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stira [4]
3 years ago
10

In binary, the second digit from the right is multiplied by the first power of two, and the _____ digit from the right is multip

lied by the fourth power of two.
A. Fourth
B. Fifth
C. Sixth
Computers and Technology
2 answers:
Mila [183]3 years ago
8 0

Answer: B. Fifth

Explanation:

aivan3 [116]3 years ago
6 0

Answer:

Your answer would be, B. Fifth digit ftom the right.

Explanation:

Binary with letters works like this,

the first 3 bits from the left, determine if the character is caps or lowercase, 010 is capital and 011 is lowercase, the next 5 are like this,

First - 2 to the power of four

Second - 2 to the power of three

Third - 2 to the power of two

Fourth - 2 to the powet of one

Fifth - 2 to the power of zero

So, given that information, the fourth digit from the right is 2 to the power of the three. The sixth digit is a bit to determine the capitalization, so the only answer left is B.

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1. Write a static method named computeOddSum that is passed an array of int's named numbers. The method must compute and return
S_A_V [24]

Answer:

The method in Java is as follows:

public static int computeOddSum(int [] myarray){

       int oddsum = 0;

       for(int i=0; i<myarray.length; i++ ) {

        if(myarray[i]%2==1) {

           oddsum+=myarray[i];

        }

     }

     return oddsum;

   }

Explanation:

This defines the static method

public static int computeOddSum(int [] myarray){

This initializes oddsum to 0

       int oddsum = 0;

This iterates through the array

       for(int i=0; i<myarray.length; i++ ) {

This checks for odd number

        if(myarray[i]%2==1) {

The odd numbers are added, here

           oddsum+=myarray[i];

        }

     }

This returns the calculated sum of odd numbers

     return oddsum;

   }

To call the method from main, use:

<em>int [] myarray = {3, 10, 11, 2, 6, 9, 5};</em>

<em>      System.out.println("Odd sum = " + computeOddSum(myarray));</em>

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Diano4ka-milaya [45]

Answer:

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Explanation:

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Answer:

Following are the responses to the given question:

Explanation:

Its machine slows for each further level of evaluation by an n/m factor. Therefore are the times for implementation for levels 2, 3, and 4 are \frac{kn}{m} \ and\  \frac{kn^2}{m^2} \ and \ \frac{kn^3}{m^3}.

So, the level values are:

level \ 2 = \frac{kn}{m}\\\\level \ 3 = \frac{kn^2}{m^2}\\\\level \ 4 = \frac{kn^2}{m^2}\\\\

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Natasha_Volkova [10]

Answer:

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