Answer:
It will always return (completely sorted) after 99 data comparisons.
It will always require at least 99 comparisons
Explanation:
Answer:
Explanation:
The following is written in Java. It continues asking the user for inputs until they enter a -1. Then it saves all the values into an array and calculates the number of values entered, the highest, and lowest, and prints all the variables to the screen. The code was tested and the output can be seen in the attached image below.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
class Brainly {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
int highest, lowest;
ArrayList<Integer> myArr = new ArrayList<>();
while (true) {
System.out.println("Enter a number [0-10] or -1 to exit");
int num = in.nextInt();
if (num != -1) {
if ((num >= 0) && (num <= 10)) {
count+= 1;
myArr.add(num);
} else {
System.out.println("Wrong Value");
}
} else {
break;
}
}
if (myArr.size() > 3) {
highest = myArr.get(0);
lowest = myArr.get(0);
for (int x: myArr) {
if (x > highest) {
highest = x;
}
if (x < lowest) {
lowest = x;
}
}
System.out.println("Number of Elements: " + count);
System.out.println("Highest: " + highest);
System.out.println("Lowest : " + lowest);
} else {
System.out.println("Number of Elements: " + count);
System.out.println("No Highest or Lowest Elements");
}
}
}
The 3 files you need to have for a successful mail merge are:
- An Excel spreadsheet works
- Outlook Contact List.
- Apple Contacts List or Text file, etc.
<h3>What is Mail Merge?</h3>
This is known to be the act of carrying out a Mail Merge and it is one where a person will need to use a Word document and a recipient list, that is an Excel workbook.
Files needed are:
- Text file,
- address files, etc.
The 3 files you need to have for a successful mail merge are:
An Excel spreadsheet worksOutlook Contact List.
Apple Contacts List or Text file, etc.
Learn more about mail merge from
brainly.com/question/20904639
#SPJ1
If 29 bits of the 32 available addressing bits are used for the subnet, then only 3 bits giving 2^3=8 combinations remain for the host addresses.
In reality, the all zeros and all ones addresses are reserved. So, 8 addresses can exist, but 6 of those are available.
The way the question is formulated it seems the answer 8 is what they're after.