View the Exhibit and examine the data in the PRODUCTStable.

Which statement would add a column called PRICE, which cannot contain NULL?
ALTER TABLE products

• A.ADD price NUMBER(8,2) DEFAULT 0 NOT NULL;
  ALTER TABLE products
• B.ADD price NUMBER(8,2) NOT NULL;
  ALTER TABLE products
• C.ADD price NUMBER(8,2) DEFAULT NOT NULL;
  ALTER TABLE products
• D.ADD price NUMBER(8,2) DEFAULT CONSTRAINT p_nn NOT NULL.

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Examine the contents of the EMP table:

What is the result?

• A.It will return the six employees earning the lowest salaries. In ascending order.
• B.It will return the five employees earning the highest salaries. In descending order.
• C.It will return the five employee's earning the lowest salaries. In ascending order.
• D.It will return the six employees earning the highest salaries. In descending order.

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What is true about non-equijoin statement performance?

• A.The Oracle join syntax performs better than the SQL:1999 compliant ANSI join syntax.
• B.The join syntax used makes no difference to performance.
• C.The BETWEEN condition always performs less well than using the >= and <= conditions.
• D.Table aliases can improve performance.
• E.The BETWEEN condition always performs better than using the >= and <= conditions.

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The EMPLOYEES table contains columns EMP_ID of data type NUMBER and HIRE_DATE of data type DATE
You want to display the date of the first Monday after the completion of six months since hiring.
The NLS_TERRITORY parameter is set to AMERICA in the session and, therefore, Sunday is the first day of the week Which query can be used?

• A.SELECT emp_id,NEXT_DAY(ADD_MONTHS(hire_date,6),1) FROM employees;
• B.SELECT emp_id,ADD_MONTHS(hire_date,6), NEXT_DAY('MONDAY') FROM employees;
• C.SELECT emp_id,NEXT_DAY(ADD_MONTHS(hite_date,6),'MONDAY') FROM employees;
• D.SELECT emp_id,NEXT_DAY(MONTHS_BETWEEN(hire_date,SYSDATE),6) FROM employees;

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You must display details of all users whose username contains the string 'ch_'. (Choose the best answer.) Which query generates the required output?

• A.SELECT * FROM users
  Where user_name LIKE '%ch\_%' ESCAPE '\';
• B.SELECT * FROM users
  Where user_name LIKE '%ch_';
• C.SELECT * FROM users
  Where user_name LIKE 'ch\_%' ESCAPE '_';
• D.SELECT * FROM users
  Where user_name LIKE '%ch_%'ESCAPE'%';

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