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rewona [7]
3 years ago
10

Samuel is working on decimal and binary conversion for his college project. He is using the binary number 111011 and wants to ex

plain its conversion to the decimal number 59. Which formula will Samuel use to explain the conversion?
A)1 x 25 + 24 + 1 x 23 + 0 x 22 + 1 x 21+ 1 x 20
B)0 x 25 + 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21+ 1 x 20
C)1 x 25 + 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21+ 1 x 20
D)20 x 1 + 21 x 1 + 22 x 0 + 23 x 1 + 24 x 1 + 25 x 1
E)1 x 25 + 1 x 25 + 22 x 0 + 23 x 1 + 24 x 1 + 25 x 1
Computers and Technology
2 answers:
Contact [7]3 years ago
4 0
<span>1 X 25 + 1 X 24 + 1 X 23 + 0 X 22 + 1 X 21 + 1 X 20 thus your Answer is C</span>
polet [3.4K]3 years ago
3 0

Answer:

C)1*2^{5} + 1*2^{4} + 1*2^{3} + 0*2^{2} + 1*2^{1} + 1*2^{0}

Explanation:

We start the conversion from the final bit, that is, the least significant.

The counter starts at zero, and we start adding, this way:

111011

D = 1*2^{0} + 1*2^{1} + 0*2^{2} + 1*2^{3} + 1*2^{4} + 1*2^{5} = 1*2^{5} + 1*2^{4} + 1*2^{3} + 0*2^{2} + 1*2^{1} + 1*2^{0}

So the correct answer is:

C)1*2^{5} + 1*2^{4} + 1*2^{3} + 0*2^{2} + 1*2^{1} + 1*2^{0}

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Help me understand why it does not loop correctly and how to make the converter a continious loop? What am I doing wrong and how
Aleksandr-060686 [28]

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If you are trying to make a loop run a certain number of time in Python, then...

To repeat something a certain number of times, you may:

1. Use range or xrange for i in range(n): # do something here.

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4 0
2 years ago
Write a static method called bothStart that allows the user to input two Strings and returns the String that is the longest subs
marishachu [46]

Answer:

  1.    public static String bothStart(String text1, String text2){
  2.        String s = "";
  3.        if(text1.length() > text2.length()) {
  4.            for (int i = 0; i < text2.length(); i++) {
  5.                if (text1.charAt(i) == text2.charAt(i)) {
  6.                    s += text1.charAt(i);
  7.                }else{
  8.                    break;
  9.                }
  10.            }
  11.            return s;
  12.        }else{
  13.            for (int i = 0; i < text1.length(); i++) {
  14.                if (text1.charAt(i) == text2.charAt(i)) {
  15.                    s += text1.charAt(i);
  16.                }else{
  17.                    break;
  18.                }
  19.            }
  20.            return s;
  21.        }
  22.    }

Explanation:

Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).  

<em />

Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).

There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).

If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.

The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>

<em />

At the end, return the s as output (Line 21).

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