11. fruits=['apple','banana','pear'],print(fruits[-1][-1])result is _______________， print(fruits.index('apple')) result is _______________

def rectangle_overlap(rect1_bl_x,rect1_bl_y,                      rect1_tr_x,rect1_tr_y,                      rect2_bl_x,rect2_bl_y,                      rect2_tr_x,rect2_tr_y):    """    (int,int,int,int,int,int,int,int) -> str        Function determines whether two rectangles overlap. When rectangles    overlap, the function checks for the following scenarios        1. The two rectangles share the same coordinates        2. The first rectangle is contained within the second        3. The second rectangle is contained within the first        4. The rectangles have overlapping area, but neither is completely           contained within the other        Function inputs represent x and y coordinates of bottom left and top right    corners of rectangles (see lab document)               The function return one of the following strings corresponding to the     scenario        "no overlap"        "identical coordinates"        "rectangle 1 is contained within rectangle 2"        "rectangle…

15. Code to Recurrence Relation What is the recurrence relation of the runtime of the following algorithm: T(n) represents the time it takes to complete func called with the parameter n value n. def func(n): if n == 0: return print(n) func(n//2) func(n//3) Pick ONE option T(n) = T(n-1) + T(n-2) + C; T(1) = C T(n) = T(n-2) + T(n-3) + C; T(1) = C T(n) = T(n/2) + T(n/3) + C; T(1) = C T(n) = T(n-5) + C; T(1) = C Clear Selection 81

Student students[3]; for(int i=0;i<3;i++) fillStruct(&students[i], records[i]); for (int i = 0; i < 3; i++) { printf("Student Name: %s, ",students[i].name); printf("Student ID: %s, ",students[i].id); printf("DOB: %s, ",students[i].ddmmyyyy); printf("City of DOB: %s\n",students[i].cityBirth); } %3D } Task3.1: the program above calls the function fillstruct and provides as arguments one record entry of a student and a pointer of a struct. The function must fill the struct with information of the record. However, for the date entry, the date is provided in a format that needs to be changed: instead of YYYY-MM-DD the date should be filled in the struct as DD- MM-ҮҮҮҮ. Define the function that does the above task void fillStruct(struct studentStruct *std, char currentStd[4][5®]);

13. While dictionaries are designed so that we can find a value based on a key, every so often we need to perform a reverse lookup. For example, imagine a dictionary where the keys are names and the values are phone numbers. Usually we want to find the phone number based on the name, but occasionally you might need to find a name based on the phone number. Write a function called reverse_lookup(dictionary, value) that returns all the keys in the dictionary associated with the specified value. Note: In dictionaries, the keys are guaranteed to be unique, but the values are not. So your function should return a list containing all of the matching keys (if any) Example: Given this dictionary, groups = {'Apple': 'Fruit', 'Spinach': 'Vegetable', 'Banana': 'Fruit'} reverse_lookup (groups. Vegetable') ["Spinach'] reverse_lookup (groups, Fruit') ['Apple'. 'Banana'] reverse_lookup (groups. 'Meat') []

What is the value of the median-of-3 for the following list. [80, 47, 41, 21, 40, 68, 78, 18, 75, 46, 30] Notes: • Your answer should be a single valid, non-negative, literal Python int value. For example, 123 is a valid int literal. • This question is asking about before any partition operation is carried out. • This is not asking for the median-of-3 index. • You can pre-check your answer (to check it's a valid, non-negative, literal int).

# dates and times with lubridateinstall.packages("nycflights13") library(tidyverse)library(lubridate)library(nycflights13) Qustion: Create a function called date_quarter that accepts any vector of dates as its input and then returns the corresponding quarter for each date Examples: “2019-01-01” should return “Q1” “2011-05-23” should return “Q2” “1978-09-30” should return “Q3” Etc. Use the flight's data set from the nycflights13 package to test your function by creating a new column called quarter using mutate()

Find the body for:def find_first_occurrence(msg: str, string_to_find: str) -> int:    """Return the index of the first occurence of string_to_find    in msg, or -1 if it does not occur, ignoring case.        >>> find_first_occurrence('October holidays: Halloween and Thanksgiving', 'H')    8    >>> find_first_occurrence('Because DEC 25 == OCT 31', 'H')    -1    >>> find_first_occurrence('hip hip hooray', 'ip')    1    """

FIX THE PROGRAM BELOW.     #Function to get the key value ind dictionary def get_key(val):     for key, value in result.items():          if val == value:              return key       return "key doesn't exist"   #Function definition to find the tallest height def tall(list_height):          return max(list_height)   #Function to find the shortest height def short(list_height):          return min(list_height)     # creating an empty list list_names list_names = []    # Get the number of baby names as input from user num1 = int(input("Enter number of baby names : "))    # iterating till the range for i in range(0, num1):     name = input()     list_names.append(name) # adding the element        print("list of baby names : ",list_names)   # creating an empty list list_height list_height = []    # iterating till the range for i in range(0, num1):     height = int(input())     list_height.append(height) # adding the element        print("list of baby heights : ", list_height)   # using…

3 4 5 5 7 3 9 3 1 2 B 4 5 7 9 0 2 3 4 5 8 0 1 def calculate_trip_time( iata_src: str, iata_dst: str, flight_walk: List[str], flights: FlightDir ) -> float: Return a float corresponding to the amount of time required to travel from the source airport to the destination airport to the destination airport, as outlined by the flight_walk. PERBE The start time of the trip should be considered zero. In other words, assuming we start the trip at 12:00am, this function should return the time it takes for the trip to finish, including all the waiting times before, and between the flights. If there is no path available, return -1.0 >>> calculate_trip_time("AA1", 2.0 >>> calculate_trip_time("AA1", >>> calculate_trip_time("AA1", >>> calculate_trip_time("AA1", 0.0 >>> calculate_trip_time("AA4", 14.0 >>> calculate_trip_time("AA1", 7.5 >>> calculate_trip_time("AA1", 2.0 *** "AA2", ["AA1", "AA2"], TEST_FLIGHTS_DIR_FOUR_CITIES) "AA7", ["AA7"] "AA1"], TEST_FLIGHTS_DIR_FOUR_CITIES) "AA7", ["AA1",…