You need to display the first names of all customers from the CUSTOMERS table that contain the character 'e' and have the character 'a' in the second last position.
Which query would give the required output?

• A.SELECT cust_first_name
  FROM customers
  WHERE INSTR(cust_first_name, 'e')<>0 AND
  SUBSTR(cust_first_name, -2, 1)='a';
• B.SELECT cust_first_name
  FROM customers
  WHERE INSTR(cust_first_name, 'e')IS NOT NULL AND
  SUBSTR(cust_first_name, 1, -2)='a';
• C.SELECT cust_first_name
  FROM customers
  WHERE INSTR(cust_first_name, 'e')<>0 AND
  SUBSTR(cust_first_name, LENGTH(cust_first_name), -2)='a';
• D.SELECT cust_first_name
  FROM customers
  WHERE INSTR(cust_first_name, 'e')<>'' AND
  SUBSTR(cust_first_name, -2, 1)='a';

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The SYSDATEfunction displays the current Oracle Server date as:
21-MAY-19
You wish to display the date as:
MONDAY, 21 MAY, 2019
Which statement will do this?
SELECT TO_DATE(SYSDATE, 'FMDAY, DD MONTH, YYYY') FROM DUAL;

• A.SELECT TO_CHAR(SYSDATE, 'FMDD, DAY MONTH, YYYY') FROM DUAL;
• B.SELECT TO_CHAR(SYSDATE, 'FMDAY, DDTH MONTH, YYYY') FROM DUAL;
• C.SELECT TO_CHAR(SYSDATE, 'FMDAY, DD MONTH, YYYY') FROM DUAL;
• D.

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Evaluate the following two queries:
SQL> SELECT cust_last_name, cust_city
FROM customers
WHERE cust_credit_limit IN (1000, 2000, 3000);
SQL> SELECT cust_last_name, cust_city
FROM customers
WHERE cust_credit_limit = 1000 or cust_credit_limit = 2000 or
cust_credit_limit = 3000
Which statement is true regarding the above two queries?

• A.Performance would improve in query 2 only if there are null values in the CUST_CREDIT_LIMIT column.
• B.There would be no change in performance.
• C.Performance would degrade in query 2.
• D.Performance would improve in query 2.

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Examine the types and examples of relationships that follow:

Which option indicates correctly matched relationships? (Choose the best answer.)

• A.1-c, 2-a, 3-b, and 4-d
• B.1-d, 2-b, 3-a, and 4-c
• C.1-c, 2-d, 3-a, and 4-b
• D.1-a, 2-b, 3-c, and 4-d

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Which two are true about the NVL, NVL2, and COALESCE functions?

• A.NVL2 can have any number of expressions in the list.
• B.NVL can have any number of expressions in the list.
• C.The first expression in NVL2 is never returned.
• D.COALESCE stops evaluating the list of expressions when it finds the first non-null value.
• E.COALESCE stops evaluating the list of expressions when it finds the first null value.
• F.NVL must have expressions of the same data type.

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