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olganol [36]
3 years ago
10

Wells drilled by a nonprofit called Water for South Sudan use a pump that can provide up to 5,500 gallons of water per day. Use

this total amount to calculate the following:
How many gallons per day per person could the well produce for a village with a population of 850 if every gallon went to domestic use?
Engineering
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

6.47 gallon per person per day

Explanation:

Given

The total amount of water available for domestic use is  5500 gallon

Let us consider that the entire available water goes to domestic use.

The total population of the village is 850.

The amount of water available to each villager is

\frac{5500}{850}\\6.47 gallon per person per day

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3 years ago
To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.
GREYUIT [131]

Income

amount budgeted

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5 0
3 years ago
computer language C++ (Connect 4 game)( this is all the info that was givin no input or solution) I used the most recent version
Mariana [72]

Answer:

C++ code explained below

Explanation:

#include "hw6.h"

//---------------------------------------------------

// Constructor function

//---------------------------------------------------

Connect4::Connect4()

{

ClearBoard();

}

//---------------------------------------------------

// Destructor function

//---------------------------------------------------

Connect4::~Connect4()

{

// Intentionally empty

}

//---------------------------------------------------

// Clear the Connect4 board

//---------------------------------------------------

void Connect4::ClearBoard()

{

// Initialize Connect4 board

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

board[r][c] = ' ';

// Initialize column counters

for (int c = 0; c < COLS; c++)

count[c] = 0;

}

//---------------------------------------------------

// Add player's piece to specified column in board

//---------------------------------------------------

bool Connect4::MakeMove(int col, char player)

{

// Error checking

if ((col < 0) || (col >= COLS) || (count[col] >= ROWS))

return false;

// Make move

int row = count[col];

board[row][col] = player;

count[col]++;

return true;

}

//---------------------------------------------------

// Check to see if player has won the game

//---------------------------------------------------

bool Connect4::CheckWin(char player)

{

// Loop over all starting positions

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

if (board[r][c] == player)

{

// Check row

int count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) &&

(board[r+d][c] == player)) count++;

if (count == WIN) return true;

 

// Check column

count = 0;

for (int d = 0; d < WIN; d++)

if ((c+d < COLS) &&

(board[r][c+d] == player)) count++;

if (count == WIN) return true;

 

// Check first diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) && (c+d < COLS) &&

(board[r+d][c+d] == player)) count++;

if (count == WIN) return true;

 

// Check second diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r-d >= 0) && (c+d < COLS) &&

(board[r-d][c+d] == player)) count++;

if (count == WIN) return true;

}

return false;

}

//---------------------------------------------------

// Print the Connect4 board

//---------------------------------------------------

void Connect4::PrintBoard()

{

// Print the Connect4 board

for (int r = ROWS-1; r >= 0; r--)

{

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw board contents

cout << "| ";

for (int c = 0; c < COLS; c++)

cout << board[r][c] << " | ";

cout << "\n";

}

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw column numbers

cout << " ";

for (int c = 0; c < COLS; c++)

cout << c << " ";

cout << "\n\n";

}

//---------------------------------------------------

// Main program to play Connect4 game

//---------------------------------------------------

int main()

{

  int choice;

  int counter = 0;

  srand (time(NULL));

  Connect4 board;

  cout << "Welcome to Connect 4!" << endl << "Your Pieces will be labeled 'H' for human. While the computer's will be labeled 'C'" << endl;

  board.PrintBoard();

  cout << "Where would you like to make your first move? (0-6)";

  cin >> choice;

  while (board.MakeMove(choice,'H') == false){

  cin >> choice;

  }

  counter++;

  while (board.CheckWin('C') == false && board.CheckWin('H') == false && counter != 21){

  while (board.MakeMove(rand() % 7, 'C') == false){}

  board.PrintBoard();

  cout << "Where would you like to make your next move?" << endl;

  cin >> choice;

  board.MakeMove(choice,'H');

  while (board.MakeMove(choice,'H') == false){

  cin >> choice;

  }

  counter++;

  }

 

  if (board.CheckWin('C')){

  cout << "Computer Wins!" << endl;}

  else if (counter == 21){cout << "Tie Game!" << endl;}

  else {cout << "Human Wins!" << endl;}

  board.PrintBoard();

}

4 0
3 years ago
If a lever operates at a mechanical disadvantage, it means that the ________.
alekssr [168]

Answer:

The correct answer is option 'B': Load is far from fulcrum and the effort is applied near the fulcrum

Explanation:

A lever works on the principle of balancing of torques. The torque about the fulcrum by the load should be equal to the torque by the applied effort. Since we know that the torque is proportional to both the force and the distance it is applied from the distance from the axis of rotation. A lever is used when we need to lift a heavy load by utilizing this effect of the lever arm.

A mechanical disadvantage occurs when we are not able to lift the weight easily due to the fact we apply effort near the fulcrum.

4 0
3 years ago
Read 2 more answers
Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change,
Doss [256]

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

6 0
3 years ago
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