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slavikrds [6]
3 years ago
14

____ documents consist of the text to be displayed on a Web page, together with a number of special characters: tags that achiev

e formatting, special effects, and references to other similar documents.
Computers and Technology
1 answer:
Y_Kistochka [10]3 years ago
3 0

Answer:

HTML

Explanation:

HyperText Markup Language ( HTML ) -

It is the foundation block of the Web page , which gives the main frame and structure to the web page .  

It is basically the type of language which is coded and generates a web page , and many web pages combine together to make up a web site in the internet .  

Many features like , formatting , color , fonts , special effects and special characters can be used in a coding of the HTML .

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How long would you need to work at McDonalds to earn enough money to buy a optiplex 3050, monitor, mouse and keyboard all from d
r-ruslan [8.4K]
So assuming you were buying this piece of trash with the mouse and keyboard it would be around $450. Average pay per hour is $9.10-$10.20 depending on were you live so we will just say $9.50. You'd need to work rougly 47 hours (without taxes) So i'd say 65-75 Hours.
8 0
3 years ago
Read 2 more answers
Draw a flowchart to accept two numbers, and then output the result as the first number
spin [16.1K]

Answer:

See attachment for flowchart

Explanation:

- The flowchart starts and ends with an oval shape.

- The flowchart accepts input for variables First and Second using the parallelogram shape.

- After both inputs have been collected;

The flowchart calculates First^Second and stores the result in variable Result.

-The value of Result is printed, afterwards.

8 0
3 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
Which of these describes, in increasing order, the file sizes of the same image in different formats? Group of answer choices
Nesterboy [21]

Answer:

If left being the shortest file size and right being the biggest file size...

It is JPEG, PNG, TIFF.

Explanation:

TIFF is used primarily for editing photos or for uncompressed high quality images for things like 4K textures. The file size will be bigger since generally it is uncompressed.

PNG reserves the maximum quality of the image and can not be compressed. (It can be compressed but the results are very minimal).

JPEG allows for the user to decide the compression and quality of the image. JPEG also doesn't hold transparency data, thus making JPEG generally use less space than PNG and definitely more than TIFF.  

8 0
3 years ago
TRUE/FALSE
algol [13]
<h2>Answer:</h2>

Following are given answers to each part with explanation:

<h2>Explanation:</h2>

1. lt takes multiple clock cycles to access data from memory.

The statement is TRUE.

Accessing data from memory acquires two steps. In first step the address is read and in second step we go to the address in order to access the data.

2.Each byte of memory has a unique address.

The statement is TRUE.

Each byte in the memory attains a unique address therefore we can access the data residing at each byte specifically by giving address.

3.The total memory used by all running programs can never be larger than the computer's physical memory.

The statement is TRUE.

The running parts are always a sub part o a system so combining them all cannot exceed the total size of computer's physical memory.

4.The lower half (or least significant half) of the EBX register is called BX.

The statement is TRUE.

BX is he least significant or lower half of the register EBX (extended BX).

5.The following data locations are in order of fastest access time to slowest access time: cache, registers, main memory.

The statement is TRUE.

Yes the order of data locations from faster to lower is given correctly:

Cache is accessed faster than registers and in the end comes main memory.

6.The ALU performs only addition, subtraction, multiplication, and division operation:s

The statement is FALSE.

ALU is an abbreviation of Arithmetic Logic unit. So it performs arithmetical operations such as addition, subtraction, multiplication, and division as well as the Logical operations that are AND OR NOT XOR.

7. The control unit (CU) coordinates the sequencing of execution steps in an instruction cycle.

The statement is TRUE.

As obvious from the name Control unit (CU) has assigned the coordination of execution of steps in sequence.

8.The EIP register is updated when an instruction is fetched.

The statement is TRUE.

Whenever the user retrieve any instruction from EIP or fetch the instruction from it. EIP gets updated.

9.The step to fetch an operand is always necessary in the instruction cycle.

The statement is TRUE.

Most of the Instructions demand an operand in the instruction cycle so that they could be operated on. While some instruction not demand the operand such as Exit. But majority demands the operand so it is true.

10.In an instruction cycle, the operands are fetched before the instruction is fetched.

The statement is FALSE.

The first part of instruction cycle is to fetch the instruction, whereas the operands are fetched after it. So the statement get FALSE.

<h3>I hope it will help you!</h3>
5 0
2 years ago
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