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NEW QUESTION: 1
Universal Containersは、プロファイルを含まない変更セットを介してカスタムタブをProduction（Enterprise Edition）にデプロイしました。カスタムタブの表示に関して正しい説明はどれですか。
A. カスタムタブはすべてのユーザーに対してデフォルトでオンになっています。
B. すべてのユーザーに対してカスタムタブがデフォルトでオフになっています。
C. カスタムタブがすべてのユーザーに公開されます。
D. すべてのユーザーに対してカスタムタブが非表示になります。
Answer: D

NEW QUESTION: 2
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10.0.0.0.255.255.255.254
D. 許可10.0.0.1.0.0.0.0
Answer: A
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's 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.

NEW QUESTION: 3
Which of the following describe design-time objects?
There are 3 correct answers to this question.
Response:
A. Catalog objects
B. Native objects
C. Repository objects
D. Active
E. Inactive
Answer: B,C,E

NEW QUESTION: 4

A. Option D
B. Option B
C. Option A
D. Option C
Answer: D