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barxatty [35]
2 years ago
11

a) In an office, it is required to take hundreds of copies of the same type written document. Which traditional method of docume

nt preparation can be used for this?
Computers and Technology
1 answer:
olga2289 [7]2 years ago
3 0

Answer:

I don't know

Explanation:

You might be interested in
Write a C program (doublecopy) that allows a user to extract some part of an existing file (fileSource) and copy it twice to a n
Slav-nsk [51]

Answer:

Check the explanation

Explanation:

#include <stdio.h>

#include <sys/stat.h>

#include <stdlib.h>

#include <fcntl.h>

#include <errno.h>

#include <unistd.h>

extern int errno;

struct stat st;

int main(int argc, char **argv){

  int num1 = atoi(argv[1]); // Getting num1 from user

  int num2 = atoi(argv[2]); // Getting num2 from user

  char *fileSource = argv[3];

  char *fileTarget = argv[4];

 

  int source_fd = open(fileSource, O_RDONLY); // opening the file in read only mode

  int target_fd = open(fileTarget, O_WRONLY | O_CREAT); // opening the target file in Writeonly mode if file is not found it will create

   

  char *ch = (char *) calloc(num2+num1, sizeof(char));

 

  stat(fileSource, &st);

  if(st.st_size < (num1 + num2)){

      printf("File Size is smaller than the specified bytes\n");

      read(source_fd, ch, st.st_size); // reading the file upto the end

      write(target_fd, ch, st.st_size); // write to the file

      write(target_fd, ch, st.st_size); // two times writing to the file

  }else{

      if(lseek(source_fd, (off_t)num1, SEEK_SET) < 0 ) // moving the cursor to after the specified bytes from the start

      {

          printf("Some Error occured while seeking the file");

          return -1;

      }

      read(source_fd, ch, num2); // reading num2 bytes from the source

      write(target_fd, ch, num2); // writing two times to the target

      write(target_fd, ch, num2);

     

  }

 

  return 0;

 

}

1 2 #include <stdio.h> #include <sys/stat.h> #include <stdlib.h> #include <fcntl.h> #include <errno.h> #include <unistd.h> 4

The code screenshot and code output are attached below.

3 0
3 years ago
We learned that an ideal multiple access protocol has FOUR desirable
Nataly [62]

Answer:

We learned that an ideal multiple access protocol has FOUR desirable

characteristics, describe them.

Explanation:

1.- Very experienced technology and easy to implement.

2.- Rigid resource management and unfit for variable traffic flows.

3.- Requires antenna duplexer for duplex transmission.

<u>Duplex</u> <em>is a term used in telecommunication to define a system that is capable of maintaining two-way communication, sending and receiving messages simultaneously. </em>

4.- Normally FDMA is combined with multiplexing FDD.

<u>FDMA</u> (acronym in English for Frequency Division Multiple Access) <em>is a multiplexing technique used in multiple communication protocols, both digital and analog, mainly radio frequency, and among them in mobile phones of GSM networks. </em>

<u>FDD</u> (Frequency Division Duplexing in English) <em>designates a duplex method in the environment of wireless telecommunications and over certain wired networks.</em>

4 0
4 years ago
To ensure rapid recovery and security of data, backup data should be ________.
Nataly [62]

Answer:B kept at a secure location at the facility

Explanation:

8 0
3 years ago
Software piracy is acceptable as it helps us obtain software cheaper or sometimes even for free.
ELEN [110]

Answer: 1. False 2. True

Explanation: Piracy in any capacity is in fact illegal and might possibly be a felony I'm not quite sure. But it's just like stealing a toy from the store, somebody worked hard to make it and you best be paying for it.

Software licenses on the other hand usually give you permission to install the software on one or many machines depending on the license you purchase, but usually, if you paid for it once, you can use it anywhere.

8 0
3 years ago
Read 2 more answers
Note: You can use a word document to write your answers and copy-paste your answer to the area specified. a. (5 points) Convert
MrMuchimi

Answer:

EA9_{16} = 3753

CB2_{16} = 3250

(1011 1110 1101 1011 1010)_2 = 781754

(1010 1000 1011 1000 1110 1101)_2 = 11057389

(1011 1110 1101 1011 1010)_2 = BEDBA

(1010 1000 1011 1000 1110 1101)_2 =  A8B8ED

74510_8= 221416

67210_8 = 203212

Explanation:

Solving (a): To base 10

(i)\ EA9_{16

We simply multiply each digit by a base of 16 to the power of their position.

i.e.

