Answer:
Explanation:
There are various questions that you can ask in this scenario, such as
What grading policies are being implemented?
How many student grades are being calculated by the program?
What is the requirements for a student to pass?
All of these questions would allow you to get an idea of how extensive the code may be and its complexity. Once you know this you would look at the code revolving around actually looping through the data and doing the necessary calculations. You can then determine how to manipulate the code and make it much more efficient.
You would also need to know how the student data is being saved, which will help determine if it is the best data structure for saving this type of data or if it can be replaced in order to maintain the data secure while increasing the speed of the program. Mainly since this information needs to be continuously used from the data structure.
<span>Financial transactions were one of the first purposes of computers. This is where "ticker tapes" came from; early computers were used to keep track of how stocks were performing. Computers in early days used large mainframe systems and punch cards to keep track of this type of data.</span>
Answer:
Byte, the basic unit of information in computer storage and processing. A byte consists of 8 adjacent binary digits (bits), each of which consists of a 0 or 1. The string of bits making up a byte is processed as a unit by a computer; bytes are the smallest operable units of storage in computer technology.
Explanation:
Can I be brainliest? TYSMMMMMM
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
