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fiasKO [112]
3 years ago
10

How can vlan hopping attacks be prevented on a network?

Computers and Technology
1 answer:
brilliants [131]3 years ago
5 0

The solution is to disable auto trunking and move native VLANs to unused VLANs.

VLAN hopping is a PC security misuse, a strategy for assaulting networked assets on a virtual LAN (VLAN). The fundamental idea driving all VLAN hopping assaults is for an assaulting host on a VLAN to access traffic on different VLANs that would ordinarily not be accessible.

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Which of these can be represented visually by a flowchart?
Mariana [72]

Answer:

i think its C.

if its wrong i'm truly sorry

Explanation:

4 0
3 years ago
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Write a program whose input is a character and a string, and whose output indicates the number of times the character appears in
Zina [86]

Answer:

Here is the C++ program:

#include <iostream> // to include input output functions

using namespace std; // to identify objects like cin cout

int counter(string userString, char character) { //function counter

   int count = 0;   // counts the no of times a character appears in the string

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

// loop to move through the string to find the occurrence of the character

       if (userString[i] == character) //if characters is found in the string

           count++;   //counts the occurrence of the character in the string

   return count; }   //returns the no of times character occurs in the string

int main() { //start of the main() function body

   string s; // stores the string entered by the user

   cout<<"Enter a string: "; //prompts user to enter the string

   cin>>s; //reads the string from user

   char ch; //stores the character entered by the user

   cout<<"Enter a character: "; //prompts user to enter a character

   cin>>ch; //reads the character from user

   cout << counter(s, ch) << endl; }  

//calls counter function to find the number of times a character occurs in the //string

Explanation:

The counter function works as following:

It has a count variable which stores the number of occurrences of a character in the userString.

It uses a for loop which loops through the entire string.

It has i position variable which starts with the first character of the string and checks if the first character of userString matches with the required character.

If it matches the character then count variable counts the first occurrence of the character and in the userString and is incremented to 1.

If the character does not match with the first character of the userString then the loops keeps traversing through the userString until the end of the userString is reached which is specified by the length() function which returns the length of the string.

After the loop ends the return count statement is used to return the number of occurrences of the character in the userString.

The main() function prompts the user to enter a string and a character. It then calls counter() function passing string s and character ch arguments to it in order to get the number of times ch appears in s.

The output is attached in a screenshot.

6 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

<em>Hence, there are 1365 ways </em>

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)

<em>From the given parameters; 3 out of 15 is detective;</em>

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;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

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
Which tools are found in the Quick Analysis feature? Check all that apply.
a_sh-v [17]

Answer:

A, C, E

Explanation:

8 0
3 years ago
Read 2 more answers
1. Digital videos look sharpest when they are displayed at a resolution that is larger than the frame size.
boyakko [2]

Answer:

(d) all of the above

Explanation:

This is because, not only is digital video a core technology for digital television, it also happens to be a core technology for video conferencing and video messaging. This could be seen in its application in messaging apps for private discussion of for holding official meetings between employees in virtual conference.

8 0
2 years ago
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