. Only the part of net income proportionate to investor share is recorded as increase in investment account. For example if investor have 30% share in investee company and the net income of investee is 100000 then only 30000 will be recorded as an increase in investment account.
SolutionThe investees net income should be recorded as an increa.pdf
SolutionThe investees net income should be recorded as an increa.pdf
asif1401
Question Answer 1 True The corpus callosum , otherwise called the callosal commissure, is a wide, level heap of neural filaments around 10 cm long underneath the cortex in the eutherian cerebrum at the longitudinal gap. It associates the left and right cerebral sides of the equator and encourages interhemispheric correspondence. It is the biggest white matter structure in the mind, comprising of 200–250 million contralateral axonal projections. The (back) bit of the corpus callosum is known as the splenium; the foremost (front) is known as the genu (or \"knee\"); between the two is the truncus, or \"body\", of the corpus callosum. The part between the body and the splenium is frequently extraordinarily contracted and along these lines alluded to as the \"isthmus\". The platform is the part of the corpus callosum that ventures posteriorly and poorly from the anteriormost genu, as can be seen on the sagittal picture of the cerebrum showed on the privilege. The platform is so named for its likeness to a winged creature\'s mouth. Corpus callosum On either side of the corpus callosum, the filaments transmit in the white matter and go to the different parts of the cerebral cortex; those bending forward from the genu into the frontal flap constitute the forceps foremost, and those bending in reverse into the occipital projection, the forceps back. Between these two sections is the fundamental body of the filaments which constitute the tapetum and expand along the side on either side into the transient projection, and cover in the focal part of the horizontal ventricle. 2 False Ramus communicans (plural rami communicantes) is the Latin expression utilized for a nerve which associates two different nerves, and can be deciphered as \"conveying branch\".When utilized without further definition, it quite often alludes to an imparting branch between a spinal nerve and the thoughtful trunk. All the more particularly, it generally alludes to one of the accompanying : Dark ramus communicans White ramus communicans The dark and white rami communicantes are in charge of passing on autonomic signs, particularly for the thoughtful sensory system. Their distinction in tinge is brought about by contrasts in myelination of the nerve filaments contained inside, i.e. there are more myelinated than unmyelinated filaments in the white rami communicantes while the opposite is valid for the dim rami communicantes. 3 False tentorium cerebelli isolates the occipital flaps of the cerebral sides of the equator from the cerebellum. Its inward, sunken, free fringe adds to the tentorial indent. (See underneath.) The outer, arched fringe encases the transverse sinus where it appends to the dura over within the occiput. Past the \"petrous edge,\" the tentorium is tied down to the foremost and back clinoid forms 4 False myelin sheath(deep) and Neurilemma (superficial) 5 False Cranial nerves are the nerves that develop straightforwardly from the cerebrum (counting the brainstem), a.
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
asif1401
No. of Moles = Given Mass/Molecular Mass =3.0 x 1013/98 = 3.06*1011 Moles Solution No. of Moles = Given Mass/Molecular Mass =3.0 x 1013/98 = 3.06*1011 Moles.
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
asif1401
Let the sample size be n , Mean of the sample = 27 (same as population mean) Standard deviation of sample = population standard deviation / sqrt(n) = 2.5/sqrt(n) Solution Let the sample size be n , Mean of the sample = 27 (same as population mean) Standard deviation of sample = population standard deviation / sqrt(n) = 2.5/sqrt(n).
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
asif1401
import java.util.Scanner; public class Assignment4 { public static void main(String args[]) { final int LEN = 5; int[] hand = new int[LEN]; Scanner input = new Scanner(System.in); //input the hand System.out.println(\"Enter five numeric cards, no face cards. Use 2-9.\"); for (int i = 0; i < hand.length; i++) { hand[i] = input.nextInt(); } //sort the collection bubbleSortCards(hand); //determine players hand type //flow of evaluation -> checking complex hands first if (containsFullHouse(hand)) { System.out.println(\"Full House!\"); } else if (containsStraight(hand)) { System.out.println(\"Straight!\"); } else if (containsFourOfaKind(hand)) { System.out.println(\"Four of a Kind!\"); } else if (containsThreeOfaKind(hand)) { System.out.println(\"Three of a Kind!\"); } else if (containsTwoPair(hand)) { System.out.println(\"Two Pair!\"); } else if (containsPair(hand)) { System.out.println(\"Pair!\"); } else System.out.println(\"High Card!\"); } //cardInSets is called with hand[] and pair to check if a pair is found. public static boolean containsPair(int hand[]) { int pair = 2; return cardInSets(hand, pair) != NOT_FOUND; } //cardInSets is called with hand[] and pair to check if a pair is found. //Then calls back again with fcard result to check for another pair. public static boolean containsTwoPair(int hand[]) { int pair = 2; int fcard = cardInSets(hand, pair); int scard = cardInSets(hand, pair, fcard); return fcard!=NOT_FOUND && scard!=NOT_FOUND; } //cardInSets is called with hand[] and threeaKind to check if three of a kind has been found. public static boolean containsThreeOfaKind(int hand[]) { int threeaKind = 3; return cardInSets(hand, threeaKind) != NOT_FOUND; } //hasStraight will resolve to true, unless the previous value++ is not equivalent. //The loop ends as soon as hasStaight is false or the end of the hand is reached. public static boolean containsStraight(int hand[]) { boolean hasStraight = true; int i=1; do { hasStraight = (hand[i] == (hand[i-1] + 1)); } while(hasStraight && ++i < hand.length); return hasStraight; } //cardInSets is called with hand[] and threeKinds to check if three of a kind is found. //Then calls back again with pair and tcard result to check for a pair of another card type. public static boolean containsFullHouse(int hand[]) { int threeKinds = 3; int pair = 2; int tcard = cardInSets(hand, threeKinds); int pcard = cardInSets(hand, pair, tcard); return tcard!=NOT_FOUND && pcard!=NOT_FOUND; } //cardInSets is called with hand[] and fouraKind to check if four of a kind is found. public static boolean containsFourOfaKind(int hand[]) { int fouraKind = 4; return cardInSets(hand, fouraKind) != NOT_FOUND; } //When ignoreCard isn\'t needed, it overloads with a sentinel 0. public static int cardInSets(int hand[], int numKinds) { return cardInSets(hand, numKinds, 0); } //The method cardInSets is a generic evaluation that returns if the parameter specified numKinds(like 2 for pair) has been found in the hand. //The ignoreCard d.
