Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
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Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
The engineering design process important to the customer layout system permits things: A shared technique on a way to discover and remedy problems. A framework to assess ideas, tell priorities and save you myopic paintings.
<h3>What is the maximum vital step withinside the engineering layout system?</h3>
The first step withinside the engineering layout system is to outline the trouble that you are solving. This step is the maximum vital, as a defective know-how of the trouble handy ends in layout failures. Engineers paintings to discover the foundation trouble via way of means of opposite engineering the signs and symptoms they perceive.
The engineering layout system, via way of means of definition, is a sequence of steps that engineers use to discover an option to a trouble. These steps include: defining the trouble, brainstorming solutions, designing and constructing a prototype of the solution, trying out the solution, and enhancing it.
Read more bout the engineering :
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Answer:
Overall, the main difference between the courses is the fact that AP classes require an AP exam, while Dual Enrollment classes just require a passing grade. This causes colleges to favor AP classes because they know exactly what you learned based on what you knew on the AP exam.
Answer:
The answer is "Option A".
Explanation:
Please find the complete program in the attached file.
The given program includes a super-class Bike having a private integer numWheels parameter. The class EBike inherits the class Bike, EBike comprises one parameter function Object() which utilizes its variable number of the private integer Battery level of the battery parameter and the wrong choice can be defined as follows:
In choice B, it is wrong since there is no constructor with a single argument in the Bike class.
In choice C, it is wrong since there no need to call the base class constructor with the super keyword.
In choice, D is wrong because there no need to create a constructor of the base class.
In choice, E is wrong because it does not require the second EBike constructor parameter.
