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alukav5142 [94]
3 years ago
12

Which of these is unused normal, unvisited link ??

Computers and Technology
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

hover

Explanation:

hope my answer help you

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What is a spreadsheet program?<br> A spreadsheet program is a computerized version of _______
STALIN [3.7K]
A spreadsheet program is a computerized - simulations of paper accounting worksheets. The program operates on data entered in cells of a table
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3 years ago
To implement a small database, a database designer must know the "1" and the "M" sides of each relationship and whether the rela
Marysya12 [62]

Answer:

True.

Explanation:

"1" to "M" or 1 to many is a type of relationship between tables in a relational database. It means that against 1 record in a table, there can be 0 to many rows in the other table. Not that the many or child table doesn't need to have any record against the 1 or parent row, so the many side is optional.

Example can be a Customer and Order relationship. Where Customer is the parent table and Order is the child table. "A customer can have 0 to many orders".

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3 years ago
Each computer on a network requires a unique way to identify itself and to refer to other computers. This is accomplished by usi
pantera1 [17]

Answer:

True

Explanation:

<em>IP Address</em>: It is used to uniquely identify each device over the network.

3 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
3 years ago
You designed a program to create a username using the first three letters from the first name and the first four letters of the
Pavel [41]

Answer:

See Explanation

Explanation:

The question would be best answered if there are options to select from; since none is provided, I will provide a general explanation.

From the question, we understand that, you are to test for Jo Wen.

Testing your program with this name will crash the program, because Jo has 2 letters (3 letters are required), and Wen has 3 letters (4 letters are required)

So, the step that needs to be revisited is when the username is generated.

Since the person's name cannot be changed and such person will not be prevented from registering on the platform, you need to create a dynamic process that handles names whose lengths are not up to the required length.

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3 years ago
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