Answer:
Risk assessment, Input validation and Output validation.
Explanation:
Software development life cycle, SDLC, is a systematic process a software being created must pass through or follow, from the stage of conception to death of the application.
There are various processes that occurs at the beginning of SDLC, a few of the activities are risk management, input and output validation.
Risk management is used to determine the feasibility, usefulness and profitability to cost of the software before development. The input and output validation is for security control access to the data of the software.
Answer: Hierarchical structure
Explanation:
The hierarchical structure is the most preferred structure for an organisation. As more and more people join an organisation with the passage of time we can clearly see the different links between the persons and the their hierarchical level to the organisation. In this process it becomes easier to manage the people at the various levels.
<span>Charts are inserted into an excel spreadsheet using the commands in the charts group on the Insert tab on the ribbon.
There, on the Insert tab, you will find many options when it comes to the things that you want to insert into your Excel spreadsheet, such as images, tables, and charts, among other things.</span>
Answer:
a. The algorithm will likely require more time since it is now being run sequentially on more computers.
Explanation:
Algorithm is a software which enables the users to solve complex problems with input of commands. The software runs the command inserted and then output is generate. The algorithms can be run parallel and sequentially on more computers.
Answer:
The function in Python is as follows:
def digitSum( n ):
if n == 0:
return 0
if n>0:
return (n % 10 + digitSum(int(n / 10)))
else:
return -1 * (abs(n) % 10 + digitSum(int(abs(n) / 10)))
Explanation:
This defines the method
def digitSum( n ):
This returns 0 if the number is 0
<em> if n == 0:
</em>
<em> return 0
</em>
If the number is greater than 0, this recursively sum up the digits
<em> if n>0:
</em>
<em> return (n % 10 + digitSum(int(n / 10)))
</em>
If the number is lesser than 0, this recursively sum up the absolute value of the digits (i.e. the positive equivalent). The result is then negated
<em> else:
</em>
<em> return -1 * (abs(n) % 10 + digitSum(int(abs(n) / 10)))</em>
