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dalvyx [7]
4 years ago
10

01000100 01101111 00100000 01111001 01101111 01110101 00100000 01101011 01101110 01101111 01110111 00100000 01111001 01101111 01

110101 01110010 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01100011 01101111 01100100 01100101 00111111 00001010 01000001 01101110 01110011 01110111 01100101 01110010 00100000 01110100 01101000 01101001 01110011 00100000 01101001 01101110 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01100011 01101111 01100100 01100101
(another 30 point easy question (Yes I do extra point easy questions a lot...that's why I'm still a helping hand despite answering so many things and yes this is my last easy question of the day NO TRANSLATING)
Computers and Technology
2 answers:
dalvyx [7]4 years ago
6 0

01011001 01100101 01100001 00100000 01001001 00100000 01101011 01101110 01101111 01110111 00100000 01101101 01111001 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00101100 00100000 01100100 01101111 00100000 01111001 01101111 01110101 00111111 00100000 01001100 01001111 01001100 00100000 01101010 01101011



^^^^ What do you think

Anna71 [15]4 years ago
5 0
This was quite tricky but, I found a program online that answers your question. The answer is (Do you know your binary code? Answer this in binary code)
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I need to know the answer to this question
RoseWind [281]

It's D because all of this are TRUE.

3 0
3 years ago
The 7-bit ASCII code for the character ‘&’ is: 0100110 An odd parity check bit is now added to this code so 8 bits are trans
Lorico [155]

Answer

First part:

The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.

Second part:

The invalid bit sequence are option a. 01001000 and d. 11100111

Explanation:

Explanation for first part:

In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.

If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.

If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.  

Explanation for second part:

A valid odd parity bit sequence will always have odd number of 1s.

Since in option a and d,  total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.

And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.

7 0
3 years ago
Which term describes the process of training a machine to do simple, repetitive tasks, and adapt or correct its performance base
yKpoI14uk [10]
Automation. ... It involves taking a machine or software that was taught to do simple repetitive tasks (traditional automation) and teaching it to intuitively adapt or correct its performance based on changing conditions, at speed and scale.
8 0
3 years ago
Write a program that removes all spaces from the given input.
sasho [114]

Explanation:

#include<iostream>

#include<string.h>

using namespace std;

char *removestring(char str[80])

{

   int i,j,len;

   len = strlen(str);

   for( i = 0; i < len; i++)

   {

       if (str[i] == ' ')

       {

           for (j = i; j < len; j++)

               str[j] = str[j+1];

           len--;

       }

   }

   return str;

}

int main ()

{  

char  str[80];

   cout << "Enter a string : ";

   cin.getline(str, 80);

   strcpy(removestring(str), str);

   cout << "Resultant string : " << str;

   return 0;

}

In this program the input is obtained as an character array using getline(). Then it is passed to the user-defined function, then each character is analyzed and if space is found, then it is not copied.

C++ does not allow to return character array. So Character pointer is returned and the content is copied to the character array using "strcpy()". This is a built in function to copy pointer to array.

5 0
4 years ago
Write a while loop that prints user_num divided by 2 until user_num is less than 1. The value of user_num changes inside of the
rewona [7]

Answer:

user_num >= 1

while true do

print "user_num/2"

Explanation:

First, you write that the user_num is equal to or is greater than 1. Then you write in the while loop that while that's true, print user_num divided by two.

6 0
2 years ago
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