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2 years ago

Which of the following best explains how algorithms that run on a computer can be used to solve problems?

1 answer:

7 0

The statement which best explains how algorithms running on a computer can be used to solve problems is; D. Some problems cannot be solved by an algorithm.

<h3>What is an algorithm?</h3>

An algorithm is simply a standard formula or procedures which is made up of a set of finite steps and instructions that must be executed on a computer, so as to proffer solutions to a problem or solve a problem under appropriate conditions.

However, it should be noted that it is not all problems that can be solved by an algorithm, especially because the required parameters and appropriate conditions are not feasible or met.

Read more on algorithm here: brainly.com/question/24793921

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This technology was developed in the 1980s and was a successful attempt to provide services for voice, video, and video traffic

sveticcg [70]

Answer:

SONET was the only one on the list created near the 1980s (1985)

Good luck!

~Awwsome

8 0

3 years ago

Hexadecimal to denary gcse method

Anuta_ua [19.1K]

There are two ways to convert from hexadecimal to denary gcse method. They are:

Conversion from hex to denary via binary.
The use of base 16 place-value columns.

<h3>How is the conversion done?</h3>

In Conversion from hex to denary via binary:

One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.

Example - Find out the denary value of hex value 2D.

It will be:

2 = 0010

D = 1101

Put them them together and then you will have:

00101101

Which is known to be:

0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1

= 45 in denary form.

Learn more about hexadecimal from

brainly.com/question/11109762

#SPJ1

3 0

2 years ago

To join a social network you create an avatar true or false

svp [43]

False because not all social media webstites provide avatars

7 0

3 years ago

Read 2 more answers

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

A malware-infected networked host under the remote control of a hacker is commonly referred to as:

natali 33 [55]

Answer:

Option a: Trojan

Explanation:

A Trojan or Trojan horse is one of the computer malware that exist in computing world. Trojan often appears as a legitimate software to deceive user to activate it by social engineering. Once the Trojan is activated in the user computer, a hacker can remote control the infected computer for malicious purposes such as removing files, sending files, displaying message or rebooting computer.

However, Trojan cannot be replicated in the infected computer.

7 0

3 years ago

Read 2 more answers