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egoroff_w [7]
3 years ago
11

What is the name of the technology that is typically implemented on switches to avoid Ethernet connectivity problems when the wr

ong type of cable (straight or crossover) is used to connect devices?
Computers and Technology
1 answer:
Andrew [12]3 years ago
4 0

Answer:  Medium Dependent Interface Crossover (MDIX)

Explanation:

The technology that is typically implemented on switches to avoid Ethernet connectivity problems is Medium Dependent Interface Crossover(MDIX) For switches with Auto (MDIX) the connection is configured automatically and a crossover or straight-through cable can be used to connect two switches. When this MDIX interface is connected it corrects the connectivity speed and the cabling for proper device functionality.

You might be interested in
Which statement is true for a career as a graphic designer?
Brrunno [24]

Answer:

The answer is B.

Explanation:

As a graphic designer you are responsible for almost anything you do! :)

Hope this helps

6 0
3 years ago
Create a security management policy that addresses the management and the separation of duties throughout the seven domains of a
klio [65]

Seven Domains of IT Infrastructure Seven domains can be found in a typical IT infrastructure. They are as follows: User Domain, Workstation Domain, LAN Domain, LAN-to-WAN Domain, Remote Access Domain, WAN Domain, and System/Application Domain.

Explanation:

User Domain : The User Domain covers all theusers (of any rank) that haveaccess to the other six domains.

RISKS:

  • User can destroy data in application(intentionally or not) and delete all
  • User can insert infected CD or USBflash drive into the work computer

Workstation Domain : A computer of an individual user where the production takes place

RISKS:

  • The workstation’s OS can have a known software vulnerability thatallows a hacker to connect remotelyand steal data.
  • A workstation’s hard drive can fail causing lost data

LAN Domain : Contains all of the workstations,hubs, switches, and routers. TheLAN is a trusted zone

RISKS:

  • A worm can spread through the LANand infect all computers in it.
  • LAN server OS can have a knownsoftware vulnerability.

WAN Domain : Stands for Wide Area Network and consists of the Internet and semi-private lines

RISKS:

  • Service provider can have a major network outage.
  • Server can receive a DOS or DDOS attack.
  • A FTP server can allow anonymously uploaded illegal software

LAN / WAN Domain : The boundary between the trusted and un-trusted zones.The zones are filtered with a firewall

RISKS:

  • Weak ingress/egress traffic filteringcan degrade performance.
  • A firewall with unnecessary portsopen can allow access from the Internet

System / Application Storage Domain : This domain is made up of user-accessed servers suchas email and database

RISKS:

  • A fire can destroy primary data
  • A DOS attack can cripple the organization’s email

Remote Access Domain  :The domain in which a mobileuser can access the local network remotely, usually through a VPN

RISKS:

  • Communication circuit outage candeny connection.
  • Remote communication from officecan be unsecured.
  • VPN tunneling between remotecomputer and ingress/egress routercan be hacked
6 0
3 years ago
I'm stuck on this Java exercise problem and don't know how to organize my code with proper formatting and it's not helping that
Volgvan

In your "count spaces method" before the for loop, you want to declare an int variable (let's call it spc)

then, each time you find that charat, spc++.


also, the string must be declared in the main method, not the spaces count method.


The actual code:


public class CountSpaces{

public static void main(String[] args){

String instring = "(your quote here)";

int spaces = calculateSpaces(instring);

System.out.println("The number of spaces is " + spaces);

}

public static int calculateSpaces(String in){

int spc = 0;

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

if(in.charAt(i)== ' '){

spc++;

}

}

return spc;

}

}


that should be it.  If you have any questions, feel free to reach out.

4 0
3 years ago
The set of specific, sequential steps that describe exactly what a computer program must do to complete the work is called a(n)
STatiana [176]

Answer:

algorithm

Explanation:

7 0
3 years ago
1)
spayn [35]

Answer:

1)

for(i = 0; i < NUM_VALS; ++i) {

  if(userValues[i] == matchValue) {

     numMatches++;  }    }

2)  

for (i = 0; i < NUM_GUESSES; i++) {

      scanf("%d", &userGuesses[i]);   }

  for (i = 0; i < NUM_GUESSES; ++i) {

         printf("%d ", userGuesses[i]);    }

3)

sumExtra = 0;

for (i = 0; i < NUM_VALS; ++i){

     if (testGrades[i] > 100){  

        sumExtra = testGrades[i] - 100 + sumExtra;    }       }

4)

for (i = 0; i < NUM_VALS; ++i) {

    if (i<(NUM_VALS-1))  

   printf( "%d,", hourlyTemp[i]);

    else  

    printf("%d",hourlyTemp[i]); }      

Explanation:

1) This loop works as follows:

1st iteration:

i = 0

As i= 0 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 0 So the statement becomes:

userValues[0] == 2

2 == 2

As the value at 0th index (1st element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 1

Now value of i is incremented to 1 so i=1

2nd iteration:

i = 1

As i= 1 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 1 So the statement becomes:

userValues[1] == 2

2 == 2

As the value at 1st index (2nd element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 2

Now value of i is incremented to 1 so i=2

The same procedure continues at each iteration.

The last iteration is shown below:

5th iteration:

i = 4

As i= 4 and NUM_VALS = 4 This means for condition i<NUM_VALS  is false so the loop breaks

Next the statement: printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches);  executes which displays the value of

numMatches = 3

2)

The first loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which reads the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration each element at i-th index is read using scanf such as element at userGuesses[0], userGuesses[1], userGuesses[2]. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

The second loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which prints the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration, each element at i-th index is printed on output screen using printf such as element at userGuesses[0], userGuesses[1], userGuesses[2] is displayed. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

So if user enters enters 9 5 2, then the output is 9 5 2

3)

The loop works as follows:

At first iteration:

i=0

i<NUM_VALS is true as NUM_VALS = 4 so 0<4. Hence the loop body executes.

if (testGrades[i] > 100 checks if the element at i-th index of testGrades array is greater than 100. As i=0 so this statement becomes:

if (testGrades[0] > 100

As testGrades[0] = 101 so this condition evaluates to true as 101>100

So the statement sumExtra = testGrades[i] - 100 + sumExtra; executes which becomes:

sumExtra = testGrades[0] - 100 + sumExtra

As sumExtra = 0

testGrades[0] = 101

So

sumExtra = 101 - 100 + 0

sumExtra = 1

The same procedure is done at each iteration until the loop breaks. The output is:

sumExtra = 8

4)

The loop works as follows:

At first iteration

i=0

i < NUM_VALS is true as  NUM_VALS = 4 so 0<4 Hence loop body executes.

if (i<(NUM_VALS-1))   checks if i is less than NUM_VALS-1 which is 4-1=3

It is also true as 0<3 Hence the statement in body of i executes

printf( "%d,", hourlyTemp[i]) statement prints the element at i-th index i.e. at 0-th index of hourlyTemp array with a comma (,) in the end. As hourlyTemp[0] = 90; So 90, is printed.

When the above IF condition evaluates to false i.e. when i = 3 then else part executes which prints the hourlyTemp[3] = 95 without comma.

Same procedure happens at each iteration unless value of i exceeds NUM_VAL.

The output is:

90, 92, 94, 95

The programs along with their output are attached.

4 0
4 years ago
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