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lana [24]
3 years ago
12

You are network administrator for an Active Directory forest with a single domain. Then network has three sites with one domain

controller at each site. You have created and configured sites in Active Directory Sites and Services, and replication is operating normally between sites. You configure two universal groups for use in securing the network. All users are members of one universal group or the other. After configuring the universal groups, users at sites 2 and 3 report slows login and slow access to the corporate database. Users at site 1 can log in and access the corporate database with acceptable performance. You want to improve login and resource access performance for users in sites 2 and 3. What should you do?
Computers and Technology
1 answer:
Alina [70]3 years ago
4 0

Answer: Enable universal group membership caching at sites 2 and 3

Configure the domain controllers at sites 2 and 3 as global catalog servers

Explanation:

Since the user wants to improve login and resource access performance for users in sites 2 and 3, then a universal group membership caching ashould be enabled at sites 2 and 3.

Also, the domain controllers at sites 2 and 3 should be configured as global catalog servers. This is required for authentication and the resolution of the user principal name. This can be used in ensuring that login is improved for the users.

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When do you use a for loop instead of a while loop? (There may be more than one answer.)
statuscvo [17]

Answer:

- You know how many times you want the loop to run.

- When there is a defined start and end.

Explanation:

Required:

When to use for instead of while loop?

First, it should be noted that both for loop and while loop are used for iteration. However, depending on the scenario; there is preference on one to the other.

The basic syntax of for loop is:

for(start;end;condition)\{\}

<em>The condition could either be increment or decrement</em>

The basic syntax of while loop is:

while(condition)\{\}

<em>The condition could be any condition whatsoever</em>

Analyzing the options one after the other.

1. How many times the loop to run.

The for loop is preferred here because of the simplicity of the for loop.

2. & 3. Both the for and while loops are used for code repetition and to perform number calculations

4. Defined start and end.

From the syntax of the for loop

for(start;end;condition)\{\}

<em />

A for loop has a definite start and end in its basic syntax and is more preferable to use instead of a while loop.

<em />

<em />

Hence, 1 & 4 answer the question

8 0
3 years ago
What does it mean by MYRIAD of data?
pentagon [3]

Answer:

The term Myriad means an indefinitely large number of things. And when we say Myriad of data, we mean an indefinitely large amount of data like in Petabytes.

Explanation:

What it does implies is that you are dealing with a large amount of data. And hence we are dealing with something like bigData. And hence, technologies related to BigData are going to operate. And this term is used as a noun as well as an adjective when we talk about English. And hence, you can use it with "a" or without the "a" as well. It has been a part of reputable English and has been repeatedly used by top authors like Milton.

5 0
3 years ago
Using a personal computer to produce high quality printed documents
alisha [4.7K]

Answer:

Desktop publishing

6 0
4 years ago
Forms open in _______, which provides a user-friendly interface for entering data.
sergey [27]
They open in pdf format

5 0
3 years ago
Consider the two computers A and B with the clock cycle times 100 ps and 150 ps respectively for some program. The number of cyc
Kipish [7]

Answer:

Option d) B is 1.33 times faster than A

Given:

Clock time, t_{A} = 100 ps

t_{A} = 150 ps

No. of cycles per instructions,  n_{A} = 2.0

n_{B} = 1.0

Solution:

Let I be the no. of instructions for the program.

CPU clock cycle, f_{A} = 2.0 I

CPU clock cycle, f_{B} = 1.0 I

Now,

CPU time for each can be calculated as:

CPU time, T = CPU clock cycle\times clock time

T_{A} = f_{A}\times t_{A} = 2.0 I\times 100 = 200 I ps

T_{B} = f_{B}\times t_{B} = 1.0 I\times 100 = 150 I ps

Thus B is faster than A

Now,

\frac{Performance of A}{Performance of B} = \frac{T_{A}}{T_{B}}

\frac{Performance of A}{Performance of B} = \frac{200}{150}

Performance of B is 1.33 times that of A

7 0
3 years ago
Read 2 more answers
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