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NEW QUESTION: 1
ネットワークエンジニアが新しいスイッチドネットワークを設定しています。 将来的にはネットワークが拡大し、多くの新しいVLANが追加されることが予想されます。 複数のスパニングツリーインスタンスに関連するスイッチリソースと管理上の負担を軽減するために使用するスパニングツリープロトコルはどれですか？
A. RSTP
B. RPVST+
C. MST
D. PVST
E. PVST+
Answer: C

NEW QUESTION: 2
Your company is performing data preprocessing for a learning algorithm in Google Cloud Dataflow.
Numerous data logs are being are being generated during this step, and the team wants to analyze them.
Due to the dynamic nature of the campaign, the data is growing exponentially every hour. The data scientists have written the following code to read the data for a new key features in the logs.
BigQueryIO.Read
.named("ReadLogData")
.from("clouddataflow-readonly:samples.log_data")
You want to improve the performance of this data read. What should you do?
A. Call a transform that returns TableRow objects, where each element in the PCollexction represents a single row in the table.
B. Use of both the Google BigQuery TableSchema and TableFieldSchema classes.
C. Specify the Tableobject in the code.
D. Use .fromQuery operation to read specific fields from the table.
Answer: D
Explanation:
BigQueryIO.read.from() directly reads the whole table from BigQuery. This function exports the whole table to temporary files in Google Cloud Storage, where it will later be read from. This requires almost no computation, as it only performs an export job, and later Dataflow reads from GCS (not from BigQuery).
BigQueryIO.read.fromQuery() executes a query and then reads the results received after the query execution. Therefore, this function is more time-consuming, given that it requires that a query is first executed (which will incur in the corresponding economic and computational costs).

NEW QUESTION: 3
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.1.0.0.0.254
B. 許可10.0.0.0.0.0.0.1
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.255.255.255.254
Answer: A
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.