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liq [111]
3 years ago
8

In what scenario should dhcp servers also be active dhcp clients?

Computers and Technology
1 answer:
mrs_skeptik [129]3 years ago
4 0
DHCP stands for Dynamic Host Configuration Protocol. It is protocol responsible for c<span>onfiguring the IP address and other TCP/IP settings on network computers. </span>There is no scenario in which DHCP servers also be active DHCP clients. They should never be both a DHCP server and DHCP client. The two sides (client and server) should be always separated.
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What name is given to any changes to the original data such as users manually modifying data, programs processing and changing d
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Modification is the name given to an attack that makes changes into original data such as users manually modifying data, programs processing and changing data, and equipment failures.

In modification attacks, only changes are made to original data without deleting it completely. Modification is basically a type of attack in the context of information security. This type of attack creates annoying situations and discords by altering existing data in data files, inserting new false information in the network and reconfiguring network topologies and system hardware.

Modification attack is mainly mounted against sensitive and historical data. Modification attacks target integrity of the original data with an intent of fabricating it.

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3 0
1 year ago
The OSI model is currently the most widely implemented network model used to develop and build networks of any size, including t
natta225 [31]

Answer:

The correct answer to the following question will be "False".

Explanation:

  • The Open Systems Interconnection model is a conceptualization that describes and vastly simplifies a telecommunication or computer system's communication features, regardless of its inner structure and technologies underlying them.
  • This model aims is to direct manufacturers and creators so that they would modularize with the digital communication devices and computer programs they build, and to promote a consistent context that defines the roles of a network or telecom device.

Therefore, the given statement is false.

7 0
2 years ago
Microsoft words spell checker
Degger [83]
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2 years ago
Applying what formatting option to your excel workbook will make it easier to read when printed out?
blagie [28]
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8 0
2 years ago
I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
3 years ago
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