Sang ayon kaba sa pahayag na walang sinuman ang nabubuhay para sa sarili lamang? bakit?

CAPTCHAs can be used as a form of signature to create a valid contract in e-commerce. True False

Len [333]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The statement "CAPTCHAs can be used as a form of signature to create a valid contract in e-commerce. " is true. It is <span>a program or system intended to distinguish human from machine input, typically as a way of thwarting spam and automated extraction of data from websites.</span>

4 0

2 years ago

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8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101

Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

= 64 + 32 + 16 + 8 + 4 + 2 + 1

= 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

= 128 + 64 + 32 + 16

= 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

= 64 + 32 + 4 + 2 + 1

= 103

01111101 = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

= 64 + 32 + 16 + 8 + 4 + 1

= 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 10101111 11000000 11111000 00011101 to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 223

10110000 = 1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 247

11000111 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0

2 years ago

#A year is considered a leap year if it abides by the #following rules: # # - Every 4th year IS a leap year, EXCEPT... # - Every

lara [203]

Answer:

To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.

Similarly check for 400th year and multiple 0f 4. The following C program describes the function.

#include<stdio.h>

#include<stdbool.h>

bool is_leap_year(int year);

void main()

{

int y;

bool b;

printf("Enter the year in yyyy format: e.g. 1999 \n");

scanf("%d", &y); // taking the input year in yyyy format.

b= is_leap_year(y); //calling the function and returning the output to b

if(b==true)

{

printf("Thae given year is a leap year \n");

}

else

{

printf("The given year is not a leap year \n");

}

bool is_leap_year(int year)

{

if(year%100==0) //every 100th year

{

if(year%400==0) //every 400th year

{

return true;

}

else

{

return false;

}

if(year%4==0) //is a multiple of 4

{

return true;

}

else

{

return false;

}

Explanation:

Output is given as image

5 0

2 years ago

Create a class named BaseballGame that contains data fields for two team names and scores for each team in each of nine innings.

m_a_m_a [10]

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

protected String[] names = new String[2];

protected int[][] scores;

protected int innings;

public BaseballGame() {

innings = 9;

scores = new int[2][9];

for(int i = 0; i < 9; i++)

scores[1][i] = scores[0][i] = -1;

}

public String getName(int team) {

return names[team];

}

public void setNames(int team, String name) {

names[team] = name;

}

public int getScore(int team, int inning) throws Exception {

if(team < 0 || team >= 2)

throw new Exception("Team is ut of bounds.");

if(inning < 0 || inning >= innings)

throw new Exception("Inning is ut of bounds.");

return scores[team][inning];

}

public void setScores(int team, int inning, int score) throws Exception {

if(team < 0 || team >= 2)

throw new Exception("Team is ut of bounds.");

if(inning < 0 || inning >= innings)

throw new Exception("Inning is ut of bounds.");

if(score < 0)

throw new Exception("Score is ut of bounds.");

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

if(scores[team][i] == -1)

throw new Exception("Previous scores are not set.");

scores[team][inning] = score;

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

public HighSchoolBaseballGame() {

innings = 7;

scores = new int[2][7];

for(int i = 0; i < 7; i++)

scores[1][i] = scores[0][i] = -1;

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

public LittleLeagueBaseballGame() {

innings = 6;

scores = new int[2][6];

for(int i = 0; i < 6; i++)

scores[1][i] = scores[0][i] = -1;

}

6 0

2 years ago

Please help!! Caleb is working on a simple logic-based program to simulate a game of tic-tac-toe. Which programming language wou

irga5000 [103]

Basic language should do it

4 0

2 years ago