In main.py define the Student class that has two attributes: name and gpa Implement the following instance methods: • A constructor that sets name to "Louie" and gpa to 1.0 set_name(self, name) - set student's name to parameter name get_name(self) - return student's name set_gpa(self, gpa) - set student's gpa to parameter gpa get_gpa(self) - return student's gpa Ex. If a new Student object is created, the default output is: ● Louie/1.0 Ex. If the student's name is set to "Felix" and the gpa is set to 3.7, the output becom Felix/3.7 345892.2051704.qx3zqy7 LAB ACTIVITY 1 class Student: N36898 2 5 1.21.1: LAB: Student class 4 if _name_ == "__main__": 7 10 # Type your code here main.py initial_student = Student () print(initial_student.get_name(),'/', initial_student.get_gpa() initial_student.set_name("Felix') initial student.set_gpa(3.7) print(initial_student.get_name(), '/', initial_student.get_gpa

Give asymptotic upper and lower bounds for each of the following recurrences. Justify your answer.

Write a procedure sum-range (from to) for finding the sum of integers in the range from to to, inclusive. Make sure that (sum-range 4 18), (sum-range 18 4), and (sum-range 7 7) all give the expected result. Note that this function should be able to count up or count down depending on the order of the paramaters. So, (sum-range 4 18) and (sum-range 18 4) should give the same result. (sum-range 7 7) should return 7. (Dr.Racket)

1. Use the substitution method to verify that the upper bound solution to the recurrence: T(n) = 3T ([n/4])+(n²) is O(n²). Otherwise, if O(n²) is not the solution, find the correct solution to the above recurrence.

1. Use the substitution method to show the recurrence: T(n) = 4T(n/2) + (n) has solution T(n) = O(n²)