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NEW QUESTION: 1
ユーザーがUS-East-1aリージョンで2つのEBSバックアップEC2インスタンスを起動しました。
ユーザーは、インスタンスの1つのゾーンを変更したいと考えています。
ユーザーはどのように変更できますか？
A. ゾーンはAWSCLIを使用してのみ変更できます
B. 実行中のインスタンスのAMIを作成し、別のAZでインスタンスを起動します
C. AWS EC2コンソールから、[Actions]-> [Change zone]を選択し、新しいゾーンを指定します
D. インスタンスの1つを停止し、アベイラビリティーゾーンを変更します
Answer: B
Explanation:
With AWS EC2, when a user is launching an instance he can select the availability zone (AZ. at the time of launch. If the zone is not selected, AWS selects it on behalf of the user. Once the instance is launched, the user cannot change the zone of that instance unless he creates an AMI of that instance and launches a new instance from it.

NEW QUESTION: 2
Which two statements represent good use cases for Wake on LAN? (Choose two.)
A. WoL can be used to power-up hosts to access the IPMI.
B. WoL can be used to power-up hosts for on-demand PXE booting.
C. WoL can be used to save electricity by powering down underused servers and desktops.
D. WoL can be used to power-up hosts for after-hours operating system updates and application patching.
Answer: B,D

NEW QUESTION: 3
ip scp server enableコマンドの効果は何ですか？
A. 特定のSCPサーバーを許可するアクセスリストを参照する
B. SCPを許可されたコピー機能のリストに追加します。
C. ルーターがSCPサーバーになることを許可します。
D. これにより、ルーターはSCPサーバーへの要求を開始できます。
Answer: C

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. 0.021
B. 0.250
C. 0.006
Answer: A
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.