Answer: testing
Explanation:
Testing an answer with multiple attachments
Answer:
Create boxes with the following content
- Initialize sum to 0, product to 1 and number to 1
- Check if number is a prime. If yes, update sum += number and product *= number
- Increase number by 1
- If number is <= 50, go back to "check" block.
The check block has a diamond shape.
At the "yes" branch of the check block, you can create separate blocks for updating the sum and the product.
Connect the blocks using arrows.
Indicate the start and the end of the flow using the appropriate symbols.
The block to check if a number is a prime you could further decompose into the steps needed to do such a check.
<u>Threat </u>is some danger that can exploit a vulnerability.
<h3>What is Threat?</h3>
Threat is seen as Anything that can be able to take advantage of a vulnerability, intentionally as well as accidentally, and take or damage, the asset of a person.
Note that a threat is that which a person or firm is protect against and Vulnerability is seen as Weaknesses.
Therefore, <u>Threat </u>is some danger that can exploit a vulnerability.
Learn more about vulnerability from
brainly.com/question/25633298
#SPJ1
Answer:
1. Hackers could get information and sell it. 2. Hackers can get a hold of your files online. 3. Hackers can look at your photos. Download a VPN
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.
