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Marina86 [1]
2 years ago
13

The population of town A is less than the population of town B. However, the population of town A is growing faster than the pop

ulation of town B. Write a program that prompts the user to enter: The population of town A The population of town B The growth rate of town A The growth rate of town B The program outputs: After how many years the population of town A will be greater than or equal to the population of town B The populations of both the towns at that time. (A sample input is: Population of town A = 5,000, growth rate of town A = 4%, population of town B = 8,000, and growth rate of town B = 2%.)
Computers and Technology
1 answer:
defon2 years ago
3 0

Answer:

#include<iostream>

using namespace std;

void main()

{

int townA_pop,townB_pop,count_years=1;

double rateA,rateB;

cout<<"please enter the population of town A"<<endl;

cin>>townA_pop;

cout<<"please enter the population of town B"<<endl;

cin>>townB_pop;

cout<<"please enter the grothw rate of town A"<<endl;

cin>>rateA;

cout<<"please enter the grothw rate of town B"<<endl;

cin>>rateB;

while(townA_pop < townB_pop)//IF town A pop is equal or greater than town B it will break

{

townA_pop = townA_pop +( townA_pop * (rateA /100) );

townB_pop = townB_pop +( townB_pop * (rateB /100) );

count_years++;

}

cout<<"after "<<count_years<<" of years the pop of town A will be graeter than or equal To the pop of town B"<<endl;

}

Explanation:

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Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
SVETLANKA909090 [29]

Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

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2 years ago
What is the Gain (dB) of a transmission if the Maximum Data Rate is 1 Gbps and the Bandwidth =7000 MHz? Group of answer choices
denpristay [2]

Answer:

15.420 dB

Explanation:

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When you connect to an unsecured wireless network, what might dishonest or unscrupulous computer users try to do?
loris [4]

Answer:

Hackers can snoop on data sent over your network.

Hackers can use your network to access your computer's files and system information.

Explanation: Unsecured Wireless connections are wireless connections which are have no passwords they are open to the general public,such networks can be very risky to use as it gives easy access to dishonest persons who can manipulate that opportunity to SNOOP ON DATA SENT OVER YOUR NETWORKS. They can use this hacking to fraudulently steal from your bank account and obtain your private information.

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A business wants to use a new platform to help data flow between third-party platforms or between a platform and its own in-hous
lawyer [7]

Answer:

Database platform as a service

Explanation:

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7 0
2 years ago
// This pseudocode is intended to display // employee net pay values. All employees have a standard // $45 deduction from their
Vikki [24]

Answer:

C++ code is given below

Explanation:

#include<iostream>

#include <cstring>

using namespace std;

int housekeeping(string EOFNAME);

int mainLoop(string name,string EOFNAME);

int finish();

void main()

{

  string name;

  string EOFNAME = "ZZZZ";

  cout << "enter the name" << endl;

  cin >> name;

  if (name != EOFNAME)

  {

      housekeeping(EOFNAME);

  }

  if (name != EOFNAME)

  {

      mainLoop(name , EOFNAME);

  }

  if (name != EOFNAME)

  {

      finish();

  }

  system("pause");

}

int housekeeping(string EOFNAME)

{  

  cout << "enter first name " << EOFNAME << " to quit " << endl;

  return 0;

}

int mainLoop(string name, string EOFNAME)

{  

  int hours;

  int rate,gross;

  int DEDUCTION = 45;

  int net;

  cout << "enter hours worked for " << name << endl;

  cin >> hours;

  cout << "enter hourly rate for " << name << endl;

  cin >> rate;

  gross = hours*rate;

  net = gross - DEDUCTION;

  if (net > 0)

  {

      cout << "net pay for " << name << " is " << net << endl;

  }

  else

  {

      cout << "dedections not covered.net is 0.";

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  cout << "enter next name or " << EOFNAME << " to quit" << endl;

  cin >> name;

  return 0;

}

int finish()

{

  cout << "end of job"<<endl;

  return 0;

}

3 0
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