In a Regressive Round Robin Scheduler, each process is assigned a time quantum and a priority. Initially the value of the time quantum is 50 milliseconds. Every time a process has been allocated the CPU and uses its entire time quantum, i. e., does not block for I/O, 10 milliseconds is added to the time quantum and its priority level is boosted. The time quantum can be increased to a maximum of 100 milliseconds. When a process blocks before using its entire time quantum, its time quantum is reduced by 5 milliseconds and the priority is left unchanged. What type of process, i. e., CPU bound or I/0 bound, does this regressive round robin scheduler favor? Explain your answer.

Solve the following javafx application: write a javafx application that analyzes a word. the user would type the word in a text field, and the application provides three buttons for the following: - one button, when clicked, displays the length of the word.- another button, when clicked, displays the number of vowels in the word.- another button, when clicked, displays the number of uppercase letters in the word(use the gridpane or hbox and vbox to organize the gui controls).

Assume the existence of a bankaccount class. define a derived class, savingsaccount that contains two instance variables: the first a double, named interestrate, and the second an integer named interesttype. the value of the interesttype variable can be 1 for simple interest and 2 for compound interest. there is also a constructor that accepts two parameters: a double that is used to initialize the interestrate variable, and a string that you may assume will contain either "simple", or "compound", and which should be used to initialize the interesttype variable appropriately. there should also be a pair of functions getinterestrate and getinteresttype that return the values of the corresponding data members (as double and int respectively).

Use a while-end loop in a script file to calculate the sum of the first n terms of the series: summation sign on the top n bottom k=1 : )^k * k^2 +5k)/3^k show the script file and the two results of n = 10 and n = 20.