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NEW QUESTION: 1
追加のサブネットがバックエンドサーバーに使用されることが多いのに、なぜ2つのサブネットでパブリックロードバランサを作成する必要があるのですか。 （2つ選んでください。）
A. バックエンドサーバー用の追加のサブネットは、これらのサーバー用の個別のセキュリティリストを可能にします。
B. バックエンドサーバー用の追加のサブネットは、これらのサーバー用に別々のルートテーブルを考慮に入れます。
C. 使用するサブネットが多いほどパフォーマンスが向上します。
D. ロードバランサーがバックエンドサーバーと同じサブネットにないと、ルーティングが簡単になります。
Answer: A,C
NEW QUESTION: 2
According to ISO/IEC 20000, what is the minimum frequency for the Service Provider and the Customer to attend a service review meeting for discussing changes to the service scope?
A. only when there is a business need to change the service
B. annually
C. monthly
D. quarterly
Answer: B
NEW QUESTION: 3
Which of the following is the lowest TCSEC class where in the system must protected against covert storage channels (but not necessarily covert timing channels)?
A. B2
B. B3
C. A1
D. B1
Answer: A
Explanation:
The B2 class referenced in the orange book is the formal security policy model based on device labels that can use DAC (Discretionary access controls) and MAC (Mandatory Access Controls). It provides functionality about covert channel control. It does not require covert timing channels. You can review the B2 section of the Orange Book.
NEW QUESTION: 4
The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to the
$ 500 a plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of 20 tickets, what is the chance that you will win 15 or more prizes?
A. 0.006
B. 0.021
C. 0.250
Answer: B
Explanation:
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by:
r (n-r)
n!(p )[q ]/r!(n-r)!. Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have
15 5
P(15) = 20!(0.5 )(0.5 )/15!5! = 0.0148
16 4
P(16) = 20!(0.5 )(0.5 )/16!4! = 0.0046 P(17) = 0.0011 P(18) = 0.0002 P(19) = 0.00002 P(20) = 0.000001
The sum adds up to 0.0207.
