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Helga [31]
3 years ago
11

Within MethodsPractice.java, write a method public static int planParty2(int f, int c, int p) that performs the same calculation

s as the previous problem, but returns the number of p-packs needed. planParty2 does not need to consider the number of extra cans, and it should not print anything. Here are some example arguments for the planParty2 method and their expected return values:

Computers and Technology
1 answer:
Vesna [10]3 years ago
3 0

Answer:

// class definition of MethodPractice

public class MethodPractice {

   // main method that begin program execution

   public static void main(String args[]) {

       // call to the planParty2 method is done using some parameter

       // and the return value is displayed

     System.out.println("The number of p-packs needed is: " + planParty2(9, 14, 6));

     System.out.println("The number of p-packs needed is: " + planParty2(4, 6, 3));

     System.out.println("The number of p-packs needed is: " + planParty2(4, 6, 4));

   }

   

   // planParty2 method with 3 parameters and it return an integer

   // the calculation needed to get the required pack is done here

   public static int planParty2(int f, int c, int p){

       // the party owner is added to list of friend

       // and the value is assigned to totalPeople

       int totalPeople = f + 1;

       // the totalPeople times the can drink

       // is assigned to totalDrinkNeeded

       int totalDrinkNeeded = totalPeople * c;

       // number of packs required is declared

       int numberOfPacks;

       // if the totalDrinkNeeded contains exact multiple of a pack

       // we return the numberOfPack by dividing totalDrinkNeeded with p

       if (totalDrinkNeeded % p == 0){

           return numberOfPacks = totalDrinkNeeded / p;

       }

       // else we compute if the totalDrinkNeeded is not a multiple of a pack

       //  first we get the remainder when the totalDrinkNeeded is divided with p

       // then the needed can is computed by subtracting remainder from p

       // the neededCan is added to totalDrinkNeeded, then divided with p

       // and assigned to numberOfPacks

       // numberOfPacks is returned.

       else {

           int remainder = totalDrinkNeeded % p;

           int neededCan = p - remainder;

           numberOfPacks = (totalDrinkNeeded + neededCan) / p;

           return numberOfPacks;

       }

   }

}

Explanation:

Image showing the detailed question is attached.

The program is written in Java and is well commented. A sample image of program output when it is executed is attached.

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Hi, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.

Answer:

import math

def rectangle(perimeter, area):

   l1_1 = (perimeter+math.sqrt((perimeter**2)-(16*area)))/4

   l1_2 = (perimeter-math.sqrt((perimeter**2)-(16*area)))/4

   l2_1 = area/l1_1

   l2_2 = area/l1_2

   print(l1_1,l2_1)

   print(l1_2,l2_2)

   if l1_1.is_integer() and l2_1.is_integer() and l1_1>0 and l2_1>0:

       return(int(max(l1_1,l2_1)))

   elif l1_2.is_integer() and l2_2.is_integer() and l1_2>0 and l2_2>0:

       return(int(max(l1_2,l2_2)))

   else:

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Explanation:

  • We import math to make basic operations
  • We define the rectangle function that receives perimeter and area
  • We calculate one of the sides (l1_1) of the rectangle using the quadratic equation to solve 2h^2 - ph + 2a = 0
  • We calculate the second root of the quadratic equation for the same side (l1_2)
  • We calculate the second side of the rectangle using the first root on w = a/h
  • We calculate the second side of the rectangle using the second root on w= a/h
  • We verify that each component of the first result (l1_1, l2_1) is an integer (using the build-in method .is_integer) and greater than 0, if True we return the maximum value between them (using the max function) as w
  • If the first pair of sides evaluate to False we check the second root of the equation and if they meet the specification we return the max value
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Answer:

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Explanation:

From the given information:

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We can conclude that the longest codeword that could possibly be for "n" symbol is n-1 bits.

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