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4 years ago

When using the “rule of thirds,” where should focal points be placed?

1 answer:

6 0

Answer: When using the “rule of thirds” you should place focal points along a vertical or horizontal line and at a point of intersection.

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Type the correct answer in the box. Spell all words correctly.

Feliz [49]

Answer:

While trying to minimize project cost, project manager Ben realized that adjusting the schedule is not completely serving the purpose. What else can ben do to minimize project cost?

In addition to adjusting the schedule, Ben can adjust Everything. For example, the Ben can hire people on contract to reduce costs

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3 years ago

Read 2 more answers

Why is it not advisable to mark tlc plates with a pen to indicate?

nikitadnepr [17]

The plate can change the function that is in it. If you were to name it wrong, you would have to debug the code and possibly rewrite it.

3 0

4 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

4 years ago

A consumer-directed digital or computer device that has a video output signal for displaying a video game is called a .......

Allushta [10]

I think the first problem is B. and the second problem is D.

4 0

4 years ago

n channels of bandwidth Bc are multiplexed together with FDM on a link. The guard band between two adjacent channels has bandwid

Andrei [34K]

Answer: The answer to the question is as follows:

Minimum BW= n*Bc + (n-1) Bg

Explanation:

FDM is a multiplexing technique that takes several baseband signals of n Khz wide (let's assume that all have the same bandwidth), and translates them in frequency so they can be transmitted together, at a higher frequency.

If we assume no guard bands between the different signals, the mimum bandwidth needed would be n times the bandwidth of a single signal.

In order to avoid crosstalk between signals, as the communication channel is not perfect, it usually leaves some room between any 2 signals, which it is called a band guard, and because there is a band guard for any 2 contiguous signals in the spectrum, the total number of bandguards (Bg) will be equal to (n-1) signals, so the total bandwidth to be used will be as follows:

BW needed: n*Bc + (n-1) Bg.

8 0

4 years ago