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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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3 years ago
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nikklg [1K]

If you want to set the sticky bit on an existing directory, subdir, without otherwise altering its permissions. to do so, you should type chmod a <u>t  <subdir>.</u>

<h3>What is Subdir?</h3>

The Definition of the term subdirectory is known to be a kind of an organizational directory that can be seen on a computer.

It is known to be one that can be found inside another directory  such as a subfolder. It is seen as the file a person is looking for and it is one that needs to have an extension.

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You want to set the sticky bit on an existing directory, subdir, without otherwise altering its permissions. To do so, you should type:

chmod a+_____ <subdir>

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6 0
2 years ago
Find the number of ways in which a committee of 4 can be chosen from six boys and
natima [27]

Answer:

465 ways

Explanation:

Atleast 1 girl and 1 boy

Possible combinations :

1 girl ; 3 boys = 6C1 ; 6C3

2 girls ; 2 boys = 6C2 ; 6C2

3 girls ; 1 boy = 6C3 ; 6C1

(6C1 * 6C3) + (6C2 * 6C2) + (6C3 * 6C1)

Combination formula:

nCr = n! ÷ (n-r)!r!

We can also use a calculator :

6C1 = 6

6C3 = 20

6C2 = 15

Hence,

(6C1 * 6C3) + (6C2 * 6C2) + (6C3 * 6C1)

(6 * 20) + (15 * 15) + (20 * 6)

120 + 225 + 120

= 465 ways

6 0
2 years ago
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