EA9_{16} = E * 16^2 + A * 16^1 + 9 * 16^0

EA9_{16} = E * 256 + A * 16 + 9 * 1

In hexadecimal

A = 10; E = 14

So:

EA9_{16} = 14 * 256 + 10 * 16 + 9 * 1

EA9_{16} = 3753

(ii)\ CB2_{16}

This gives:

CB2_{16} = C * 16^2 + B * 16^1 + 2 * 16^0

CB2_{16} = C * 256 + B * 16 + 2 * 1

In hexadecimal

C = 12; B =11

So:

CB2_{16} = 12 * 256 + 11 * 16 + 2 * 1

CB2_{16} = 3250

Solving (b): To base 10

(i)\ (1011 1110 1101 1011 1010)_2

We simply multiply each digit by a base of 2 to the power of their position.

i.e.

(1011 1110 1101 1011 1010)_2 = 1 * 2^{19} + 0 * 2^{18} + 1 * 2^{17} + 1 * 2^{16} +1 * 2^{15} + 1 * 2^{14} + 1 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 1 * 2^{10} + 0 * 2^9 + 1 * 2^8 +1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0

(1011 1110 1101 1011 1010)_2 = 781754

(ii)\ (1010 1000 1011 1000 1110 1101)_2

(1010 1000 1011 1000 1110 1101)_2 = 1 * 2^{23} + 0 * 2^{22} + 1 * 2^{21} + 0 * 2^{20} +1 * 2^{19} + 0 * 2^{18} + 0 * 2^{17} + 0 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 1 * 2^{13} + 1 * 2^{12} +1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 0 * 2^8 + 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 1*2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0

(1010 1000 1011 1000 1110 1101)_2 = 11057389

Solving (c): To base 16

i.\ (1011 1110 1101 1011 1010)_2

First, convert to base 10

In (b)

(1011 1110 1101 1011 1010)_2 = 781754

Next, is to divide 781754 by 16 and keep track of the remainder

781754/16\ |\ 48859\ R\ 10

48859/16\ |\ 3053\ R\ 11

3053/16\ |\ 190\ R\ 13

190/16\ |\ 11\ R\ 14

11/16\ |\ 0\ R\ 11

Write out the remainder from bottom to top

(11)(14)(13)(11)(10)

In hexadecimal

A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.

(11)(14)(13)(11)(10)=BEDBA

So:

(1011 1110 1101 1011 1010)_2 = BEDBA

ii.\ (1010 1000 1011 1000 1110 1101)_2

In b

(1010 1000 1011 1000 1110 1101)_2 = 11057389

Next, is to divide 11057389 by 16 and keep track of the remainder

11057389/16\ |\ 691086\ R\ 13

691086/16\ |\ 43192\ R\ 14

43192/16\ |\ 2699\ R\ 8

2699/16\ |\ 168\ R\ 11

168/16\ |\ 10\ R\ 8

10/16\ |\ 0\ R\ 10

Write out the remainder from bottom to top

(10)8(11)8(14)(13)

In hexadecimal

A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.

(10)8(11)8(14)(13) = A8B8ED

So:

(1010 1000 1011 1000 1110 1101)_2 =  A8B8ED

Solving (d): To octal

(i.)\ 74510

Divide 74510 by 8 and keep track of the remainder

74510/8\ |\ 9313\ R\ 6

9313/8\ |\ 1164\ R\ 1

1164/8\ |\ 145\ R\ 4

145/8\ |\ 18\ R\ 1

18/8\ |\ 2\ R\ 2

2/8\ |\ 0\ R\ 2

Write out the remainder from bottom to top

74510_8= 221416

(ii.)\ 67210

Divide 67210 by 8 and keep track of the remainder

67210/8\ |\ 8401\ R\ 2

8401/8\ |\ 1050\ R\ 1

1050/8\ |\ 131\ R\ 2

131/8\ |\ 16\ R\ 3

16/8\ |\ 2\ R\ 0

2/8\ |\ 0\ R\ 2

Write out the remainder from bottom to top

67210_8 = 203212

8 0
3 years ago
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