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2 years ago

Which of formula contains an absolute cell reference?

Computers and Technology

1 answer:

svet-max [94.6K]2 years ago

3 0

Which formula contains an absolute cell reference? =SUM($B$7:$B$9)

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To give your app users the ability to open your app directly from other apps by clicking a link, you should use:.

pickupchik [31]

To give your app users the ability to open your app directly from other apps by clicking a link, you should use: deep link. With the deep link and its URL functionality, existing app users are driven directly inside the mobile app itself.
Deep links are usually made up of two parts: a scheme (part of the link that identifies which app to open).and a host and path (the unique location in the app where your content exists).

7 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

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

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 $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
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

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

The resulting run has length 6000 records.

Part b

For step 1

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

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

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

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

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.

For step 2

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

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

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

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

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)

Now the total time for merging is given as

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

Total time in merging is 249 seconds seconds

5 0

3 years ago

A computing company is running a set of processes every day. Each process has its own start time and finish time, during which i

Helga [31]

Answer:

Above all else, let us show that this issue has the greedy choice property.

This implies that worldwide ideal arrangement can be gotten by choosing local ideal arrangements. Presently notice the following:

The global optimal solution of the issue requests us to track down the minimum number of times process_check module should be run.

It is likewise referenced that during each running interaction, process_check module should be called atleast once. Along these lines, the minimum possible number of process_check calls happens when process_check is made close to the quantity of the processes that are run.

Along these lines, we see that the global optimal solution is shaped by choosing optimal answer for the nearby advances.

Along these lines, the issue shows greedy choice property. Thus, we can utilize a greedy algorithm for this issue.

The greedy step in this calculation is to postpone the process_check as far as might be feasible. Thus, it should be run towards the finish of each interaction. So the following algorithm can be utilized:

Sort the cycles by utilizing the completion times.

Towards the finish of the 1st process in the arranged list, run the process_check module. Right now any process that is now running is removed the arranged rundown. The main interaction is additionally removed from the sorted list.

Now repeat stage 2, till all processes are finished.

In the above mentioned algorithm, the costliest advance is the step of sorting. By utilizing the optimal sorting algorithm, for example, merge sort, we can achieve it in O(n log n) asymptotic time.

Along these lines, this greedy algorithm additionally takes O(n log n) asymptotic time complexity.

5 0

3 years ago

Please solve in 5 mins very fast

Dmitriy789 [7]

Answer:

a. virtual reality

b. Master Boot Records

c. Primary function of a router

d. zoom

Explanation:

4 0

3 years ago

Read 2 more answers

What term is the ability to restore service quickly and without lost data if a disaster makes your components unavailable or it

oksian1 [2.3K]

Answer:

"Recoverability" is the appropriate solution.

Explanation:

The above refers to either the DBMS work timetables where only certain processes are conducted out only when all transactions where such modifications are learned by the submission are implemented or operated.
This technique is used to reinstate make informed even without system failures.

8 0

3 years ago