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NEW QUESTION: 1
進行中のプロジェクト中に、顧客はスケジュールを変更せずに新しい要件を提供することを要求します。証明されたマネージャーは次に何をすべきですか？
A. 専門家の判断を求める
B. プロジェクト管理計画を更新する
C. コストへの影響について顧客に通知する
D. 変更管理手順に従う
Answer: A

NEW QUESTION: 2
Which of the following commands can be used to bind a NIC to a single processor when using a Performance Pack onSecure Platform?

A. fw fat path nic
B. splat proc
C. sim affinity
D. set proc
Answer: C

NEW QUESTION: 3
Refer to the exhibit.

A packet with a source IP address of 192.168.2.4 and a destination IP address of 10.1.1.4 arrives at the AcmeB router. What action does the router take?
A. forwards a packet containing an ARP request out the FastEthemet0/1 interface
B. forwards a packet containing an EIGRP advertisement out the Serial0/1 interface
C. forwards a packet containing an ICMP message out the FastEthemet0/0 interface
D. forwards the received packet out the Serial0/0 interface
Answer: C
Explanation:
CCNA - EIGRP Common Question
http://www.orbitco-ccna-pastquestions.com/CCNA---EIGRP-Common-Question.php
Looking at the output above, there is no IP route for 10.1.1.4 address on AcmeB routing table. If the router can no find a specific path in its routing table to a particular route,( In this case no path is found so AcmeB) the router will inform the source host with an ICMP message that the destination is unreachable and this will be through the same interface it has received the packet (interface Fa0/0 network 192.168.3.0/28 from the exhibit).

NEW QUESTION: 4
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.1.0.0.0.254
B. 許可10.0.0.0.0.0.0.1
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.255.255.255.254
Answer: A
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 0.0.15.255 172.23.31.255.
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 000 ) 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.