Solution

Fourth proportional to 4, 14, 24 be x.

Then, 4:14:24:x

4x=24*14

x=84

Solution

The numbers are 31 x 13 and 31 x 16

∴ Larger number =31 x 16=496

Solution

LCM of 2, 3, 4, 5, 6 and 8 is 120 (2 minutes)

In 44 minutes, they will toll together =44/2+1=23 times

Solution

Solution

Solution

Solution

Succulent means tender and Juicy. Cactus is a type of succulent plant.

The opposite of succulent is dry and hard, which can be seen in wood.

Solution

Offender is a person who commits an illegal act.

Bystander is a person who is present at an event but does not take part.

Defender is a person who protects.

Mender is a person who repairs things.

Solution

“Upon” and “on” are both used as prepositions. Prepositions are used to express the relationship of a noun and a pronoun in the sentence with another word. They can be used interchangeably in many cases.

Solution

‘Meerabai stopped the practice of animal sacrifice and challenged the custom of ‘Sati’. Despite being a young princess who grew up in a male dominated society, she showed how she could resolve matters.’

Solution

Explanation:

Step 1:

Preorder:

Inorder:

Step 2:

Preorder:

Inorder:

Step 3:

Preorder:

Inorder:

Step 4:

Preorder:

Inorder:

Solution

Hashing is an important Data Structure which is designed to use a special function called the Hash function which is used to map a given value with a particular key for faster access of elements. The efficiency of mapping depends on the efficiency of the hash function used.

Solution

Due to the nature of these algorithms, in the case of a graph with moderate number of edges, we should use Kruskal's algorithm. Prim's algorithm runs faster in a graph with many edges as it only compares limited number of edges per loop, where as Kruskal's start by sorting all the edges in the list then going through them again to check if the edge is part of the minimal spanning tree (MST) or not. Hence although both algorithms have similar running time, number of edges in the graph should be the main component when you are deciding between the algorithms. More edges compared to vertices, use Prim's, otherwise, use Kruskal's.

Solution

After adding all edge weights of every given circuit, The option C forms a negative weight cycle.

Option A: B-A, A-F, F-D, A-D, A-C, C-B → 30+(-17)+16+12+(-15)+(-16) → 10

Option B: B-A, A-C, C-B → 30+(-15)+(-16) → -1

Option C: B-A, A-F, F-D, D-C, C-B → 30+(-17)+16+(-16)+(-16) → -3

Option D: B-A, A-D, D-C, C-B → 30+12+(-16)+(-16) → 10

Solution

The Floyd Warshall algorithm is used to find all the pairs of shortest distances in a graph using dynamic programming.

Solution

Two possible Shortest paths:

Solution

Given elements : 16, 25, 28, 37

(x+1) mod 6

(16+1) mod 6 = 17 mod 6 → Slot 5 [ No collision ]

(25+1) mod 6 = 26 mod 6 → Slot 2 [ No collision ]

(28+1) mod 6 = 29 mod 6 → Slot 5 [ Collision ]

(37+1) mod 6 = 38 mod 6 → slot 2 [ Collision ]

2x mod 6

2*16 mod 6 → 32 mod 6 → Slot 2 [ No collision ]

2*25 mod 6 → 50 mod 6 → Slot 2 [ Collision ]

2*28 mod 6 → 56 mod 6 → Slot 2 [ Collision ]

2*37 mod 6 → 74 mod 6 → slot 2 [ Collision ]

x mod 6

16 mod 6 → slot 4 [ No collision ]

25 mod 6 → slot 1 [ No collision ]

28 mod 6 → slot 4 [ Collision ]

37 mod 6 → slot 1 [ Collision ]

(2x+1) mod 6

(2*16+1) mod 6 → 33 mod 6 → Slot 2 [ No collision ]

(2*25+1) mod 6 → 51 mod 6 → Slot 2 [ Collision ]

(2*28+1) mod 6 → 57 mod 6 → Slot 2 [ Collision ]

(2*37+1) mod 6 → 75 mod 6 → slot 2 [ Collision ]

Solution

Given elements : 32,56,87,23,65,26,93

x mod 7

32 mod 7 → 4

56 mod 7 → 0

87 mod 7 → 3

23 mod 7 → 2

65 mod 7 → 2 [ collision ]

(65 mod 7)+1 → 3 [ collision ]

(65 mod 7)+2 → 4 [ collision ]

(65 mod 7)+3 → 5 [ No collision ]

26 mod 7 → 5 [ collision ]

(26 mod 7)+1 → 6 [ No collision ]

The final content is given below table and index starts from 0.

