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Step2247 [10]
3 years ago
10

Object-oriented systems have three general types of cohesion: _____, _____, and _____. A. method, class, inheritance B. method,

generalization/specialization, inheritance C. generalization/specialization, class, object D. method, class, generalization/specialization E. functional, sequential, procedural
Computers and Technology
1 answer:
Verdich [7]3 years ago
4 0

Answer:

D. method, class, generalization/specialization

Explanation:

The Cohesion is said to be the level to what an element of a module is related to others. Generally, the cohesion can be understood as an internal adhesive that holds together the modules. If the software is good then it is going to have high cohesion. Method and class are type of cohesion, as they hold the modules together. And generalization/specialization also is a general type of cohesion as generalization tracks the common features among the class, and binds them into one superclass. However, the specialization means creating subclasses out of the classes, and which is the meaning of the specialization. In an ideal situation, there should be high cohesion or single responsibility. However, in the case of inheritance, there can be multiple responsibilities. And hence composition like a generalization, specialization, association, aggregation are used to make use of the composition for reuse rather than inheritance. Have you seen language supporting multiple inheritances. You will never see it though there is some way. And its never allowed to ensure high cohesion. And hence three general types of cohesion are method, class and generalization/specialization. And this is option D.

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Which of the following is not a benefit of introducing an e-commerce solution to an organisation?
emmainna [20.7K]

Answer:

A. Improving human resource management by offering more information to employees about available services, from annual leave arrangements to retirement plans.

Explanation:

e-commerce is a short for electronic commerce and it can be defined as a marketing strategy that deals with meeting the needs of consumers, by selling products or services to the consumers over the internet.

This ultimately implies that, e-commerce is strictly based on the buying and selling of goods or services electronically, over the internet or through a digital platform. Also, the payment for such goods or services are typically done over the internet such as online payment services.

Some of the benefit of introducing an e-commerce solution to an organisation's includes;

I. Improving internal communication by offering various means for exchanging information with minimum effort and cost.

II. Reducing the personalisation of services.

III. Supporting purchasing functions by offering responsive pricing models based on the analysis of market status.

7 0
3 years ago
Virtualization:
Sedaia [141]

Answer:

C. can boost server utilization rates to 70% or higher.

Explanation:

Virtualization -

It is the process of running virtual instance of the computer in the abstracted layer from the very actual hardware , is known as virtualization .

In this process multiple operating system can be operated on the single computer setup , it is useful for increasing the speed of the program.

Hence, from the question,

The correct statement with respect to virtualization is C. can boost server utilization rates to 70% or higher .

8 0
3 years ago
Given an array A of size N, and a number K. Task is to find out if it is possible to partition the array A into K contiguous sub
Dennis_Churaev [7]

Answer: First lets solve the Prerequisite part

Lets say we have an input array of N numbers {3,2,5,0,5}. We have to  find number of ways to divide this array into 3 contiguous parts having equal sum. So the output for the above input array will be 2 as there are 2 ways to divide the array. One is (3,2),(5),(0,5) and the other is (3,2),(5,0),(5).

Following are the steps to achieve the above outcome.

  • Let p and q point to the index of array such that sum of array elements from 0 to p-1 is equal to sum of array elements from p to q which is equal to the sum of array elements from q+1 to N-1.  
  • If we see the array we can tell that the sum of 3 contiguous parts is 5. So the condition would be that sum of all array elements should be equal to 5 or sum of each contiguous part is equal to sum of all array elements divided by 5.
  • Now create 2 arrays prefix and postfix of size of input array. Index p of prefix array carries sum of input array elements from index 0 to index p. Index q of postfix array carries sum of input array elements from index p to index N-1
  • Next move through prefix array suppose at the index p of prefix array : value of prefix array == (sum of all input array elements)/5.
  • Search the postfix array for p index found above. Search it starting from p+2 index. Increment the count variable by 1 when the value of postfix array =(sum of all input array elements)/5 and push that index of postfix array into a new array. Use searching algorithm on new array to calculate number of values in postfix array.

Now lets solve the main task

We have an array A of size N and a number K. where A[]= {1,6,3,4,7} N=5 and K=3. We have to find if its possible to partition A into 3 contiguous subarrays such that sum of elements in each subarray is the same. It is possible in this example. Here we have 3 partitions (1,6),(3,4),(7) and sum of each of subarrays is same (1+6) (3+4) (7) which is 7.

