trình bày ngắn gọn hiểu biết của bản thân về mạng máy tính theo mô hình OSI hoặc TCP/IP

Answer:

String words[]=str.split("\\s");  

String capitalizeWord="";  

for(String w:words){  

   String first=w.substring(0,1);  

   String afterfirst=w.substring(1);  

   capitalizeWord+=first.toUpperCase()+afterfirst+" ";  

}  

return capitalizeWord.trim();  

Explanation:

Define the word you are trying to capitalize and split it into an array.

String words[]=str.split("\\s");

Create a string for the capital output to be created as

String capitalizeWord="";  

Capitalize the word using the "first.toUpperCase()" and "afterfirst" functions

for(String w:words){  

   String first=w.substring(0,1);  

   String afterfirst=w.substring(1);  

   capitalizeWord+=first.toUpperCase()+afterfirst+" ";  

}  

Answer:

Below is the required JAVA Code:

Explanation:

public static int largestDigit(int n) {

   if (n < 0) {

       return largestDigit(-1 * n);

   } else if (n == 0) {

       return 0;

   } else {

       int digit = n % 10;

       int maxDigit = largestDigit(n / 10);

       if (digit > maxDigit)

           maxDigit = digit;

       return maxDigit;

   }

}

\color{red}Method\;tested\;in\;a\;complete\;java\;program\;if\;you\;are\;interested.

public class LargestDigitRecursive {

   public static int largestDigit(int n) {

       if (n < 0) {

           return largestDigit(-1 * n);

       } else if (n == 0) {

           return 0;

       } else {

           int digit = n % 10;

           int maxDigit = largestDigit(n / 10);

           if (digit > maxDigit)

               maxDigit = digit;

           return maxDigit;

       }

   }

   public static void main(String[] args) {

       System.out.println(largestDigit(14263203));

       System.out.println(largestDigit(845));

       System.out.println(largestDigit(52649));

       System.out.println(largestDigit(3));

       System.out.println(largestDigit(0));

       System.out.println(largestDigit(-573026));

       System.out.println(largestDigit(-2));

   }

}

OUTPUT:

6

8

9

3

0

7

2

Answer:

Three cups would be needed.

Explanation:

2 2/3 × 3/4 = 8/2 × 3/4 = 24/8 = 3

Answer:

The statement is as follows:

print("{0:,.1f}".format(number))

Explanation:

Required

Statement to print 1234567.456 as 1,234,567.5

To do this, we make use of the format keyword, and we set the print format in the process.

To round up number to 1 decimal place, we use the following format:

"{0:,.1f}"

To include comma in the thousand place, we simply include a comma sign before the number of decimal place of the output; i.e. before 1

"{0:,.1f}"

So, the print statement is:

print("{0:,.1f}".format(number))

Answer:

Although this word was made up in order to be a contender for the longest word in English, it can be broken down into smaller chunks in order to understand it.

Anti- means that you are against something; monopoly means the exclusive control over something; geographic is related to geography; and the remaining part has to do with nationalism.

So this word means something like 'a nationalistic feeling of being against geographic monopoly,' but you can see that it doesn't have much sense.

(answer copied from  Kalahira just to save time)

Answer:

computer

Explanation:

is an ICT device that help and makes things or work easy

Answer:

Blurring can provide the optical illusion of a fast movement. It is often used in videos to hide sensitive or age-inappropriate information, the protection of identities (hence security) or just to provide decency.

Explanation:

Mostly used during animated videos where unlike (real videos) the blur does not occur naturally, the blur is used to make speed believable. Speed invigorates, and highlights intensity. Blurring still images can help to make more realistics motions of speed. An example where lots of blurring is used as prostproduction effects is in Flash.

Cheers

Answer:

hi

Explanation:

i did not known answers

Answer:

A.Extreme animation

hope it helpss

Answer:

b. Joyce recruits users to test new software by explaining to them how it will benefit the organization.

Explanation:

Skills can be classified as hard or soft. Hard skills include skills such as technical skills that are acquired through technical knowledge and training.

Soft skills on the other hand are skills exhibited as a result of personality traits, character and behaviours such as time management, leadership, communication and organizational skills.

Some hard skills include;

i. Ability to design banners and posters

ii. Proficiency in using programming languages such as Java to build software applications

iii. Evaluating the result of a test or research.

