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3 years ago

Consider the following code segment - 3

Computers and Technology

1 answer:

abruzzese [7]3 years ago

8 0

Answer:

The answer is "Choice C".

Explanation:

In the above-given code, a 1D array "list" is declared that holds integer values, and it uses three print which can be defined as follows:

In a first print method, it uses the length method that holds the array length that is "9".

In a second print method, it uses the array that prints the first element value of the array that is "22".

In the last print method, it uses the array that holds the length of the array, which will give an error message that is "ArrayIndexOutOfBoundsException".
That's why only the choice C is correct.

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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

What is one pass of a coding sequence called?

Oksi-84 [34.3K]

Answer:

In computer programming, a one-pass compiler is a compiler that passes through the parts of each compilation unit only once, immediately translating each part into its final machine code. ... This refers to the logical functioning of the compiler, not to the actual reading of the source file once only.

Explanation:

8 0

3 years ago

Read 2 more answers

In a system with paging as the address translation mechanism, using multi-level paging always causes less memory usage for stori

Zarrin [17]

Answer:

The given statement is "True". A further explanation is given below.

Explanation:

Multi level paging can conserve storage capacity as opposed to sequential paging, including some of its hierarchical method Multi-level paging completing such.
Rather precisely, with either the aid of the number of co-association, clients will determine which unique category of 2²⁰ pages whose data are stored as well as the pick itself.

3 0

3 years ago

2. How do upgrading and retraining help you cope with change?

Luden [163]

<h2>Upgrading and retraining are mandatory to move along with the world.</h2>

Explanation:

Let us understand the term deeply,

Upgrading - Updating yourself with the latest

Retraining - learning new skills

Let me give you a real-life example which is nothing but "mobiles". If you are not updated then:

you will sit with mobile to make calls and
do money transactions only by stepping into the bank,
connect with people only through calls or directly visiting them,
distance break up the relationship,
booking tickets in classical way, etc.

These could be done in one touch if you have latest mobile with necessary applications.

In a similar way, we need to get retrained to get to learn new skills, technologies so that we can do our job the best, to be on track, be productive, convert your valuable knowledge in terms of money, to be peaceful in day today transactions, etc.

7 0

3 years ago

Which term describes a number in base 16, using the digits from zero to nine along with A, B, C, D, E, and

cupoosta [38]

Answer:

It is Hexadecimal

Explanation:

Hexadecimal EX: A3 B8

8 0

3 years ago