Deck 9: Advanced Query Formulation With SQL

Question 1

Outer join problems cannot be formulated in SQL at all using the cross product style.

Answer 1

False
Explanation: Outer join problems can be formulated using the cross product style with proprietary SQL extensions.

Question 2

To be union compatible each corresponding column from both temporary tables must have compatible data types.

Answer 2

True

Question 3

Using the join operator style,the result set of a one-sided outer join depends on the direction and position of the table names in the query.

Answer 3

True

Question 4

The following SQL statement is an example of a Type II nested query:
SELECT StdSSN,StdLastName,StdMajor
FROM Student
WHERE Student.StdSSN IN
(SELECT StdSSN FROM Enrollment
WHERE EnrGrade > 3.5)

Question 5

A nested query cannot have multiple levels,i.e.you cannot have a nested query inside another nested query.

Question 6

Nested queries can be used as part of a condition in WHERE or HAVING clauses,but they cannot be used in the FROM clause of a query.

Question 7

Type II nested queries are very useful for difference problems but should not be used for join problems.

Question 8

A good table design usually eliminates the use of null values,so their effects on query results is not generally a concern for database developers.

Question 9

Typically,a full outer join is used to combine two tables that are similar but are not union compatible.

Question 10

The following SQL statement is an example of a Type II nested query:
SELECT StdSSN,StdFirstName,StdLastName,StdMajor
FROM Student
WHERE EXISTS
(SELECT StdSSN FROM Enrollment
WHERE Student.StdSSN = Enrollment.StdSSN
AND EnrGrade > 3.5)

Question 11

One-Sided Outer Join: an operator that generates the join result (the nonmatching rows)from one of the input tables.SQL supports the one-sided outer join operator through the LEFT JOIN and RIGHT JOIN keywords.

Question 12

The following SQL statement would execute in both Access and Oracle 8i:
SELECT FacSSN,FacLastName,StdSSN,StdLastName
FROM Faculty FULL JOIN Student
ON Faculty.SSN = Student.SSN

Question 13

Using a Type I nested query is the standard way to reference related tables in SQL DELETE statements.

Question 14

One of the reasons Type II nested queries are not used for join problems is that their repetitive execution makes them inefficient for this purpose.

Question 15

Type II nested queries should not be used for a join operation when the result table contains any columns from the inner query.

Question 16

The following SQL statement would return only the non-matching rows of the left table:
SELECT FacSSN,LastName
FROM Faculty LEFT JOIN Student
ON Faculty.SSN = Student.SSN
WHERE Student.StdSSN IS NULL

Question 17

Type I nested queries can be used like loops and Type II nested queries can be used like procedures in a programming language.

Question 18

In simple conditions,only rows in which the condition evaluates to false are excluded from the result set.

Question 19

Type I nested queries are useful for some join problems but should not be used for difference problems.

Question 20

A full outer join generates a table with the matching rows plus the nonmatching rows from both tables.

Question 21

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-(ORACLE)SELECT CNAME,DATE_OUT,RTN FROM CUSTOMER,RENTALS
WHERE CUSTOMER.CID = RENTALS.CID
AND BIRTHPLACE IN ('ERIE','CARY')AND EXISTS
(SELECT * FROM RENTCOST
WHERE COST < 40 AND RENTALS.MAKE= RENTCOST.MAKE)
The CNAMEs shown by the execution of this query are:

A) SIMON
B) GREEN, BLACK, SIMON
C) GREEN
D) GREEN, SIMON

Question 22

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT DISTINCT CNAME FROM CUSTOMER
WHERE CID IN
(SELECT CID FROM RENTALS WHERE RETURN IN
(SELECT BIRTHPLACE FROM CUSTOMER))
What is the meaning of this query?

