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Bond [772]
3 years ago
11

Write an enhanced for loop that multiplies all elements in an int[] array named factors, accumulating the result in a variable n

amed product. Numbers.java 123456789101112 public class Numbers{ public int multiply(int[] factors) { int product
Computers and Technology
1 answer:
IgorC [24]3 years ago
8 0

public class Numbers {

   public int multiply(int[] factors){

       int product = 1;

       for (int i : factors){

           product *= i;

       }

       return product;

   }

   public static void main(String[] args) {

       int [] factors = {1,2,3,4,5,6,7,8,9,10,11,12};

       Numbers num = new Numbers();

       System.out.println(num.multiply(factors));

   }

   

}

I hope this helps!

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Suppose we perform a sequence of n operations on a data structure such that if some condition C(k) holds then the kth operation
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Discuss the advantages of using analogue multi-tester or digital multi-tester over the other.​
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Answer:

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2 years ago
Find the double word-length 2's complement representation of each of the following decimal numbers:a. 3874
miss Akunina [59]

Answer:

-3874₁₀ = 1111 1111 1111 1111 1111 1111 1101 1110₂

Explanation:

2's complement is a way for us to represent negative numbers in binary.

To get 2's complement:

1. Invert all the bits

2. Add 1 to the inverted bits

Summary: 2's complement = -N = ~N + 1

1. Inverting the number

3874₁₀ = 1111 0010 0010₂

~3874₁₀ = 0000 1101 1101₂

2. Add 1 to your inverted bits

~3874₁₀ + 1 = 0000 1101 1101₂ + 1

= 0000 1101 1110₂

You can pad the most signigicant bits with 1's if you're planning on using more bits.

so,

12 bits                          16 bits

0000 1101 1110₂  = 1111 0000 1101 1110₂

They asked for double word-length (a fancy term for 32-bits), so pad the left-most side with 1s' until you get a total of 32 bits.

           32 bits

= 1111 1111 1111 1111 1111 1111 1101 1110

7 0
3 years ago
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