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ryzh [129]
4 years ago
12

Which of the following is true of how the Internet has responded to the increasing number of devices now using the network? a) T

he Internet protocols are changed every year to adapt to the new devices that have been connected to the network b) While the number of devices connected to the Internet has grown, the network itself has not grown c) The protocols of the Internet were designed to scale as new devices are added d) The protocols of the Internet are no longer necessary thanks to the large number of devices now connected to the network
Computers and Technology
2 answers:
zalisa [80]4 years ago
8 0

Answer:

A

Explanation:

The internet protocols are changed every year to adapt to the new devices that have been connected to the network. Back in the 1990s, most traffic used a few protocols.  Pv4 routed packets, TCP turned those packets into connections, SSL (later TLS) encrypted those connections, DNS named hosts to connect to, and HTTP was often the application protocol using it all.

For many years, there were negligible changes to these core Internet protocols; HTTP added a few new headers and methods, TLS slowly went through minor revisions, TCP adapted congestion control, and DNS introduced features like DNSSEC. The protocols themselves looked about the same ‘on the wire’ for a very long time (excepting IPv6, which already gets its fair amount of attention in the network operator community.)

As a result, network operators, vendors, and policymakers that want to understand (and sometimes, control) the Internet have adopted a number of practices based upon these protocols’ wire ‘footprint’ — whether intended to debug issues, improve quality of service, or impose policy.

Now, significant changes to the core Internet protocols are underway. While they are intended to be compatible with the Internet at large (since they won’t get adoption otherwise), they might be disruptive to those who have taken liberties with undocumented aspects of protocols or made an assumption that things won’t change.

USPshnik [31]4 years ago
5 0

Answer:

Its C, they were designed to scale as new devices are added,

Explanation:

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What security weaknesses/vulnerabilities exist in Wireless local area network device hardware and software?
AfilCa [17]

Answer:

Explanation:

There are many weaknesses/vulnerabilities, some of which are the following...

Default Network Hardware, many individuals will go out and buy a new router and install it in their home. These devices come with a preset configuration including a preset security password which is many times a default password used for every router of the same model. This can be easily obtained by anyone who can then access the network.

Access Point hacking, an experienced individual can use a packet sniffer to detect the SSID that is frequently sent from the router in order to create an access point to be able to access the network.

WEP encryption is another vulnerability. These are very low security passwords that can be cracked using different software in a short period of time.

These are some of many vulnerabilities that wireless local networks have, but there are also many precautions and security measures that can be taken to prevent them.

7 0
3 years ago
Integral 3t+ 1 / (t + 1)^2​
Verizon [17]

Answer:

3ln|t+1|+\frac{2}{t+1} +C

Explanation:

We'll be using u-substitution for this problem.

Let

u=t+1\\du=dt

Substitute

\int\limits {\frac{3u-2}{u^2}} \, du

Split the fraction

\int\limits {\frac{3u}{u^2} } \, du -\int\limits {\frac{2}{u^2} } \, du

Move the constants out

3\int\limits {\frac{u}{u^2}du -2\int\limits {u^{-2}} \, du

Simplify

3\int\limits {\frac{1}{u}du -2\int\limits {u^{-2}} \, du

Integrate

3ln|u|+\frac{2}{u} +C

Substitute

3ln|t+1|+\frac{2}{t+1} +C

5 0
3 years ago
Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
zalisa [80]

Answer:

Explanation:

#include<iostream>

#include<ctime>

#include<bits/stdc++.h>

using namespace std;

double calculate(double arr[], int l)

{

double avg=0.0;

int x;

for(x=0;x<l;x++)

{

avg+=arr[x];

}

avg/=l;

return avg;

}

int biggest(int arr[], int n)

{

int x,idx,big=-1;

for(x=0;x<n;x++)

{

if(arr[x]>big)

{

big=arr[x];

idx=x;

}

}

return idx;

}

int main()

{

vector<pair<int,double> >result;

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

int choice;

cin>>choice;

while(choice!=2)

{

int n,m;

cout<<"Enter N"<<endl;

cin>>n;

cout<<"Enter M"<<endl;

cin>>m;

int c=m;

double running_time[c];

while(c>0)

{

int arr[n];

int x;

for(x=0;x<n;x++)

{

arr[x] = rand();

}

clock_t start = clock();

int pos = biggest(arr,n);

clock_t t_end = clock();

c--;

running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;

}

double avg_running_time = calculate(running_time,m);

result.push_back(make_pair(n,avg_running_time));

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

cin>>choice;

}

for(int x=0;x<result.size();x++)

{

cout<<result[x].first<<" "<<result[x].second<<endl;

}

}

8 0
3 years ago
A loop within another loop is known as a(n) ____ loop.
Natali [406]
Nested






..................................
4 0
3 years ago
Consider the following algorithms. Each algorithm operates on a list containing n elements, where n is a very large integer.
DochEvi [55]

We have that the appropriate statement will be

  • An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

From the question we are told

Consider the following <u>algorithms</u>. Each <u>algorithm</u> operates on a list containing n <em>elements</em>, where n is a very large <u>integer</u>.

I. An algorithm that accesses each <u>element</u> in the list twice.

II. An <em>algorithm </em>that accesses each <u>element </u>in the list n times.

III. An <u>algorithm</u> that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the <em>algorithms </em>run in <em>reasonable </em>time?

<h3>Algorithm  </h3>

Generally In order to get <em>admission </em>to every thing in the list twice, the algorithm has to traverse the listing twice,

which leads to 2*n entry to operations.

And if the every factor is accessed n times, the listing will be traversed n time, which leads to n^2 get right of entry to operations.

If n is a very giant <em>integer</em>, each 2*n and n^2 are plenty larger.

So, there will be <em>solely </em>ten entry to operations and this algorithm will have a sensible jogging time.

Therefore

An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

For more information on  algorithm  visit

brainly.com/question/950632

4 0
2 years ago
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