Following are the steps to achieve the above outcome.

  • In order create K contiguous subarrays where each subarray has equal sum, first check the condition that sum of all elements in the given array should be divisible by K. Lets name another array as arrsum that will be the size of array A. Traverse A from first to  last index and keep adding current element of A with previous value in arrsum. Example A contains (1,6,3,4,7} and arrsum has {1,7,10,14,21}
  • If the above condition holds, now check the condition that each subarray or partition has equal sum. Suppose we represent sum1 to sum of all element in given array and sum2 of sum of each partition then: sum2 = sum2 / K.
  • Compare arrsum to subarray, begining from index 0 and when it becomes equal to sum2 this means that end of one subarray is reached. Lets say index q is pointing to that subarray.
  • Now from q+1 index find p index in which following condition holds: (arrsum[p] - arrsum[q])=sum2
  • Continue the above step untill K contigous subarrays are found. This loop will break if, at some index, sum2 of any subarray gets greater than required sum2 (as we know that every contiguous subarray should contain equal sum).

A easier function Partition for this task:

int Partition(int A[], int N, int k) // A arra y of size N and number k

{      int sum = 0;    int count = 0;  //variables initialization    

   for(int j = 0; j < N; j++)  //Loop that calculates sum of A

  sum = sum + A[j];        

  if(sum % k != 0) //checks condition that sum of all elements of A should be //divisible by k

   return 0;        

   sum = sum / k;  

   int sum2 = 0;  //represents sum of subarray

  for(int j = 0; j < N; j++) // Loop on subarrays

  {      sum2=sum2 + A[j];  

   if(sum2 == sum)    { //these lines locates subarrays and sum of elements //of subarrays should be equal

       sum2 = 0;  

       count++;  }  }  

/*calculate count of subarrays whose

sum is equal to (sum of complete array/ k.)

if count == k print Yes else print No*/

if(count == k)    

return 1;  

   else

   return 0;  }

6 0
4 years ago
The Yuba College Library would like a program to calculate patron fines for overdue books. Fines are determined as follows: Pape
mestny [16]

Answer:

Explanation:

The following Python code creates all of the necessary functions in order to request and validate all of the inputs from the user. Once everything is entered and validated it calculates the total fee and outputs it to the user. Also calls the entire class at the end using a test variable...

class Customer:

   library_card = 0

   patrons_name = ""

   patrons_address = ""

   type_of_book = ""

   book_title = ""

   days_overdue = 0

   fee = 0.0

   book_list = ['paperback regular', 'paperback best seller', 'magazine', 'hardcover']

   def __init__(self):

       self.library_card = self.get_library_card()

       self.patrons_address = input("Patron's Address: ")

       self.book_title = input("Book Title: ")

       self.type_of_book = self.get_type_of_book()

       self.days_overdue = float(self.get_days_overdue())

       self.calculate_total()

       print("Your total Fee is: " + str(self.fee))

   def get_library_card(self):

       library_card = int(input("Enter 4-digit library card number: "))

       if (type(library_card) == type(0)) and (len(str(library_card)) == 4):

           return library_card

       else:

           print("Invalid Card number:")

           self.get_library_card()

   def get_type_of_book(self):

       type_of_book = input("Type of Book 1-4: ")

       if (int(type_of_book) > 0) and (int(type_of_book) <= 4):

           return int(type_of_book)

       else:

           print("Invalid Type")

           self.get_type_of_book()

   def get_days_overdue(self):

       days_overdue = input("Number of Days Overdue: ")

       if int(days_overdue) >= 0:

           return days_overdue

       else:

           print("Invalid number of days")

           self.get_days_overdue()

   def calculate_total(self):

       if self.type_of_book == 1:

           self.fee = 0.20 * self.days_overdue

           if self.fee > 8:

               self.fee = 8

       elif self.type_of_book == 2:

           self.fee = 0.50 * self.days_overdue

           if self.fee > 15:

               self.fee = 15

       elif self.type_of_book == 3:

           self.fee = 0.25 * self.days_overdue

           if self.fee > 5:

               self.fee = 5

       else:

           self.fee = 0.30 * self.days_overdue

           if self.fee > 23:

               self.fee = 23

test = Customer()

5 0
3 years ago
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