In this activity, you'll decide how a computer (in this case, a smartphone)

Which of the following is considered a modern method of communication?

Vika [28.1K]

The answer is C. Tablet, because that is the quickest, easiest, and most recent way of communication.

4 0

3 years ago

The term packet is used fairly generically to refer to protocol data unit (PDU). There are PDU equivalent names in the different

Westkost [7]

Answer:

Bits

Explanation:

The protocol data unit is the representative unit of data in the OSI layer of a network. The OSI system has seven layers.

The physical layer is the first layer of the system and the protocol data unit is represented as bits of data.

Note that the term packet is the PDU for data in the network layer of the OSI network system.

4 0

2 years ago

Sdq1fdszgdSHTGafeges'

kumpel [21]

Answer:

iooooooooooooooooooooo

Explanation:

ooooooooooooooo

oooo9ooooo9oooooooooo

ooooooooooooooppppooooooooooo

oooooooooooo

ooooooooooooooooooooo

6 0

3 years ago

Assuming that the message variable contains the string "Happy holidays", the statement message.find("days"); returns a. -1b. 10c

katrin [286]

Answer:

Answer is b. 10

Explanation:

The find() method is used to search the string for a specified value and searches the first occurrence of that value. It returns the position of that specified value. If the value is not found, this method returns -1.

Lets say we have a message variable that contains the string Happy holidays.

message = "Happy holidays"

We use the method find() to find days word from the Happy holidays string in the message variable.

The find() method will find the position of day.

If we look at the string Happy holidays we see that the word day is at the 10th position of the string.

H 1

a 2

p 3

p 4

y 5

h 6

o 7

l 8

i 9

d 10

a 11

y 12

s 13

So it is at 10th position so the statement

message.find("days")

returns 10

7 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