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vovikov84 [41]
3 years ago
6

Kristen wants to view the records of her database in ascending order. What should she do?

Computers and Technology
2 answers:
Veseljchak [2.6K]3 years ago
7 0
The answer is OC. Sort the table
Elanso [62]3 years ago
6 0

the answer is Sort the table

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Describe Format Painter by ordering the steps Jemima should follow to make the format of her subheadings more
borishaifa [10]

Answer:

Step 1: Select the subheading with the desired format.

Step 2:

Navigate to the ribbon area.

Step 3:

Go to the Clipboard command group.

Step 4:

Click the Format Painter icon.

Step 5:

Click and drag the icon on the second subheading.

Step 6: Use Reveal Formatting to make sure formatting was copied.

Explanation:

I just did this one hope it helps

8 0
3 years ago
____ are model building techniques where computers examine many potential solutions to a problem, iteratively modifying various
gregori [183]

Answer:

E. Genetic algorithms

Explanation:

In Computer science, Genetic algorithms are model building techniques where computers examine many potential solutions to a problem, iteratively modifying various mathematical models, and comparing the mutated models to search for a best alternative.

6 0
3 years ago
This software system shall consist of four source files (.cpp) and four header files (.h) defining four classes. The first class
salantis [7]

Answer:

Please see the attached file for the complete answer.

Explanation:

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6 0
4 years ago
Consider a disk with the following characteristics (these are not parameters of any particular disk unit): block size B 512 byte
telo118 [61]

Answer:

a) total capacity = (Block size + I. Gap size)*N

total capacity = (512+128 Bytes)* 20= 12800 Bytes = 12.8 Kb

Useful.cap = Block size * N = 512*20 = 10240 Bytes = 10.24 Hb

b) Cylinders c) [tex] T = (512+128 Gb) *20 *15*2 = 384000 Bytes= 384Kb

U = 512Gb*20*15*2= 307200 Bytes= 307.2 Kb

d) TS= 384000 Bytes *400 =153600000 Bytes= 153500 Kb= 153. 6Mb

UT = 307200 Bytes*400=122880000 Bytes = 122880 Kb = 122.88 Mb

e) RS= 2400 rpm * \frac{60 s}{1 min}= 144000 rps

Tr= \frac{total capacity}{RS} = \frac{12800 By}{144000 rps}= 0.0889 bytes/s

And if we convert using 1 s = 1000 ms we have"

0.0889 bytes/s * 1000 =88.89 bytes/ms

The block transfer time btt would be given by:

btt= \frac{512 bytes}{88.9 bytes/ms}=5.76 ms

And the average rotational delay would be given by:

rd= \frac{1}{2} (\frac{1}{2400 rpm}) 60 s  \frac{1000 ms}{1 s}= 12.5 ms

f) For this case we can calculate the average time to locate and transfer adding the following time:

TT= btt +rd+ st = 5.76 +12.5 +30 ms= 48.26 ms

g) For this case we can calculate the time to transfer 20 random blocks like this:

t_1 = 20*(s+rd+btt) = 20*(30+12.5+5.76) =965.2 ms

And the time to transfer 20 consecutive blocks using double buffering would be:

t_2 = s+ rd+ 20 btt = 30 +12.5 + 20 (5.76)=157.7 ms

Explanation:

Part a

For this case we need to calculate first the total capcity like this:

total capacity = (Block size + I. Gap size)*N

Where N represent the number of blocs per track, and if we replace we got:

total capacity = (512+128 Bytes)* 20= 12800 Bytes = 12.8 Kb

And the useful capacity is given by:

Useful.cap = Block size * N = 512*20 = 10240 Bytes = 10.24 Hb

Part b

For this case the number of cylinders correspond to the number of tracks.

Cylinders = tracks= 400

Part c

First we can calculate the total cylinder capacity like this:

T = (512+128 Gb) *20 *15*2 = 384000 Bytes= 384Kb

And the useful capacity is:

U = 512Gb*20*15*2= 307200 Bytes= 307.2 Kb

Part d

We can calculate the totals like on part d but we just need to multiply by 400 since that represent the number of tracks per surface

TS= 384000 Bytes *400 =153600000 Bytes= 153500 Kb= 153. 6Mb

UT = 307200 Bytes*400=122880000 Bytes = 122880 Kb = 122.88 Mb

Part e

For this case we can convert the revolution per minute in revolutions per second like this:

RS= 2400 rpm * \frac{60 s}{1 min}= 144000 rps

And we can calculate the transfer rate like this:

Tr= \frac{total capacity}{RS} = \frac{12800 By}{144000 rps}= 0.0889 bytes/s

And if we convert using 1 s = 1000 ms we have"

0.0889 bytes/s * 1000 =88.89 bytes/ms

The block transfer time btt would be given by:

btt= \frac{512 bytes}{88.9 bytes/ms}=5.76 ms

And the average rotational delay would be given by:

rd= \frac{1}{2} (\frac{1}{2400 rpm}) 60 s \frac{1000 ms}{1 s}= 12.5 ms

Part f

For this case we can calculate the average time to locate and transfer adding the following time:

TT= btt +rd+ st = 5.76 +12.5 +30 ms= 48.26 ms

Part g

For this case we can calculate the time to transfer 20 random blocks like this:

t_1 = 20*(s+rd+btt) = 20*(30+12.5+5.76) =965.2 ms

And the time to transfer 20 consecutive blocks using double buffering would be:

t_2 = s+ rd+ 20 btt = 30 +12.5 + 20 (5.76)=157.7 ms

7 0
3 years ago
It is desirable to provide a degree of __________ __________ among classes so that one class is not adversely affected by anothe
alexandr402 [8]

Answer:

The right approach is "Traffic isolation ".

Explanation:

  • A significant amount of those same traffic insulation is necessary. Therefore one class isn't influenced by yet another traffic class which makes a mistake. Then maybe the packages throughout the traffic could collapse.
  • This also eliminates uncertainty with people operating the infrastructure. If you don't need the VLANs to speak to one another because you could implement anything about this illustration as well as add it to certain VLANs.
4 0
3 years ago
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