Answer:
cinematographer
Explanation:
Kate handles the technical aspects of the camerawork. The keyword is technical aspects. Therefore, she is more of a cinematographer than a camera operator. They sound like they can be interchanged, but not necessarily. The difference between the two roles is that a camera operator is more familiar with the camera equipment and is responsible for creating a plan of execution that requires technical skills. A Cinematographer, on the other hand, relies on the camera operator to film the shots. He/she is in charge of the actual shooting and is the one who decides on the aspects of how to light the scene or how the shot will be composed.
Answer:
The correct answer is:
C. ndx = 0;
while (ndx < 3) {
ar[ndx] = 0;
ndx++;
}
Explanation:
The declaration given is:
int ar[3];
This means the array consists of three locations and is named as ar.
We know that the indexes are used to address the locations of an array and the index starts from 0 and goes upto to 1 less than the size of the array which means the indexes of array of 3 elements will start from 0 and end at 2.
Now in the given options we are using ndx variable to run the while loop.
So the code to assign zero to all elements of array will be
ndx = 0;
while(ndx<3)
{
ar[ndx] = 0;
ndx++;
}
Hence, the correct answer is:
C. ndx = 0;
while (ndx < 3) {
ar[ndx] = 0;
ndx++;
}
The answer here is Blind carbon copy (bcc).
The original recipients of the letter are unable to see (blind) who else receives a copy if those parties are on the bcc list.
Answer:
Explanation:
The following code is written in Java and is a function/method that takes in an int array as a parameter. The type of array can be changed. The function then creates a counter and loops through each element in the array comparing each one, whenever one element is found to be a duplicate it increases the counter by 1 and moves on to the next element in the array. Finally, it prints out the number of duplicates.
public static int countDuplicate (int[] arr) {
int count = 0;
for(int i = 0; i < arr.length; i++) {
for(int j = i + 1; j < arr.length; j++) {
if(arr[i] == arr[j])
count++;
}
}
return count;
}
