2014年7月2日星期三

1Z0-027 1z1-061 dernières questions d'examen certification Oracle et réponses publiés

Pass4Test est un site de vous ramener au succès. Pass4Test peut vous aider à promouvoir les connaissances essentielles pour le test Oracle 1Z0-027 et passer le test à la première fois.

Est-ce que vous vous souciez encore pour passer le test Oracle 1z1-061? Pourquoi pas choisir la formation en Internet dans une société de l'informatique. Un bon choix de l'outil formation peut résoudre le problème de prendre grande quantité de connaissances demandées par le test Oracle 1z1-061, et vous permet de préparer mieux avant le test. Les experts de Pass4Test travaillent avec tous efforts à produire une bonne Q&A ciblée au test Oracle 1z1-061. La Q&A est un bon choix pour vous. Vous pouvez télécharger le démo grantuit tout d'abord en Internet.

Si vous vous inscriez le test Oracle 1Z0-027, vous devez choisir une bonne Q&A. Le test Oracle 1Z0-027 est un test Certification très important dans l'Industrie IT. C'est essentielle d'une bonne préparation avant le test.

Code d'Examen: 1Z0-027
Nom d'Examen: Oracle (Oracle Exadata Database Machine Administration, Software Release 11.x)
Questions et réponses: 72 Q&As

Code d'Examen: 1z1-061
Nom d'Examen: Oracle (Oracle Database 12c: SQL Fundamentals)
Questions et réponses: 75 Q&As

Est-ce que vous vous souciez encore de réussir le test Oracle 1Z0-027? Est-ce que vous attendez plus le guide de formation plus nouveaux? Le guide de formation vient de lancer par Pass4Test peut vous donner la solution. Vous pouvez télécharger la partie de guide gratuite pour prendre un essai, et vous allez découvrir que le test n'est pas aussi dur que l'imaginer. Pass4Test vous permet à réussir 100% le test. Votre argent sera tout rendu si vous échouez le test.

Vous pouvez comparer un peu les Q&As dans les autres sites web que lesquelles de Pass4Test, c'est pas difficile à trouver que la Q&A Oracle 1Z0-027 est plus complète. Vous pouvez télécharger le démo gratuit à prendre un essai de la qualité de Pass4Test. La raison de la grande couverture des questions et la haute qualité des réponses vient de l'expérience riche et la connaissances professionnelles des experts de Pass4Test. La nouvelle Q&A de Oracle 1Z0-027 lancée par l'équipe de Pass4Test sont bien populaire par les candidats.

Beaucoup de gens trouvent difficile à passer le test Oracle 1z1-061, c'est juste parce que ils n'ont pas bien choisi une bonne Q&A. Vous penserez que le test Oracle 1z1-061 n'est pas du tout autant dur que l'imaginer. Le produit de Pass4Test non seulement comprend les Q&As qui sont impressionnées par sa grande couverture des Questions, mais aussi le service en ligne et le service après vendre.

1z1-061 Démo gratuit à télécharger: http://www.pass4test.fr/1z1-061.html

NO.1 Which normal form is a table in if it has no multi-valued attributes and no partial
dependencies?
A. First normal form
B. Second normal form
C. Third normal form
D. Fourth normal form
Answer: B

certification Oracle   certification 1z1-061   1z1-061 examen   1z1-061 examen   1z1-061 examen

NO.2 View the Exhibit and examine the structure of the product, component, and PDT_COMP
tables.
In product table, PDTNO is the primary key.
In component table, COMPNO is the primary key.
In PDT_COMP table, <PDTNO, COMPNO) is the primary key, PDTNO is the foreign key referencing
PDTNO in product table and COMPNO is the foreign key referencing the COMPNO in component
table.
You want to generate a report listing the product names and their corresponding component names,
if the component names and product names exist.
Evaluate the following query:
SQL>SELECT pdtno, pdtname, compno, compname
FROM product _____________ pdt_comp
USING (pdtno) ____________ component USING (compno)
WHERE compname IS NOT NULL;
Which combination of joins used in the blanks in the above query gives the correct output?
A. JOIN; JOIN
B. FULL OUTER JOIN; FULL OUTER JOIN
C. RIGHT OUTER JOIN; LEFT OUTER JOIN
D. LEFT OUTER JOIN; RIGHT OUTER JOIN
Answer: C

