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NEW QUESTION: 1
次の展示に示す構成を持つVNet1という名前の仮想ネットワークがあります。

ドロップダウンメニューを使用して、図に示されている情報に基づいて各ステートメントを完成させる回答の選択肢を選択します。
注：それぞれの正しい選択には1ポイントの価値があります。

Answer:
Explanation:

説明

ボックス1：アドレススペースを追加する
仮想ネットワーク内のIaaS仮想マシン（VM）およびPaaSロールインスタンスは、接続先のサブネットのアドレス空間に基づいて、指定した範囲からプライベートIPアドレスを自動的に受け取ります。 192.168.1.0/24アドレス空間を追加する必要があります。
ボックス2：ネットワークインターフェイスを追加する
10.2.1.0/24ネットワークが存在します。ネットワークインターフェイスを追加する必要があります。
参照：
https://docs.microsoft.com/en-us/office365/enterprise/designing-networking-for-microsoft-azure-iaas
https://docs.microsoft.com/en-us/azure/virtual-network/virtual-networks-static-private-ip-arm-pportal

NEW QUESTION: 2
ネットワーク管理者は、SSH経由でリモートデバイスに接続したいと考えています。プロトコルが使用しているOSIモデルの次のレイヤーはどれですか？
A. 0
B. 1
C. 2
D. 3
E. 4
Answer: E

NEW QUESTION: 3
A customer's goal is to optimize the existing backup infrastructure containing Symantec Backup Exec software and an IBM TS3310 (3576-L5B) tape library.
Currently, the two LTO-5 drives are running at 80 % utilization. Because the customer expects a 30 % capacity increase over the next three years, the backup window will be missed in the future.
Which budget-sensitive solution should be offered to this customer?
A. An expansion unit (E9U) for the existing TS3310 library to gain more cartridge slots
B. Add a second TS3310 (3576-L5B) with LTO-5 drives installed at a secondary site for disaster recovery
C. Upgrade the TS3310 (3576-L5B) by replacing the LTO-5 drives and cartridges with LTO-6 drives and cartridges
D. Add two IBM TS7620 appliances with VTL licenses and replication software for redundancy
Answer: A
Explanation:
(http://www-01.ibm.com/common/ssi/cgibin/ssialias?infotype=OC&subtype=NA&htmlfid=897/ENUS3576E9U&appname=System%20Storage)

NEW QUESTION: 4
An individual has a 10% probability of suffering from a headache when he has a cold versus 1% when he does not have a cold. If the individual has a probability of developing a cold equal to 5%, what is the probability that this individual suffers from a headache?
A. 1.45%
B. 11%
C. 2.23%
Answer: A
Explanation:
Assume event A is the event in which the individual suffers from a headache. Using the Total
Probability Rule, P(A) = 0.1 x 0.05 + 0.01 x 0.95 = 1.45%.