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Alenkasestr [34]

2 years ago

12

Did this technological advancement take away or create jobs? Overall, what was the contribution of this technological advancemen

t to society?

Computers and Technology

1 answer:

mars1129 [50]2 years ago

3 0

It took away jobs and made day to day tasks easier

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Which of the following would be better communicated with an image than with text?

slega [8]

Answer:

A decor to a wedding

Explanation:

its bc who would want to know how its described when u could look of how it looks u might be speechless

8 0

2 years ago

One of the most common uses of spreadsheet are

otez555 [7]

I think it is Index Match?

6 0

2 years ago

Suppose users share a 1-Gbps link. Also, suppose each user requires 200 Mbps when transmitting, but each user only transmits 30

V125BC [204]

Answer:

The answer is "5 users and 1 block".

Explanation:

In Option a:

Bandwidth total $= 1-Gbps \times 1000$

$= 1,000 \ Mbps$

Any User Requirement $= 200 \ Mbps$

The method for calculating the number of approved users also is:

Now, calculate the price of each person for overall bandwidth and demands,

$\text{Sponsored user amount} = \frac{\text{Bandwidth total}}{\text{Each user's requirement}}$

$=\frac{1000}{200}\\\\=\frac{10}{2}\\\\= 5 \ users$

In Option b:

$\text{blocking probability} = \frac{link}{\text{1-Gbps} = 10^9 \frac{bits}{sec}}$

$\ let = 0.05 = \frac{100}{20} \\\\\text{blocking probability} = \frac{ 200 \times 10^6}{\frac{100}{20}}$

$= \frac{ 200 \times 10^6 \times 20 }{100}\\\\= \frac{ 2 \times 10^6 \times 20 }{1}\\\\= 40 \times 10^6 \\\\$

mean user $= 25 \times \frac{1}{20} \\\\$

$= 1.25$

max user $= \frac{10^9}{40 \times 10^6} \\\\$

$= \frac{10^9}{4 \times 10^7} \\\\ = \frac{10^2}{4} \\\\ = \frac{100}{4} \\\\= 25 \\\\ =\ \ 1 \ \ block \\\\$

3 0

2 years ago

Which type of electronic community displays discussions on different topics in threads?

LekaFEV [45]

Science had been done eating junk food and eating a lot more then a week after

7 0

2 years ago

Suppose there are 69 packets entering a queue at the same time. Each packet is of size 7 MiB. The link transmission rate is 1.7

svp [43]

Answer:

69.08265412 milliseconds

Explanation:

Lets first convert 7 MiB to bits

$7*1024*1024*8=58720256$ bits

Now convert bits to Gbits

$58720256/10^{9} =0.058720256$ Gbits

Queuing Delay = Total size/transmission link rate

Queuing Delay= $\frac{0.058720256}{1.7} =0.03454132706$ seconds

Delay of packet number 3 = $0.03454132706*2=0.06908265412$ seconds

or $0.06908265412= 69.08265412$ milliseconds

7 0

2 years ago