Oracle   1z1-061 examen   1z1-061 examen   certification 1z1-061

NO.3 View the Exhibit for the structure of the student and faculty tables.
You need to display the faculty name followed by the number of students handled by the faculty at
the base location.
Examine the following two SQL statements:
Which statement is true regarding the outcome?
A. Only statement 1 executes successfully and gives the required result.
B. Only statement 2 executes successfully and gives the required result.
C. Both statements 1 and 2 execute successfully and give different results.
D. Both statements 1 and 2 execute successfully and give the same required result.
Answer: D

Oracle examen   certification 1z1-061   certification 1z1-061   1z1-061   certification 1z1-061   1z1-061 examen

NO.4 Examine the structure proposed for the transactions table:
Which two statements are true regarding the creation and storage of data in the above table
structure?
A. The CUST_STATUS column would give an error.
B. The TRANS_VALIDITY column would give an error.
C. The CUST_STATUS column would store exactly one character.
D. The CUST_CREDIT_LIMIT column would not be able to store decimal values.
E. The TRANS_VALIDITY column would have a maximum size of one character.
F. The TRANS_DATE column would be able to store day, month, century, year, hour, minutes,
seconds, and fractions of seconds
Answer: B,C

Oracle examen   1z1-061 examen   certification 1z1-061   1z1-061
Explanation:
VARCHAR2(size)Variable-length character data (A maximum size must be specified:
minimum size is 1; maximum size is 4, 000.)
CHAR [(size)] Fixed-length character data of length size bytes (Default and minimum size
is 1; maximum size is 2, 000.)
NUMBER [(p, s)] Number having precision p and scale s (Precision is the total number of
decimal digits and scale is the number of digits to the right of the decimal point; precision
can range from 1 to 38, and scale can range from -84 to 127.)
DATE Date and time values to the nearest second between January 1, 4712 B.C., and
December 31, 9999 A.D.

NO.5 Examine the types and examples of relationships that follow:
1.One-to-one a) Teacher to students
2.One-to-many b) Employees to Manager
3.Many-to-one c) Person to SSN
4.Many-to-many d) Customers to products
Which option indicates the correctly matched relationships?
A. 1-a, 2-b, 3-c, and 4-d
B. 1-c, 2-d, 3-a, and 4-b
C. 1-c, 2-a, 3-b, and 4-d
D. 1-d, 2-b, 3-a, and 4-c
Answer: C

Oracle examen   certification 1z1-061   certification 1z1-061   1z1-061 examen   1z1-061

NO.6 In the customers table, the CUST_CITY column contains the value 'Paris' for the
CUST_FIRST_NAME 'Abigail'.
Evaluate the following query:
What would be the outcome?
A. Abigail PA
B. Abigail Pa
C. Abigail IS
D. An error message
Answer: B

certification Oracle   certification 1z1-061   certification 1z1-061   certification 1z1-061

NO.7 You need to create a table for a banking application. One of the columns in the table has the
following requirements:
1. You want a column in the table to store the duration of the credit period.
2) The data in the column should be stored in a format such that it can be easily added and
subtracted with date data type without using conversion functions.
3) The maximum period of the credit provision in the application is 30 days.
4) The interest has to be calculated for the number of days an individual has taken a credit for.
Which data type would you use for such a column in the table?
A. DATE
B. NUMBER
C. TIMESTAMP
D. INTERVAL DAY TO SECOND
E. INTERVAL YEAR TO MONTH
Answer: D

certification Oracle   1z1-061   certification 1z1-061

NO.8 View the Exhibit and evaluate the structure and data in the CUST_STATUS table.
You issue the following SQL statement:
Which statement is true regarding the execution of the above query?
A. It produces an error because the AMT_SPENT column contains a null value.
B. It displays a bonus of 1000 for all customers whose AMT_SPENT is less than CREDIT_LIMIT.
C. It displays a bonus of 1000 for all customers whose AMT_SPENT equals CREDIT_LIMIT, or
AMT_SPENT is null.
D. It produces an error because the TO_NUMBER function must be used to convert the result of the
NULLIF function before it can be used by the NVL2 function.
Answer: C

Oracle examen   1z1-061 examen   1z1-061 examen   1z1-061   1z1-061 examen   1z1-061 examen
Explanation:
The NULLIF Function The NULLIF function tests two terms for equality. If they are equal the function
returns a null, else it returns the first of the two terms tested. The NULLIF function takes two
mandatory parameters of any data type. The syntax is NULLIF(ifunequal, comparison_term), where
the parameters ifunequal and comparison_term are compared. If they are identical, then NULL is
returned. If they differ, the ifunequal parameter is returned.

没有评论:

发表评论