1. Who was able to complete the puzzle the fastest in Trial 1? Person 2 did it the fastest on Trial 1

2. Who was able to complete the puzzle the fastest in Trial 5?

Person 5 did it the fastest on Trial 5

3. How does each person’s time in Trial 1 compare with their time in Trial 5? (Describe the general trend.)

Because in Trial 1 they did not know the puzzle but in Trial 5 they know the puzzle.

4. Why do you think some were able to complete the puzzle faster than others?

Because they are older.

5. How does this experiment demonstrate learning by trial and error? Be specific.
In Trial 1 we did not know how to solve it and made a lot of errors, but after 2-3 try's we got to know where all the puzzles go, and at Trial 5 we did it in less time.

Can you make money in why are you guys so

You have a small company and want to keep your costs low, but it is important your employees share data. which network would provide you with the most economical solution?

This question involves a class named textfile that represents a text file. public class textfile { private string filename; private string filename; private arraylist words; // constructors not shown // postcondition: returns the number of bytes in this file public int filesize() { } // precondition: 0 < = index < words. size() // postcondition: removes numwords words from the words arraylist beginning at // index. public void deletewords(int index, int numwords) { } // precondition: 0 < = index < = words. size() // postcondition: adds elements from newwords array to words arraylist beginning // at index. pub lic voidaddwords(int index, string[] newwords) { } // other methods not shown } complete the filesize() method. the filesize() is computed in bytes. in a text file, each character in each word counts as one byte. in addition, there is a space in between each word in the words arraylist, and each of those spaces also counts as one byte. for example, suppose the words arraylist stores the following words: { mary had a little lamb; its fleece was white as snow. } the filesize() method would compute 4 + 3 + 1 + 6 + 5 + 4 + 6 + 3 + 5 + 2 + 5 as the sum of the lengths of each string in the arraylist. the value returned would be this sum plus 10, because there would also be 10 spaces in between the 11 words. complete the filesize() method below: // postcondition: returns the number of bytes in this file public int filesize() { }