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kirza4 [7]
3 years ago
12

What is the difference between a design pattern and a DLL?

Computers and Technology
1 answer:
jeka57 [31]3 years ago
6 0

Explanation:

Design Pattern

Design Pattern is defined as how the structure is coded across design - functionality , data abstraction , and leaving space for the future inclusions.

The future inclusions are the business logic which are subjected to change .

DLL - is the interface similar named in java language which specifies the contract of what the product or code is going to perform without exposing the underlying complexities of the implementation .

Example :  

Calculator is an example of  DLL class which contains - add , multiply, subtract , divide methods. The implementation class implements the calculator and write the implementation of the add, multiply methods.

There can be another class that implements this calculator contract and implement scientific calculation logic. In this way you have different choices or flavors of the same contract .

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Describe the difference between the circumscribed and inscribed options when using the AutoCAD Polygon command
marysya [2.9K]

Answer: Describe the difference between circumscribed and inscribed options when using the autocad polygon tool. Circumscribed draws the object around the circle while inscribed draws the object inside the circle. The Length is equal to 5.3151 and the Angle is equal to 41 degrees.

Explanation:

3 0
3 years ago
Which of the following is true about sorting functions?
il63 [147K]

Answer: Option D -- Sorting an already sorted array of size n with quicksort takes O(n log n) time.

Explanation:

Sorting an already sorted array of size n with quicksort takes O(n log n) time is true about sorting functions while other options are wrong.

8 0
3 years ago
A group of statisticians at a local college has asked you to create a set of functions that compute the median and mode of a set
Vanyuwa [196]

Answer:

from functools import reduce

def mean(mylist):

   score = reduce(lambda x,y: x + y, mylist)/ len(mylist)

   return score

def median(mylist):

   sorted(mylist)

   list_len = len(mylist) % 2

   i = round(len(mylist)/2)

   x = len(mylist)//2

   if list_len == 0:

       median = (mylist[x] + mylist[x+1]) / 2  

   else:

       median = mylist[i]

   return median

def mode(mylist):

   unique = set(mylist)

   unique = list(unique)

   collector = [mylist.count(key) for key in unique]

   maxi = max(collector)

   loc = collector.index(maxi)

   return unique[loc]

def main():

   scores = input( 'Enter list of numbers: ').split(",")

   scores = [int(score) for score in scores]

   

   operation = input('Enter operation: ')

   operator = ['mean', 'median', 'mode']

   

   for x in iter(list, 0):

       if operation in operator:

           break

       print("Invalid operation: ")

       operation = input('Enter operation')

   

   index_loc = operator.index(operation)

   

   if index_loc == 0:

       return mean(scores)

   elif index_loc == 1:

       return median(scores)

       #return np.median(scores)  can be used of the defined function

   elif index_loc == 2:

       #return stats.mode(scores)[0]  can be used of the defined function

       return mode(scores)

print( main( ) )

Explanation:

The main python function calls conditionally three statistical functions namely mean, median and mode. It prompts for user input for a list of integer numbers and a function name name to return the corresponding result.

8 0
3 years ago
Write a program to read a list of exam scores given as integer percentages in the range 0-100. Display the total number of grade
Tju [1.3M]

Answer:

import java.util.Scanner;

public class Program

{

  public static void main(String [] Args)

  {

      int totalAGrades = 0, totalBGrades = 0, totalCGrades = 0, totalDGrades = 0, totalFGrades = 0, counter=0,maximum = 0, minimum = 9999, num, total = 0,smallest = 0,largest = 0;

       Scanner in = new Scanner(System.in);

       System.out.println("Enter exam percentage: ");

       System.out.println("Enter a negative examScore: ");

       int examScore = in.nextInt();

       while(examScore > 0)

       {

         counter++;

         if(examScore < 0){          

             break;}  

         else if(examScore > maximum){      

              maximum = examScore;}

         else if(examScore < minimum)   {  

              minimum = examScore;}

         

          total = total + examScore;  

     

          if(examScore <= 50 && examScore>0)

              smallest = examScore;

              if(examScore > 90 && examScore <=100)

              largest = examScore;

         

      if(examScore>=90 && examScore<=100)

          totalAGrades++;

      else if(examScore>=80 && examScore<=89)

          totalBGrades++;

      else if(examScore>=70 && examScore<=79)

          totalCGrades++;

      else if(examScore>=60 && examScore<=69)

          totalDGrades++;

      else if(examScore>=0 && examScore<=59)

          totalFGrades++;

      examScore = in.nextInt();

     

   }

      System.out.println("Total number of scores is = " + counter );

      System.out.println("Total Number of each Letter grade : " + counter);

      System.out.println("Percentage of total for each letter grade : ");

          System.out.println("Total number of A grades = "+ totalAGrades);

          System.out.println("Total number of B grades = "+ totalBGrades);

          System.out.println("Total number of C grades = "+ totalCGrades);

          System.out.println("Total number of D grades = "+ totalDGrades);

          System.out.println("Total number of F grades = "+ totalFGrades);

     

      System.out.println("Lowest exam Score is :"+smallest);

      System.out.println("Highest exam Score is :"+largest);

      System.out.println("Average exam Score = "+ (total / counter));

  }

}  

Explanation:

  • Get the exam information from user as input and run a while loop until examScore is greater than zero.
  • Use conditional statement to check the scores of students and increment the relevant grade accordingly.
  • Finally display all the information including grades and scores.
3 0
3 years ago
Which formula is a simple formula?<br> O =E10/5+1<br> 0 -5+A1*B1<br> O =C3+10-E3<br> 0 =10-(A3-D13)
steposvetlana [31]

Answer:

O =C3+10-E3

Explanation:

Only addition and subtraction :D

8 0
2 years ago
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