This is little bit tricky question.
Step-1:
n= number of vertices
n-1 = number of edges
Example: n=5 vertices and n-1=4 edges

Step-2: The above graph T won’t have cycle then we are calling as tree. Here, every pairs of vertices in T is connected by exactly one path.
Note: The above properties is nothing but minimum spanning tree properties.

We can also write (¬p → q) into (p ∨ q)

Here, we can minimize the boolean form into
¬p ∨ (p → q)
= ¬p ∨ (¬p ∨ q)
= ¬p ∨ q
In question, they are not discussed about the truth values of p, it implies that ¬p ∨ q also be True or False. So, we cannot be determined.

Option-A is happens when A and B are independent.
Option-B is happens when A and B are mutually exclusive.
Option-C is not happens.
Option-D is P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).

Here, there is no terminal symbol ‘a’, so the answer never be option B,C,D.
Option-A is also wrong because the string generated by grammar is bccccd.

The regular expression a+b denotes the set {a, b}.

Option-A: (100101100)​ ₂​ = (300)​ ₁₀
300 is divisible by 4
Option-B: (1110001110001)​ ₂​ = (7281)_{​ 10}
7281 is not divisible by 4
Option-C: (11110011)​ ₂​ = (243)​ ₁₀
243 is not divisible by 4
Option-D: (10101010101010)​ ₂​ = (10,922)_{​ 10}
10,922 is not divisible by 4.

→ The half adder adds two single binary digits A and B. It has two outputs, sum (S) and carry (C). The carry signal represents an overflow into the next digit of a multi-digit addition. The value of the sum is 2C + S. The simplest half-adder design incorporates an XOR gate for S and an AND gate for C. The Boolean logic for the sum (in this case S) will be A′B + AB′ whereas for the carry (C) will be AB.

⇒ a=a⊕b
⇒ b=a ⊕b
= a⊕b ⊕b (Substitute (1))
= 0 ⊕ b
= b
⇒ a = b⊕a
= b⊕a⊕b (Substitute (1))
= a⊕b⊕b
= a⊕0
= a

Method 1:
(X+Y) + (XY)’
= X+Y+X’+Y’
= (X+X’)+(Y+Y’)
= 1
Method 2:

Given question is ambiguous. It is not mentioned how the inverter is connected. Placing NOT gate in different positions gives different solutions.

The sizeof operator will print number of bytes of a data type.
sizeof(3.14f) → It will consider as float data type. The float data type size is 4 bytes.
sizeof(3.14) → It will consider as double data type. The double data type size is 8 bytes.
sizeof(3.141) → It will consider as double data type. The double data type size is 8 bytes.
Output= 4 8 8
Note: The exact size of each of these 3 types depends on the C compiler implementation (or) platform

Step-1: First we have to convert into binary numbers
(35)​ ₁₀​ = (​ 0010 0011​ ) ​ ₂
(7)​ ₁₀​ = (0000 0111)​ ₂
Step-2: Perform OR operation on both binary numbers.
0010 0011
0000 0111
---------------
0010 0111
----------------
Step-3: Convert result into Decimal number
(0010 0111)​ ₂​ = (39)_{​ 10}