import java.util.Scanner;public class Assignment4 { public st.pdf
import java.util.Scanner;public class Assignment4 { public st.pdf
asif1401
In computer graphics are geometric functions which used to design various graphics acoording to the user requirements output primitives: some of the important output primitive are 1)text 2)line 3)tone 4)interpolation the above primitive consists of two components they index and types of line and tone consists of pattern index line: i)line indexes: the width or thickness of line and color together to be considered with nnm is called as lineindex ii)line type: it indicates type of line (dashed,dotted,solid) these will be allocated by numbers tone: the component inside it is tone pattern index. it consists of color and pattern information 1)horizontal hatch 2)vertical hatch 3)cross hatch 4)right up hacth 5)leftup hatch here the first three didgits is for pattern and last three digits is for color interpolation : i)line interpolation:it uses coordinates system of user ii)great circle interpolation:it uses map projection coordinates text: it deals with textual data .it conatain following components i)size ii)width iii)type iv)font v)height input primtives: input primitives are nothing but input devices ,generally the input devices are consists of both physical and logical 1)mouse 2)tablet 3)electronic pens all the above device contains the similar input primitive it is coordinate position 4)Keyboard: the keyboard is example of string device its input primitive is characters 5)sensing devices ; these devices comes under vector devices its input primitive values is between 0 and 1 Solution In computer graphics are geometric functions which used to design various graphics acoording to the user requirements output primitives: some of the important output primitive are 1)text 2)line 3)tone 4)interpolation the above primitive consists of two components they index and types of line and tone consists of pattern index line: i)line indexes: the width or thickness of line and color together to be considered with nnm is called as lineindex ii)line type: it indicates type of line (dashed,dotted,solid) these will be allocated by numbers tone: the component inside it is tone pattern index. it consists of color and pattern information 1)horizontal hatch 2)vertical hatch 3)cross hatch 4)right up hacth 5)leftup hatch here the first three didgits is for pattern and last three digits is for color interpolation : i)line interpolation:it uses coordinates system of user ii)great circle interpolation:it uses map projection coordinates text: it deals with textual data .it conatain following components i)size ii)width iii)type iv)font v)height input primtives: input primitives are nothing but input devices ,generally the input devices are consists of both physical and logical 1)mouse 2)tablet 3)electronic pens all the above device contains the similar input primitive it is coordinate position 4)Keyboard: the keyboard is example of string device its input primitive is characters 5)sensing devices ; these devices comes under vector devices its input primitive values is betwe.
In computer graphics are geometric functions which used to design va.pdf
In computer graphics are geometric functions which used to design va.pdf
asif1401
I =2 [sint] put value I = 2 Solution I =2 [sint] put value I = 2.
I =2 [sint]put valueI = 2SolutionI =2 [sint]put valueI.pdf
I =2 [sint]put valueI = 2SolutionI =2 [sint]put valueI.pdf
asif1401
f(x)=sin(x) + c Solution f(x)=sin(x) + c.
f(x)=sin(x) + cSolutionf(x)=sin(x) + c.pdf
f(x)=sin(x) + cSolutionf(x)=sin(x) + c.pdf
asif1401
Dipteran insects and their positive role in environment. The word Diptera is derived from Greek di meaning two and ptera meaning wings.Though these insects are known to have two wing they efficiently use only there for flight and the hind wings being reduced to balanced organs called \"Halterers\". The order diptera includes a large number of species most of them are known for their ability to cause diseases in human beings but they are of considerable ecological and human importance.They are known to cause dreadful diseases like malaria,dengue, yellow fever by acting as vectors but their beneficial role in ecosystem cannot be ignored. Some of the positive aspects of the order can be listed as follows: 1. We know that pollination is an essential mechanism for fruit production and we are aware that bees are important pollinators .Dipterans are the second largest pollinators after the hymenopterans.In fact they are said to be the earliest pollinating agents. Many crop plants are dependent on these insect for pollination without which the fruit bearing process is not completed. The chocolate which we enjoy would have not been there had there been no flies. It is a known fact that the chocolate plant Theobroma cocoa produces very small flowers and are self incompatable for various reasons very small midges of the families Ceratopogonidae and Cecidomyiidae pollinate the small white flowers emerging from the stems. so next time we curse a fly remember that we will go without chocolate if we eradicate them. 2. The study of genetics has made many things impossible possible now , we should not forget the contribution of our fruit fly here . Most of the genetic studies are based upon the experiments conducted on fruit flies,because of there smaller genome and easy to rear they serve as excellent genetic model organisms. 3. They play an important role at various trophic level both as consumers and as prey. In many aquatic ecosystems they form the main food source for birds and fishes.As herbivores in wetlands, flies can be very beneficial in controlling potential eutrophication.Many of the aquatic flies are known to reduce algal proliferation, despite very high algal productivity. 4. They form a dominant taxa in temperate ecosystems. 5.Maggots are reared and are used as fishing baits. 6. Some of the maggots which selectvely feed on the necrotic tissue are used in medicine in debraidment to clear wounds. 7.Some members belonging to the families Muscidae or the Sphaeroceridae are detritivores, meaning they feed on decaying material. . These are beneficial in that they speed up nutrient cycling and thus lead to a richer soil (indirectly).Larve of some of the diptran insects are known to act as excellent scavengers , decomposers. 8 .Some flies play an important role in bilogical control of weeds and pests. 9. They are used in forensic labs also . some of the maggots that feed on corpses provide the evidence for the time of death in that they are known to fe.
