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3 years ago

Output a brief description of the game of hangman and how to play.

Computers and Technology

1 answer:

musickatia [10]3 years ago

3 0

Answer:

Hangman is game which can be played with a pencil or pen and a paper. It is a multiplayer game which requires minimum two players. It is a guessing game, in which a player thinks of a word or a phrase and other player or players tries to guess it correctly, letter by letter or whole word or phrase at once, in a certain number of guesses.

In this game, the word to be guessed is represented by dashes. If the guessed word or letter is correct, the player fills the blank spaces. But if the guessed word or letter is wrong, the player draws an element of a hanged man stick figure one by one as a tally mark.

This game ends if the player who is guessing guesses the word or the phrase correctly and the guesser wins; or if the player guesses enough wrong letters of phrases that the other player finishes the drawing of the hangman and the guesser loses.

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Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,

Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary denary or decimal

11111111 = 2⁸ = 256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s = /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1  1  1 1 1 1

128 64 32 16 8 4 2 1

step 3

in the ip network address: 198.16.0.0/19 (subnet representation) we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts+16: 198.16.16.0 - 198.16.31.255

+16: 198.16.32.0- 198.16.47.255 etc

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11111000.000000 /21

now we have added 5 1s in the third field to reserve 11 0s

subnet mask: 255.255.8.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1  1 1 1

128 64 32 16 8  4 2 1

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (subnet representation) we increment this using 8

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts +16: 198.16.24.0- 198.16.31.255

+16: 198.16.32.0- 198.16.112.255 etc

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1 1 1 1

128 64 32 16 8 4 2 1

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts +16: 198.16.40.0- 198.16.55.255

+16: 198.16.56.0- 198.16.71.255 etc

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11100000.000000 /19

now we have added 3 1s in the 3rd field to reserve 13 0s

subnet mask: 255.255.32.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1  1  1 1 1 1 1

128 64 32  16 8 4 2 1

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts +16: 198.16.72.0- 198.16.103.255

+16: 198.16.104.0- 198.16.136.255 etc

5 0

3 years ago

Get user input as a boolean and store it in the variable tallEnough. Also, get user input as a boolean and store it in the varia

frozen [14]

Answer:

I will code in JAVA.

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

boolean tallEnough;

boolean oldEnough;

Scanner input = new Scanner(System.in);

tallEnough = input.nextBoolean(); //wait the input for tallEnough

oldEnough = input.nextBoolean(); //wait the input for OldEnough

if(tallEnough && oldEnough){

System.out.print(true);

} else {

System.out.print(false);

}

Explanation:

First, to accept user inputs you have to import the class Scanner. Then declare both variables before allowing the user to set input values for both boolean variables.

In the if-else statement checks if both variables are true, then prints true. Another case prints always false.

8 0

3 years ago