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Snezhnost [94]
3 years ago
13

Write a function PrintShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, pr

int "Too few.". If more than 4, print "Too many.". Else, print "N: Lather and rinse." numCycles times, where N is the cycle number, followed by "Done.". End with a newline. Example output for numCycles = 2:
1: Lather and rinse.
2: Lather and rinse.
Done.
Hint: Define and use a loop variable.
Sample program:
#include
using namespace std;

int main() {
PrintShampooInstructions(2);
return 0;
}
Computers and Technology
2 answers:
Finger [1]3 years ago
8 0

Answer:

public static void PrintShampooInstructions(int numberOfCycles){

       if (numberOfCycles<1){

           System.out.println("Too Few");

       }

       else if(numberOfCycles>4){

           System.out.println("Too many");

       }

       else

           for(int i = 1; i<=numberOfCycles; i++){

               System.out.println(i +": Lather and rinse");

           }

       System.out.println("Done");

   }

Explanation:

I have used Java Programming language to solve this

Use if...elseif and else statement to determine and print "Too Few" or "Too Many".

If within range use a for loop to print the number of times

Nady [450]3 years ago
4 0

Answer:

#In Python

def shampoo_instructions(num_cycles):

   if num_cycles < 1: ///

       print ('Too few.') ///

   elif num_cycles > 4:

       print ('Too many.')

   else:

       i = 0

       while i<num_cycles:

           print (i+1,": Lather and rinse.")

           i = i + 1

       print('Done.')

user_cycles = int(input())

shampoo_instructions(user_cycles)

Explanation:

def shampoo_instructions(num_cycles): #def function with loop

   if num_cycles < 1:  #using 1st if statement

       print('Too few.')

   elif num_cycles > 4:

       print ('Too many.')

   else:

       i = 0

       while i<num_cycles:

           print (i+1,": Lather and rinse.")

           i = i + 1

       print('Done.')

user_cycles = int(input())

shampoo_instructions(user_cycles)

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Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr
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Answer:

Here is the C++ program:

#include <iostream>  // to use input output functions

#include <cmath>  // to use math functions like sqrt()

#include <iomanip>  //to use setprecision method

using namespace std;   //to access objects like cin cout

int main ()  {  //start of main function

  double speedA;  //double type variable to store average speed of car A

  double speedB;  //double type variable to store average speed of car B

  int hour;  //int type variable to hold hour part of elapsed time

  int minutes;  //int type variable to hold minutes part of elapsed time

  double shortDistance;  // double type variable to store the result of shortest distance between car A and B

  double distanceA;  //stores the distance of carA

  double distanceB;  //stores the distance of carB

  double mins,hours;   //used to convert the elapsed time

cout << "Enter average speed of car A: " << endl;  //prompt user to enter the average speed of car A

cin >> speedA;   //reads the input value of average speed of car A from user

cout << "Enter average speed of car B: " << endl ;  //prompt user to enter the average speed of car B

cin >> speedB;   //reads the input value of average speed of car A from user

cout << "Enter elapsed time (in hours and minutes, separated by a space): " << endl;  //prompts user to enter elapsed time

cin>> hour >> minutes;    //reads elapsed time in hours and minutes

  mins = hour * 60;  //computes the minutes using value of hour

  hours = (minutes+mins)/60;     //computes hours using minutes and mins

distanceA = speedA * (hours);  // computes distance of car A

distanceB = speedB * (hours);   //computes distance of car B

   shortDistance =sqrt((distanceA * distanceA) + (distanceB * distanceB));   //computes shortest distance using formula √[(distanceA)² + (distanceB)²)]

cout << "The (shortest) distance between the cars is: "<<fixed<<setprecision(2)<<shortDistance;

//display the resultant value of shortDistance up to 2 decimal places

Explanation:

I will explain the program with an examples:

Let us suppose that the average speeds of cars are:

speedA = 70

speedB = 55

Elapsed time in hours and minutes:

hour = 2

minutes = 30

After taking these input values the program control moves to the statement:

mins = hour * 60;  

This becomes

mins = 2 * 60

mins = 120

Next

hours = (minutes+mins)/60;

hours = (30 + 120) / 60

         = 150/60

hours = 2.5

Now the next two statements compute distance of the cars:

distanceA = speedA * (hours);  

this becomes

distanceA = 70 * (2.5)

distanceA = 175

distanceB = speedB * (hours);

distanceB = 55 * (2.5)

distanceB = 137.5

Next the shortest distance between car A and car B is computed:

shortDistance = sqrt((distanceA * distanceA) + (distanceB * distanceB));

shortDistance = sqrt((175 * 175) + (137.5 * 137.5))

                        = sqrt(30625 + 18906.25)

                        = sqrt(49531.25)

                        =  222.556173

shortDistance =  222.56

 

Hence the output is:

The (shortest) distance between the cars is: 222.56        

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