Dipteran insects and their positive role in environment.The word D.pdf
Dipteran insects and their positive role in environment.The word D.pdf
asif1401
creating dynamic web project in eclipse 1.) create Survey servlet class and update code with below class code 2) update web.xml with below content 3) create register,html file //Survey servlet class import java.io.*; import javax.servlet.ServletException; import javax.servlet.http.*; public class Survey extends HttpServlet { public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { //setting content type response.setContentType(\"text/html\"); //getting printwrite object PrintWriter out = response.getWriter(); // accessing Form values from request String firstName=request.getParameter(\"fName\"); String lastName=request.getParameter(\"lName\"); String emaiId=request.getParameter(\"userEmail\"); String dob=request.getParameter(\"dob\"); //printing values out.println(\"Thanks for completing our Customer Survey form!\"); out.println(\"The data you entered was: First name:\"+firstName+\"Last name:\"+lastName+\"Email address:\"+emailId+\"Birth date:\"+dob); //closing connection out object out.close(); } } //register.html FirstName: LatName: Email Id: Date of Birth: //web.xml SurveySurveySurvey/servlet/Registerregister.html Solution creating dynamic web project in eclipse 1.) create Survey servlet class and update code with below class code 2) update web.xml with below content 3) create register,html file //Survey servlet class import java.io.*; import javax.servlet.ServletException; import javax.servlet.http.*; public class Survey extends HttpServlet { public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { //setting content type response.setContentType(\"text/html\"); //getting printwrite object PrintWriter out = response.getWriter(); // accessing Form values from request String firstName=request.getParameter(\"fName\"); String lastName=request.getParameter(\"lName\"); String emaiId=request.getParameter(\"userEmail\"); String dob=request.getParameter(\"dob\"); //printing values out.println(\"Thanks for completing our Customer Survey form!\"); out.println(\"The data you entered was: First name:\"+firstName+\"Last name:\"+lastName+\"Email address:\"+emailId+\"Birth date:\"+dob); //closing connection out object out.close(); } } //register.html FirstName: LatName: Email Id: Date of Birth: //web.xml SurveySurveySurvey/servlet/Registerregister.html.
creating dynamic web project in eclipse1.) create Survey servlet c.pdf
creating dynamic web project in eclipse1.) create Survey servlet c.pdf
asif1401
// Accepts number and base and returns result in integer import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Base { public static void main(String[] args) throws IOException { System.out.println(\"Enter number: \"); BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String st=br.readLine(); System.out.println(\"Enter base\"); String st1=br.readLine(); int n=Integer.parseInt(st); int b=Integer.parseInt(st1); int sum=0; int i=0; while(n!=0) { sum=(int) (sum+(n%10)*Math.pow(b,i++)); n=n/10; } System.out.println(sum); } } Solution // Accepts number and base and returns result in integer import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Base { public static void main(String[] args) throws IOException { System.out.println(\"Enter number: \"); BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String st=br.readLine(); System.out.println(\"Enter base\"); String st1=br.readLine(); int n=Integer.parseInt(st); int b=Integer.parseInt(st1); int sum=0; int i=0; while(n!=0) { sum=(int) (sum+(n%10)*Math.pow(b,i++)); n=n/10; } System.out.println(sum); } }.
