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NEW QUESTION: 1
Refer to the exhibit.
R2 is unable to access the 172.16.1.0/30 network between R1 and R3. Which option is a possible reason for the failure?
A. The subnet mask on the link between R2 and R3 is smaller than /30.
B. The wildcard mask on R3 is misconfigured.
C. The seed metric for redistributing into RIP on R3 is missing.
D. The OSPF processes on R2 and R3 are different.
E. Auto-summary is misconfigured under the RIP process of R3.
Answer: C
Explanation:
The problem is that RIP requires a seed metric to be specified when redistributing routes into that protocol.
A seed metric is a "starter metric" that gives the RIP process a metric it can work with. The OSPF metric of cost is incomprehensible to RIP, since RIP's sole metric is hop count.
Reference.
http://www.thebryantadvantage.com/CCNP%20Certification%20BSCI%20Exam%20Tutorial%20R oute%
20Redistribution%20Seed%20Metric.htm

NEW QUESTION: 2
A Fast Path Upgrade of a cluster:
A. Upgrades all cluster members except one at the same time.
B. Treats each individual cluster member as an individual gateway.
C. Is not a valid upgrade method in R76.
D. Is only supported in major releases (R70 to R71, R75 to R76).
Answer: C

NEW QUESTION: 3
ある会社が独立した著者向けに電子書籍を発行しています。
同社は、Power Appsポータルソリューションを実装して、今後の本に関する発表を一般に公開したいと考えています。
会社の機能を推奨する必要があります。
どの機能を使用する必要がありますか？回答するには、回答領域で適切なオプションを選択します。
注：それぞれの正しい選択は1ポイントの価値があります。

Answer:
Explanation:

Explanation

Reference:
https://www.onactuate.com/upgrades/what-is-the-new-powerapps-portal/
https://docs.microsoft.com/en-us/powerapps/maker/portals/configure/page-templates

NEW QUESTION: 4
Examine the structure of the EMPLOYEEStable.

There is a parent/child relationship between EMPLOYEE_IDand MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_IDis 123.
Which query provides the correct output?
SELECT e.last_name, m.manager_id
A. FROM employees e LEFT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123;
SELECT e.last_name, e.manager_id
B. FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.employee_id)
WHERE e.employee_id = 123;
SELECT m.last_name, e.manager_id
C. FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123;
D. FROM employees e RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123;
SELECT e.last_name, m.manager_id
Answer: C