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Allushta [10]
3 years ago
12

People who connect one network to another within the company or even across organizations are A. central connectors. B. boundary

spanners. C. external specialists. D. peripheral specialists.
Computers and Technology
1 answer:
Lunna [17]3 years ago
5 0

People who connect one network to another within the company or even across organizations are boundary spanners.  

People who serve as ambassadors to linking information and communications on informal networks within different department in an organisation and with other organisations or personnel  far and near for the good of their organisation are called Boundary Spanners.

The importance of Boundary  spanning in organisation includes

  • Exchange of expertise information
  • Nurturing of connections with people from different part s of the world.
  • Improved innovation in businesses

See more here: brainly.com/question/14728967

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If you are referencing cell (C2)in Excel and want to be able to copy the formula and keep using the data in cell C2 in every pla
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Read 2 more answers
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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.

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
3 years ago
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