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3 years ago

Which device is similar to a hub but can isolate groups of devices on a network?

Computers and Technology

1 answer:

alexdok [17]3 years ago

5 0

Answer:

Router

Explanation:

A hub is so similar like a switch, but a switch is smarter than the hub because can be to know what port is connected, but is this particular question we have a router like options, this is the smartest of the three options because we can segregate and isolate groups of the device on a network, this is the most complex to use of the three option because we have to make some settings like the DHCP.

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Suppose you have a program P and 90 percent of P can be parallelized, but 10 percent of P is inherently sequential and cannot be

EleoNora [17]

Answer:

Means no matter how many processors you use, speed up never increase from 10 times.

Explanation:

If a problem of size W has a serial component Ws,then performance using parallelism:

Using Amdahl's Law:

Tp = (W - Ws )/ N + Ws

Here, Ws = .1,

W - Ws = .9

Performance Tp = (.9 / N) + .1

---------------------------------------------------------

Speed Up = 1 / ( (.9 / N) + .1)

If N -> infinity, Speed Up <= 10

Means no matter how many processors you use, speed up never increase from 10 times.

5 0

3 years ago

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

3 years ago

How do you fix peer answer bad gateway 502. Please help

kondaur [170]

Answer:

Reload the page.

Look for server connectivity issues.

Check for any DNS changes.

Sift through your logs.

Fix faulty firewall configurations.

Comb through your website's code to find bugs.

Contact your host.

Explanation:

7 0

4 years ago

When the program generates or populates reports after getting the data and processinng,it is called

TiliK225 [7]

Answer:

In computing, extract, transform, load (ETL) is the general procedure of copying data from one or more sources into a destination system which represents the data differently from the source(s) or in a different context than the source(s).

Data extraction involves extracting data from homogeneous or heterogeneous sources; data transformation processes data by data cleansing and transforming them into a proper storage format/structure for the purposes of querying and analysis; finally, data loading describes the insertion of data into the final target database such as an operational data store, a data mart, data lake or a data warehouse. Hope this helps! PLEASE GIVEE ME BRAINLIST!!! =)

6 0

3 years ago

Consider an error-free 64 kbps satellite channel used to send 512 byte data frames in one direction, with very short acknowledge

Vladimir [108]

Answer:

The answer is "2".

Explanation:

In the given question some information is missing, that is "The propagation time for satellite to earth" which is "270 milliseconds" so, the description to this question can be defined as follows:

Given values:

Bandwidth = 64 kbps

Data frames = 512 bytes

Propagation Time (tp ) =270 ms

Change Bandwidth kbps to bps:

1 kb= 1024 bytes

calculated bandwidth= 64 kbps = 64×1024 bps = 65536 bps

1 bytes = 8 bits

512 bytes = 512 × 8 = 4096 bits

Frame length = 4096 bits

Formula

Transmission time (Tt) = Frame length /Bandwidth

Window size = 1+2a

where a = Propagation time/Transmission time

Calculate Transmission time:

Transmission time (Tt) = 4096 / 65536

Transmission time (Tt)= 625 m.sec

Calculate Window size:

Window size = 1+2(270/625)

Window size = 1+2(0.432)

Window size = 1+0.864

Window size = 1.864

Window size = 2

3 0

3 years ago