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Nana76 [90]
3 years ago
12

Classification, affinity grouping, and estimation are______methods used by managers to make their data more meaningful.

Computers and Technology
1 answer:
Arisa [49]3 years ago
6 0
Answer: data minding
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1. When the Cisco 2950 switch is first turned on, it behaves like a hub. Why is this so, and what must happen before the switch
Sav [38]

Answer:

The switch sends broadcast traffic to all ports like a hub to get the number of ports used and build its MAC address table.

Explanation:

A network switch is a device used in a LAN to connect all available workstations. Unlike a network hub, it is able to send frames from a host or workstation to another using its MAC address table.

An example of a switch is the Cisco 2950 switch. When it is first turned on, it acts as a hub, broadcasting traffic to all its ports. This process is used to create a MAC address table to get the IP address of workstations and the ports they relate to, enabling it to send unicast traffic to a specific port.

7 0
3 years ago
A combination lock has the following basic properties:
fomenos

Answer:

public class CombinationLock {

   

    private int combinationNumber1 = 0;

    private int combinationNumber2 = 0;

    private int combinationNumber3 = 0;

   

   CombinationLock(int combinationNumber1, int combinationNumber2, int combinationNumber3){

       this.combinationNumber1 = combinationNumber1;

       this.combinationNumber2 = combinationNumber2;

       this.combinationNumber3 = combinationNumber3;

   }

   

   public boolean open(int number1, int number2, int number3){

       if(number1 == combinationNumber1 && number2 == combinationNumber2 && number3 == combinationNumber3)

           return true;

       else

           return false;

   }

   

   public boolean changeCombo(int number1, int number2, int number3, int newNumber1, int newNumber2, int newNumber3){

       

       if (open(number1, number2, number3)){

           combinationNumber1 = newNumber1;

           combinationNumber2 = newNumber2;

           combinationNumber3 = newNumber3;

           return true;

       }else

           return false;

   }

}

Explanation:

- <em>Three variables</em> are created to hold the combination.

- A <em>constructor</em> is created to set the combination.

- A boolean method called <em>open</em> is created to check if the given numbers are correct, and <u>returns true</u> (<u>otherwise returns false</u>).

- A boolean method method called <em>changeCombo</em> is created to check if given numbers are correct. If they are correct, it assigns new values for the combination and <u>returns true</u> (<u>otherwise returns false</u>).

5 0
3 years ago
Which mitigation technique would prevent rogue servers from providing false ip configuration parameters to clients?
IrinaVladis [17]
DHCP snooping is a mitigation technique<span> to </span>prevent rogue<span> DHCP </span>servers  <span>from </span>providing false IP configuration parameters to clients.  DHCP snooping<span> is a security feature that acts like a firewall between untrusted hosts and trusted </span>DHCP<span> servers. It is</span><span> a series of techniques applied to improve the security of a </span>DHCP<span> infrastructure. </span>
6 0
3 years ago
Read 2 more answers
Brainly is the stupidest homework site. The past like 10 questions ive asked ive had no answer to. Thanks a lot!
Irina18 [472]

oof. I can help send me the questions.

7 0
3 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

  • Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW  = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

5 0
3 years ago
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