Answer:
Only
Option: void f1(float array[], int size);
is valid.
Explanation:
To pass an array as argument in a function, the syntax should be as follows:
functionName (type arrayName[ ] )
We can't place the size of the array inside the array bracket (arrayName[100]) as this will give a syntax error. The empty bracket [] is required to tell the program that the value that passed as the argument is an array and differentiate it from other type of value.
Answer:
#include <iostream>
using namespace std;
int main()
{
int arr[]={3,-9,9,33,-4,-5, 100,4,-23};
int pos;
int n=sizeof(arr)/sizeof(arr[0]);
for(int i=0;i<n;i++){
if(arr[i]>=0){
pos++;
}
}
cout<<"Number of positive integers is "<<pos<<endl;
return 0;
}
Explanation:
create the main function in the c++ programming and declare the array with the element. Then, store the size of array by using the formula:
int n=sizeof(arr)/sizeof(arr[0]);
after that, take a for loop for traversing the array and then check condition for positive element using if statement,
condition is array element greater than or equal to zero.
if condition true then increment the count by 1.
this process happen until the condition true
and finally print the count.
<h2>Answer:</h2>
<u>The correct option is</u><u> (B) hang up and call back using the banks official phone number</u>
<h2>Explanation:</h2>
There are a lot of cases where people pretend to call from the banks where the receivers have the account. The caller tries to take the information from the receiver and pretends to be the bank official. If there is any doubt then the receiver should hang up the call and call back the official number of the bank to confirm that whether somebody has called from the bank to get the information.
Answer:
import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.print("Input a word: ");
String userinput = input.nextLine();
for(int i =0;i<userinput.length();i+=2) {
System.out.print(userinput.charAt(i));
}
}
}
Explanation:
This line prompts user for input
System.out.print("Input a word: ");
This declares a string variable named userinput and also gets input from the user
String userinput = input.nextLine();
The following iterates through every other character of userinput from the first using iteration variable i and i is incremented by 2
for(int i =0;i<userinput.length();i+=2) {
This prints characters at i-th position
System.out.print(userinput.charAt(i));
Answer:
A
Issues can lead reoccuring costomers to not want to come back, resulting in this issue.