A) List customers whose rental was returned to their birthplace
B) List customers whose rental was returned to someone else's birthplace
C) List customers whose rental was returned to a city that is also a birthplace
D) None of the above

Question 23

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT DISTINCT CID,CNAME FROM CUSTOMER
WHERE CID IN
(SELECT CID FROM RENTALS WHERE MAKE IN ('FORD','TOYOTA'))
The CNAMEs shown by the execution of this query are:

A) BLACK
B) BLACK, JONES
C) BLACK, JONES, MARTIN
D) BLACK, JONES, MARTIN, VERNON

Question 24

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT CID,CNAME FROM CUSTOMER HERE EXISTS
(SELECT CID FROM RENTALS
WHERE CUSTOMER.CID = RENTALS.CID
AND PICKUP = 'CARY')
The execution of this query lists the following CNAMEs:

A) BLACK, GREEN, JONES, MARTIN, SIMON
B) BLACK, JONES, MARTIN
C) BLACK, GREEN, JONES, MARTIN, SIMON, VERNON
D) BLACK

Question 25

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-(ACCESS)SELECT CNAME,DATE_OUT FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
WHERE AGE < 40 AND RTN IN
(SELECT RTN FROM RENTLENGTH
WHERE DAYS > 1)
The CNAMEs shown by the execution of this query are:

A) JONES, MARTIN
B) GREEN, JONES, MARTIN
C) BLACK, JONES, MARTIN
D) BLACK, GREEN, JONES, MARTIN

Question 26

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT DISTINCT CNAME,AGE FROM CUSTOMER,RENTALS
WHERE CUSTOMER.CID <> RENTALS.CID
The execution of this query produces the following number of rows:

A) 0
B) 1
C) 2
D) 6

Question 27

When null values exist in one of the columns in a simple condition,it is possible that those rows will not appear in the result of either the simple condition or its negation.

Question 28

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT CID,CNAME FROM CUSTOMER WHERE 0 =
(SELECT COUNT(*)FROM RENTALS
WHERE CUSTOMER.CID = RENTALS.CID)
What is the interpretation of this query?

A) List the customers who do not have rentals
B) List the customers who have one rental
C) List the customers who have 0 or more rentals
D) List the customers who have 1 or more rentals

Question 29

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-DELETE FROM CUSTOMER WHERE CID IN
(SELECT CID FROM RENTALS WHERE RETURN IN
(SELECT BIRTHPLACE FROM CUSTOMER))
The number of rows deleted from CUSTOMER by the execution of this query is:

A) 6
B) 5
C) 4
D) 3

Question 30

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT DISTINCT CID,CNAME FROM CUSTOMER
WHERE CID IN
(SELECT CID FROM RENTALS WHERE MAKE IN ('FORD','TOYOTA')
AND RTN IN
(SELECT RTN FROM RENTLENGTH
WHERE DAYS < 3))
The CNAMEs shown by the execution of this query are:

A) BLACK
B) BLACK, JONES
C) BLACK, JONES, MARTIN
D) JONES, MARTIN

Question 31

When performing a grouping operation,the SQL standard specifies that all rows with null values are grouped together.

Question 32

Null values affect:

A) simple conditions involving comparison operators
B) compound conditions involving logical operators
C) aggregate calculations and grouping
D) all of the above

Question 33

Nested queries can be used like a procedure (Type I nested query)in which the nested query is executed one time or like a loop (Type II nested query)in which the nested query is executed repeatedly.

A) Type I nested query, Type III nested query
B) Type I nested query, Type II nested query
C) Type II nested query, Type I nested query
D) None of the above

Question 34

Select the statement that is correct about the outer join operator:

A) The outer join operator generates the join result (the matching rows).
B) The outer join operator generates the join result (the nonmatching rows).
C) The outer join operator generates the join result (the matching rows) plus the nonmatching rows.
D) The outer join operator generates the join result (the nonmatching rows) but excludes the matching rows.