Accepts number and base and returns result in integerimport jav.pdf
Accepts number and base and returns result in integerimport jav.pdf
asif1401
a. steering committee An advisory committee usually made up of high level stakeholders and/or experts who provide guidance on key issues such as company policy and objectives, budgetary control, marketing strategy, resource allocation, and decisions involving large expenditures. b. Role of steerig commitee Control over the IT process of Define the IT processes, organisation and relationships that satisfies the business requirement for IT of being agile in responding to the business strategy whilst complying with governance requirements and providing defined and competent points of contact by focusing on establishing transparent, flexible and responsive IT organisational structures and defining and implementing IT processes with owners, roles and responsibilities integrated into business and decision processes is achieved by: and is measured by c) Role of CEO Decisionmaking may be simply defined as choosing a course of action from among alternatives. If there are no alternatives, then no decision is required. A basis assumption is that the best decision is the one that involves the most revenue or the least amount of cost. The task of management with the help of the management accountant is to find the best alternative. The process of making decisions is generally considered to involve the following steps: 1 Identify the various alternatives for a given type of decision. 2. Obtain the necessary data necessary to evaluate the various alternatives. 3. Analyze and determine the consequences of each alternative. 4. Select the alternative that appears to best achieve the desired goals or objectives. 5. Implement the chosen alternative. 6. At an appropriate time, evaluate the results of the decisions against standards or other desired result Role of Accounting manager - Planning and control techniques which management may use include business budgeting, costvolumeprofit analysis, incremental analysis, flexible budgeting, segmental contribution reporting, inventory models, and capital budgeting models. Nature of Accounting Information - Information for acounting manager -In order for the accounting department to make meaningful analysis of data, it is necessary to distinguish between fixed and variable costs and other types of costs that are not important in the recording of business transactions. CEO - Some but not all of the information needed by management can be provided from financial statements and historical accounting records. In addition to historical data, management will expect the management accountant to provide other types of data, such as estimates, forecasts, future data, and standards. Management Accounting and Decision-Making managerial technique requires an identifiable type of information. The accounting department will be expected to provide the information required by a specific tool. In order for the accounting department to make many types of analysis, a separation of costs into fixed and variable will be required. The managem.
a. steering committeeAn advisory committee usually made up of high.pdf
a. steering committeeAn advisory committee usually made up of high.pdf
asif1401
Answer. Treponema ,Borrelia and leptospira Family :Leptospiraceae-Genus:Leptospira Family -Spirochaetaceae-Genus:Treponema,Borrelia Spirochetes are gram neative bacteria that are long,thin,helcal and motile Axial filaments (a form of flagella ) is found between the peptidoglcan layer and outer membrane and running parallel to them are the locomotory organelles C) cause the disease relapsing fever Clinical significance :its a vector borne disease. Relapsing fever (Borrelia recurrentis)-lice,ticks Clinical aspect :sudden fever,chills,headache,nausea for 2-9 days,symptoms reappear after 3-10 days ;evolution continueswith similar cycles Laboratory diagnosis: Detection of spirochetes in blood smear ELISA test Genus Leptospira have the general charateristics of axial filamemts which provide high motility Solution Answer. Treponema ,Borrelia and leptospira Family :Leptospiraceae-Genus:Leptospira Family -Spirochaetaceae-Genus:Treponema,Borrelia Spirochetes are gram neative bacteria that are long,thin,helcal and motile Axial filaments (a form of flagella ) is found between the peptidoglcan layer and outer membrane and running parallel to them are the locomotory organelles C) cause the disease relapsing fever Clinical significance :its a vector borne disease. Relapsing fever (Borrelia recurrentis)-lice,ticks Clinical aspect :sudden fever,chills,headache,nausea for 2-9 days,symptoms reappear after 3-10 days ;evolution continueswith similar cycles Laboratory diagnosis: Detection of spirochetes in blood smear ELISA test Genus Leptospira have the general charateristics of axial filamemts which provide high motility.
Answer. Treponema ,Borrelia and leptospiraFamily Leptospiraceae-G.pdf
Answer. Treponema ,Borrelia and leptospiraFamily Leptospiraceae-G.pdf
asif1401
help system
male presentation...pdf.................
male presentation...pdf.................
MirzaAbrarBaig5
diagnosting testing help to determine / monitoring the patient condition and find out disease, and evaluate progress of patient.
diagnosting testing bsc 2nd sem.pptx....
diagnosting testing bsc 2nd sem.pptx....
Ritu480198
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Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii Solution Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii.
Fe+3 solution is more acidic because it produces .pdf
Fe+3 solution is more acidic because it produces .pdf
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Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions. Solution Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions..
Copper chloride. Copper ions will form precipitat.pdf
Copper chloride. Copper ions will form precipitat.pdf
asif1401
CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks. Solution CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks..
CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf
CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf
asif1401
User interface design process Navigation Design Input Design Output Design Solution User interface design process Navigation Design Input Design Output Design.
User interface design processNavigation DesignInput DesignOu.pdf
User interface design processNavigation DesignInput DesignOu.pdf
asif1401
True.it is an example of an independent-measures design Solution True.it is an example of an independent-measures design.
