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Cloud and Storage Services Engineer Exam

  • Exam Number/Code : NS0-701
  • Exam Name : Cloud and Storage Services Engineer Exam
  • Questions and Answers : 213 Q&As
  • Update Time: 2019-01-10
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NEW QUESTION: 1
親が削除された場合に子レコードを残したい場合、どのようなタイプの関係が存在する必要がありますか?
Answer:
Explanation:
1-1

NEW QUESTION: 2

A. Option E
B. Option C
C. Option B
D. Option A
E. Option D
Answer: D

NEW QUESTION: 3
ユーザーがCIDR 20.0.0.0/24のVPCを作成しました。
ユーザーはCIDR 20.0.0.0/25のパブリックサブネットを作成しました。
ユーザーはCIDR 20.0.0.128/25でプライベートサブネットを作成しようとしています。
このシナリオに当てはまるのは、以下の記述のどれですか。
A. ユーザーはCIDRが20.0.0.128 / 25のプライベートサブネットを作成できます。
B. CIDR範囲が間違っているため、ユーザーはプライベートサブネットを作成できません
C. AWSではCIDR 20.0.0.0/25が許可されていないため、この文は誤りです。
D. CIDRが重複しているため、ユーザーはプライベートサブネットを作成できません
Answer: A
Explanation:
ユーザーがVPCでサブネットを作成するとき、彼はサブネットのCIDRブロックを指定します。
サブネットのCIDRブロックは、VPCのCIDRブロック(VPC内の単一のサブネットの場合)、またはサブセット(複数のサブネットを有効にする場合)と同じにすることができます。ユーザーがVPC内に複数のサブネットを作成する場合サブネットのCIDRブロックが重なってはいけないので、この場合、ユーザーはCIDRブロック20.0.0.0/24を持つVPCを作成し、256のIPアドレス(20.0.0.0〜
20.0.0.255 ..ユーザーはこのCIDRブロックを2つのサブネットに分割し、それぞれ128のIPアドレスをサポートすることができます。
1つのサブネットはCIDRブロック20.0.0.0/25(アドレス20.0.0.0-20.0.0.127用)を使用し、他のサブネットはCIDRブロック20.0.0.128/25(アドレス20.0.0.128-20.0.0.255用)を使用します。

NEW QUESTION: 4
Refer to the exhibit.

Routers R1 and R2 are IPv6 BGP peers that have been configured to support a neighbor relationship over an IPv4 internet work. Which three neighbor IP addresses are valid choices to use in the highlighted section of the exhibit? (Choose three.)
A. 10.67.0.2:0:0:0:0:0:0
B. ::10.67.0.2
C. ::0A43:0002
D. 10.67.0.2::
E. 0:0:0:0:0:0:10.67.0.2
F. 0A43:0002::
Answer: B,C,E
Explanation:
Explanation/Reference:
Explanation:
The automatic tunneling mechanism uses a special type of IPv6 address, termed an "IPv4- compatible" address. An IPv4-compatible address is identified by an all-zeros 96-bit prefix, and holds an IPv4 address in the low-order 32-bits. IPv4-compatible addresses are structured as follows:

Therefore, an IPv4 address of 10.67.0.2 will be written as ::10.67.0.2 or
0:0:0:0:0:0:10.67.0.2 or ::0A43:0002 (with 10[decimal] = 0A[hexa] ; 67[decimal] = 43[hexa] ; 0[hexa] = 0
[decimal] ; 2[hexa] = 2[decimal])