Input array for Bubble sort 56, 93, 23, 87, 32, 65, 26

After first pass: 56, 23, 87, 32, 65, 26, 93

Except 56 elements every element has changed their order.

So, the final answer is 6.

Solution

If D_1,D_2,…,D_n are domains in a relation model, then the relation is a table, which is a subset of D_1×D_2×…×D_n.

Solution

Minimum no. of tables needed in 2 as the relationship between and is , it is and total participation. So, we can merge A and B into 1 task and to entity C, 1 table is required. Hence, minimum no. of tables required is 2.

Solution

FD’s from given diagram:

Emp_ID→Name,Dept_Name,Salary

Emp_ID,Course_Title→Data_Completed

(Emp_ID, Course_Title)+ = Emp_ID, Course_Title, Name, Dept_Name, Salary, Date_Completed. So, the Candidate key is (Emp_ID, Course_Title).

Part of the key (Emp_ID) is determining Non-prime attributes, Name, Dept_Name, Salary. So, it is a partial dependency.

Hence, relation EMPLOYEE is in 1NF but not in 2NF.

Solution

The schema T(P, Q, R, S) is decomposed into T1 (P, Q) and T2(R, S) and there are only two FDs P -> Q and R -> S. So, the decomposition is dependency preserving

T1 ∩ T2 is empty. So, the decomposition is not lossless.

Solution

Ordinal data has a defined order (e.g., Poor < Fair < Good < Excellent), whereas nominal data like country names or employee IDs do not have a meaningful order.

Solution

The z-score formula is: z = (X - µ) / σ = (40 - 30) / 14.14 = 10 / 14.14 ≈ 0.7.

Solution

If the first 5 components explain 85% of the variance, the remaining 95 components explain 100% - 85% = 15%.

Solution

The main advantage of hierarchical clustering is that it allows for the determination of the number of clusters after the analysis, as it constructs a dendrogram without requiring a predefined number of clusters.

Solution

Finding the shortest path between a pair of nodes in a graph

Solution

Increasing the number of hidden units in a fully-connected deep network with one hidden layer allows the model to learn more complex patterns in the training data, which typically reduces bias. However, this increased capacity also makes the model more sensitive to fluctuations in the training data, leading to higher variance and a greater risk of overfitting.

Solution

Classifying Spotify users based on their listening history

Weather forecast using data collected by a satellite

Predicting tuberculosis using patient's chest X-Ray

Solution

Number of weights from input to hidden layer = 3×4 = 12 Bias term for the four neurons in hidden layer = 4 Number of weights from hidden to output layer = 4 Bias term in the final output layer = 1 Summing the above = 21

Solution

Models which overfit have a high bias

Models which underfit have a high variance

Solution

LDA reduces the data to C−1 dimensions, where C is the number of classes.

Solution

As the number of evidence variables increases in rejection sampling, the efficiency typically decreases exponentially. This is because the probability of generating a sample that satisfies all the evidence constraints becomes significantly smaller, leading to a dramatic reduction in the number of accepted samples. Consequently, more samples need to be generated to obtain a sufficient number of valid samples, making the process less efficient as the number of evidence variables grows.

Solution

These are networks that contain cycles. In the context of graphical models, loops make inference more complex because variables can influence each other through multiple paths.

Solution

In the likelihood weighting method, when evidence is consistent with a given sample, the weight of that sample is adjusted based on the likelihood of the evidence given the sample. As the number of evidence variables increases, the likelihood of randomly generating a sample that aligns with all the evidence decreases. This means more samples will be assigned low or zero weights. Consequently, the variance among the weights increases.

