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Aleks [24]
3 years ago
8

Two DHCP servers, Server1 and Server2, are running Windows Server 2016. As the administrator, you create a scope called Scope1.

Server1 is the primary DHCP server. Which of the following best describes the easiest way to assign 80% of the addresses to Server1 and 20% to Server2?
a. On Scope1, run the DHCP Split-Scope Configuration Wizard.b. Create a multicast scope.c. Create a DHCP policy.d. Create a superscope.
Computers and Technology
1 answer:
kumpel [21]3 years ago
5 0

Answer:

The answer is "Option a"

Explanation:

Split-scope is also an easy and simple approach to deliver DHCP consistency and workload management into your system. Server 2008 R2 provides a convenient divide-scope guide which removes several operational efforts but can only be to use if all databases run on R2, and wrong choices can be described as follows:

  • In option b, It uses the Ip address for multicast, that's why it is wrong.
  • In option c, It is wrong because it uses a windows interface, that works on policies.  
  • In option d, It is wrong because it is an administrative feature.

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Time in transit would be correct
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3 years ago
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Suppose list1 is an MyArrayList and list2 is a MyLinkedList. Both contains 1 million double values. Analyze the following code:
hjlf

Answer:

The correct answer for the given question is " The Code fragment A runs fastly than the code fragment of B".

Explanation:

In this question there are some information is missing i. e options. The question does not give any options. The options for the given question is given below

(A.) The Code fragment A runs fastly than the code fragment of B.

(B.) The Code fragment B runs fastly than code fragment of  A.

(C)  The Code fragment A runs as fastly as code fragment of B.  

So we conclude the answer i.e option(A) because As given in the question list1  is a MyArrayList and list2 is a MyLinkedList.  , in list1 we fetching the data easily and fastly means that it remove the data easily as compare to list2 As MyArrayList is storing the list only and also we can fetch the data easily manner.

The list2 is an object of MyLinkedList means that it manipulating the data fastly as compared to MyArrayList but if we compared the fetching of data then  MyArrayList is a better option so the code fragment runs fastly then code fragmented B.

7 0
3 years ago
A technique for identifying and tracking assets using technologies such as RFID and smart cards is known as electronic tagging.
Mademuasel [1]

Answer: True

Explanation: Electronic tagging is defined as the activity in which the monitoring of the things and people are done through the electronic markers or device .This process is usually done by attaching a tag sized component on the ankle of the human to monitor them.

The tracking of the good and human is done by production of the time alerts through the tag .Thus the statement given in the question is true.

4 0
3 years ago
How many possible password of length four to eight symbols can be formed using English alphabets both upper and lower case (A-Z
Fynjy0 [20]

Answer:

In a password, symbol/characters can be repeated. first calculate the total

symbols which can be used in a password.

So there are total 26(A-Z),26(a-z),10(0-9) and 2(_,$) symbols.

that is equal to 26+26+10+2=64.

Total number of password of length 4:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64=16777216

Total number of password of length 5:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64*64=1073741824

Similarly,

Total number of password of length 6:

64*64*64*64*64*64=68719476736

Total number of password of length 7:

64*64*64*64*64*64*64=4398046511104

Total number of password of length 8:

64*64*64*64*64*64*64=281474976710656

Hence the total number of password possible is:285,942,833,217,536

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