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kumpel [21]
3 years ago
5

What scheme is usually used today for encoding signed integers?

Computers and Technology
1 answer:
dusya [7]3 years ago
6 0

Answer:

 For encoding signed integer the two's complement scheme are used as, the number can be positive, with no fraction part and it can be negative. The computer can stored information in the form of bits and the value with the zero and one. Encoding of signed number in the two's complement with the positive integer it can be done by binary number.

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What is the value of the variable result after these lines of code are executed?
zhuklara [117]

Answer:

result = 6

Explanation:

Just substitute the numbers for the variables:

(3*2) - (0*2) = 6

so result = 6

4 0
3 years ago
Read 2 more answers
In the following four questions, we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the pa
pantera1 [17]

Answer:

1)we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Initially suppose there is only one link between source and destination. Also suppose that the entire MP3 file is sent as one packet. The TRANSMISSION DELAY is:

3 Seconds

2)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. The END TO END DELAY(transmission delay plus propagation delay) is

3.05 seconds

3)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. how many bits will the source have transmitted when the first bit arrives at the destination.

500,000 bits

4)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP3 file is sent as one packet. Suppose there is no congestion, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. The end-to-end delay is

6.1 seconds

5)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose that the MP3 file is broken into 3 packets, each of 10 Mbits. Ignore headers that may be added to these packets. Also ignore router processing delays. Assuming store and forward packet switching at the router, the total delay is

4.05 seconds

6)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 TDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

7)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

6 0
3 years ago
Write your code to define and use two functions. The Get User Values function reads in num Values integers from the input and as
ZanzabumX [31]

Answer:

#include <iostream>

#include <vector>

using namespace std;

/* Define your function here */

vector<int> GetUserValues(vector<int>& userValues, int numValues) {

  int tmp = 0;

  vector<int> newVec;

 

  for(int i = 0; i < numValues; i++) {

     cin >> tmp;

     newVec.push_back(tmp);

     

  }

 

  return newVec;

}

void OutputIntsLessThanOrEqualToThreshold(vector<int> userValues, int upperThreshold) {

  for (int i = 0; i < userValues.size(); ++i) {

     if(userValues.at(i) < upperThreshold) {

          cout << userValues.at(i) << " ";

     }

  }

 

  cout << endl;

}

int main() {

  vector<int> userValues;

  int upperThreshold;

  int numValues;

 

  cin >> numValues;

  userValues = GetUserValues(userValues, numValues);

  cin >> upperThreshold;

  OutputIntsLessThanOrEqualToThreshold(userValues, upperThreshold);

  return 0;

}

Explanation:

Perhaps their is a better way to code this, but I couldn't figure out what to do with the pointer in the first function.

6 0
3 years ago
Write a program that asks for the user's first, middle, and last name as one input. The names must be stored in three different
nydimaria [60]

Answer:

Here is the C++ program:

#include <iostream>  //to use input output functions  

#include <cstring>  // used to manipulate c strings  

using namespace std;  //to identify objects cin cout

const int SIZE = 20;  //fixes constant size for firstName, middleName and lastName arrays  

const int FULL_SIZE = 60;  //constant size for fullName array  

int main() {  //start of main method  

 char firstName[SIZE]; //declares a char type array to hold first name  

 char middleName[SIZE];//declares a char type array to hold middle name  

 char lastName[SIZE]; //declares a char type array to hold last name  

 char fullName[FULL_SIZE]; //declares a char type array to hold full name

 cout << "Enter first, middle, and last name:  ";  //prompts user to enter first name

 cin>>firstName>>middleName>>lastName;  //reads last name from user, stores it in lastName array

 strcpy(fullName, lastName);  //copies last name from lastName to fullName array using strcpy method which copies a character string from source to destination

 strcat(fullName, ", "); //appends comma "," and empty space " " after last name in fullName using strcat method which appends a string at the end of another string

 strcat(fullName, firstName);  //appends first name stored in firstName to the last of fullName

 strcat(fullName, " ");  //appends an empty space to fullName

  char temp[2];  //creates a temporary array to hold first initial of middle name

temp[0] = middleName[0];  //holds the first initial of middle name

strcat(fullName, temp);  //appends first initial of middle name stored in temp to the last of fullName

strcat(fullName, ".");   //appends period

    cout<<fullName;  } //displays full name

Explanation:

I will explain the program with an example

Lets say user enters Neil as first name, Patrick as middle and Harris as last name. So firstName char array contains Neil, middleName char array contains Patrick and lastName char array contains Harris.  

Next the fullName array is declared to store Neil Patrick Harris in a specific format:

strcpy(fullName, lastName);

This copies lastName to fullName which means fullName now contains Harris.

 strcat(fullName, ", ");

This appends comma and an empty space to fullName which means fullName now contains Harris with a comma and empty space i.e. Harris,

strcat(fullName, firstName);

This appends firstName to fullName which means Neil is appended to fullName which now contains Harris, Neil

 strcat(fullName, " ");  

This appends an empty space to fullName which means fullName now contains Harris, Neil with an empty space i.e. Harris, Neil

char temp[2];

temp[0] = middleName[0];

This creates an array to hold the first initial of middle name i.e. P of Patrick

strcat(fullName, temp);

This appends an the first initial of middleName to fullName which means fullName now contains Harris, Neil P

strcat(fullName, ".");

This appends a period to fullName which means fullName now contains Harris, Neil P.

Now    cout <<"Full name is:\n"<<fullName<< endl; prints the fullName so the output of this entire program is:

Full name is:                                                                                                                                  Harris, Neil P.

The screenshot of the program along with its output is attached.

8 0
3 years ago
1.Give single word
EastWind [94]

Answer:

A. Versatile

B. Accuracy

C. High speed

D. Animation.

5 0
3 years ago
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