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2 years ago

If you said the bottom margin, you are talking about what?

Computers and Technology

2 answers:

Mama L [17]2 years ago

6 0

The space between the last line of words and the bottom of the page

user100 [1]2 years ago

3 0

The space between the words and the edge of the paper

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RAM IS often referred

miss Akunina [59]

3

thank me later :))))))))

7 0

2 years ago

Why is formatting important with a table?

Crazy boy [7]

Answer:

Explanation:

it is important to be succinct, and help the reader to find the information that they need quickly, and without wading through reams of superfluous information. formatting of table plays an important role in presenting the data clearly and scuccinctly.

7 0

3 years ago

Which options can you choose to determine the width and height of data or cells in a spreadsheet? You can choose the or rulers t

Tju [1.3M]

Answer :

Double click on boarder width an height will automatically adjust.

Manual procedure are following:

1.Open desire file of spreadsheet

2. Click on home tab in cell group and then click format.

3. Under cell size click on default width.

4. In standered column width box type new measurements and then click OK.

4 0

3 years ago

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.

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.

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

For what purpose are high-level programming languages used for

Mars2501 [29]

Answer:

To speed up the compiler during run time

Explanation:

5 0

3 years ago