Help please?

It's a holiday poop in my bitt

Zina [86]

Answer:

bitt

Explanation:

pp

4 0

3 years ago

Read 2 more answers

You are going to write a program for Computer test which will read 10 questions from a file, order them randomly and provide the

ollegr [7]

Answer:

from pprint import pprint

import random

match = {

"mcq1": "a",

"mcq2": "c",

"mcq3": "c",

"mcq4": "d",

"mcq5": "a",

"mcq6": "a",

"mcq7": "a",

"mcq8": "b",

"mcq9": "d",

"mcq10": "a"

}

file = open("test","r")

questions = set()

for i in range(10):

ques = []

for line in file:

if "-" in line:

break

else:

ques.append(line)

questions.add(tuple(ques))

def quiz(questions, match):

questions = list(questions)

marks = 0

for i in range(10):

pprint(questions[i])

mcq = "mcq",str(i+1)

mcq = list(mcq)

answer = input("enter your answer: ")

if answer == match["".join(mcq)]:

marks += 1

return marks

print("Your score is: ",quiz(questions, match))

Explanation:

the file which has been loaded is test, and its data is provided below...

mcq1. The device which converts analog signals to digtital signals and vice versa is called.

a) mother board

b) TAP

c) Modem

d) I/O device

-

mcq2. The main components of a computer system are.

a) TAP, CPU, Printer

b) CPU, Input device

c) CPU, ALU, CU

d) CPU , Output device , Memory unit, Control unit

-

mcq3. A source program is a program.

a) writter in machine laguage

b) translated in machine langaue

c) written in high level language

d) required to boot a computer

-

mcq4.Which image format supports transparency in images.

a) PNG

b) GIF

c) JPG

d) A & B

-

mcq5. (10111) 2 = (?) 10.

a) 23

b) 50

c) 24

d) 89

-

mcq6. UNICODE is an example of.

a) character encoding set

b) driver

c) software

d) database

-

mcq7. (10111) 2 = (?) 10.

a) 23

b) 50

c) 24

d) 89

-

mcq8. NTFS stand for.

a) Network File Saving

b) New Technology File System

c) Newt Trend File Saving

d) Non Technology File System

-

mcq9. FF is example of.

a) Octal number system

b) Binary Number System

c) Decimal Number System

d) Hexadecimal number system

-

mcq10. Emails are sent with the help of ?

a) SMTP

b) FTP

c) HTTP

d) UDP

-

3 0

3 years ago

Components that enhance the computing experience, such as computer keyboards, speakers, and webcams, are known as

never [62]

Answer:

- Peripheral devices

Explanation:

Peripheral devices are defined as computer devices which are not the element of the essential/basic computer function. These devices can be internal as well as external and are primarily connected to the computer for entering or getting information from the computer. For example, the keyboards or mouse functions to enter data into the computer for processing and receiving information while the output devices like speakers, projectors, printers, etc. are used to get the information out of the computer.

3 0

3 years ago

Is a computer network that spans a relatively small area, allowing all computer users to connect with each other to share data a

Natalija [7]

The answer is a Local Area Network

5 0

3 years ago

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