a. Query for the given instruction: select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name. Explanation: Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join. Use the natural outer join between instructor and teaches tables. Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Question

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Answer 1

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

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Année scolaire: 2024/2025 Collège SMARA Devoir à domicile 1 semestre 2 Prof: NILAJ PHYSIQUE CHIMIE Nom et Prénom: Niveau: 3APIC Exercice 1: Durée: 1 heure 1) Donner la définition des termes suivantes? Le Mouvement : La vitesse moyenne: 2) Répondre par « Vrai » ou «> Un objet peut être en mouvement et au repos Si la vitesse est constante, le mouvement sera uniforme. 3) Compléter les phrases par les termes qui conviennent : Pour déterminer le mouvement ou le repos d'un objet, vous devez choisir un autre objet appelé le d'un point d'un corps mobile est l'ensemble des positions qu'il occupe durant son mouvement. Il y a trois types de trajectoires: circulaire. La distance de freinage dépend de 4) Cocher la bonne réponse? > La relation entre la vitesse, la distance et le temps: d Vm = dxt > L'unité internationale de la vitesse moyenne est : m/s 5) On considère la figure suivante : Compléter le tableau par les mots suivants : En mouvement au repos A C Km/h et et l' m. s-1 15 A B C Sol…

5. Here is a tree corresponding to T(64) for an unknown recurrence relation: 32 + lg 64 8+ lg 16 2+ lg 4 2+lg4 2+lg4 8+ lg 16 2+lg4 7 7 7 7 7 7 7 7 Fill in the details for the corresponding recurrence relation: T(n) = T(n/4)+ T(1)= Put scratch work below. Scratch work is not graded but may be used for regrade partial credit.

Top screenshot is question, the rest are examples and explinations.

Answer 2

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 8

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Answer 9

Chapter 4, Problem 2PE

Explanation of Solution

a.

Query for the given instruction:

select ID, name, count (course_id, sec_id, year, semester) as “Sections taught” from instructor natural left outer join teaches group by ID, name.

Explanation:

Here, get a list with instructor’s ID, name, and the number of sections that they have taught using an outer join.

Use the natural outer join between instructor and teaches tables.

Get the instructor’s ID, name, and the number of sections taught by the instructor by using a group by clause on the ID and name attribute...

Answer 10

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Answer 11

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Answer 12

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Answer 13

Explanation of Solution

c.

Query for the given instruction:

select course_id, sec_id, ID; decode (name, NULL, ‘-’, name) from (section natural left outer join teaches) natural left outer join instructor where semester = ‘Spring’ and year = 2018.

Explanation:

Initially perform a natural left outer between section, teaches and instructor table...

Answer 14

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 15

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

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Check out a sample textbook solution

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Answer 19

Ch. 4 - Prob. 1PE Ch. 4 - Prob. 2PE Ch. 4 - Prob. 3PE Ch. 4 - Prob. 4PE Ch. 4 - Prob. 5PE Ch. 4 - Prob. 6PE Ch. 4 - Prob. 7PE Ch. 4 - Prob. 8PE Ch. 4 - SQL allows a foreign-key dependency to refer to...Ch. 4 - Given the relations a(name, address, title) and...

Concept explainers

Explanation of Solution

Explanation of Solution

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

Explanation of Solution

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...

Want to see the full answer?

Chapter 4 Solutions

Concept explainers

Explanation of Solution

Explanation of Solution

b. Query without using outer join: Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I. Explanation: Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

Explanation of Solution

Explanation of Solution

d. Query for the given instruction: select dept_name, count (ID) from department natural left outer join instructor group by dept_name Explanation: Initially perform a natural left outer join between department and instructor table...

Want to see the full answer?

Chapter 4 Solutions

b.

Query without using outer join:

Select ID, name, (where count (*) as “Section taught” from teaches T where T.id = I.id) from instructor I.

Explanation:

Display a list with instructor’s ID, name, and the number of sections that they have taught using a sub-query...

d.

Query for the given instruction:

select dept_name, count (ID) from department natural left outer join instructor group by dept_name

Explanation:

Initially perform a natural left outer join between department and instructor table...