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.
I believe the answer is A. Add animation to clip art and text
Answer:
A,B
Explanation:
It is A,B because therefore it is true A,B if attribute A determines both attribute B and C, A,B if attribute A and B determine attribute C, A,B is a composite determinant.
Is this java or python pls explain or else i can’t answer
Answer:
a.When inspecting an instrument, Technicians apply specific Inspection Tests; more than one Test can be applied for an inspection. Each piano is inspected by at least one technician, and a technician can inspect more than one instrument.
Explanation:
An Inspection test is a formal approach used to test a system or product such as machines, package, software. This can be done by dimension inspection, visual inspection, welding inspection, function test, factory acceptance test. Three major factors to be considered in the test plan include:
Test Coverage,
Test Methods, and
Test Responsibilities
There are specific inspection tests that should be applied when inspecting an instrument. A technician can apply more than one test type to assess the authenticity of a product. More than one technician is needed to ascertain the working mehanism of a machine to ensure it has no fault. A technician can inspect more than one instrument depending on his diversity of specialization.
