code of create table and insert table only please Create the following tables in your database with the following schema:Table: Countries+-| Column NameТуре| country_id| country_name| int| varcharI-------+country_id is the primary key for this table.Each row of this table contains the ID and the name of one country.Table: Weather+-----| Column Name| Type |country_id| weather state| day| intint| date |+-+------+(country_id,Each row of this table indicates the weather state in a country for one day.day) is the primary key for this table.Add the following data to your tables:Input:Countries table:+------| country_id | country_name |+| 2| 3| USA| Australia| PeruI China| Morocco| Spain| 5+-Weather table:+--| country_id | weather_state | day| 2| 2| 2| 3| 3| 3| 5| 5| 5| 7| 7| 3| 16| 18| 21| 25| 22| 20| 25| 27| 31| 15| 12| 12| -2| 2019-11-07 || 2019-11-09 I| 2019-11-23 || 2019-11-28| 2019-12-01 || 2019-12-02 || 2019-11-05 || 2019-11-15 || 2019-11-25 || 2019-10-23 || 2019-11-01| 2019-10-28| 2019-10-27| 2019-11-10 || 2019-11-11| 2019-11-12| 8| 9| 32019-12-23 Write an SQL query to find the type of weather in each country for November 2019.The type of weather is:• Cold if the average weather_state is less than or equal 15,• Hot if the average weather_state is greater than or equal to 25, and• Warm otherwise.Return result table in any order.The results should be:Output:+-I country_name | weather_type| USA| Australia| Peru| Morocco| China| ColdI Cold| Hot| HotI WarmExplanation:Average weather_state in USA in November is (15) /1= 15 so weather type is Cold.Average weather_state in Austraila in November is (-2 + 0 + 3) /3 = 0.333 so weather type is Cold.Average weather_state in Peru in November is (25) / 1 = 25 so the weather type is Hot.Average weather_state in China in November is (16 + 18 + 21)/3 = 18.333 so weather type is Warm.Average weather_state in Morocco in November is (25 + 27+31)/3 = 27.667 so weather type is Hot.We know nothing about the average weather_state in Spain in November so we do not include it inthe result table.

We need to write code of create table and insert table only.

Create the following tables in your database with the following schema: Table: Countries | Column Name | Туре | country_id I country_name +---- | int | varcharI -----+---------+ country_id is the primary key for this table. Each row of this table contains the ID and the name of one country. Table: Weather ------ | Column Name 1 Туре I ---- I country_id | weather state | int I day | int | date | +- +------+ (country_id, day) is the primary key for this table. Each row of this table indicates the weather state in a country for one day. Add the following data to your tables: Input: Countries table: +------------ I country_id | country_name | | USA | Australia | Peru I China | Morocco | Spain | 2 Weather table: +---------- | country_id | weather_state | day | 2019-11-01 | | 2019-10-28 | | 2019-10-27 | | 2019-11-10 | | 2019-11-11 | | 2019-11-12 | | 2019-11-07 I | 2019-11-09 | | 2019-11-23 | | 2019-11-28 | | 2019-12-01 | | 2019-12-02 | | 2019-11-05 | | 2019-11-15 | | 2019-11-25 | | 2019-10-23 | | 2019-12-23 | | 2 | 2 | 2 | 15 | 12 | 12 | -2 | 3 | 16 | 18 | 21 | 25 | 22 | 20 | 25 | 27 | 31 | 7 1 3

Create the following tables in your database with the following schema: Table: Countries | Column Name | Туре | country_id I country_name +---- | int | varcharI -----+---------+ country_id is the primary key for this table. Each row of this table contains the ID and the name of one country. Table: Weather ------ | Column Name 1 Туре I ---- I country_id | weather state | int I day | int | date | +- +------+ (country_id, day) is the primary key for this table. Each row of this table indicates the weather state in a country for one day. Add the following data to your tables: Input: Countries table: +------------ I country_id | country_name | | USA | Australia | Peru I China | Morocco | Spain | 2 Weather table: +---------- | country_id | weather_state | day | 2019-11-01 | | 2019-10-28 | | 2019-10-27 | | 2019-11-10 | | 2019-11-11 | | 2019-11-12 | | 2019-11-07 I | 2019-11-09 | | 2019-11-23 | | 2019-11-28 | | 2019-12-01 | | 2019-12-02 | | 2019-11-05 | | 2019-11-15 | | 2019-11-25 | | 2019-10-23 | | 2019-12-23 | | 2 | 2 | 2 | 15 | 12 | 12 | -2 | 3 | 16 | 18 | 21 | 25 | 22 | 20 | 25 | 27 | 31 | 7 1 3

Oracle 12c: SQL

3rd Edition

ISBN:9781305251038

Author:Joan Casteel

Publisher:Joan Casteel

Chapter13: Views

Section: Chapter Questions

Problem 4MC

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