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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 seed metric for redistributing into RIP on R3 is missing.
C. Auto-summary is misconfigured under the RIP process of R3.
D. The wildcard mask on R3 is misconfigured.
E. The OSPF processes on R2 and R3 are different.
Answer: B
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. Is only supported in major releases (R70 to R71, R75 to R76).
C. Is not a valid upgrade method in R76.
D. Treats each individual cluster member as an individual gateway.
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 RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123;
SELECT e.last_name, m.manager_id
B. 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
C. 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
D. FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123;
Answer: D