True.it is an example of an independent-measures designSolution.pdf
True.it is an example of an independent-measures designSolution.pdf
asif1401
The main class of the tictoe game looks like. public class Main { public void play() { TicTacToe game = new TicTacToe(); System.out.println(\"Welcome! Tic Tac Toe is a two player game.\"); System.out.print(\"Enter player one\'s name: \"); game.setPlayer1(game.getPrompt()); System.out.print(\"Enter player two\'s name: \"); game.setPlayer2(game.getPrompt()); boolean markerOk = false; while (!markerOk) { System.out.print(\"Select any letter as \" + game.getPlayer1() + \"\'s marker: \"); String marker = game.getPrompt(); if (marker.length() == 1 && Character.isLetter(marker.toCharArray()[0])) { markerOk = true; game.setMarker1(marker.toCharArray()[0]); } else { System.out.println(\"Invalid marker, try again\"); } } markerOk = false; while (!markerOk) { System.out.print(\"Select any letter as \" + game.getPlayer2() + \"\'s marker: \"); String marker = game.getPrompt(); if (marker.length() == 1 && Character.isLetter(marker.toCharArray()[0])) { markerOk = true; game.setMarker2(marker.toCharArray()[0]); } else { System.out.println(\"Invalid marker, try again\"); } } boolean continuePlaying = true; while (continuePlaying) { game.init(); System.out.println(); System.out.println(game.getRules()); System.out.println(); System.out.println(game.drawBoard()); System.out.println(); String player = null; while (!game.winner() && game.getPlays() < 9) { player = game.getCurrentPlayer() == 1 ? game.getPlayer1() : game.getPlayer2(); boolean validPick = false; while (!validPick) { System.out.print(\"It is \" + player + \"\'s turn. Pick a square: \"); String square = game.getPrompt(); if (square.length() == 1 && Character.isDigit(square.toCharArray()[0])) { int pick = 0; try { pick = Integer.parseInt(square); } catch (NumberFormatException e) { //Do nothing here, it\'ll evaluate as an invalid pick on the next row. } validPick = game.placeMarker(pick); } if (!validPick) { System.out.println(\"Square can not be selected. Retry\"); } } game.switchPlayers(); System.out.println(); System.out.println(game.drawBoard()); System.out.println(); } if (game.winner()) { System.out.println(\"Game Over - \" + player + \" WINS!!!\"); } else { System.out.println(\"Game Over - Draw\"); } System.out.println(); System.out.print(\"Play again? (Y/N): \"); String choice = game.getPrompt(); if (!choice.equalsIgnoreCase(\"Y\")) { continuePlaying = false; } } } public static void main(String[] args) { Main main = new Main(); main.play(); } } The TicTacToe class code will be like import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /** * * author Iamavinashkotina */ public class TicTacToe { private char[][] board = new char[3][3]; private String player1; private String player2; private int currentPlayer; private char marker1; private char marker2; private int plays; private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); protected void init() { int counter = 0; for (int i = 0; i < 3; i++) { for (int i1 = 0; i1 < 3; i1++) {.
The main class of the tictoe game looks like.public class Main {.pdf
The main class of the tictoe game looks like.public class Main {.pdf
asif1401
the inner electrons Solution the inner electrons.
the inner electronsSolutionthe inner electrons.pdf
the inner electronsSolutionthe inner electrons.pdf
asif1401
Solution The investee\'s net income should be recorded as an increase to the investment account. False Explanation Not all the net income of investee is recorded as an increase to investment account. Only the part of net income proportionate to investor share is recorded as increase in investment account. For example if investor have 30% share in investee company and the net income of investee is 100000 then only 30000 will be recorded as an increase in investment account.
SolutionThe investees net income should be recorded as an increa.pdf
SolutionThe investees net income should be recorded as an increa.pdf
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Question Answer 1 True The corpus callosum , otherwise called the callosal commissure, is a wide, level heap of neural filaments around 10 cm long underneath the cortex in the eutherian cerebrum at the longitudinal gap. It associates the left and right cerebral sides of the equator and encourages interhemispheric correspondence. It is the biggest white matter structure in the mind, comprising of 200–250 million contralateral axonal projections. The (back) bit of the corpus callosum is known as the splenium; the foremost (front) is known as the genu (or \"knee\"); between the two is the truncus, or \"body\", of the corpus callosum. The part between the body and the splenium is frequently extraordinarily contracted and along these lines alluded to as the \"isthmus\". The platform is the part of the corpus callosum that ventures posteriorly and poorly from the anteriormost genu, as can be seen on the sagittal picture of the cerebrum showed on the privilege. The platform is so named for its likeness to a winged creature\'s mouth. Corpus callosum On either side of the corpus callosum, the filaments transmit in the white matter and go to the different parts of the cerebral cortex; those bending forward from the genu into the frontal flap constitute the forceps foremost, and those bending in reverse into the occipital projection, the forceps back. Between these two sections is the fundamental body of the filaments which constitute the tapetum and expand along the side on either side into the transient projection, and cover in the focal part of the horizontal ventricle. 2 False Ramus communicans (plural rami communicantes) is the Latin expression utilized for a nerve which associates two different nerves, and can be deciphered as \"conveying branch\".When utilized without further definition, it quite often alludes to an imparting branch between a spinal nerve and the thoughtful trunk. All the more particularly, it generally alludes to one of the accompanying : Dark ramus communicans White ramus communicans The dark and white rami communicantes are in charge of passing on autonomic signs, particularly for the thoughtful sensory system. Their distinction in tinge is brought about by contrasts in myelination of the nerve filaments contained inside, i.e. there are more myelinated than unmyelinated filaments in the white rami communicantes while the opposite is valid for the dim rami communicantes. 3 False tentorium cerebelli isolates the occipital flaps of the cerebral sides of the equator from the cerebellum. Its inward, sunken, free fringe adds to the tentorial indent. (See underneath.) The outer, arched fringe encases the transverse sinus where it appends to the dura over within the occiput. Past the \"petrous edge,\" the tentorium is tied down to the foremost and back clinoid forms 4 False myelin sheath(deep) and Neurilemma (superficial) 5 False Cranial nerves are the nerves that develop straightforwardly from the cerebrum (counting the brainstem), a.
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
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No. of Moles = Given Mass/Molecular Mass =3.0 x 1013/98 = 3.06*1011 Moles Solution No. of Moles = Given Mass/Molecular Mass =3.0 x 1013/98 = 3.06*1011 Moles.