Question 35

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-(ACCESS)SELECT CNAME,DATE_OUT,RTN FROM CUSTOMER INNER JOIN RENTALS
ON CUSTOMER.CID = RENTALS.CID
WHERE AGE < 40 AND EXISTS
(SELECT * FROM RENTLENGTH
WHERE RENTALS.RTN = RENTLENGTH.RTN
AND DAYS > 1)
The CNAMEs shown by the execution of this query are:

A) JONES, MARTIN
B) GREEN, JONES, MARTIN
C) BLACK, JONES, MARTIN
D) BLACK, GREEN, JONES, MARTIN

Question 36

When using the SUM function with two columns,a row containing a null value for either column would produce a null value for that row.

Question 37

The following two functions will always produce the same value when performed in a SQL statement:
SUM(Column1)+ SUM(Column2)
SUM(Column1 + Column2)

Question 38

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT CID,CNAME FROM CUSTOMER WHERE CID NOT IN
(SELECT CID FROM RENTALS)
The execution of this query produces the following number of rows:

A) 0
B) 1
C) 2
D) 3

Question 39

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-(ACCESS)DELETE RENTALS.* FROM RENTALS INNER JOIN CUSTOMER
ON CUSTOMER.CID = RENTALS.CID
AND CNAME IN ('BLACK','SIMON'))
The execution of this query deletes the following number of rows from the table RENTALS:

A) 1
B) 2
C) 3
D) 4

Question 40

The COUNT(*)function always returns the same value as the COUNT(column)function.

Question 41

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT * FROM VOTE WHERE Vote2 >=10
The execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 42

In outer join queries,non-matched rows in the result table would contain ____________ values in certain columns.

Question 43

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT * FROM VOTE WHERE Vote1>10 OR VOTE2 >10
The execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 44

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT * FROM VOTE WHERE Vote2 <10
The execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 45

The SQL keyword(s)____________ in a SELECT query generate a result table that contains the matching rows and the non-matching rows of both of the input tables.

Question 46

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT MAKE FROM RENTALS,CUSTOMER
WHERE RENTALS.CID = CUSTOMER.CID
AND RESID_CITY = 'HEMET'
GROUP BY MAKE
HAVING COUNT (DISTINCT RENTALS.CID)=
(SELECT COUNT(*)FROM CUSTOMER
WHERE RESID_CITY = 'HEMET')
The execution of this query produces the following number of rows:

A) 0
B) 1
C) 2
D) 3

Question 47

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT * FROM VOTE WHERE Vote2 >=10
UNION
SELECT * FROM VOTE
WHERE Vote2 <10
The execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 48

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT * FROM VOTE WHERE Vote1>10 AND VOTE2 >10
The execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 49

The SQL keyword(s)____________ in a SELECT query generate a result table that contains the matching rows and the non-matching rows of the left table.

Question 50

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT CID FROM RENTALS GROUP BY CID
HAVING COUNT (DISTINCT MAKE)= (SELECT COUNT(*)FROM RENTCOST)
The operation performed by this query is:

A) A difference
B) A division
C) A union
D) A join

Question 51

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT MAKE FROM RENTALS,CUSTOMER
WHERE RENTALS.CID = CUSTOMER.CID
AND RESID_CITY = 'HEMET '
GROUP BY MAKE
HAVING COUNT (DISTINCT RENTALS.CID)=
(SELECT COUNT(*)FROM CUSTOMER
WHERE RESID_CITY = 'HEMET')
The meaning of this query is the following:

A) List all makes of cars rented to customers residing in Hemet
B) List all makes of cars rented to at least one customer residing in Hemet
C) List all makes of cars rented to all customers residing in Hemet
D) None of the above

Question 52

An operator that generates a result table that contains the matching rows of both tables plus the non-matching rows of both of the input tables is a(n)_____________________.

Question 53

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID).
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned.
RENTCOST $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below.
RENTLENGTH $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.

