StayWell has increased the monthly rent of each large property by $150. Update the monthly rents in the LARGE_PROPERTY table accordingly.
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StayWell has increased the monthly rent of each large property by $150. Update the monthly rents in the LARGE_PROPERTY table accordingly.
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- CUSTOMERColumns = 7, Rows = 10 CustomerID LastName FirstName Address ZIP Phone Email 1 Shire Robert 6225 Evanston Ave N 98103 206-524-2433 Robert.Shire@somewhere.com 2 Goodyear Katherine 7335 11th Ave NE 98105 206-524-3544 Katherine.Goodyear@somewhere.com 3 Bancroft Chris 12605 NE 6th Street 98005 425-635-9788 Chris.Bancroft@somewhere.com 4 Griffith John 335 Aloha Street 98109 206-524-4655 John.Griffith@somewhere.com 5 Tierney Doris 14510 NE 4th Street 98005 425-635-8677 Doris.Tierney@somewhere.com 6 Anderson Donna 1410 Hillcrest Parkway 98273 360-538-7566 Donna.Anderson@elsewhere.com 7 Svane Jack 3211 42nd Street 98115 206-524-5766 Jack.Svane@somewhere.com 8 Walsh Denesha 6712 24th Avenue NE 98053 425-635-7566 Denesha.Walsh@somewhere.com 9 Enquist Craig 534 15th Street 98225 360-538-6455 Craig.Enquist@elsewhere.com 10 Anderson Rose 6823 17th Ave NE 98105 206-524-6877 Rose.Anderson@elsewhere.com EMPLOYEEColumns = 5, Rows = 5 EmployeeID LastName FirstName…IF POSSIBLE PLEASE TYPE OR WRITE ELIGIBLY illustrate the process of normalizing the data shown in this table to third normal form (3NF). officeNo officeAddress telNo mgrStaffNo name B001 8 Jefferson Way, Portland, ME 97201 503-555-3618 S1500 Tom Daniels B002 City Center Plaza, Seattle, WA 98122 206-555-6756 S0010 Mary Martinez B003 14 – 8th Avenue, New York, NY 10012 212-371-3000 S0145 Art Peters B004 16 – 14th Avenue, Seattle, WA 98128 206-555-3131 S2250 Sally Stern Primary key: officeNo Why the above table is not in 3NF? Describe and illustrate the process of normalizing the data shown in this table to third normal form (3NF). c) Identify the primary and foreign keys in your 3NF relations.what is these codes insert into values()... CREATE TABLE [Car] ([CarID] varchar(30),[SerialNumber] varchar(50),[Model] varchar(30),[Colour] varchar(30),[Year] varchar(20),PRIMARY KEY ([CarID])); CREATE TABLE [ServiceTicket] ([ServiceticketID] varchar(40),[Serviceticketnumber] varchar(40),[CarID] varchar(30),[CustomerID] int,[Datarecieved] varchar(30),[Comments] varchar(100),[Datareturnedcustomer] varchar(100),PRIMARY KEY ([ServiceticketID]),CONSTRAINT [FK_ServiceTicket.CustomerID]FOREIGN KEY ([CustomerID])REFERENCES [Customer]([CustomerID]),CONSTRAINT [FK_ServiceTicket.CarID]FOREIGN KEY ([CarID])REFERENCES [Car]([CarID])); CREATE TABLE [ServiceMechanic] ([ServicemechanicID] varchar(40),[ServiceticketID] varchar(40),[ServiceID] int,[MechanicID] int,[Hours] varchar(10),[Comment] varchar(100),[Rate] varchar(40),PRIMARY KEY ([ServicemechanicID]),CONSTRAINT [FK_ServiceMechanic.MechanicID]FOREIGN KEY ([MechanicID])REFERENCES [Mechanic]([MechanicID]),CONSTRAINT…
- Problem: JMS TechWizards is a local company that provides technical services to several small businesses in the area. The company currently keeps its technicians and clients’ records on papers. The manager requests you to create a database to store the technician and clients’ information. The following table contains the clients’ information. Client Number Client Name Street City State Postal Code Telephone Number Billed Paid Technician Number AM53 Ashton-Mills 216 Rivard Anderson TX 78077 512-555-4070 $315.50 $255.00 22 AR76 The Artshop 722 Fisher Liberty Corner TX 78080 254-555-0200 $535.00 $565.00 23 BE29 Bert's Supply 5752 Maumee Liberty Corner TX 78080 254-555-2024 $229.50 $0.00 23 DE76 D & E Grocery 464 Linnell Anderson TX 78077 512-555-6050 $485.70 $400.00 29 GR56 Grant Cleaners 737 Allard Kingston TX 78084 512-555-1231 $215.00 $225.00 22…CREATE TABLE employees8(employee_id NUMBER(10) PRIMARY KEY,first_name VARCHAR(25) UNIQUE,last_name VARCHAR(25) UNIQUE,job_id NUMBER(3),salary NUMBER(6) NOT NULL,department_name VARCHAR2(20));CREATE