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
asif1401
Let the sample size be n , Mean of the sample = 27 (same as population mean) Standard deviation of sample = population standard deviation / sqrt(n) = 2.5/sqrt(n) Solution Let the sample size be n , Mean of the sample = 27 (same as population mean) Standard deviation of sample = population standard deviation / sqrt(n) = 2.5/sqrt(n).
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
asif1401
import java.util.Scanner; public class Assignment4 { public static void main(String args[]) { final int LEN = 5; int[] hand = new int[LEN]; Scanner input = new Scanner(System.in); //input the hand System.out.println(\"Enter five numeric cards, no face cards. Use 2-9.\"); for (int i = 0; i < hand.length; i++) { hand[i] = input.nextInt(); } //sort the collection bubbleSortCards(hand); //determine players hand type //flow of evaluation -> checking complex hands first if (containsFullHouse(hand)) { System.out.println(\"Full House!\"); } else if (containsStraight(hand)) { System.out.println(\"Straight!\"); } else if (containsFourOfaKind(hand)) { System.out.println(\"Four of a Kind!\"); } else if (containsThreeOfaKind(hand)) { System.out.println(\"Three of a Kind!\"); } else if (containsTwoPair(hand)) { System.out.println(\"Two Pair!\"); } else if (containsPair(hand)) { System.out.println(\"Pair!\"); } else System.out.println(\"High Card!\"); } //cardInSets is called with hand[] and pair to check if a pair is found. public static boolean containsPair(int hand[]) { int pair = 2; return cardInSets(hand, pair) != NOT_FOUND; } //cardInSets is called with hand[] and pair to check if a pair is found. //Then calls back again with fcard result to check for another pair. public static boolean containsTwoPair(int hand[]) { int pair = 2; int fcard = cardInSets(hand, pair); int scard = cardInSets(hand, pair, fcard); return fcard!=NOT_FOUND && scard!=NOT_FOUND; } //cardInSets is called with hand[] and threeaKind to check if three of a kind has been found. public static boolean containsThreeOfaKind(int hand[]) { int threeaKind = 3; return cardInSets(hand, threeaKind) != NOT_FOUND; } //hasStraight will resolve to true, unless the previous value++ is not equivalent. //The loop ends as soon as hasStaight is false or the end of the hand is reached. public static boolean containsStraight(int hand[]) { boolean hasStraight = true; int i=1; do { hasStraight = (hand[i] == (hand[i-1] + 1)); } while(hasStraight && ++i < hand.length); return hasStraight; } //cardInSets is called with hand[] and threeKinds to check if three of a kind is found. //Then calls back again with pair and tcard result to check for a pair of another card type. public static boolean containsFullHouse(int hand[]) { int threeKinds = 3; int pair = 2; int tcard = cardInSets(hand, threeKinds); int pcard = cardInSets(hand, pair, tcard); return tcard!=NOT_FOUND && pcard!=NOT_FOUND; } //cardInSets is called with hand[] and fouraKind to check if four of a kind is found. public static boolean containsFourOfaKind(int hand[]) { int fouraKind = 4; return cardInSets(hand, fouraKind) != NOT_FOUND; } //When ignoreCard isn\'t needed, it overloads with a sentinel 0. public static int cardInSets(int hand[], int numKinds) { return cardInSets(hand, numKinds, 0); } //The method cardInSets is a generic evaluation that returns if the parameter specified numKinds(like 2 for pair) has been found in the hand. //The ignoreCard d.
import java.util.Scanner;public class Assignment4 { public st.pdf
import java.util.Scanner;public class Assignment4 { public st.pdf
asif1401
In computer graphics are geometric functions which used to design various graphics acoording to the user requirements output primitives: some of the important output primitive are 1)text 2)line 3)tone 4)interpolation the above primitive consists of two components they index and types of line and tone consists of pattern index line: i)line indexes: the width or thickness of line and color together to be considered with nnm is called as lineindex ii)line type: it indicates type of line (dashed,dotted,solid) these will be allocated by numbers tone: the component inside it is tone pattern index. it consists of color and pattern information 1)horizontal hatch 2)vertical hatch 3)cross hatch 4)right up hacth 5)leftup hatch here the first three didgits is for pattern and last three digits is for color interpolation : i)line interpolation:it uses coordinates system of user ii)great circle interpolation:it uses map projection coordinates text: it deals with textual data .it conatain following components i)size ii)width iii)type iv)font v)height input primtives: input primitives are nothing but input devices ,generally the input devices are consists of both physical and logical 1)mouse 2)tablet 3)electronic pens all the above device contains the similar input primitive it is coordinate position 4)Keyboard: the keyboard is example of string device its input primitive is characters 5)sensing devices ; these devices comes under vector devices its input primitive values is between 0 and 1 Solution In computer graphics are geometric functions which used to design various graphics acoording to the user requirements output primitives: some of the important output primitive are 1)text 2)line 3)tone 4)interpolation the above primitive consists of two components they index and types of line and tone consists of pattern index line: i)line indexes: the width or thickness of line and color together to be considered with nnm is called as lineindex ii)line type: it indicates type of line (dashed,dotted,solid) these will be allocated by numbers tone: the component inside it is tone pattern index. it consists of color and pattern information 1)horizontal hatch 2)vertical hatch 3)cross hatch 4)right up hacth 5)leftup hatch here the first three didgits is for pattern and last three digits is for color interpolation : i)line interpolation:it uses coordinates system of user ii)great circle interpolation:it uses map projection coordinates text: it deals with textual data .it conatain following components i)size ii)width iii)type iv)font v)height input primtives: input primitives are nothing but input devices ,generally the input devices are consists of both physical and logical 1)mouse 2)tablet 3)electronic pens all the above device contains the similar input primitive it is coordinate position 4)Keyboard: the keyboard is example of string device its input primitive is characters 5)sensing devices ; these devices comes under vector devices its input primitive values is betwe.