Some soft skills include;

i. Making public speeches that captivate the audience.

ii. Ability to recruit and test capabilities in others.

iii. Leading a team

So, Joyce recruiting users to test new software by explaining to them how it will benefit the organization is a form of soft skill since it consists, at minimum, of leadership and communication skills

Answer:

Yup

Explanation:

Answer:

Explanation:

Given that:

The no of flow pairs = 80

Each link's capacity = 10 Gbps

Link capacity amongst TOR switches and hosts = 1 Gbps

Because each connection is distributed among all flows, each flow crossing the same link distributes the link's capacity with the other flows. The connection capacity needed by each flow in the network is determined by the highest flow rate.

Maximum rate flow = [tex]\dfrac{link's \ capacity}{no \ of \ flow \ pairs}[/tex]

[tex]= \dfrac{10 \ Gbps}{80}[/tex]

Since 1 Gbps = 1000 Mbps

[tex]= \dfrac{10000}{80}[/tex]

= 125 Mbps

Hence, maximum rate flow = 125 Mbps

b.

Each switch at Tier-1 is connected to each and every switch at Tier-2 in the highly integrated structural topology. As a result, there are n-disjoint routes between tier-2 and tier-1 switches.

Number of paths between tier-1 and tier-2 is 4, thus, there exist four paths from tier-1 switch to tier 2.

The capacity whereby all data routes convey is determined by the maximum rate of flow.  

number of available paths = 4  

Each link's capacity = 10 Gbps

Maximum flow rate = 4 × 10 Gbps

Maximum flow rate = 40 Gbps

c.

Because each connection is distributed between all flows, each flow crossing a similar link distributes the link's capacity with other flows.

number of flow pairs = 160

Each link's capacity = 10 Gbps

maximum rate of flow = 10000 Mbps / 160

= 62.5 Mbps

For 160 flow pairs, maximum flow rate = 62.5 Mbps

Each switch in Tier-1 communicates with each switch in Tier-2 in the highly linked topology network. As a result, n-disjoint routes between tier-2 and tier-1 switches are possible. As there exist 20 hosts engaged in the network around each host, the number of routes from either a tier-1 switch or tier-2 switches is 20.

here;

no of paths = 20

each link's capacity = 10 Gbps

Maximum flow rate = 20 × 10  

= 200 Gbps

As such, with 160 pairs of flow, maximum flow rate = 200 Gbps

Answer:

Step 1 : Start

Step 2 : Set target= 0

Step 3 : Use a While loop

Step 4 : input donation

Step 5 : Sum - - > target += donation

Step 6 :, check loop condition

Step 7 : End

Explanation:

Algorithm is a sequence of written instructions or steps which we want the computer to in other to solve a particular problem :

Counter is set at 0

Total amount is initially set to := 0

Using a while loop ;

While target amout is less than 200 ; user inputs the amout to donate and it is added to target amount. This loop continues until target amount is greater than 200. Then the loop terminates.

Answer:

b i just guessed

Explanation:

Answer:

Explanation:

Pink: Down 5 then left 2.

Yellow: Left 3 and down 2.

Green: Right 7, down 4 and left 1.

Purple: Up 6 and left 9.

Red: Left 7, down 5 and left 1.

You can do the last one, blue :)

Answer:

Explanation:

u=up, d=down, r=right, l=left

yellow: l3d2

pink: d5l2

green: r7d4l1

purple: u6l9

red: l7d5l1

blue: r2u7l5

Answer:

//In java

for (int i = 0; i <= 100; i++)

{

if ( i % 2 == 0 )

{

System.out.println( i + " ");

}

}

Answer:

"Portrait orientation" would be the correct answer.

Explanation:

Thus the above is the correct answer.

Answer:

1)PERFORMANCE

2)RANGE

Explanation:

A mesh network can be regarded as local network topology whereby infrastructure nodes connect dynamically and directly, with other different nodes ,cooperate with one another so that data can be efficiently route from/to clients. It could be a Wireless mesh network or wired one.

Wireless mesh network, utilize

only one node which is physically wired to a network connection such as DSL internet modem. Then the one wired node will now be responsible for sharing of its internet connection in wireless firm with all other nodes arround the vicinity. Then the nodes will now share the connection in wireless firm to nodes which are closest to them, and with this wireless connection wide range of area can be convered which is one advantage of wireless mesh network, other one is performance, wireless has greater performance than wired one.

Some of the benefits derived from configuring a wireless mesh network is

1)PERFORMANCE

2)RANGE

Based on the information given, the correct options are performance and range.

