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NEW QUESTION: 1
インスタンス管理計画を定義するとき、何を考慮しなければなりませんか？
注：この質問には2つの正解があります。
A. インスタンスを保持する期間
B. 保持されるインスタンスの最大数
C. インスタンスを生成するJob Serverの数
D. 入力ファイルリポジトリサーバーの使用可能なストレージスペース
Answer: A,B
NEW QUESTION: 2
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.254
B. Permit 10.0.0.0.0.0.0.1
C. Permit 10.0.0.0.255.255.255.254
D. Permit 10.0.0.1.0.0.0.0
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 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
Where in balance sheet certain account increases value and another decreases value?
choose the correct answer
Response:
A. Ledger entry
B. Debit entry
C. Journal entry
D. Credit entry
Answer: C
