Answer:
C: Security
Explanation:
Communications and information systems principles need to be, among other things, secure. They need to be able to protect sensitive information from those who intentionally not need to know. Some incident information like voice, networks, and data, are very sensitive and thus, should be secure to the right levels and should comply with privacy laws and data protection.
When drilling, students in a study session use flashcards to quiz one another. thus, Option B is the correct statement.
<h3>
What do you mean by drill and practice?</h3>
The term drill and practice can be described as a way of practice characterized by systematic repetition of concepts, examples, and exercise problems.
Drill and exercise is a disciplined and repetitious exercise, used as an average of coaching and perfecting an ability or procedure.
Thus, When drilling, students in a study session use flashcards to quiz one another. Option B is the correct statement.
Learn more about drill and practice:
brainly.com/question/27587324
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Answer:
The program in Python is as follows:
num1 = int(input())
num2 = int(input())
if num1 >=0 and num2 >= 0:
print(num1+num2)
elif num1 <0 and num2 < 0:
print(num1*num2)
else:
if num1>=0:
print(num1**2)
else:
print(num2**2)
Explanation:
This gets input for both numbers
num1 = int(input())
num2 = int(input())
If both are positive, the sum is calculated and printed
<em>if num1 >=0 and num2 >= 0:</em>
<em> print(num1+num2)</em>
If both are negative, the products is calculated and printed
<em>elif num1 <0 and num2 < 0:</em>
<em> print(num1*num2)</em>
If only one of them is positive
else:
Calculate and print the square of num1 if positive
<em> if num1>=0:</em>
<em> print(num1**2)</em>
Calculate and print the square of num2 if positive
<em> else:</em>
<em> print(num2**2)</em>
Answer:
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
Explanation:
The Worst case will happen when f(a) > 2*f(b) ; f(b) > 2*f(c) ; f(c) > 2*f(d) ; f(d) > 2*f(e) and f(e) > 2*f(f).
Where f(x) is frequency of character x.
Lets consider the scenario when
f(a) = 0.555, f(b) = 0.25, f(c) = 0.12, f(d) = 0.05, f(e) = 0.02 and f(f) = 0.005
Please see attachment for image showing the steps of construction of Huffman tree:- see attachment
From the Huffman tree created, we can see that endcoding e() of each character are as follows:-
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
e(e) = 11110
e(f) = 11111
So we can see that maximum length of encoding is 5 in this case.
