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NEW QUESTION: 1
Note: This question is part of a series of questions that use the same set of answer choices. An answer choice may be correct for more than one question in the series.
You use a portable computer that has Windows 7 installed. The computer has a single hard disk drive and a dual-core CPU.
You need to analyze the performance of the computer to meet the following requirements:
*Include the CPU, Memory, and Disk subsystems.
*Rate the subsystems and find out the lowest-rated component.
What should you do?
A. Opern Performance Monitor, and customize the System Monitor view.
B. Open System Properties, and configure use profiles settings.
C. Open System Properties, and configure the page file settings.
D. Opern Performance Monitor, and configure a Data Collector Set.
E. Run powercfg.exe.
F. Open Task Manager.
G. Open System Properties, and configure environment variables.
H. Configure Event Viewer subscriptions.
I. Run the Windows Experience Index.
J. Open System Properties, and configure performance settings.
Answer: I

NEW QUESTION: 2
The system administrator notices that many employees are using passwords that can be easily guessed or
are susceptible to brute force attacks.
Which of the following would BEST mitigate this risk?
A. Increase the minimum password length.
B. Enforce account lockout policies.
C. Shorten the maximum life of account passwords.
D. Enforce password rules requiring complexity.
Answer: D
Explanation:
Section: Access Control and Identity Management
Explanation/Reference:
Explanation:
Password complexity often requires the use of a minimum of three out of four standard character types for
a password. The more characters in a password that includes some character complexity, the more
resistant it is to brute force attacks.

NEW QUESTION: 3
Instructions
-Refer to the exhibits to answer the questions related to this task.
-THIS TASK DOES NOT REQUIRE DEVICE CONFIGURATION.
-To access the multiple-choice question, click on the numbered box on the left of the top panel.
-There is one multiple-choice question with this task. Be sure to answer the question before
selecting the Next button.
Scenario You are deploying a small wireless test network in a lab. The network is made up of a wireless controller (WLC1), a dual radio access point (AP1) and a switch (SW1) that is configured as a DHCP server. The IP subnet being used for this network is 10.10.10.0/24.


Cisco 640-722 Exam

Cisco 640-722 Exam


Cisco 640-722 Exam
In setting up the wireless network, it is desired that the DHCP server that is configured in WLC1 provides IP addressing to the WLAN. When devices attempt to join the network, they fail to receive an IP address in the range provided by the DHCP server in WLC1, but they do receive an IP address from the DHCP server in SW1. What is causing this issue?
A. The primary DHCP server address in WLC1 is incorrect.
B. The primary DHCP server configured in WLC1 is not enabled.
C. The external DHCP server in SW1 will always be the preferred DHCP server and WLC1 will always be the secondary DHCP server in the WLAN.
D. The DHCP server address for SW1 needs to be removed from the DHCP server configuration in WLC1.
Answer: B
Explanation:
This can be verified by the following screen shot, showing the DHCP server is disabled at the bottom:

NEW QUESTION: 4
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.0
B. Permit 10.0.0.1.0.0.0.254
C. Permit 10.0.0.0.0.0.0.1
D. Permit 10.0.0.0.255.255.255.254
Answer: B
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.

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