a. Query: select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’); Explanation:

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Answer 1

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Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

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i have attached my 2 java questions . please answer them correctly, add all comments etc . thank you.

Can you help me solve this problem using Master's Theorem:Solve the recurrence relation f(n) = 3af(n/a) + (n + a)2 with f(1) = 1 and a > 1 byfinding an expression for f(n) in big-Oh notation.

here is example 7.6## Example 7.6 Suppose the sample population is χ 2 (2), which is non-normal but with same variance 4. ▶ Repeat the simulation, but replacing the N(0, 4) samples with χ 2 (2) samples. ▶ Calculate the empirical confidence level.(Empirical confidence level) n <- 20 alpha <- 0.05 UCL <- replicate(1000, expr = { x <- rchisq(n,df=2) (n-1)*var(x)/qchisq(alpha,df=n-1) }) sum(UCL >4) mean(UCL > 4) ## t.test function n <- 20 x <- rnorm(n,mean=2) result <- t.test(x,mu=1) result$statistic result$parameter result$p.value result$conf.int result$estimate

Answer 2

Expert Solution & Answer

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

Expert Solution & Answer

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Answer 3

Expert Solution & Answer

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

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Answer 4

Expert Solution & Answer

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

Expert Solution & Answer

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Answer 5

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Answer 6

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Answer 7

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

Answer 8

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Answer 9

Chapter 3, Problem 21E

Explanation of Solution

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

Answer 10

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Answer 11

Explanation of Solution

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

Answer 12

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Answer 13

Explanation of Solution

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Answer 14

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Answer 15

Explanation of Solution

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Answer 16

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

Answer 17

Explanation of Solution

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...

Answer 18

Expert Solution & Answer

Answer 19

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Answer 20

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Answer 21

Ch. 3 - Prob. 1PE Ch. 3 - Prob. 2PE Ch. 3 - Prob. 3PE Ch. 3 - Suppose that we have a relation marks(ID, score)...Ch. 3 - Suppose that we have a relation marks(ID, score)...Ch. 3 - The SQL like operator is case sensitive (in most...Ch. 3 - Prob. 7PE Ch. 3 - Prob. 8PE Ch. 3 - Prob. 9PE Ch. 3 - Prob. 10PE

a. Query: select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’); Explanation:

Explanation of Solution

a. Query: select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’); Explanation:

Explanation of Solution

b. Query: Select memb_no, name from member m where not exists (select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no)); Explanation: The above query is get the “memb_n...

Explanation of Solution

c. Query: Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

Explanation of Solution

d. Query: select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

Explanation of Solution

e. Query: create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) ) Explanation: The above query is used creat...

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Chapter 3 Solutions

a.

Query:

select memb_no, name from member natural join borrowed where isbn in (select isbn from book where publisher=’McGraw-Hill’);

Explanation:

b.

Query:

Select memb_no, name from member m where not exists

(select isbn from book where publisher=’McGraw-Hill’ except (select isbn from member natural join borrowed where m.memb_no=memb_no));

Explanation:

The above query is get the “memb_n...

c.

Query:

Select memb_no, name from member natural join borrowed natural join book group by memb_no, name, publisher having count (isbn)>5...

d.

Query:

select sum(num_of_books)/ count(memb_no) from (select memb_no, (select count(*) from borrowed where member.memdber_no=borrowed...

e.

Query:

create table participated (report_number char(20) primary key, license char(20) references car, driver_id char(20) references driver, damage_amount dec(6,2) )

Explanation:

The above query is used creat...