Answer:
def find_max(num_1, num_2):
max_val = 0.0
if (num_1 > num_2): # if num1 is greater than num2,
max_val = num_1 # then num1 is the maxVal.
else: # Otherwise,
max_val = num_2 # num2 is the maxVal
return max_val
max_sum = 0.0
num_a = float(input())
num_b = float(input())
num_y = float(input())
num_z = float(input())
max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)
print('max_sum is:', max_sum)
Explanation:
I added the missing part. Also, you forgot the put parentheses. I highlighted all.
To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)
Answer:
Sensitivity Levels
Explanation:
Sensitivity Level is option use in email to inform the recipient that they should exercise discretion in accordance with sharing the content of the message.
In data presentation of computing
systems and applications, when a user click the submit button on the form, the
name-value pair of each form is sent because it is an open-ended data structure
that allows future extension without altering existing code or data.
Answer:
- public class FindDuplicate{
-
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
-
- int n = 5;
- int arr[] = new int[n];
-
- for(int i=0; i < arr.length; i++){
- int inputNum = input.nextInt();
- if(inputNum >=1 && inputNum <=n) {
- arr[i] = inputNum;
- }
- }
-
- for(int j =0; j < arr.length; j++){
- for(int k = 0; k < arr.length; k++){
- if(j == k){
- continue;
- }else{
- if(arr[j] == arr[k]){
- System.out.println("True");
- return;
- }
- }
- }
- }
- System.out.println("False");
- }
- }
Explanation:
Firstly, create a Scanner object to get user input (Line 4).
Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).
Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.
If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".
Answer:
See explaination
Explanation:
StackExample.java
public class StackExample<T> {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);
/**
* Returns a reference to the element at the top of this stack.
* The element is not removed from the stack.
* atreturn element on top of stack
* atthrows EmptyCollectionException if stack is empty
*/
public T peek() throws EmptyCollectionException
{
if (isEmpty())
throw new EmptyCollectionException("stack");
return stack[top-1];
}
/**
* Returns true if this stack is empty and false otherwise.
* atreturn true if this stack is empty
*/
public boolean isEmpty()
{
return top < 0;
}
}
//please replace "at" with the at symbol
Note:
peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.
