Answer:
public static void print_popcorn_time(int bag_ounces){
if(bag_ounces<3){
System.out.println("Too Small");
}
else if(bag_ounces>10){
System.out.println("Too Large");
}
else{
bag_ounces*=6;
System.out.println(bag_ounces+" seconds");
}
}
Explanation:
Using Java prograamming Language.
The Method (function) print_popcorn_time is defined to accept a single parameter of type int
Using if...else if ....else statements it prints the expected output given in the question
A complete java program calling the method is given below
public class num6 {
public static void main(String[] args) {
int bagOunces = 7;
print_popcorn_time(bagOunces);
}
public static void print_popcorn_time(int bag_ounces){
if(bag_ounces<3){
System.out.println("Too Small");
}
else if(bag_ounces>10){
System.out.println("Too Large");
}
else{
bag_ounces*=6;
System.out.println(bag_ounces+" seconds");
}
}
}
<span>If a thread is not finished running, perhaps because it had to wait or it was preempted, it is typically restarted on the same processor that previously ran it. This is this known as </span>processor affinity.
Answer:
Explanation:
The following code is written in Java. It creates the raiseToPower method that takes in two int parameters. It then uses recursion to calculate the value of the first parameter raised to the power of the second parameter. Three test cases have been provided in the main method of the program and the output can be seen in the attached image below.
class Brainly {
public static void main(String[] args) {
System.out.println("Base 5, Exponent 3: " + raiseToPower(5,3));
System.out.println("Base 2, Exponent 7: " + raiseToPower(2,7));
System.out.println("Base 5, Exponent 9: " + raiseToPower(5,9));
}
public static int raiseToPower(int base, int exponent) {
if (exponent == 0) {
return 1;
} else if (exponent == 1) {
return base;
} else {
return (base * raiseToPower(base, exponent-1));
}
}
}
ioq8oy because of the haoss
Explanation:
Answer:
Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list
Explanation:
Assuming that both lists have firs_t and last_ pointers.
For a singly-linked list ; when locating a kth element, you have iterate through a number of k-1 elements which means that locating an element will be done only in one ( 1 ) direction
For a Doubly-linked list : To locate the Kth element can be done from two ( directions ) i.e. if the Kth element can found either by traversing the number of elements before it or after it . This makes finding the Kth element faster because the shortest route can be taken.
<em>Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list </em>
