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NEW QUESTION: 1
Your network contains a System Center 2012 Configuration Manager environment.
You need to create a report that lists all of the client computers that do not have an Application named App1 installed.
Which query should you use to create the report?
A. SELECT SYS.Netbios_Name0 from v_R_System SYS WHERE SYS.ResourceID NOT IN (SELECT SYS.ResourceID FROM v_R_System AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP.ResourceID WHERE ARP.DisplayName0 = 'App1')
B. SELECT SYS.Netbios_Name0, ARP.DisplayName0 FROM v_R_System AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP. ResourceID WHERE ARP.DisplayName0 NOT IN ('App1')
C. SELECT SYS.Netbios_Name0 from v_R_System SYS WHERE SYS.ResourceID IN (SELECT SYS.ResourceID FROM v_R_System AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID - ARP.ResourceID. WHERE ARP.DisplayName0 = 'App1')
D. SELECT SYS.Netbios_Name0, ARP.DisplayName0 FROM v_R_System AS SYS INNER JOIN V_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID - ARP.ResourceID WHERE ARP.DisplayName0 NOT LIKE 'App1'
Answer: A
Explanation:
SELECT computer name FROM table v_R_System (SQL Syntax)
WHERE resource ResourceID is NOT IN
Add Remove Programs WHERE DisplayName0 = 'App1'
SQL INNER JOIN Keyword
The INNER JOIN keyword returns rows when there is at least one match in both tables.
Reference: http://www.w3schools.com/sql/sql_join_inner.asp
NEW QUESTION: 2
どのタイプの分析により、悪用がネットワークに影響を与える可能性があるかを確認できますか？
A. カジュアル
B. 確率的
C. 説明的
D. 推測
Answer: B
Explanation:
決定論的分析では、分析に使用されるすべてのデータが事前にわかっています。一方、確率的分析は、何かが発生する可能性を想定して行われますが、いつ、どのように発生するか正確にはわかりません。
NEW QUESTION: 3
You want to configure a device to select an OSPF-learned route as the preferred path overan EBGP-learned route Which action must you take?
A. Decrease the OsPF administrative distance
B. Increase the oSPF administrative distance
C. Decrease the OSPF cost
D. Increase the OSPF cost
Answer: A