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Svetradugi [14.3K]

3 years ago

6

What are scape codes and the effectes on out put

Computers and Technology

1 answer:

Nina [5.8K]3 years ago

6 0

They are called escape (not scape) codes. They are special characters used to send commands to the output device such as "new line" or "next page", etc.

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Which of the following is not a valid SQL command? (Points : 2) UPDATE acctmanager SET amedate = SYSDATE WHERE amid = 'J500';

Volgvan

Answer:

UPDATE acctmanager WHERE amid = 'J500';

Explanation:

The statement written above is not valid SQL statement because there is no SET after UPDATE. Update is always used with SET.If you are updating something then you need to specify the value with which that value should be replaced.

UPDATE acctmanager SET amname = UPPER(amname);

This statement does not contains WHERE clause but it will run all the values of amname column will get updated in the table acctmanager.

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2 years ago

You are going to visit a national park, and have never been there before. You are using a map to try and make the distance trave

Firdavs [7]

Answer:

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8 0

2 years ago

The _____ feature will allow users to view nonprinting formatting marks to aid in editing a document. View Alignment Show/Hide I

Annette [7]

The __Show/Hide Insert___ feature will allow users to view non printing formatting marks to aid in editing a document.

The answer is D. Show/Hide Insert

Let me know if this is correct

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2 years ago

Is “Annabelle” a true story? I mean it is fiction right?

Julli [10]

Annabelle is indeed a true story making it fiction

5 0

2 years ago

Read 2 more answers

Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the

bekas [8.4K]

Answer:

The answer to this question can be described as follows:

Explanation:

Given data:

Performance of the CPU:

The Fastest Factor Fraction of Work:

$f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65$

Current Feature Speedup:

$K_1=$ 1.5

CPU upgrade=6000

Disk activity:

The quickest part is the proportion of the work performed:

Current Feature Speedup:

$k_2=3$

Disk upgrade=8000

System speedup formula:

$s=\frac{1}{(1-f)+(\frac{f}{k})}$

Finding the CPU activity and disk activity by above formula:

CPU activity:

$S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1) \\$

Disk activity:

$S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)$

CPU:

Formula for CPU upgrade:

$= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)$

DISK:

Formula for DISK upgrade:

$=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)$

equation (3) and (4),

Thus, for the least money the CPU alternative is the best performance upgrade.

b)

From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

c)

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095

4 0

2 years ago