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VMariaS [17]
2 years ago
8

If two devices simultaneously transmit data on an Ethernet network and a collision occurs, what does each station do in an attem

pt to resend the data and avoid another collision
Computers and Technology
1 answer:
Arada [10]2 years ago
8 0

Explanation:

Both station will retract their data. They will wait a random amount of time before sending out data again. This lessens the chance of collision again.

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My mom works from home selling her craft work online to people all over the world. She can do this from home because we have acc
8_murik_8 [283]
I would go with A) The Internet 


I hope that turns out well for you. Good luck! (:
6 0
3 years ago
Which unique address is a 128-bit address written in hexadecimal?
pentagon [3]
Every hexadecimal digit represents 4 bits, so the address has 128/4 = 32 digits.

A GUID (Globally Unique IDentifier) has 128 bits. They are usually written like this:

{38a52be4-9352-4<span>53e-af97-5c3b448652f0}.</span>

There are different types of guids, depending on how they are generated. The first digit of the third group reveals the type. In the example above it is 4. A type 4 guid is fully random (except of course for the 4).

3 0
3 years ago
Can any software run on any processor
jekas [21]

Answer:

Yes.

Explanation:

Your software requires CPU instruction if CPU doesn't provide that instructions the software won't work/run.

hope this helps you

have a great day:)

6 0
3 years ago
WILL GIVE BRAINLIEST!! NO LINKS!!!
anyanavicka [17]

Answer:

Maybe try a different charger, and if that works, your phone may have an issue..

Explanation:

Sometimes cords just.. don't work right if you use 'em long enough.

7 0
3 years ago
Read 2 more answers
) Consider the ambiguous CFG problem: given a context free grammar, does there exist a string that can be generated by two diffe
vodomira [7]

Answer:

See explaination

Explanation:

The post correspondence problem is: Given a_1, \ldots, a_n and b_1, \ldots, b_n , is there a sequence of indices 11. ..., im such that a_{i_1}\ldots a_{i_m} = b_{i_1}\ldots b_{i_m} .

To reduce from the PCP to the ambiguous CFG problem, do the following. Add some more symbols c_1, \ldots, c_n and define the following languages:

L_{A} = \{a_{i_1}\ldots a_{i_k}c_{i_k}\ldots c_{i_1} \mid k \geq 1\}

L_{B} = \{b_{i_1}\ldots b_{i_k}c_{i_k}\ldots c_{i_1} \mid k \geq 1\}

L_{AB} = L_A \cup L_B .

Define the following grammar for the language L_{AB} :

S \to S_{A} \mid S_{B}

S_A \to a_iS_Ac_i \mid a_ic_i, 1 \leq i \leq n

S_B \to b_iS_Bc_i \mid b_ic_i, 1 \leq i \leq n

It is easy to see that the given grammar generates the langauge L_{AB} . The reduction outputs the grammar thus constructed.

To see why this is a valid reduction, let there be a solution to the post correspondence problem, i.e. 11. ..., im such that a_{i_1}\ldots a_{i_m} = b_{i_1}\ldots b_{i_m} . Then the string a_{i_1}\ldots a_{i_m}c_{i_m}\ldots c_{i_1} = b_{i_1}\ldots b_{i_m}c_{i_m}\ldots c_{i_1} has two difference derivations in the given grammar, one using S_A and the other using S_B .

Conversely, let a string generated by the given grammar have two different derivations. Note that S_A and S_B can only generate one derivation tree for any string, hence the only way for two different derivations to happen is for one of them to go to S_A and the other to S_B .

Let the string for which ambiguity appears be of the form a_{i_1}\ldots a_{i_m}c_{i_m}\ldots c_{i_1} . Then the other derivation must produce a string of the form b_{i_1}\ldots b_{i_m}c_{i_m}\ldots c_{i_1} , this is because both derivations must produce the same string for ambiguity to occur, hence the part which uses c_i must be the same. Note that this further implies that a_{i_1}\ldots a_{i_m} = b_{i_1}\ldots b_{i_m} , which is equivalent to saying that the post correspondence problem has a solution.

Hence the reduction works as expected. This proves that the ambiguous CFG problem is undecidable, otherwise post correspondence problem would be decidable, which is a contradiction.

4 0
3 years ago
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