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padilas [110]
3 years ago
6

A que generacion pertenecen los gadgets?

Computers and Technology
1 answer:
alukav5142 [94]3 years ago
8 0

Answer:

English please

Explanation:

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Which command should you enter to configure a single port to discard inferior bpdus 200-125?
Airida [17]

Answer:

spanning-tree portfast bpduguard

Explanation:

spanning- tree protocol (STP) is a layer 2 protocol in the OSI model. It is automatically configured in a switch to prevent continual looping of BPDUs, to avoid traffic congestion. The fastport bpduguard is only applicable in non-trunking access in a switch. It is more secure to configure the fastport mode in switch port connected directly to a node, because there are still bpdus transfer in a switch to switch connection.

BPDUs Guard ensures that inferior bpdus are blocked, allowing STP to shut an access port in this regard.

8 0
3 years ago
Users in a corporation currently authenticate with a username and password. A security administrator wishes to implement two-fac
Katen [24]

Answer:

Option (C) is the correct answer of this question.

Explanation:

Smart card is the security administrator wishes to implement two-factor authentication to improve security.Normally this data is affiliated with either meaning, information, or both, and will be stored and transmitted within the chip of the card.A smart card, usually a Chip Card type.

  • It is a flexible card that holds an integral computer chip that preserves and sends a signal data, either database or semiconductor form.
  • Used for controlling access to a resource.
  • Data authentication, encryption, cloud storage, and software processing can be established by smart cards.

Other options are incorrect.

4 0
3 years ago
Write a program that takes a point (x,y) from theuser and find where does the point lies. The pointcan
agasfer [191]

Answer:

C++ Program .

#include<bits/stdc++.h>

using namespace std;

int main()

{

int x,y;//declaring two variables x and y.

string s;//declaring string s..

cout<<"enter x and y"<<endl;

cin>>x>>y;//taking input of x and y..

if(x>=0 &&y>=0) //condition for 1st quadrant..

cout<<"the point lies in 1st Quadrant"<<endl;

else if(x<=0 &&y>=0)//condition for 2nd quadrant..

cout<<"the point lies in 2nd Quadrant"<<endl;

else if(x>=0 &&y<=0)//condition for 3rd quadrant..

cout<<"the point lies in 3rd Quadrant"<<endl;

else //else it is in  4th quadrant..

cout<<"the point lies in 4th Quadrant"<<endl;

cout<<"enter n to terminate the program"<<endl;

while(cin>>s)//if the user has not entered n the program will not terminate..

{

   if(s=="n")

   {

       cout<<"the program is terminated"<<endl;

       exit(0);

   }

   cout<<"you have not entered n please enter n to terminate the program<<endl;

}

}

Explanation:

The above written program is for telling the point lies in which quadrant.I am first declaring two variables x and y.Then after that taking input of x and y after that checking in which quadrant the point lies.

Taking input of the string s declared earlier for program termination the program will keep running until the user enters n.

4 0
3 years ago
Easy coding question, please help.
borishaifa [10]

Answers:

What is the index of the last element in the array? stArr1.length()-1

This prints the names in order. How would I print every other value? Change line 4 to: index = index +2

Change line 7 to: i < names.length

5 0
3 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
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