Which of the following is a benefit, as well as a risk, associated with client/server networks?

The _______________ domain refers to any endpoint device used by end users, which includes but is not limited to mean any smart

viva [34]

The answer is <span>workstation. </span>The workstation domain refers to any endpoint device used by end users, which includes but is not limited to mean any smart device in the end user's physical possession and any device accessed by the end user, such as a smartphone, laptop, workstation, or mobile device.

5 0

2 years ago

Together with some anthropologists, you’re studying a sparsely populated region of a rainforest, where 50 farmers live along a 5

Varvara68 [4.7K]

Answer:

Explanation:

Farmers are always both directly and indirectly connected to each other

Their network is mostly strong

Networks become weak only on the edges (ends) of the river but doesn't completely dimnish

With the available network length, the center of river bank forms the strongest network of all and becomes a key player in defining the balance property of overall network

The network is very well structurally balanced and we can see that through the below image

20 miles 10 20 30 40 50

See attachment file for diagram

Considering the total length of river as 50miles and and the center of the whole length will be at 25th mile. From that point, if we consider a farmer will be be having friends for a length of 20miles both along upstream and downstream.

By this he'll be in friend with people who are around 80% of the total population. As me move from this point the integrity increases and this results in a highly balanced structural network.

6 0

2 years ago

What might be one reason why a stock becomes more valuable over time

Nikolay [14]

1. The General Market is Down This is for the most part the most evident reason that a stock is underestimated and happens when the large scale perspective of the economy is poor. It is helpful for financial specialists to have some fundamental apparatuses to esteem the general market so they can get ready as the market progresses toward becoming underestimated.

7 0

2 years ago

Using Amdahl’s Law, calculate the speedup gain of an application that has a 60 percent parallel component for (a) two processing

Nina [5.8K]

Answer:

a) Speedup gain is 1.428 times.

b) Speedup gain is 1.81 times.

Explanation:

in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:

$t=\frac{1 }{(S + (1- S)/N) }$

Where S is the portion of the application that must be performed serially, and N is the number of processing cores.

(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t = \frac{1}{(0.4 + (1-0.4)/2)} =1428671$

Speedup gain is 1.428 times.

(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t=\frac{1}{(0.4 + (1-0.4)/4)} = 1.8181$

Speedup gain is 1.81 times.

8 0

2 years ago

The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided

poizon [28]

Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

Random randNum = new Random();

int people = 5;

int[] birth_Day = new int[365+1];

// setting up birthsdays

for (int i = 0; i < birth_Day.length; i++)

birth_Day[i] = i + 1;

int iteration;

// varying number n

while (people <= 100)

{

System.out.println("Number of people: " + people);

// creating new birth day array

int[] newbirth_Day = new int[people];

int count = 0;

iteration = 100000;

while(iteration != 0)

{

count = 0;

for (int i = 0; i < newbirth_Day.length; i++)

{

// generating random birth day

int day = randNum.nextInt(365);

newbirth_Day[i] = birth_Day[day];

}

// check for same birthdays

for (int i = 0; i < newbirth_Day.length; i++)

{

int bday = newbirth_Day[i];

for (int j = i+1; j < newbirth_Day.length; j++)

{

if (bday == newbirth_Day[j])

{

count++;

break;

}

iteration = iteration - 1;

}

System.out.println("Probability: " + count + "/" + 100000);

System.out.println();

people += 5;

}

4 0

2 years ago