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bogdanovich [222]
3 years ago
13

Write a for loop to populate array userGuesses with NUM GUESSES integers. Read integers using Scanner. Ex If NUM GUESSES is 3 an

d user enters 9 5 2, then userGuesses is 19, 5, 2) 1 import java til. Sca nner 3 public class Store Guesses 4 public static void main (string args) Scanner SCnr new Scanner (System in final int NUM GUESSES int[] user Guesses new int[NUM. GUESSES] int 0; Your solution goes here 10 for (i 03 i NUM GUESSES ++i){ 11 user Guesses [i] scnr.nextInt 12 13 14 15 for (i 0; i NUM GUESSES ++i) 16 System.out print (u 17 18 19 for (i 0; i NUM GUESSES ++i)t 20 System Ou print (userGuesses[i] 21 Run X Testing for 12, 4, 6) 2 4 6 Expected output Your output 2 4 6 2 4 6 Tests aborted
Computers and Technology
1 answer:
inn [45]3 years ago
7 0

Answer:  

import java.util.Scanner;  

public class StoreGuesses {  

  public static void main (String [] args) {  

      Scanner scnr = new Scanner(System.in);  

      final int NUM_GUESSES = 3;  

      int[] userGuesses = new int[NUM_GUESSES];  

      int i = 0;  

     

      for (i = 0; i < NUM_GUESSES; ++i){  

          userGuesses[i] = scnr.nextInt();  

      }  

      for (i = 0; i < NUM_GUESSES; ++i){  

          System.out.print(userGuesses[i] + " ");  

      }  

  }  

}  

/*

Output:

2 4 6  

2 4 6  

*/

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NAND is logically complete. Use only NAND gates to constructgate-level circuits that compute the
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Answer:

Hi, for this exercise we have two laws to bear in mind:

Morgan's laws

NOT(А).NOT(В) = NOT(A) + NOT (B)

NOT(A) + NOT (B) = NOT(А).NOT(В)

And the table of the Nand

INPUT OUTPUT

A B A NAND B

0 0         1

0 1         1

1 0         1

1 1         0

Let's start!

a.

Input            OUTPUT

A       A     A NAND A

1         1             0

0        0            1

b.

Input            OUTPUT

A       B     (A NAND B ) NAND (A NAND B )

0         0            0

0         1            0

1          0             0

1          1             1

C.

Input            OUTPUT

A       B     (A NAND A ) NAND (B NAND B )

0        0           0

0        1             1

1         0            1

1          1            1

Explanation:

In the first one, we only need one input in this case A and comparing with the truth table we have the not gate

In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.

In the third case we have that the OR is A + B and we know in base of the morgan's law that:

A + B = NOT(NOT(А).NOT(В))

So, we have to negate the two inputs and after make nand with the two inputs negated.

I hope it's help you.

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