It should be noted that a mesh network can be regarded as the local network topology whereby infrastructure nodes connect directly, with other different nodes ,cooperate with one another so that data can be efficiently route from/to clients.

The advantage of configuring a wireless mesh network is that it enhances performance and range.

Learn more about network on:

https://brainly.com/question/1167985

Answer:

Test if L(M1-2) is empty.

Construct FA M2-1 from M1 and M2 which recognizes the language L(>M2) - L(>M1).

COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.

Test if L(M2-1) is empty.

Answer yes if and only if both answers were yes.

Explanation:

An algorithm must be guaranteed to halt after a finite number of steps.

Each step of the algorithm must be well specified (deterministic rather than non-deterministic).

Three basic problems:

Given an FA M and an input x, does M accept x?

Is x in L(M)

Given an FA M, is there a string that it accepts?

Is L(M) the empty set?

Given an FA M, is L(M) finite?

Algorithm for determining if M accepts x.

Simply execute M on x.

Output yes if we end up at an accepting state.

This algorithm clearly halts after a finite number of steps, and it is well specified.

This algorithm is also clearly correct.

Testing if L(M) is empty.

Incorrect "Algorithm"

Simulate M on all strings x.

Output yes if and only if all strings are rejected.

The "algorithm" is well specified, and it is also clearly correct.

However, this is not an algorithm because there are an infinite number of strings to simulate M on, and thus it is not guaranteed to halt in a finite amount of time.

COMMENT: Note we use the algorithm for the first problem as a subroutine; you must think in this fashion to solve the problems we will ask.

Correct Algorithm

Simulate M on all strings of length between 0 and n-1 where M has n states.

Output no if and only if all strings are rejected.

Otherwise output yes.

This algorithm clearly halts after a finite number of steps, and it is well specified.

The correctness of the algorithm follows from the fact that if M accepts any strings, it must accept one of length at most n-1.

Suppose this is not true; that is, L(M) is not empty but the shortest string accepted by M has a length of at least n.

Let x be the shortest string accepted by M where |x| > n-1.

Using the Pumping Lemma, we know that there must be a "loop" in x which can be pumped 0 times to create a shorter string in L.

This is a contradiction and the result follows.

COMMENT: There are more efficient algorithms, but we won't get into that.

Testing if L(M) is finite

Incorrect "Algorithm"

Simulate M on all strings x.

Output yes if and only if there are a finite number of yes answers.

This "algorithm" is well specified and correct.

However, this is not an algorithm because there are an infinite number of strings to simulate M on, and thus it is not guaranteed to halt in a finite amount of time.

COMMENT: Note we again use the algorithm for the first problem as a subroutine.

Correct Algorithm

Simulate M on all strings of length between n and 2n-1 where M has n states.

Output yes if and only if no string is accepted.

Otherwise output no.

This algorithm clearly halts after a finite number of steps, and it is well specified.

The correctness of the algorithm follows from the fact that if M accepts an infinite number of strings, it must accept one of length between n and 2n-1.

This builds on the idea that if M accepts an infinite number of strings, there must be a "loop" that can be pumped.

This loop must have length at most n.

When we pump it 0 times, we have a string of length less than n.

When we pump it once, we increase the length of the string by at most n so we cannot exceed 2n-1. The problem is we might not exceed n-1 yet.

The key is we can keep pumping it and at some point, its length must exceed n-1, and in the step it does, it cannot jump past 2n-1 since the size of the loop is at most n.

 This proof is not totally correct, but it captures the key idea.

COMMENT: There again are more efficient algorithms, but we won't get into that.

   Other problems we can solve using these basic algorithms (and other algorithms we've seen earlier this chapter) as subroutines.

       COMMENT: many of these algorithms depend on your understanding of basic set operations such as set complement, set difference, set union, etc.

Given a regular expression r, is Lr finite?

Convert r to an equivalent FA M.

COMMENT: note we use the two algorithms for converting a regular expression to an NFA and then an NFA to an FA.

Test if L(M) is finite.

Output the answer to the above test.

Given two FAs M1 and M2, is L(M1) = L(M2)?

Construct FA M1-2 from M1 and M2 which recognizes the language L(>M1) - L(>M2).

COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.

Test if L(M1-2) is empty.

Construct FA M2-1 from M1 and M2 which recognizes the language L(>M2) - L(>M1).

COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.

