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NEW QUESTION: 1
R2にはローカルで生成されたプレフィックス192.168.130.0/24があり、次の構成があります。
neighbor 1.1.1.1 route-map OUT outコマンドを使用して、route-map OUTコマンドをeBGPネイバーR1（1.1.1.1）に適用した場合の結果は何ですか。
A. R1は、192.168.130.0 / 24を1つのASホップではなく2つのASホップとして認識します。
B. R1は192.168.130.0/24以外のルートを受け入れません
C. ネットワーク192.168.130.0/24はR1テーブルでは許可されていません
D. R1は、192.168.30.0 / 24宛てのトラフィックを転送しません
Answer: A
NEW QUESTION: 2
What percentage of the HP Latex ink vehicle is water (by weight)?
A. 95% (by weight)
B. 10% (by weight)
C. 25% (by weight)
D. 66% (by weight)
Answer: D
Explanation:
Reference:
http://www.hp.com/hpinfo/newsroom/press_kits/2011/HPLatexSummit/LatexTechnology.pd f
NEW QUESTION: 3
Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
A. 192.168.1.32/27
B. 192.168.1.48/29
C. 192.168.1.8/29
D. 192.168.1.16/28
E. 192.168.1.40/29
Answer: E
Explanation:
Explanation/Reference:
248 mask uses 5 bits (1111 1000)
42 IP in binary is (0010 1010)
The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP ...
1111 1000 AND
0010 1010 equals
0010 1000 - which is .40
/24 is standard class C mask.
adding the 5 bits from the .248 mask gives /29