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NEW QUESTION: 1
次のうち、データ所有者の責任を説明しているベストはどれですか？
A. 適切なユーザーへのアクセシビリティを確保し、適切なレベルのデータセキュリティを維持する
B. データストレージやアーカイブなど、基本的なデータの可用性を維持する
C. 継続的なデータ整合性のための定期的な監査による品質と検証の実施
D. 情報が組織の使命に与える影響を判断する
Answer: D

NEW QUESTION: 2

A. Option D
B. Option A
C. Option C
D. Option B
Answer: B
Explanation:
When the user has configured Transmission Control Protocol (TCP) or Secure Sockets Layer (SSL) for
both front-end and back-end connections of the Elastic Load Balancer, the load balancer forwards the
request to the back-end instances without modifying the request headers unless the proxy header is
enabled. SSL does not support sticky sessions. If the user has enabled a proxy protocol it adds the
source and destination IP to the header.
Reference:
http://docs.aws.amazon.com/ElasticLoadBalancing/latest/DeveloperGuide/enable-proxy-protocol.html

NEW QUESTION: 3
1.0より大きいCPI値は何を示していますか？
A. 推定値を超えるコスト
B. 実際の値でのコスト
C. 推定値以下のコスト
D. 推定値でのコスト
Answer: C

NEW QUESTION: 4
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.0.0.0.1
B. Permit 10.0.0.1.0.0.0.0
C. Permit 10.0.0.1.0.0.0.254
D. Permit 10.0.0.0.255.255.255.254
Answer: C
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.