Microsoft Access is a database management system that is part of the Microsoft Office Suite.
The answer & explanation for this question is given in the attachment below.
Answer:
Option C i.e., Interoperability is the correct option
Explanation:
Interoperability performs for computers or its components for the purpose to communicate and it is important to improve the development of the Internet of Things. It performs the communication as well as share their services between computer or its components.
It also contains challenges occurred at the time of developing and implementing the protocols by which they communicate.
The answer is the team. Teams are self-establishing which no one not even the scrum master tells the team how to turn product backlog into augmentations of stoppable functionality. Each team member put on his or her knowledge to all of the complications. The interaction that outcomes progresses the entire scrum team’s general competence and usefulness. The optimal size for a scrum team is seven people, plus or minus two. When there are fewer than five team members, there is less interface and as a result less productivity improvement. The product owner and scrum master roles are not comprised in this count. Team arrangement may variation at the end of a sprint. Every time Team membership is altered, the productivity increased from self-organization is reduced.
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