In computer graphics are geometric functions which used to design va.pdf
In computer graphics are geometric functions which used to design va.pdf
asif1401
I =2 [sint] put value I = 2 Solution I =2 [sint] put value I = 2.
I =2 [sint]put valueI = 2SolutionI =2 [sint]put valueI.pdf
I =2 [sint]put valueI = 2SolutionI =2 [sint]put valueI.pdf
asif1401
f(x)=sin(x) + c Solution f(x)=sin(x) + c.
f(x)=sin(x) + cSolutionf(x)=sin(x) + c.pdf
f(x)=sin(x) + cSolutionf(x)=sin(x) + c.pdf
asif1401
Dipteran insects and their positive role in environment. The word Diptera is derived from Greek di meaning two and ptera meaning wings.Though these insects are known to have two wing they efficiently use only there for flight and the hind wings being reduced to balanced organs called \"Halterers\". The order diptera includes a large number of species most of them are known for their ability to cause diseases in human beings but they are of considerable ecological and human importance.They are known to cause dreadful diseases like malaria,dengue, yellow fever by acting as vectors but their beneficial role in ecosystem cannot be ignored. Some of the positive aspects of the order can be listed as follows: 1. We know that pollination is an essential mechanism for fruit production and we are aware that bees are important pollinators .Dipterans are the second largest pollinators after the hymenopterans.In fact they are said to be the earliest pollinating agents. Many crop plants are dependent on these insect for pollination without which the fruit bearing process is not completed. The chocolate which we enjoy would have not been there had there been no flies. It is a known fact that the chocolate plant Theobroma cocoa produces very small flowers and are self incompatable for various reasons very small midges of the families Ceratopogonidae and Cecidomyiidae pollinate the small white flowers emerging from the stems. so next time we curse a fly remember that we will go without chocolate if we eradicate them. 2. The study of genetics has made many things impossible possible now , we should not forget the contribution of our fruit fly here . Most of the genetic studies are based upon the experiments conducted on fruit flies,because of there smaller genome and easy to rear they serve as excellent genetic model organisms. 3. They play an important role at various trophic level both as consumers and as prey. In many aquatic ecosystems they form the main food source for birds and fishes.As herbivores in wetlands, flies can be very beneficial in controlling potential eutrophication.Many of the aquatic flies are known to reduce algal proliferation, despite very high algal productivity. 4. They form a dominant taxa in temperate ecosystems. 5.Maggots are reared and are used as fishing baits. 6. Some of the maggots which selectvely feed on the necrotic tissue are used in medicine in debraidment to clear wounds. 7.Some members belonging to the families Muscidae or the Sphaeroceridae are detritivores, meaning they feed on decaying material. . These are beneficial in that they speed up nutrient cycling and thus lead to a richer soil (indirectly).Larve of some of the diptran insects are known to act as excellent scavengers , decomposers. 8 .Some flies play an important role in bilogical control of weeds and pests. 9. They are used in forensic labs also . some of the maggots that feed on corpses provide the evidence for the time of death in that they are known to fe.
Dipteran insects and their positive role in environment.The word D.pdf
Dipteran insects and their positive role in environment.The word D.pdf
asif1401
creating dynamic web project in eclipse 1.) create Survey servlet class and update code with below class code 2) update web.xml with below content 3) create register,html file //Survey servlet class import java.io.*; import javax.servlet.ServletException; import javax.servlet.http.*; public class Survey extends HttpServlet { public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { //setting content type response.setContentType(\"text/html\"); //getting printwrite object PrintWriter out = response.getWriter(); // accessing Form values from request String firstName=request.getParameter(\"fName\"); String lastName=request.getParameter(\"lName\"); String emaiId=request.getParameter(\"userEmail\"); String dob=request.getParameter(\"dob\"); //printing values out.println(\"Thanks for completing our Customer Survey form!\"); out.println(\"The data you entered was: First name:\"+firstName+\"Last name:\"+lastName+\"Email address:\"+emailId+\"Birth date:\"+dob); //closing connection out object out.close(); } } //register.html FirstName: LatName: Email Id: Date of Birth: //web.xml SurveySurveySurvey/servlet/Registerregister.html Solution creating dynamic web project in eclipse 1.) create Survey servlet class and update code with below class code 2) update web.xml with below content 3) create register,html file //Survey servlet class import java.io.*; import javax.servlet.ServletException; import javax.servlet.http.*; public class Survey extends HttpServlet { public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { //setting content type response.setContentType(\"text/html\"); //getting printwrite object PrintWriter out = response.getWriter(); // accessing Form values from request String firstName=request.getParameter(\"fName\"); String lastName=request.getParameter(\"lName\"); String emaiId=request.getParameter(\"userEmail\"); String dob=request.getParameter(\"dob\"); //printing values out.println(\"Thanks for completing our Customer Survey form!\"); out.println(\"The data you entered was: First name:\"+firstName+\"Last name:\"+lastName+\"Email address:\"+emailId+\"Birth date:\"+dob); //closing connection out object out.close(); } } //register.html FirstName: LatName: Email Id: Date of Birth: //web.xml SurveySurveySurvey/servlet/Registerregister.html.
