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NEW QUESTION: 1
Which of the followings can only be used for BGP route selection? (Multiple Choice)
A. IP-PREFIX
B. AS-PATH-FILTER
C. COMMUNITY-FILTER
D. ACL
Answer: B,C
NEW QUESTION: 2
You need to grant appropriate permissions to the SSISOwners SQL Server login. What should you do?
A. Map the login to the SSISDB database. Assign the user to the ssis_admin role.
B. Map the login to the msdb database. Assign the user to the db_ssisadmin role.
C. Map the login to the msdb database. Assign the user to the db_owner role.
D. Map the login to the msdb database. Assign the user to the ssis_admin role.
E. Map the login to the SSISDB database. Assign the user to the db_owner role.
F. Map the login to the SSISDB database. Assign the user to the db_ssisadmin role.
Answer: F
Explanation:
Correct role is db_ssisadmin http://msdn.microsoft.com/en-us/library/ms141053.aspx
NEW QUESTION: 3
どのような一般的な多層防御方法が攻撃対象を減らすのに役立ちますか?
A. インターネットに面したDMZリンクにシステムを配置して、トラフィックフローを制御します
B. UDPプロトコルを使用して、帯域幅とプロトコルのオーバーヘッドを保持します
C. IPS、ファイアウォール、およびAAAベースのプラットフォームとサービスを展開します
D. 銅線接続をファイバーベースの接続に置き換えます
E. 8文字のパスワードを使用
Answer: C
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10.0.0.0.255.255.255.254
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.0.0.0.1
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.
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