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lutik1710 [3]
1 year ago
14

Which part of the ethernet address is assigned to vendors to identify the equipment?

Computers and Technology
1 answer:
Nitella [24]1 year ago
4 0

First three bytes of the ethernet address exists assigned to vendors to identify the equipment.

<h3>What is Ethernet address?</h3>

A media access control address stands for a unique identifier allocated to a network interface controller for usage as a network address in communications within a network segment. This use exists as standard in most IEEE 802 networking technologies, including Ethernet, Wi-Fi, and Bluetooth.

This address exists as the identifying mark that allows a networked computer to communicate with the Internet. The ethernet address exists how we can associate an “IP address” to an individual computer — without this– number, there's no form to access any servers, websites, email, etc.

The first three bytes (pairs of hexadecimal characters) of any unicast address include that vendor address component of the MAC address. The staying three bytes carry the serial number of that vendor's interface card. Dell, Inc. Apple, Inc.

Hence, first three bytes of the ethernet address exists assigned to vendors to identify the equipment.

To learn more about Ethernet address refer to:

brainly.com/question/7284219

#SPJ4

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Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)
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(1)Minterms complement = XYZ (2) Compliment of Minterms = Σm(0,1, 2 , 4 , 6) (3) (X+Y+Z) (4) Minimized SOP = Z + XY

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Explanation:

Solution

Recall that:

Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)

(1) Canonical Disjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as minterms

In Minterm we assign 'I' to each uncomplimented variable and '0' to each complemented/complementary variable

For the given question stated we ave the following:

Minterms = XYZ, XYZ, XYZ, XYZ, XYZ.

(2) Canonical Conjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as maxterms.

In Maxterms we assign '0' to each uncomplimented variable and '1' to each complemented/complementary variable

Compliment of Minterms = Σm(0,1, 2 , 4 , 6)

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