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NEW QUESTION: 1
Regarding the IP routing, the following correct ()? (multiple choice)
A. Routing is the third layer concept in the OSI model
B. In the LAN, the route includes the following two parts: IP address and MAC address.
C. IP routing is the path information that guides IP packet forwarding.
D. Any route must include the following three parts: source address, destination address, and next hop.
Answer: A,C

NEW QUESTION: 2
Which two types of mailboxes are supported by Cisco Unity Express? (Choose two.)
A. general delivery
B. personal
C. temporary
D. voice
E. mailing list
F. group
Answer: A,B

NEW QUESTION: 3
開発者は、アカウントのSOQLステートメントの結果セットを反復処理することにより、メモリ内でリソースを大量に消費するアクションを実行するApexクラスのメソッドを特定する必要があります。このメソッドは、DMLステートメントを実行して、変更をデータベースに保存します。トランザクション制御を確保し、ガバナーの制限を超えないようにするためのベストプラクティスとして、開発者が実装する必要がある2つの手法はどれですか。選ぶ
2つの答え
A. System.Limitクラスを使用して、現在のCPUガバナー制限の消費を監視します。 （逃した）
B. 部分的なDMLステートメントを使用して、有効なデータのみがコミットされるようにします。
C. @ ReadOnlyアノテーションを使用して、SOQLによって返される行数をバイパスします。
D. Database.Savepointメソッドを使用して、データベースの整合性を強制します。 （逃した）
Answer: A,D

NEW QUESTION: 4
Note: This question is part of a series of questions that use the same or similar answer choices. An answer choice may be correct for more than one question in the series. Each question is independent of the other questions in this series. Information and details provided in a question apply to that question.
You have a database for a banking system. The database has two tables named tblDepositAcct and tblLoanAcct that store deposit and loan accounts, respectively. Both tables contain the following columns:

You need to determine the total number of deposit and loan accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN
tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
B. CustNo
C. CustNo = L.CustNo
D. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL OR
E. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT
CustNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
I. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
J. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
K. CustNo IS NULL
Answer: F
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.