Answer:
- public class Main {
-
- public static void main(String[] args) {
- int myArray[] = {3, 7, 2, 5, 9,11, 24, 6, 10, 12};
- int myArray2 [] = {1, 2, 3, 4 ,5};
-
- displayValue(myArray);
- reverseDisplay(myArray);
- displaySum(myArray);
- displayLess(myArray, 10);
- displayHighAvg(myArray);
- displayBoth(myArray, myArray2);
- }
-
- public static void displayValue(int arr[]){
- for(int i = 0; i < arr.length; i++){
- System.out.print(arr[i] + " ");
- }
-
- System.out.println();
- }
-
- public static void reverseDisplay(int arr[]){
- for(int i = arr.length-1; i >= 0; i--){
- System.out.print(arr[i] + " ");
- }
-
- System.out.println();
- }
-
- public static void displaySum(int arr[]){
- int sum = 0;
- for(int i = 0; i < arr.length; i++){
- sum += arr[i];
- }
- System.out.println(sum);
- }
-
- public static void displayLess(int arr[], int limit){
-
- for(int i = 0; i < arr.length; i++){
- if(arr[i] < limit){
- System.out.print(arr[i] + " ");
- }
- }
- System.out.println();
- }
-
-
- public static void displayHighAvg(int arr[]){
- int sum = 0;
- for(int i = 0; i < arr.length; i++){
- sum += arr[i];
- }
-
- double avg = sum / arr.length;
-
- for(int i = 0; i < arr.length; i++){
- if(arr[i] > avg){
- System.out.print(arr[i] + " ");
- }
- }
- System.out.println();
- }
-
- public static void displayBoth(int arr1[], int arr2 []){
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr1[j] == arr2[i]){
- System.out.print(arr1[j] + " ");
- }
- }
- }
- System.out.println();
- }
- }
Explanation:
There are five methods written to solve all the problems stated in the question.
Method 1 : displayValue (Line 15 - 21)
This is the method that take one input array and use the print() method to display the all the elements in the array.
Method 2: reverseDisplay (Line 23 - 26)
This method will take one input array and print the value in the reverse order. We just need to start with the last index when running the for-loop to print the value.
Method 3: displaySum (Line 31 - 37)
This method will take one input array and use a for-loop to calculate the total of the values in the array.
Method 4: displayLess (Line 39 - 47)
This method will take one two inputs, a array and a limit. We use the limit as the condition to check if any value less than the limit, then the value will only be printed.
Method 5: displayHighAvg (Line 50 - 64)
This method will take one input array and calculate the average. The average will be used to check if any element in the array higher than it, then the value will only be printed.
Method 6: displayBoth (Line 66 - 75)
This method will take two input arrays and compare both of them to find out if any value appears in both input arrays and print it out.
<span>It is false that a thin red border indicates the active cell. As a matter of fact, an active cell is indicated by a thin green border. That means that you have clicked that particular cell, and that it is activated, so that you can use it and write whatever you want there. Usually, even in the real world, green means that something is active, and red that it is inactive.</span>
Answer:
client
Explanation:
client is someone that needs help
Answer:
B) managing a database for a large department store
C)managing the computer network for a large department store
Explanation:
An individual who majors in Information Technology is considered to be a professional with varying knowledge in the area of computers, networking, databases, and security. Such an individual is considered to be an expert in the area of database management, integration of software, and computer network optimization.
Hence, the most appropriate jobs for someone who majors in information technology is both options B and C