Please answer the question as soon as possible . The informations are already given . You must put another information to answer the user views question . Do not forget to screenshot the sql queries part . Thank you .

SQL> CREATE TABLE Customer 2 (Customer_No Char(5) NOT NULL, 3 Customer_Name Varchar2(30) NOT NULL, 4 Gender Char(1) NOT NULL, 5 Address Varchar2(60) NOT NULL, 6 Email_Address Varchar2(60) NOT NULL, 7 Telephone_No Number(10) NOT NULL, 8 No_of_Clothes Number(2NOT NULL, 9 Cust_Size Char(5) NOT NULL, 10 Receipt_Date Varchar2(15) NOT NULL, 11 Return_Date Varchar2(15) NOT NULL, 12 CONSTRAINT Customer_No_pk PRIMARY KEY (Customer_No)); Table created. SQL> DESC Customer Name Null? Type CUSTOMER_NO CUSTOMER_NAME GENDER ADDRESS EMAIL_ADDRESS TELEPHONE_NO NO_OF_CLOTHES CUST_SIZE RECEIPT_DATE RETURN_DATE NOT NULL CHAR(5) NOT NULL VARCHAR2(30) NOT NULL CHAR(1) NOT NULL VARCHAR2(60) NOT NULL VARCHAR2(60) NOT NULL NUMBER (10) NOT NULL NUMBER(2) NOT NULL CHAR(5) NOT NULL VARCHAR2(15) NOT NULL VARCHAR2(15) SQL> CREATE TABLE Designer 2 ( Designer_No Char(5) NOT NULL, 3 Designer_Name Varchar2(30) NOT NULL, 4 Address Varchar2(60) NOT NULL, 5 Email Address Varchar2(60) NOT NULL, 6 Telephone_No Number(10) NOT NULL, 7 Social_Med_Plat Varchar2(30) NOT NULL, 8 Salary Number (4) NOT NULL, 9 CONSTRAINT Designer_No_pk PRIMARY KEY (Designer_No)); Table created. SQL> DESC Designer Name Null? Type DESIGNER_NO DESIGNER_NAME ADDRESS EMAIL_ADDRESS TELEPHONE_NO SOCIAL_MED_PLAT SALARY NOT NULL CHAR(5) NOT NULL VARCHAR2(30) NOT NULL VARCHAR2(60) NOT NULL VARCHAR2(60) NOT NULL NUMBER (10) NOT NULL VARCHAR2(30) NOT NULL NUMBER (4) SQL> CREATE TABLE Clothes 2 ( Clothes_No Char(5) NOT NULL, 3 Customer_No Char(5) NOT NULL, 4 Designer_No Char(5) NOT NULL, 5 Type_Clothes Varchar2(30) NOT NULL, 6 Cond_Clothes Varchar2(20) NOT NULL, 7 Type_Fabric Varchar2(20) NOT NULL, 8 Colour_Clothes Varchar2(10) NOT NULL, 9 Price Number(3) NOT NULL, 10 CONSTRAINT Clothes_No_pk PRIMARY KEY (Clothes_No), 11 CONSTRAINT Customer_No_fk FOREIGN KEY (Customer_No) REFERENCES Customer, 12 CONSTRAINT Designer_No_fk FOREIGN KEY (Designer_No) REFERENCES Designer); Table created. SQL> DESC Clothes Name Null? Type CLOTHES_NO CUSTOMER_NO DESIGNER_NO TYPE_CLOTHES COND_CLOTHES TYPE_FABRIC COLOUR_CLOTHES PRICE NOT NULL CHAR(5) NOT NULL CHAR(5) NOT NULL CHAR(5) NOT NULL VARCHAR2(30) NOT NULL VARCHAR2(20) NOT NULL VARCHAR2(20) NOT NULL VARCHAR2(10) NOT NULL NUMBER(3) 5.4 User Views Note: List down 3 DBA SQL queries asked by top management. You may use join relations, aggregate or/and grouping function for the queries