creating dynamic web project in eclipse1.) create Survey servlet c.pdf
creating dynamic web project in eclipse1.) create Survey servlet c.pdf
asif1401
// Accepts number and base and returns result in integer import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Base { public static void main(String[] args) throws IOException { System.out.println(\"Enter number: \"); BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String st=br.readLine(); System.out.println(\"Enter base\"); String st1=br.readLine(); int n=Integer.parseInt(st); int b=Integer.parseInt(st1); int sum=0; int i=0; while(n!=0) { sum=(int) (sum+(n%10)*Math.pow(b,i++)); n=n/10; } System.out.println(sum); } } Solution // Accepts number and base and returns result in integer import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Base { public static void main(String[] args) throws IOException { System.out.println(\"Enter number: \"); BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String st=br.readLine(); System.out.println(\"Enter base\"); String st1=br.readLine(); int n=Integer.parseInt(st); int b=Integer.parseInt(st1); int sum=0; int i=0; while(n!=0) { sum=(int) (sum+(n%10)*Math.pow(b,i++)); n=n/10; } System.out.println(sum); } }.
Accepts number and base and returns result in integerimport jav.pdf
Accepts number and base and returns result in integerimport jav.pdf
asif1401
a. steering committee An advisory committee usually made up of high level stakeholders and/or experts who provide guidance on key issues such as company policy and objectives, budgetary control, marketing strategy, resource allocation, and decisions involving large expenditures. b. Role of steerig commitee Control over the IT process of Define the IT processes, organisation and relationships that satisfies the business requirement for IT of being agile in responding to the business strategy whilst complying with governance requirements and providing defined and competent points of contact by focusing on establishing transparent, flexible and responsive IT organisational structures and defining and implementing IT processes with owners, roles and responsibilities integrated into business and decision processes is achieved by: and is measured by c) Role of CEO Decisionmaking may be simply defined as choosing a course of action from among alternatives. If there are no alternatives, then no decision is required. A basis assumption is that the best decision is the one that involves the most revenue or the least amount of cost. The task of management with the help of the management accountant is to find the best alternative. The process of making decisions is generally considered to involve the following steps: 1 Identify the various alternatives for a given type of decision. 2. Obtain the necessary data necessary to evaluate the various alternatives. 3. Analyze and determine the consequences of each alternative. 4. Select the alternative that appears to best achieve the desired goals or objectives. 5. Implement the chosen alternative. 6. At an appropriate time, evaluate the results of the decisions against standards or other desired result Role of Accounting manager - Planning and control techniques which management may use include business budgeting, costvolumeprofit analysis, incremental analysis, flexible budgeting, segmental contribution reporting, inventory models, and capital budgeting models. Nature of Accounting Information - Information for acounting manager -In order for the accounting department to make meaningful analysis of data, it is necessary to distinguish between fixed and variable costs and other types of costs that are not important in the recording of business transactions. CEO - Some but not all of the information needed by management can be provided from financial statements and historical accounting records. In addition to historical data, management will expect the management accountant to provide other types of data, such as estimates, forecasts, future data, and standards. Management Accounting and Decision-Making managerial technique requires an identifiable type of information. The accounting department will be expected to provide the information required by a specific tool. In order for the accounting department to make many types of analysis, a separation of costs into fixed and variable will be required. The managem.
a. steering committeeAn advisory committee usually made up of high.pdf
a. steering committeeAn advisory committee usually made up of high.pdf
asif1401
Answer. Treponema ,Borrelia and leptospira Family :Leptospiraceae-Genus:Leptospira Family -Spirochaetaceae-Genus:Treponema,Borrelia Spirochetes are gram neative bacteria that are long,thin,helcal and motile Axial filaments (a form of flagella ) is found between the peptidoglcan layer and outer membrane and running parallel to them are the locomotory organelles C) cause the disease relapsing fever Clinical significance :its a vector borne disease. Relapsing fever (Borrelia recurrentis)-lice,ticks Clinical aspect :sudden fever,chills,headache,nausea for 2-9 days,symptoms reappear after 3-10 days ;evolution continueswith similar cycles Laboratory diagnosis: Detection of spirochetes in blood smear ELISA test Genus Leptospira have the general charateristics of axial filamemts which provide high motility Solution Answer. Treponema ,Borrelia and leptospira Family :Leptospiraceae-Genus:Leptospira Family -Spirochaetaceae-Genus:Treponema,Borrelia Spirochetes are gram neative bacteria that are long,thin,helcal and motile Axial filaments (a form of flagella ) is found between the peptidoglcan layer and outer membrane and running parallel to them are the locomotory organelles C) cause the disease relapsing fever Clinical significance :its a vector borne disease. Relapsing fever (Borrelia recurrentis)-lice,ticks Clinical aspect :sudden fever,chills,headache,nausea for 2-9 days,symptoms reappear after 3-10 days ;evolution continueswith similar cycles Laboratory diagnosis: Detection of spirochetes in blood smear ELISA test Genus Leptospira have the general charateristics of axial filamemts which provide high motility.
Answer. Treponema ,Borrelia and leptospiraFamily Leptospiraceae-G.pdf
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asif1401
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