What type of constructor appears below? public Student(int s_id, int s_age, double s_gavg) {
id = s_id;
age = s_age;
gradeAverage = s_gavg;
}

A. default
B. void
C. parameterized
D. public

How to print: number is equal to: 1 and it is odd number number is equal to: 2 and it is even number number is equal to: 3 and it is odd number number is equal to: 4 and it is even number in the console using java using 1 if statement, 1 while loop, 1 else loop also using % to check odds and evens

This question involves a class named textfile that represents a text file. public class textfile { private string filename; private string filename; private arraylist words; // constructors not shown // postcondition: returns the number of bytes in this file public int filesize() { } // precondition: 0 < = index < words.size() // postcondition: removes numwords words from the words arraylist beginning at // index. public void deletewords(int index, int numwords) { } // precondition: 0 < = index < = words.size() // postcondition: adds elements from newwords array to words arraylist beginning // at index. pub lic voidaddwords(int index, string[] newwords) { } // other methods not shown } complete the filesize() method. the filesize() is computed in bytes. in a text file, each character in each word counts as one byte. in addition, there is a space in between each word in the words arraylist, and each of those spaces also counts as one byte. for example, suppose the words arraylist stores the following words: { mary had a little lamb; its fleece was white as snow. } the filesize() method would compute 4 + 3 + 1 + 6 + 5 + 4 + 6 + 3 + 5 + 2 + 5 as the sum of the lengths of each string in the arraylist. the value returned would be this sum plus 10, because there would also be 10 spaces in between the 11 words. complete the filesize() method below: // postcondition: returns the number of bytes in this file public int filesize() { }

Create a function (prob3_6) that will do the following: input a positive scalar integer x. if x is odd, multiply it by 3 and add 1. if the given x is even, divide it by 2. repeat this rule on the new value until you get 1, if ever. your program will output how many operations it had to perform to get to 1 and the largest number along the way. for example, start with the number 3: because 3 is odd, we multiply by 3 and add 1 giving us 10. 10 is even so we divide it by 2, giving us 5. 5 is odd so we multiply by 3 and add one, giving us 16. we divide 16 (even) by two giving 8. we divide 8 (even) by two giving 4. we divide 4 (even) by two giving 2. we divide 2 (even) by 2 to give us 1. once we have one, we stop. this example took seven operations to get to one. the largest number we had along the way was 16. every value of n that anyone has ever checked eventually leads to 1, but it is an open mathematical problem (known as the collatz conjectureopens in new tab) whether every value of n eventually leads to 1. your program should include a while loop and an if-statement.