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Levart [38]
2 years ago
5

2. what are the advantages of breaking up a single logical message into a number of fixed-sized packets and then sending each on

e of those packets independently through the network?
Computers and Technology
1 answer:
Ludmilka [50]2 years ago
8 0

The advantages of breaking up a single logical message into a number of fixed-sized packets is that it makes it more flexible and also it increases the performance and also the reliability of the internet.

There is an Easy lost data recovery Internet as it is an heterogeneous network.

<h3>Are packets made of fixed size?</h3>

Packets are known to come in different sizes. They can also be made of  a fixed size.

The examples are 1,000 or 1,500 bytes, or other kinds of variable sizes, based on the system. Its advantages is one where errors can be corrected easily and data recovery is fast.

Learn more about fixed-sized packets from

brainly.com/question/3991989

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Write a program that helps a young student learn to make change. Use the random number generator to generate change amounts betw
erastovalidia [21]

Answer:

#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

 

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

 

int quarters = coins/25;

 

coins = coins%25;

 

int dimes = coins/10;

 

coins = coins%10;

 

int nickles = coins/5;

coins = coins%5;

 

int pennies = coins/1;

 

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

 

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

 

}

 

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

 

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

 

int quarters = coins/25;

 

coins = coins%25;

 

int dimes = coins/10;

 

coins = coins%10;

 

int nickles = coins/5;

coins = coins%5;

 

int pennies = coins/1;

 

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

 

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

 

}

 

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}

Explanation:

  • Use the built-in rand function to generate a random coin.
  • Get all the required values for quarters, dimes, nickles and pennies and store them in variables.
  • Check if the coins give a value of temp, then answer is correct and the values of user will be added.
  • Optimum solution is displayed  If it does not matches with the optimum solution. Display that the answer is wrong along with solution.
6 0
2 years ago
What used to produce protein under the direction of DNA?​
Leya [2.2K]

Genes

A gene is the section of DNA required to produce one protein.
7 0
3 years ago
If the Mona Lisa was in a word document, what would be the orientation
skelet666 [1.2K]
Portrait orientation
8 0
3 years ago
Once you turn 18, if you choose one of the 4 options for accessing credit, how will you be sure you use that credit responsibly?
dmitriy555 [2]

Answer:

hm not sure man

Explanation:

sorry, give me crown and 5 stars  tho

6 0
2 years ago
The function below takes two arguments: a dictionary of strings (keys) to integers (values) a_dict and a list of strings key_lis
vitfil [10]

Answer:

  1. def getData(a_dict, key_list):
  2.    result = []
  3.    for key in key_list:
  4.        result.append(a_dict[key])
  5.    
  6.    return result  
  7. result = getData( {"puffin": 5, "corgi": 2, "three": 3} , ["three", "corgi"])
  8. print(result)

Explanation:

Let's define a function <em>getData() </em>with two parameters,<em> a_dict </em>and <em>key_list</em> as required by the question (Line 1).

Since the function is to return a list of associated values of dictionaries, a new list,<em> result</em>,  is declared and initialized with empty values (Line 2).

Next, use for-loop to traverse through every string in the input <em>key_list </em>(Line 4) and use the traversed key to address the value in the<em> a_dict </em>and add it to the <em>result</em> list (Line 5)

At last, return the <em>result </em>list as output (Line 7)

We can test the function using the test case from the question and we shall see the output as follows:

[3, 2]

8 0
3 years ago
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