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OMG-Certified Systems Modeling Professional - Model User
- Exam Number/Code : OMG-OCSMP-MU100
- Exam Name : OMG-Certified Systems Modeling Professional - Model User
- Questions and Answers : 213 Q&As
- Update Time: 2019-01-10
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NEW QUESTION: 1
What is a table domain?
A. Static sets of values based on the values in the table window
B. Dynamic sets of values based on the values of another object
C. Static sets of values based on the values of another object
D. Dynamic sets of values based on the values in the table window
Answer: B
NEW QUESTION: 2
Which Orange Book evaluation level is described as "Verified Design"?
A. B1.
B. B2.
C. A1.
D. B3.
Answer: C
Explanation:
Level A1 is described as Verified Design.
The following answers are incorrect:
B3. This is incorrect because level B3 is described as Security Domains.
B2. This is incorrect because level B2 is described as Structured Protection. B1. This is incorrect because level B1 is described as Labeled Security.
NEW QUESTION: 3
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 customers who have only loan accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
B. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL OR
C. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
E. CustNo IS NULL
F. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
G. CustNo = L.CustNo
H. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN
tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
I. CustNo
J. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
K. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
Answer: H
Explanation:
The RIGHT JOIN keyword returns all records from the right table (table2), and the matched records from the left table (table1). The result is NULL from the left side, when there is no match.
References: https://www.w3schools.com/sql/sql_join_right.asp
NEW QUESTION: 4
Ein Arbeitsprozess ...
Bitte wählen Sie die richtige Antwort.
A. Verwendet einen gemeinsamen Speicherbereich, der als gemeinsamer Speicher bezeichnet wird.
B. Wird inaktiv, während auf einen Benutzer gewartet wird.
C. Bleibt über den Dispatcher mit einem Bildschirm verbunden.
D. Wird aktiv, während auf einen Benutzer gewartet wird.
Answer: A