Test if L(M2-1) is empty.

Answer yes if and only if both answers were yes.

Answer:

We can use APC to help with dynamic tracks by :-

Answer:

The algorithm is as follows:

1. Input sentence

2. Input character

3. Length = len(sentence)

4. count = 0

5. For i = 0 to Length-1

 5.1 If sentence[i] == character

    5.1.1 count++

6. Print count

7. Stop

The program in C++ is as follows:

#include <iostream>

#include <string.h>

using namespace std;

int main(){

char str[100];   char chr;

cin.getline(str, 100);

cin>>chr;

int count = 0;

for (int i=0;i<strlen(str);i++){

 if (str[i] == chr){

  count++;} }

cout << count;

return 0;}

Explanation:

I will provide explanation for the c++ program. The explanation can also be extended to the algorithm

This declares the string and the character

char str[100];   char chr;

This gets input for the string variable

cin.getline(str, 100);

This gets input for the character variable

cin>>chr;

This initializes count to 0

int count = 0;

This iterates through the characters of the string

for (int i=0;i<strlen(str);i++){

If the current character equals the search character, count is incremented by 1

 if (str[i] == chr){  count++;} }

This prints the count

cout << count;

Answer:

The function is as follows:

def addingAllTheWeirdStuff(array1,array2):

   sumOdd = 0

   for i in array2:

       if i % 2 == 1:

           sumOdd+=i

   for i in array1:

       if i < 10:

           sumOdd+=i

   print("Sum:",sumOdd)

   sumEven = 0

   for i in array1:

       if i % 2 == 0:

           sumEven+=i

   for i in array2:

       if i > 10:

           sumEven+=i

   print("Sum:",sumEven)

Explanation:

This declares the function

def addingAllTheWeirdStuff(array1,array2):

This initializes the sum of odd numbers in array 2 to 0

   sumOdd = 0

This iterates through array 2

   for i in array2:

This adds up all odd numbers in it

       if i % 2 == 1:

           sumOdd+=i

This iterates through array 1

   for i in array1:

This adds up all elements less than 10 to sumOdd

       if i < 10:

           sumOdd+=i

This prints the calculated sum

   print("Sum:",sumOdd)

This initializes the sum of even numbers in array 1 to 0

   sumEven = 0

This iterates through array 1

   for i in array1:

This adds up all even numbers in it

       if i % 2 == 0:

           sumEven+=i

This iterates through array 2

   for i in array2:

This adds up all elements greater than 10 to sumEven

       if i > 10:

           sumEven+=i

This prints the calculated sum

   print("Sum:",sumEven)

Answer:

b. False

Explanation:

False, unchecked exceptions do not cause compilation errors and do not require try/catch blocks or throws statements. However, although they are not required it is good programming to include them in order to handle any exceptions that may arise. If you do not include a proper way to handle such an exception the program will still run but may run into an exception during runtime. If this occurs then the entire program will crash because it does not know how to handle the exception since you did not provide instructions for such a scenario.

Answer:

The function is as follows:

def get_win_percentage(self):

       return self.team_wins / (self.team_wins + self.team_losses)

Explanation:

This defines the function

def get_win_percentage(self):

This calculates the win percentage and returns it to main

       return self.team_wins / (self.team_wins + self.team_losses)

See attachment for complete (and modified) program.

Answer: def get_win_percentage(self):

Explanation: got it right on edgen

Answer:

chosen ciphertext

Explanation:

Chosen ciphertext attack is a scenario in which the attacker has the ability to choose ciphertexts C i and to view their corresponding decryptions – plaintexts P i . It is essentially the same scenario as a chosen plaintext attack but applied to a decryption function, instead of the encryption function.

Cyber attack usually associated with obtaining decryption keys that do not run in fixed time is called the chosen ciphertext attack.

Hence, chosen ciphertext attempts the use of modular exponentiation.

Learn more : https://brainly.com/question/14826269

Answer: hi mom

Explanation: got it right on edgen

Answer:

here you go. wish you a great day tomorrow.

and in fact,computer science is somewhat the right category

Abstract art may be - and may seem like - almost anything. This because, unlike the painter or artist who can consider how best they can convey their mind using colour or sculptural materials and techniques. The conceptual artist uses whatever materials and whatever form is most suited to putting their mind across - that would be anything from the presentation to a written statement. Although there is no one kind or structure employed by abstract artists, from the late 1960s specific tendencies emerged.