TABLE jobs8(job_id NUMBER(3) PRIMARY KEY,job_title VARCHAR2(25),min_salary NUMBER(4)check(min_salary >=2900),max_salary NUMBER(6)check(max_salary<=100000)); CREATE TABLE departments8(department_id NUMBER(2) PRIMARY KEY,department_name VARCHAR2(30),location_name VARCHAR2(15) DEFAULT 'Istanbul'); how can I do this question according to these 3 tables ? Create a complex view that joining with three tables.Please help me to transcribe SQL codes to python codes create table ab asselect *from aliWHERE itemprice IS NOT NULL; SELECT COUNT(MerchandiseCode), UserIDFROM abGROUP BY useridORDER BY COUNT(MerchandiseCode) DESC; select sum(totalsales), useridfrom abgroup by useridorder by sum(totalsales) desc; SELECT COUNT(transactiondate), UserIDFROM abGROUP BY useridORDER BY COUNT(transactiondate) DESC;
- MySQL CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...6. Find the last name of all employees that were not hired on Tuesday *hint: (UseDATE_FORMAT() function amd the format you need is %W, also use UPPER()7. Find the number of employees in each department who have a manager.8. Find the number of employees for each manager whose employees' minimum salary is greaterthan 5000.9. Find the number of…Some rows of a User table are shown below: User ucode name phone scode 7 Alex 847 - 3902 UX 8 Tony 203 - 3902 PX 9 Charles BD 11 Mary 877 - 3333 BD Which of the following queries retrieves the rows where there is no phone? a. SELECT * FROM User WHERE phone IS NULL b. SELECT * FROM User WHERE phone = “ ” c. SELECT * FROM User WHERE phone NOT IS NULL d. SELECT * FROM User WHERE phone > 0Database Q1: Which one of the following INSERT statements will successfully insert a single row? The following table has been created: CREATE TABLE student_table ( stud_id NUMBER (6), last_name VARCHAR2 (20), first_name VARCHAR2 (20), lunch_num NUMBER (4) ; a) INSERT VALUES INTO student_table (143354, ‘Roberts’, ‘Cameron’, 6543); b) INSERT INTO student_table COLMUNS (stud_id, last_name, lunch_num) VALUES (143352, ‘Roberts’, 5543, ‘Cameron’); c) INSERT TO student_table (stud_id, lunch_num, first_name, last_name) VALUES (143352, 6543, ‘Cameron’, ‘Roberts’); d) INSERT INTO student_table VALUES (143354, ‘Roberts’, ‘Cameron’, 6543); Q2: What does the following SQL statement display: SELECT EMP_ID, LAST_NAME, SALARY FROM EMPLOYEES WHERE SALARY > (SELECT MIN (E. SALARY) FROM EMPLOYEES E) ORDER BY SALARY DESC; a) Employees information who earn less money than the maximum salary b) Employees information who earn more money than the highest employee salary c) Employees information who earn…
- Please see image for information SQL Code: https://gist.githubusercontent.com/GistMasterPro/89cad4b0f93db12893c238b40d1d1d94/raw/707e8a95b879a58e4eafe21672bae0e1358975eb/sqlcode.sqlBelow are some rows of the table INVOICE COD PROV_COD DATE TYPE LOC TOTAL 2910 192 2022-03-11 90 TX 1928 9301 384 2022-05-03 90 NY 2800 Overdue invoices are those whose date plus TYPE days have passed. Which of the following shows all invoices with overdue dates? a. SELECT * FROM INVOICE WHERE CURDATE() - DATE > TYPE b. SELECT * FROM INVOICE WHERE CURDATE()-TYPE >DATE c. SELECT * FROM INVOICE WHERE DATE+TYPE < CURDATE() d. SELECT * FROM INVOICE WHERE DATE+TYPE > CURDATE()The view V_PAT_ADT_LOCATION_HX returns one row of information per bed stay. The column ADT_DEPARTMENT_ID is the ID of the bed's department. The column ADT_DEPARTMENT_NAME is that department's name. Two distinct departments can potentially share the same name. Which of the following queries would return exactly one row per department and display the name of the department? A. SELECT ADT_DEPARTMENT_ID, MIN(ADT_DEPARTMENT_NAME) FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID B. SELECT MIN(ADT_DEPARTMENT_ID), ADT_DEPARTMENT_NAME FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID C. SELECT 'ADT_DEPARTMENT_ID', ADT_DEPARTMENT_NAME FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID D. SELECT ADT_DEPARTMENT_ID, 'ADT_DEPARTMENT_NAME' FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID