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NEW QUESTION: 1
Refer to the exhibit.

The network is converged. After link-state advertisements are received from Router_A, what information will
Router_E contain in its routing table for the subnets 208.149.23.64 and 208.149.23.96?
A. O 208.149.23.64 [110/3] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.172.23.10, 00:00:16, Serial 1/0
B. O 208.149.23.64 [110/1] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.173.23.10, 00:00:16, FastEthernet 0/0
C. O 208.149.23.64 [110/13] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/13] via 190.172.23.10, 00:00:16, Serial 1/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
D. O 208.149.23.64 [110/13] via 190.173.23.10, 00:00:07, FastEthernet 0/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
Answer: D
Explanation:
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface.
The interface cost is calculated with the formula 108 / Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 /
100,000,000 = 1. Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets - B is not correct.
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 64; T3 cost = 108 / 45 Mbps
= 2) so surely OSPF will choose the path through T3 link -> Router_E will choose the path from Router_A through
FastEthernet0/0, not Serial1/0 - C & D are not correct.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -
they are surely incorrect.

NEW QUESTION: 2
RMANリカバリカタログの使用が必要な3つの機能はどれですか？ （3つ選択してください。）
A. 単一のデータベースのバックアップに関するカスタマイズされたレポートの作成
B. REPORT SCHEMAコマンドを使用して、過去のデータベースのデータファイルとテーブルスペースを一覧表示する
C. REPORT SCHEMAコマンドを使用して、データベースの現在のデータファイルとテーブルスペースを一覧表示します
D. RMANに保存されたスクリプトを使用する
E. 暗号化されたバックアップの作成
F. BACKUPコマンドでKEEPFOREVER句を使用する
Answer: C,D,F

NEW QUESTION: 3
In the third grade of Windblown School there are 108 students, one third of them failed the math test and
1/6 failed that literature test. At least how many students failed both tests?
A. 6.
B. 8.
C. 12.
D. 10.
E. 0.
Answer: E
Explanation:
Explanation/Reference:
Explanation:
(1/3 x 108 = 36) failed the math test.
(1/6 x 108 = 18) failed that literature test.
The least amount of people that failed both tests is (18 + 36 -108 = -54), there can't be an negative Overlapping between the groups so the least amount of people who failed both tests is zero.