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IrinaK [193]
3 years ago
9

The platform in E-Learning refers to:

Computers and Technology
1 answer:
ollegr [7]3 years ago
8 0
A.

The type of computer system that the course can be taken on.
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Google Analytics can collect behavioral data from which systems?
ankoles [38]

Answer:

E.A,B and C

Explanation:

The behavioral data can be collected from E-commerce platforms,Mobile applications,Online point-of-sales system  by Google Analytics because these platform a person always searches what he is interested in.It describes their behavior.What type of lifestyle the person is living,the persons daily schedules etc.So the answer is option E.

6 0
3 years ago
Al and Bill are arguing about the performance of their sorting algorithms. Al claims that his O(N log N)-time algorithm is alway
lisabon 2012 [21]

Answer:

See in Explanation

Explanation:

All you know is that an O(n log n) algorithm is eventually (a lot) faster than an O(n^{2} ) algorithm, but for any specific value of n, you can't say anything. Indeed, the asymptotic growth of the function doesn't depend on its value on n. This means that if  f(n)=O(n^{2} ) then for all c_{1} ,....\ , c_{99}, the following function is also O(n^{2} ):

f^{'} (n)=[\ cnf(n)\  if \ n

So asymptotic notation is completely uninformative regarding the performance of algorithms on specific values of n.

You could say that functions such as f^{'}(n) are artificial and never really occur in practice. This is not quite true (you can hard-code some auxiliary information for small n, for example), but even if you consider "natural" functions, asymptotic notation doesn't help you to determine anything other than asymptotic's: consider for example n^{2} against 10^{100}n\  log \ n (such examples definitely happen in practice, for example in fast matrix multiplication).

7 0
3 years ago
The title element in the head section of an HTML document specifies the text
bagirrra123 [75]

Answer: c. that's displayed in the title bar of the browser

Explanation:

The title element in the html write up is used to describe the title of the work that is been carried out.

It has the format <title></title>, the title name is indicated between the opening and closing title tag.

The title does not display on the main page of the Web user's display, but it can be seen on the title bar of the Web browser.

The title element help Web user to have an idea of what a Web page is about.

7 0
3 years ago
In a digital computer system, the data is transferred between the CPU and the other components
FinnZ [79.3K]

Answer:

The data transmission or information exchange from the CPU to either the computer's peripherals is accomplished via "Computer ports". The further explanation is given below.

Explanation:

To link a display, camera, microphones, or other devices, computer ports have several purposes. The Processor (CPU) also interacts through a bus to peripherals. Several categories of buses you must have got to hear of would be universal servo controller, PCI, or Compulsive-ATA.

  • A peripheral device has always been defined as any auxiliary appliance including a mouse as well as a keyboard, which somehow helps connect to either the PC but instead operates with it.
  • Computer systems were composed of various hardware. This would include the CPU, storage, buses, devices, etc.

So that the above is the right answer to the given scenario.

8 0
3 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

5 0
4 years ago
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