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slega [8]
3 years ago
13

1. Perform the following Binary Additions 101101 + 10100 *

Computers and Technology
2 answers:
daser333 [38]3 years ago
5 0

Answer:

Solutions:

Ans#1:

   101101

<u>+   10100 </u>

<u>1000001  </u>

<u />

Ans#2:

01010011

<u>+01110110</u>

<u>11001001 </u>

Ans#3:

101101

<u>- 10100 </u>

<u>011001  </u>

<u />

<u>Ans#4:</u>

<u />

11001011

<u>- 1010110 </u>

<u>10010100</u>

<u />

Explanation:

The given answers are in Binary.

In binary we add the digits in 1 and 0 form.

Addition in Binary:

1 + 0 = 1

0 + 1 = 1

0 + 0 = 0

1 + 1 = 10

Subtraction in Binary:

1 - 0 = 1

1 - 1 = 0

0 - 1 = 1 (it's because the 0 borrows a 1)

0 - 0 = 0

USPshnik [31]3 years ago
3 0

Answer:

1.         carry   1  1  1

                      1  0 1 1 0 1

                      +  1 0 1 0 0

                       ------------------------

   Answer: 1  0 0 0  0 0 1  

2.         carry  1 1  1 0 1  1  

                     0 1 0 1 0 0 1 1

                  + 0 1 1  1 0  1  1 0

                  --------------------------

    Answer: 0 1  1 0 0 1 0 0 1

3.                   1 0 1 1 0 1

                    -   1 0 1 0 0

                   -------------------------

       Answer: 0 1 1 0 0 1

4.                      1 1 0 0 1 0 1 1

                      -    1 0 1 0 1 1 0

                     -----------------------------

        Answer:   01 1 1 0 1 0 1

Explanation:

In Binary addition when you add 1 + 1 = 0, carry over the 1, i.e 10

in Subtraction when you subtract 0-1 = 1,  brow 1 from left side, resulting in -1 carried over.

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7 0
3 years ago
Write a program that asks the user to enter the amount s/he has budgeted for a month. The amount should be between 1000 and 2000
Elena-2011 [213]

Answer:

The program to this question can be given as:

Program:

  //import package for user input.

import java.util.*;      

//define class

public class Budget                      

{

public static void main(String [] a)            //define main method.

{

//creating Scanner class object.

Scanner ob =new Scanner(System.in);    

//define variable.      

double budgetamount=0, amountspent=0 ,difference=0,total=0,num=0 ;        

int count = 0;                            

System.out.println("How much have you budgeted for the month? :");        //print message.

budgetamount=ob.nextDouble();                             //taking input

       while(budgetamount != 0)           //checkig number greater then 4 digite.

       {

           budgetamount= budgetamount/10;        

           ++count;

       }

       if(count>4)                 //condition

       {

       System.out.println("enter each expense, then type -999 to quit: ");

       while(num!=-999)               //taking expense  

       {

       total=total+num;                 //totaling expense

       num=scan.nextDouble();                  

       }

       if(total<=budgetamount)           //condtion for over budget.

       {

       System.out.print("under budget:");         //print message

       System.out.print(budgetamount-total);

       }

       else

       {

       System.out.print("over budget:");              //for under budget

       System.out.print(total-budgetamount);

       }

       }

       else

       {

       System.out.println("not valid");        //message for number lessthen 4 digit.    

       }                                      

}

}

output:

How much have you budgeted for the month? : 1200.55.

enter each expense, then type -999 to quit: 365.89

556.90

339.98

-999

over budget:1262.77

Explanation:

The explanation of this program can be given as:

In the above program we import the package in that is used for scanner class. This class is used for the input from the user after input we use the while loop and if-else statement. The while loop is the entry control loop It is used for input validation and if-else is used for the checking condition. Then we insert expense that is inserted by the user. Then we calculate under-budgeted and over-budgeted by conditional statement that is if-else. and at the last we print it.

5 0
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Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working​ alone, one machine would take one hour long
Alborosie

<em><u>Answer</u></em>

5 hours

<em><u>Explanation</u></em>

The two working together can finish a job in

\frac{20}{9}  \: hours

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

\frac{1}{x}    +  \frac{1}{x - 1}

This should be equal to 20/9 hours.

\frac{1}{x}  +  \frac{1}{x - 1}  =  \frac{9}{20}

Multiply through by;

20x(x - 1) \times \frac{1}{x}  +20x(x - 1) \times   \frac{1}{x - 1}  =  20x(x - 1) \times \frac{9}{0}

20(x - 1)  +20x = 9x(x - 1)

20x - 20+20x = 9{x}^{2}  - 9x

9{x}^{2}  - 9x - 20x - 20x + 20= 0

9{x}^{2}  - 49x  + 20= 0

Factor to get:

(9x - 4)(x - 5) = 0

x =  \frac{4}{9}  \: or \: x = 5

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

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Answer:

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Explanation:

Please check the answer section.

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