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NEW QUESTION: 1
ホットスポットに関する質問-RIPv2のトラブルシューティング
シナリオ:
マネージドネットワークサービスを提供する会社で働いており、小規模オフィスを運営している不動産クライアントでネットワークの問題が発生しています。ネットワークの問題のトラブルシューティングを行います。
-ルーターR1は本社をインターネットに接続し、ルーターR2とR3は内部ルーターです
-ルーターR1でNATが有効になっている
-R1、R2、R3の間で有効になっているルーティングプロトコルはRIPv2です
-R1は、内部ルーターが転送するためのデフォルトルートをRIPv2に送信します
R1へのインターネットトラフィック
-サーバー1とサーバー2はそれぞれVLAN 100と200に配置され、ルーターR2とのスティック構成で引き続き実行されています。
R1、R2、R3、およびL2SW1デバイスでコンソールにアクセスできます。
問題をトラブルシューティングするには、showコマンドのみを使用してください。






R1、R2、R3、およびL2SW1デバイスでコンソールにアクセスできます。
問題をトラブルシューティングするには、showコマンドのみを使用してください。
R2の構成を調べます。ルーターR2をソースとするR3 LANネットワークを宛先とするトラフィックは、R3ではなくR1に転送されます。何が問題になるのでしょうか?

A. 特定された問題はありません。デフォルトルートはルーターR1によってRIPv2ドメインに伝播されるため、この動作は正常です。
B. R3でRIPv2が有効になっていません。
C. パッシブインターフェイス機能を使用して、R2とR3の間でRIPv2ルーティング更新が抑制されます。
D. R3でRIPv2が有効になっていますが、RIPv2ドメインにアドバタイズされていないR3 LANネットワーク。
Answer: B
Explanation:
Explanation:
First we should check the routing table of R2 with the "show ip route" command.

In this table we cannot find the subnet "10.10.12.0/24" (R3 LAN network) so R2 will use the default route advertised from R1 (with the command "default-information originate" on R1) to reach unknown destination, in this case subnet 10.10.12.0/24 -> R2 will send traffic to
10.10.12.0/24 to R1.
Next we need to find out why R3 did not advertise this subnet to R2. A quick check with the "show running-config" on R3 we will see that R3 was not configured with RIP ( no "router rip" section).
Therefore we can conclude RIPv2 was not enabled on R3.

NEW QUESTION: 2
다중 스레드 응용 프로그램은 단일 스레드 응용 프로그램보다 더 위험합니다.
A. 경쟁 조건.
B. 데이터베이스 삽입.
C. 바이러스 감염.
D. 패킷 스니핑.
Answer: A

NEW QUESTION: 3
A researcher has a sample of 700 observations from a population whose standard deviation is known to be
1,235.6. The mean of the sample is calculated to be 219.2. The null hypothesis is stated as Ho: mean
150 and the alternative is non-directional. The p-value in this case equals ________.
A. 12.82%
B. 13.88%
C. 6.94%
D. 10.16%
Answer: B
Explanation:
Explanation/Reference:
Explanation:
The z-statistic under the null equals (219.2 - 150)/(1235.6/(700

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.5)) = 1.48. Since the alternative is non- directional, you must use a two-tailed test and ask the question, "What is the probability of observing a z- statistic at least as large in magnitude as 1.48?" The answer to this will give you the p- value for this sample under the two-tailed test. The right-tailed probability of observing a z-statistic which is at least as big as 1.48 equals 1.0 - 0.9306 = 0.0694 = 6.94%. The left-tailed probability of observing a z-statistic which is at least as small as - 1.48 also equals 6.94% since the standard normal distribution is symmetrical about zero. Therefore, the p-value of this two-tailed test in this sample equals 6.94%*2 = 13.88%.