→ Sign bit S = 0 (It means positive number)
→ E=1000 0000B = 128D (in normalized form)
→ Fraction is 1.11B (with an implicit leading 1) = 1 + 1×2^-1 + 1×2^-2
= 1.75D
→ The number is +1.75 × 2^(128-127)
= +3.5D

In 2’s complement numbers, the range of integers are from -2^n-1 to 2^n-1 – 1

Number of RAM chips required = Memory capacity / RAM chip capacity.
=(256K * 8) / (32K x 1)
=(256*1024*8) / (32*1024*1)
=64

A ring counter is a type of counter composed of flip-flops connected into a shift register, with the output of the last flip-flop fed to the input of the first, making a "circular" or "ring" structure.
The “MODULO” or “MODULUS” of a counter is the number of states the counter counts or sequences through before repeating itself and a ring counter can be made to output any modulo number. A “mod-n” ring counter will require “n” number of flip-flops connected together to circulate a single data bit providing “n” different output states
So, modulus-12 requires , 12 flip-flops.

→ Finding complement boolean expression, we require Demorgan’s law.
→ DeMorgan’s law and after complementing:
⇒ ( AB ( B’C + AC ))’
⇒ (A + B)’ + (B’C + AC)’
⇒ (A’ + B’) + ( B’C)’ (AC)’
⇒ (A’ + B’) + ( B + C’)(A’ + C’)

→ ISO/IEC 646, like ASCII, is a 7-bit character set. It does not make any additional codes available, so the same code points encoded different characters in different countries. Escape codes were defined to indicate which national variant applied to a piece of text, but they were rarely used, so it was often impossible to know what variant to work with and, therefore, which character a code represented, and in general, text-processing systems could cope with only one variant anyway.
→ Extended Binary Coded Decimal Interchange Code (EBCDIC) is an 8-bit binary code for numeric and alphanumeric characters.
→ BCD encoding uses 4 bits to represent each digit from the range 0 to 9 in its binary form.
→ In case of Gray codes, any number of bits can be used to represent a character, according to the requirement.

1. An adder is a digital circuit that performs addition of numbers. The half adder adds two binary digits called as augend and addend and produces two outputs as sum and carry; XOR is applied to both inputs to produce sum and AND gate is applied to both inputs to produce carry.
2. The full adder adds 3 one bit numbers, where two can be referred to as operands and one can be referred to as bit carried in. And produces 2-bit output, and these can be referred to as output carry and sum.
Half adder is used to add two numbers of the least significant bits, so one half adder is required. In order to add remaining 16 bits of two numbers , we require 16 full adders

Given function f = A’B’C + A’B’C’
=A’B’(C+C’) (C+C’=1)
=A’B’
=(A+B)’
So we can implement with minimum Two number of 2x1 multiplexers

Binary expression of (3*4096 + 15*256 + 5*16 + 3)
=(12,288+3840+80+3)
=(16211)₁₀
=‭(0011111101010011‬)₂
Total number of 1’s in binary representation is 10.

→ AB + AB' + A'C + AC
= A(B + B') + C(A + A')
= A+C
As the expression is independent of 'B'

Final Pop sequence: 22112

We need to find the minimum time to sort the names by using quick sort.
The Worst case time complexity is O(n²)
Average and best case time complexity is O(n logn)
Step-1: Time taken to sort 1000 names by quicksort
100= c*nlogn
= c*1000log1000
= 100
c=1/10log1000
Now, time taken to sort 100 names by quicksort,
=c*nlogn
= (1/10log1000)*100*log100
= 6.7

Optimal merge pattern will give the optimal result after performing sorted order. Every time it will take least frequency elements and performing merging.
→ Final order is F3, F4, F1, F5, F6, F2

In this, maximum size of cluster = 4 (S6, S3, S7, S1)
→ Worst case of finding a number is equal to maximum size of cluster + 1(after searching all the cluster it enters into empty cluster)
→ Maximum no. of comparisons = 4+1
= 5

Step-1: Pop elements from Stack-1 and push into Stack-2.
For this,
int x=element=stack-1,pop();
stack-2.push(element);

Step-2: Once the complete stack-1 gets copied to Stack-2, then we can simply call pop() on s2, it will remove the element-1.
So, A queue can be implemented using 2 stacks.

Given diagram is ER diagram , So any object, for example, entities, attributes of an entity, relationship sets, and attributes of relationship sets, can be represented with the help of an ER diagram.
We need to convert ER diagram to relational model.
There are two entities in the ER diagram named Person and Exam with attributes which are represented in ellipse.
Table name: Person

Now Create table for a relationship
Add the primary keys of all participating Entities as fields of table with their respective data types.
If relationship has any attribute, add each attribute as field of table.
Declare a primary key composing all the primary keys of participating entities.
Declare all foreign key constraints.
Table for “Qualification” relationship

→ When the database system crashes after the transactions have committed then we need to redo the log records.
→ And if the database system crashes before the transactions have committed then we need to undo the log records.
So from above theory we can say that option (B) is the correct Solution.

Total number of records in the data base=n
Given block will hold either 3 records or 10 key pointers.
Number of blocks to store “n” records = n/3
Number of blocks to store dense file index = n/10
Total blocks to hold data file and dense index = n/3 + n/10 = 13n/30 blocks.

JavaScript can "display" data in different ways:
Writing into an HTML element, using innerHTML.
Writing into the HTML output using document.write().
Writing into an alert box, using window.alert().
Writing into the browser console, using console.log().

Here, simple way to solve this question is,
First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.