answer:Lisa has 10 friends and X marbles. She wants to give each friend at least one marble and no two friends should receive the same number of marbles. To find the minimum number of additional marbles Lisa needs, we can consider the worst-case scenario where each friend receives a different number of marbles. In this scenario, the first friend receives 1 marble, the second friend receives 2 marbles, the third friend receives 3 marbles, and so on, up to the tenth friend who receives 10 marbles. The sum of the marbles given to each friend in this scenario can be calculated using the formula for the sum of an arithmetic series: Sum = (First term + Last term) * Number of terms / 2 In this case, the first term is 1, the last term is 10, and the number of terms is 10. So, the sum of the marbles given to each friend is: (1 + 10) * 10 / 2 = 55 To ensure that Lisa can give each friend at least one marble and no two friends receive the same number of marbles, she needs to have at least 55 marbles. Given that the minimum number of additional marbles she needs is 21, we can write: X + 21 = 55 Subtracting 21 from both sides of the equation, we find: X = 34 The value of X is 34. The answer is: 34

answer:```python import random def roll_dice(): Simulate the rolling of two dice and return the outcome. return random.randint(1, 6) + random.randint(1, 6) def estimate_probabilities(num_rolls): Roll the dice many times and estimate the probability of each outcome. outcomes = [0] * 11 # Possible outcomes are 2 through 12 for _ in range(num_rolls): outcome = roll_dice() outcomes[outcome - 2] += 1 # Adjust index to 0-based probabilities = [count / num_rolls for count in outcomes] return probabilities def print_probabilities(probabilities): Print the estimated probability of each outcome. for i, probability in enumerate(probabilities): print(fOutcome {i + 2}: {probability:.3f}) def main(): num_rolls = 1000 probabilities = estimate_probabilities(num_rolls) print_probabilities(probabilities) if __name__ == __main__: main() ``` This program defines a function `roll_dice` to simulate the rolling of two dice, a function `estimate_probabilities` to estimate the probability of each outcome by rolling the dice many times, and a function `print_probabilities` to print the estimated probabilities. The `main` function sets the number of rolls to 1000, estimates the probabilities, and prints the results. You can adjust the number of rolls by changing the `num_rolls` variable in the `main` function. The more rolls you perform, the closer the estimated probabilities will be to the actual probabilities.

answer:```python def longest_word(s): Return the longest word in a given string. words = s.split() longest = max(words, key=len) return longest # Test cases print(longest_word(I am the longest word)) # Output: longest print(longest_word(one two three four five six seven eight nine ten)) # Output: Not seven, but actually eight and seven have the same length, so the output depends on the order. ``` This function works by splitting the input string into words using the `split` method, then using the `max` function with the `key` argument set to `len` to find the longest word. If there are multiple words with the same maximum length, the `max` function will return the first one it encounters. However, this function will return the first word with the maximum length, not necessarily the first word that appears in the original string if there are multiple words with the same maximum length. To fix this, we can use the following function: ```python def longest_word(s): Return the longest word in a given string. words = s.split() max_len = max(len(word) for word in words) for word in words: if len(word) == max_len: return word # Test cases print(longest_word(I am the longest word)) # Output: longest print(longest_word(one two three four five six seven eight nine ten)) # Output: seven ``` This function first calculates the maximum length of any word in the string, then iterates over the words in the string and returns the first word that has the maximum length. This ensures that the function returns the first word with the maximum length, as required.

answer:Here is a Python implementation of the Tower of Hanoi problem using recursion: ```python def tower_of_hanoi(n, source, auxiliary, destination): if n == 1: print(fMove disc 1 from {source} to {destination}) return tower_of_hanoi(n - 1, source, destination, auxiliary) print(fMove disc {n} from {source} to {destination}) tower_of_hanoi(n - 1, auxiliary, source, destination) def main(): n = 3 tower_of_hanoi(n, 'rod1', 'rod2', 'rod3') if __name__ == __main__: main() ``` This implementation uses recursion to solve the problem. The function `tower_of_hanoi` takes four parameters: the number of discs `n`, and the names of the source, auxiliary, and destination rods. The base case for the recursion is when there is only one disc to move, in which case we simply print the move. For more than one disc, we recursively move `n - 1` discs from the source to the auxiliary rod, using the destination rod as a temporary storage. Then we move the nth disc from the source to the destination rod. Finally, we recursively move the `n - 1` discs from the auxiliary rod to the destination rod, using the source rod as a temporary storage. The `main` function sets the number of discs to 3 and calls the `tower_of_hanoi` function with the names of the rods. When you run this program, it will print the moves required to solve the Tower of Hanoi problem for 3 discs. You can adjust the number of discs by changing the value of `n` in the `main` function. Note that the number of moves required to solve the Tower of Hanoi problem is exponential in the number of discs. Specifically, it is `2^n - 1`, where `n` is the number of discs. Therefore, this program will print a large number of moves for larger values of `n`.