-SELECT DISTINCT CNAME,AGE FROM CUSTOMER WHERE CID NOT IN
(SELECT CID FROM RENTALS)
This query is NOT equivalent (order ignored)to the following query:

A) SELECT DISTINCT CNAME, AGE FROM CUSTOMER, RENTALS WHERE CUSTOMER.CID <> RENTALS.CID
B) SELECT DISTINCT CNAME, AGE FROM CUSTOMER WHERE 0 IN (SELECT COUNT(*) FROM RENTALS
C) SELECT DISTINCT CNAME, AGE FROM CUSTOMER WHERE 0 = (SELECT COUNT(*) FROM RENTALS
WHERE CUSTOMER.CID = RENTALS.CID)
D) SELECT DISTINCT CNAME, AGE FROM CUSTOMER WHERE NOT EXISTS (SELECT CID FROM RENTALS

Question 54

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT SUM(VOTE1)AS S1,SUM(VOTE2)AS S2, SUM(VOTE1)-SUM(VOTE2)AS S3,SUM(VOTE1-VOTE2)AS S4
FROM VOTE
The execution of this query produces the following result (shown in order of occurrence in the query):

A) 75 40 35 35
B) 75 40 75 35
C) 75 40 35 5
D) 75 40 40 40

Question 55

The SQL keyword(s)____________ in a SELECT query generate a result table that contains the matching rows and the non-matching rows of the right table.

Question 56

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT COUNT(*)AS Nrows,COUNT(VOTE2)AS NVOTE2 FROM VOTE
The execution of this query produces the following result (Nrows shown before NVOTE2):

A) 4, 4
B) 3, 3
C) 4, 3
D) 4, null

Question 57

An operator that generates a result table that contains the matching rows of both tables plus the non-matching rows of one of the input tables is a(n)_____________________.

Question 58

Figuer:
VOTE $$\begin{array} { | l | l | l | } \hline \text { Name } & \text { votel } & \text { vote2 } \\\hline \text { Bill } & 20 & 15 \\\hline \text { Mary } & 10 & 25 \\\hline \text { Jill } & 15 & 0 \\\hline \text { Bob } & 30 & \\\hline\end{array}$$

-SELECT Vote2,COUNT(*) FROM VOTE
GROUP BY Vote2
In SQL2,the execution of this query produces the following number of rows:

A) 1
B) 2
C) 3
D) 4

Question 59

A SELECT statement that appears inside another query is called a(n)___________________.

Question 60

A nested query in which the inner query does not reference any tables in the outer query is a(n)__________________.

Question 61

In Access,a nested query in the FROM clause would be used to overcome the inability of using the keyword ____________ inside aggregate functions.

Question 62

In a difference problem using a Type II nested query,a NOT EXISTS condition in a WHERE clause would evaluate to ________ if the nested query returned zero rows.

Question 63

The following SQL statement could not be executed because what column is included in the SELECT? _____________
SELECT StdSSN,StdLastName,StdMajor,EnrGrade
FROM Student
WHERE Student.StdSSN IN
(SELECT StdSSN FROM Enrollment
WHERE EnrGrade > 3.5)

Question 64

When a problem narrative statement contains the word "not" in it,this is usually an indication that the problem requires the use of a(n)_______________ operation.

Question 65

A(n)_______________ nested query executes one time for each row in the outer query.

Question 66

If a problem narrative specifies a requirement to list the values of one column that match any of the values in another column,this problem would require the use of a(n)_________________ operation.

Question 67

In the following SQL statement,which column would be blank if null values were present in one or more of the rows? _______________
SELECT FacSSN,COUNT(*)AS NumRows
FROM Offering
GROUP BY FacSSN

Question 68

__________________ nested queries are also known as correlated queries because they reference columns in the outer queries.

Question 69

A relational algebra operator used to combine rows from two tables is the ______________ operator.

Question 70

A(n)_________________ nested query executes one time and produces a result table.

Question 71

To evaluate the result of a compound condition,SQL:1999 uses _________________,which show how combinations of the values true,false and null combine with Boolean operators.

Question 72

In SQL statements,________________ conditions involve one or more simple conditions connected by a Boolean operator.

Question 73

If a problem narrative specifies a requirement to list the values of one column that match all of the values in another column,this problem would require the use of a(n)_________________ operation.

Question 74

In SQL statements,________________ conditions involve a column or column expression,a comparison operator,and another column,column expression or constant.