QUESTION: 17

A company wants to award a 3% increase in salary to all employees who meet these two criteria:

* They have a valid department assignment.
* Their current salary is less than $50, 000.

Which two statements will accomplish this objective? (Choose two.)

UPDATEemp FROM department AS dept, employee AS emp
SET salary_amount = salary_amount * 1.03
WHERE emp.department_number = dept.department_number
AND salary_amount < 50000;
UPDATE employee ASemp FROM department AS dept
SET salary_amount = salary_amount * 1.03
WHERE emp.department_number = dept.department_number
AND salary_amount < 50000;
UPDATE employee
SET salary_amount = salary_amount * 1.03
WHERE employee.department_number IN
(SELECT department_number FROM department
AND salary_amount < 50000);
UPDATE employee
SET salary_amount = salary_amount * 1.03
WHERE employee.department_number IN
(SELECT department_number FROM department)
AND salary_amount < 50000;

Answer(s): B,D

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QUESTION: 18

Given the following CREATE TABLE statement:

CREATE SET TABLE SPA_APPT_AVAILABILITY
( SPA_ID INTEGER NOT NULL
, APPT_DATE DATE NOT NULL
, APPT_TIME CHAR(5) NOT NULL
, TREATMENT_ID INTEGER NOT NULL
, TREATMENT_NAME CHAR(50) NOT NULL
, AESTHETICIAN_ID INTEGER NOT NULL
, COMMENTS VARCHAR(100)
, PRIMARY KEY (SPA_ID, APPT_DATE))
;

Which column(s) will be defined as the primary index?

SPA_ID
TREATMENT_ID
SPA_ID, APPT_DATE
SPA_ID, TREATMENT_ID

Answer(s): D

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QUESTION: 19

The date '2009-04-10' is represented by StartDate and the date '2008-03-07' is represented by
EndDate.
Which calculation produces the number of days between the two dates?

EndDate - StartDate
DIFF(EndDate, StartDate)
DAYS(StartDate, EndDate)
EXTRACT(DAY FROMEndDate) - EXTRACT(DAY FROM StartDate)

Answer(s): B

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QUESTION: 20

A mobile provider wants to increase the price on a specific feature included in several of their service offers. The feature name contains the string "walkabout".
Which query modifies the correct rows from the Phone_Packages table?

UPDATEPhone_Packages
SET price = price * 1.25
WHERE Features.feature_name LIKE '%walkabout%';
UPDATEPhone_Packages
SET price = price * 1.25
WHERE feature_code IN (SELECT feature_code
FROM Features
WHERE feature_name LIKE '%walkabout%');
UPDATEPhone_Packages FROM Features
SET price = price * 1.25
WHERE Features.feature_name LIKE '%walkabout%';
UPDATEPhone_Packages FROM Features
SET price = price * 1.25
WHERE feature_code IN (SELECT feature_code
FROM Features
WHERE feature_name LIKE '%walkabout%');

Answer(s): D

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QUESTION: 20

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