Solution

We have Boolean function: [~(~p ∧ q) ∧ ~( ~p ∧ ~q)] ∨ (p ∧ r)] = [(p ∨ ~q) ∧ (p ∨ q ) ∨ (p ∧ r)] = [p ∨ (p ∧ q) ∨ (p ∧ ~q) ∨(p ∧ r)] = p[1 ∨ q ∨ ~q ∨ r] = p

Solution

Statement S1: True. A heuristic is considered admissible if it never overestimates the true cost to reach the goal from any given state. This ensures that the heuristic is optimistic and will lead to an optimal solution in algorithms like A*.

Statement S2: False. A heuristic is monotonic (or consistent) if it satisfies the triangle inequality, meaning that the estimated cost from the current node to the goal should not exceed the cost from the current node to a neighboring node plus the estimated cost from the neighboring node to the goal. While monotonicity implies admissibility, it is not accurate to define monotonicity solely based on the triangle inequality without mentioning the relationship to the heuristic values at neighboring nodes.

Solution

i.e. the mean value is 21.9 ohms,

Solution

P(A) = 0.5 and P(B) = 0.8

Solution

Solution

Given:

‘X’ is a normal random variable with mean 1 and

variance 4 ⇒ X is N(1, 4)

⇒ X = 0 lies the left of the mean X=1,

P(X) for the given variable is as shown below

⇒ P (X < 0) = 0.5 – P (0 < X < 1 )

⇒ 0 < P (X < 0) < 0.5

Solution

Solution

The probability of 1 breakdown per week is 0.2572

The probability of less than 3 breakdowns per week is the sum of the probabilities of 0, 1, and 2 breakdowns per week,

i.e. 0.1225 + 0.2572+ 0.2700, i.e. 0.6497

Solution

The probability of having a 4 upwards 3 times is 0.1302

Solution

Given:

Solution

Solution

Solution

Solution

For the inverse of matrix to exist, the matrix should be

non-singular

(i.e.) determinant is non-zero

∴ I and II are equivalent statements

Solution

By Cramer’s Rule

Solution

Solution

Given:

Add -1 times row 1 to row 2 and add -2 times row 1 to row 3

Add row 2 to row 3

In terms of elementary matrices, we have just shown that

Rearrange the equation to get

Combine the matrices as in the proof of the LU decomposition to find

Solution

We have T : R3 → R3 . We write vectors x ∈ R3 as columns

Recall the standard basis

We have

So, the standard matrix of T is

Solution

Solution

Solution

Solution

Solution

Hence, C is the correct option.

Solution

Solution

The correct mode for opening a file for reading in Python is “r”

Solution

The finally block is executed even though there is a return statement in the try block.

Solution

In this approach, we perform an inorder traversal of the BST, which gives us the elements in sorted order.

The function inorder() traverses the left subtree, then appends the node’s value to the result list, and then traverses the right subtree. Once the traversal is completed, the elements are in increasing order.

The k-th smallest element is simply the (k-1)-th index in the result list (as the list is 0-indexed).

Option B is incorrect because it doesn't handle the return value properly for the k-th smallest element; the count is not updating as intended.

Option C uses a stack and performs an iterative inorder traversal, but it doesn't handle the logic for finding the k-th smallest element correctly.

Option D uses the wrong approach, performing the traversal in the right direction, which doesn't guarantee the elements will be ordered in increasing fashion.

Solution

The find_anagrams function uses two lists p_count and s_count to store the frequency of characters in p and the sliding window of s.

The sliding window moves through the string s, updating s_count to track the frequencies of characters in the current window.

When the window size reaches the size of p, it compares the s_count with p_count. If they are equal, it means that the substring is an anagram of p, and the start index is added to the result.

For Option A, the string s = "cbaebabacd" contains anagrams of p = "abc" starting at indices [0, 6], which is the correct output.

Solution

The longest_increasing_subsequence function uses dynamic programming to find the LIS.

It initializes a DP array where each index represents the length of the longest subsequence ending at that index.

For each pair of indices, if nums[i] > nums[j], it updates the DP array to reflect the longer subsequence.

The final answer is the maximum value in the DP array. In this case, for the input [10, 9, 2, 5, 3, 7, 101, 18], the LIS is [2, 5, 7, 101], which has a length of 4.

Hence, Option A is correct.

Solution

Finding the shortest path between a pair of nodes in a graph