Fare: The money paid for a journey. Fair: treating people equally without favouritism or discrimination (or) without cheating (or) trying to achieve unjust advantage The fare is too high for it to be considered fair.

We know that area of a rectangle = length * breadth = L * B
Area of rectangle (1) = L*B = (1/3) * 2 = (2/3)
Area of rectangle (2) = L*B = (3/4 - 1/3 ) * 3 = (5/4 )
Area of rectangle (3) = L*B = (1 - 3/4) * 1 = 1/4
Therefore Total area = Area(1+2+3) = (2/3) + (5/4) + (1/4) = 26/12 = 13/6

Explanation: Statement 1 & Statement 3 cannot be true simultaneously.
Statement 1: All students are inquisitive. Statement 3: No student is inquisitive

From the question we know that letters can rotate in the plane, now we will verify the options one by one. Option 1: A,D,D,D,J from these letters we can't form a palindrome.
Option 2: A,D,R,A,R from these we can form RADAR which is a palindrome.
Option 3: A,D,Z,E,D from these letters we can't form a palindrome.
Option 4: A,D,I,L,Y from these letters we can't form a palindrome.

From the given passage consider the sentence “In organizations with good work culture, metrics help improve outcomes” Option A: Metrics are useful in organizations with poor work culture. Which is incorrect based on the sentence.
Option B: Metrics are useful in organizations with good work culture. Which is correct and relevant to the considered sentence.
Option C: Metrics are always counterproductive in organizations with good work culture. Which is incorrect because we can't always guarantee good work culture.
Option D: Metrics are never useful in organizations with a good work culture. Incorrect metrics will help to make a good work culture.

The girls accounted for 60% of the qualified candidates.
60% can be written as ⅗. Which means among 5 qualified persons three of them are girls.
So, The number of boys qualified is less than the number of girls qualified.

Case 1: The first ball is Green
Probability of first ball is green = 3/5
If the first ball is green then we are not replacing so the probability of the second ball being orange is = 2/4.
Probability of first ball green and second ball orange is =3/5 * 2/4 = 6/20
Case 2: The first ball is orange
Probability of first ball is orange =2/5
If the first ball is orange then we are replacing it with another orange ball/ so the number balls will remain the same.
Probability of second ball is orange = 2/5
Probability of first ball is orange and second ball is also orange = 2/5*2/5 = 4/25
Therefore, Total probability = (6/20 + 4/25) = 23/50.

From the given information draw the image, such that
The Image is satisfying all given conditions. From this we can say that S cannot be placed at the corner. By satisfying all given conditions we can draw in different ways but in all cases S cannot be placed at the corner.

Given Conditions are Shape is to be similar, area need not to be same, A rope can pass through three poles.

Therefore least number of additional ropes required = 3.

constant < logarithmic < linear < polynomial < exponential < factorial
f(n)=n^c where c is a constant
f(n^2) = (n^2)^c = n^2c
(f(n))^2 = (n^c)^2 = n^2c
f(n^2) = (f(n))^2 is TRUE asymptotically.
Option-B: FALSE: The small omega function indicates the tightest upper bound.
f(n)^2 < o(f(n)^2) is FALSE asymptotically.
Option-C: FALSE: Consider f(n)=logn
f(n^2)=logn^2 = 2*logn
(f(n))^2 = (logn)^2 = logn * logn
f(n^2) <=Ω(f(n))^2 is FALSE asymptotically.
Option-D: FALSE:
Exponential values(Eg: 2^n, 3^n,..,k^n where k is a constant ”
f(n)=3^n
f(n^2)=f(3^(n^2)
(f(n))^2 = (3^n)^2 = 3^2n
f(n^2) >= O(f(n))^2 is FALSE asymptotically.

Regular expression ab*bab* +ba*aba* does not generate the string “abb” which is accepted by FA hence this is not true.
(ab*b)*ab*+(ba*a)*ba* does not generate the string “abbaa” hence this is also wrong.
(ab*b+ba*a)*(a*+b*) generates epsilon hence this is also wrong.
(ab*b+ba*a)*(ab*+ba*) is the correct regular expression.
GATE 2022 Computer Science and Information Technology (CS)

Even though there is no reduce-reduce conflict in CLR(1) but in LALR(1) while merging the states differ in only lookahead may get reduce-reduce conflict. So the given statement is not true.
Symbol table is accessed in all the phases of compiler and not only in lexical analysis phase.
Data flow analysis is done in the control flow graph, in the code optimization phase. If LR(1) parses a grammar then definitely it is DCFL, so LR(1) parsing is sufficient for deterministic context-free languages.

A relation with only two attributes will always be in BCNF.
Example:
R(A, B).
Two functional dependencies possible for the relation: (1) A->B and (2) B->A
If there is no functional dependency, we can assume trivial functional dependencies like AB->A and AB->B.
In all cases, functional dependencies like A->B, A must be a key.
So they all will be in BCNF irrespective of the functional depencies set.

/* Link list node */
struct Node {
int data;
struct Node* next;
{
this->data = data;
next = NULL;
}
};
struct LinkedList {
Node* head;
LinkedList() { head = NULL; }
/* Function to reverse the linked list */
void reverse()
{
// Initialize current, previous and
// next pointers
Node* current = head;
Node *prev = NULL, *next = NULL;
while (current != NULL) {
// Store next
next = current->next;
// Reverse current node's pointer
current->next = prev;
// Move pointers one position ahead.
prev = current;
current = next;
}
head = prev;
}
For the above algorithm :
Time Complexity: O(n)
Space Complexity: O(1)

For n keys and m elements without any condition the expected number of elements in a slot are n/m.
How?
For 1st element the probability of key1 ends up in slot 1 is 1/m.
For 2nd element the probability of key2 ends up in slot 2 is 1/m
..
..
For nth element the probability of keyn ends up in slot n is 1/m
Hence expected number of elements in a slot is:
1/m+1/m+... (n times)= n/m
In the given question h1 is for elements at even sequence position and h2 is for odd number of sequence positions. This will not affect the overall probability.
Hence here also it is n/m