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NEW QUESTION: 1
Siehe Ausstellung. Wie verwaltet der Router den Datenverkehr zu 192.168.12.16?

A. Der Datenverkehr zwischen allen drei Routen wird ausgeglichen.
B. Es wählt die EIGRP-Route, da sie die geringste administrative Entfernung aufweist.
C. Wählt die RIP-Route aus, da sie das längste Präfix einschließlich der Zieladresse enthält.
D. Es wählt die OSPF-Route, da es das längste Präfix einschließlich der Zieladresse hat.
Answer: D

NEW QUESTION: 2
프로젝트 관리자는 위험으로 등록되지 않은 문제가 나타 났을 때 몇 달 동안 프로젝트를 관리하고 있습니다. 이 문제는 프로젝트에 큰 영향을 미칠 수 있습니다. 프로젝트 관리자는 어떻게 해야 합니까?
A. 스폰서에게 문제가 발생했으며 프로젝트의 성공 여부가 불확실 할 수 있음을 알립니다.
B. 프로젝트에 대한 위험으로 등록되지 않았고 계획된 대응도 없기 때문에 문제를 관리하지 마십시오.
C. 프로젝트 팀 및 관련 이해 관계자와 회의를 열어 문제를 관리하는 최선의 방법에 대해 합의합니다.
D. 문제가 해결되어 더 이상 프로젝트에 대한 위험으로 나타나지 않을 때까지 프로젝트를 연기하십시오.
Answer: C

NEW QUESTION: 3
This transaction is typically used in two modes: update and full replacement:
A. Premium Payment.
B. Health Care Claim.
C. First Report of Injury.
D. Health Plan Enrollment and Dis-enrollment.
E. Coordination of Benefits.
Answer: D

NEW QUESTION: 4
Examine the data in the PROMO_BEGIN_DATE column of the PROMOTIONS table:
PROMO_BEGIN _DATE 04-jan-00
10-jan-00
15-dec-99
18-oct-98
22-aug-99
You want to display the number of promotions started in 1999 and 2000.
Which query gives the correct output?
A. SELECT SUM(DECODE(SUBSTR(promo_begin_date,8),'00',1,0)) "2000",
SUM(DECODE(SUBSTR
(promo_begin_date,8),'99',1,0)) "1999"
FROM promotions;
B. SELECT COUNT(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '99' THEN 1 ELSE 0 END) "1999", COUNT(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '00' THEN 1 ELSE 0 END) "2000" FROM promotions;
C. SELECT SUM(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '99' THEN 1 ELSE 0 END)
"1999",SUM(CASE TO_CHAR(promo_begin_date,'yyyy') WHEN '00' THEN 1 ELSE 0 END)
"2000"
FROM promotions;
D. SELECT COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,'yyyy'), 8), '1999', 1, 0))
"1999", COUNT(DECODE(SUBSTR(TO_CHAR(promo_begin_date,'yyyy'), 8),'2000', 1,
0)) "2000"
FROM promotions;
Answer: A