Answer:
int[] a = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
Explanation:
//written in java
The left side <u>int[] a</u> Declare a variable(a) which is of type Integer array(<u>int[]</u>).
The equal sign(=) assign the right side to the left side.
The right side initialize the integer array(a) with ten element.
Answer:
A. planning and gathering financial information
It could be old ... could be your connection to the internet... could be a virus. You can go to walmart or your super store and buy a usb for your computer. There are ones that can speed up your computer. I had to buy it for mine and it worked. It might be over 20 dollars depends on where you go.
Answer:
- import java.util.Scanner;
-
- public class Main {
- public static void main(String[] args) {
- int num[] = new int[51];
- Scanner input = new Scanner(System.in);
- System.out.print("Number of input: ");
- int limit = input.nextInt();
- for(int i=0; i < limit; i++){
- System.out.print("Input a number (1-50): ");
- int k = input.nextInt();
- num[k]++;
- }
- for(int j=1; j < 51; j++){
- if(num[j] > 0){
- System.out.println("Number of occurrence of " + j + ": " + num[j]);
- }
- }
- }
- }
Explanation:
The solution is written in Java.
Firstly, create an integer array with size 51. The array will have 51 items with initial value 0 each (Line 5).
Create a Scanner object and get user entry the number of input (Line 6-7).
Use the input number as the limit to control the number of the for loop iteration to repeatedly get integer input from user (Line 9-13). Whenever user input an integer, use that integer, k, as the index to address the corresponding items in the array and increment it by one (LINE 11-12).
At last, create another for loop to iterate through each item in the array and check if there is any item with value above zero (this means with occurrence at least one). If so, print the item value as number of occurrence (Line 14-17).
C. is very thin, thus allowing harmful gases to escape from it
