Question 1

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.

Accepted Answer

A) True 
 B)False  False

Question 2

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)_____________________.

Accepted Answer

full outer join

Question 3

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:

Accepted Answer

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 
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 C

Question 4

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

Accepted Answer

A) True 
 B)False

Question 5

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

Accepted Answer

The answer of A(n)_______________ nested query executes one time for...

Question 6

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:

Accepted Answer

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

Question 7

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

Accepted Answer

A) True 
 B)False

Question 8

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

Accepted Answer

A) True 
 B)False

Question 9

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.

Accepted Answer

A) True 
 B)False

Question 10

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  1)
The CNAMEs shown by the execution of this query are:

Accepted Answer

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

Question 11

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:

Accepted Answer

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

Question 12

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.

Accepted Answer

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 
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 13

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:

Accepted Answer

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

Question 14

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.

Accepted Answer

A) True 
 B)False

Question 15

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.

Accepted Answer

A) True 
 B)False

Question 16

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

Accepted Answer

The answer of A nested query in which the inner...

Question 17

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

Accepted Answer

A) True 
 B)False

Question 18

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:

Accepted Answer

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

Question 19

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

Accepted Answer

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. 
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 20

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

Accepted Answer

The answer of In Access,a nested query in the FROM...

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.

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)_____________________.

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

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

Figuer: VOTE Name votel vote2 Bill 20 15 Mary 10 25 Jill 15 0 Bob 30 -SELECT * FROM VOTE WHERE Vote2 >=10 The execution of this query produces the following number of rows:

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

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

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.

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.

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.

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.

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

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

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

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

Introduction to Database Management

Introduction to Database Development

The Relational Data Model

Query Formulation With SQL

Understanding Entity Relations

Developing Data Models for Business Databases

Normalization of Relational Tables

Physical Database Design

Application Development With Views

Stored Procedures and Triggers

View Design and Integration

Data and Database Administration

Transaction Management

Data Warehouse Technology and Management

Client-Server Processing and Distributed Databases

Object Database Management Systems

Filters

Exam 9: Advanced Query Formulation With SQL

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.

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)_____________________.

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

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

Figuer: VOTE Name votel vote2 Bill 20 15 Mary 10 25 Jill 15 0 Bob 30 -SELECT * FROM VOTE WHERE Vote2 >=10 The execution of this query produces the following number of rows:

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

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

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.

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.

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.

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.

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

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

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

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

Introduction to Database Management

Introduction to Database Development

The Relational Data Model

Query Formulation With SQL

Understanding Entity Relations

Developing Data Models for Business Databases

Normalization of Relational Tables

Physical Database Design

Application Development With Views

Stored Procedures and Triggers

View Design and Integration

Data and Database Administration

Transaction Management

Data Warehouse Technology and Management

Client-Server Processing and Distributed Databases

Object Database Management Systems

Filters