Process to Complete Registration On Learning Management System - Canvas Free for Teacher

 Process to Complete Registration On Learning Management System 

 Please use the following process step by step to complete registration for the Canvas Free for Teacher 

 Step 1 : After Teacher add you in any particular course -  You will get a similar email for Course Invitation/Canvas. Click on Get Started

Step 2: Click On Accept as shown in the snapshot below.  

Step 3: Click on Create My Account. This step is required once for the foundation course, even if you already have a Canvas Account.

Step 4 : Create a new Password and click on Register 

Step 5 : You shall login to the Canvas Homepage. You are ready to start your course! 

 

Google Map Direction - User Guide

  Introduction

 Maps shows you directions and uses real-time traffic information to find the best route to your destination. you can hear traffic alerts, where to turn, which lane to use, and if there's a better route. This guide helps you to use google map direction to get your places and its direction. 

Stepwise pocedure to use google map direction: 

Step 1: Open your browser in your system. You will see google app over the page. Click on that option. Next of this you will see google map option, by clicking this option will reach on google map.

Step 2: You will get google home page 

Step 3 : Insert your destination in search option.

Step 4 : Map will take you at inset location, you will now see your destination on map

Que 4

4]
Consider the following Entities and Relationships                [30 Marks]
Room (roomno, desc, rate)
Guest (gno, gname, no_of_days)
Relation between Room and Guest is One to One.
         Constraint: Primary key, no of days should be > 0.
Create a Database in 3NF & write queries for following.
•    Display room details according to its rates  in ascending order.
•    Find the names of guest who has allocated room for more than 3 days.
•    Find no. of AC rooms.
•    Display total amount for NON-AC rooms.
•    Find names of guest with maximum room charges.
Ans.

create table room(
roomno int primary key,
desc1 varchar2(10),
rate int
);

create table guest(
gno int primary key,
roomno int unique references room (roomno),
gname varchar2(20),
nday int
);

insert into room values(101,'Ac',2000);
insert into room values(102,'Non-Ac',3000);
insert into room values(103,'Ac',4000);
insert into room values(104,'Ac',3000);
insert into room values(105,'Ac',7000);
insert into room values(106,'Ac',3000);
insert into room values(107,'Ac',4000);
insert into room values(108,'Ac',5000);
insert into room values(109,'Non-Ac',5000);

insert into guest values(21,101,'Yuvraj-Singh',5);
insert into guest values(22,102,'Yusuf-Pathan',13);
insert into guest values(23,103,'Gautam-Gambhir',10);
insert into guest values(24,104,'Raina',10);
insert into guest values(25,105,'Sachin-Tendulkar',11);
insert into guest values(26,106,'Praveen-Kumar',11);
insert into guest values(27,107,'Ravindra-Jadeja',14);

select *from room;
select *from guest;

1]Display room details according to its rates  in ascending order.

select desc1,rate from room order by  rate asc ;

2]    Find the names of guest who has allocated room for more than 3 days.

select gname,nday from guest where nday>3;

select gname,nday from guest where nday>10;

Que7

Q. 10)    Consider the following Entities and Relationships                [30 Marks]
Sailor (sid, sname, age)
Boats (bid, bname, color)
Relation between Sailer and Boats is Many to Many with day as descriptive attribute.
Constraint: Primary key, age should be > 0.
Create a Database in 3NF & write queries for following.
•    Display details of all boats sailed by sailor ‘Ram’.
•    Display Sailor names working on blue boat.
•    Count number of boats sailed by each sailor.
•    Find the name of sailor who sailed the boat on both Tuesday & Friday.
•    Display details of the boats which is sailed maximum times on Sundays.

Ans.
create table sailor(
sid int primary key,
sname varchar(10),
age int check(age>0)
);
create table boats(
bid int primary key,
bname varchar(10),
color varchar(10)
);
create table sailor_boats(
sid int references sailor(sid),
bid int references boats(bid),
day varchar(10),
primary key(sid,bid)
);
insert into sailor values(501,'Ram',25);
insert into sailor values(502,'Sam',25);
insert into sailor values(503,'Gansam',25);
insert into sailor values(504,'Rima',25);

insert into boats values(601,'Godavari','blue');
insert into boats values(602,'Andaman','blue');
insert into boats values(603,'Albert','Yellow');
insert into boats values(604,'Areil','Weight');
insert into boats values(605,'Arizona','Black');

insert into sailor_boats values(501,601,'Monday');
insert into sailor_boats values(502,602,'Friday');
insert into sailor_boats values(503,603,'Sunday');
insert into sailor_boats values(504,605,'Tueday');
insert into sailor_boats values(501,604,'Sunday');
select *from sailor;
select *from boats;
select *from sailor_boats;

1]
select bid from sailor_boats  where sid=50;
2]selet
***************************************************
Q. 11)Consider the following Entities and Relationships                [30 Marks]
Supplier (sid, sname, addr)
Parts (pid, pname, pdesc)
Relation between Supplier and Parts is Many to Many with cost as descriptive attribute.
Constraint: Primary key, cost should be > 0.
Create a Database in 3NF & write queries for following.
•    Display Supplier details from 'Mumbai' city.
•    Update cost by 25 % for all parts supplied by supplier ‘Mr. Pawar’.
•    Display all parts supplied by each supplier.
•    Display details of parts which are supplied at maximum price by each supplier.
•    Display all suppliers who supply part ‘wheel’ and also display its cost.
Ans.
create table supplier(
sid int primary key,
sname varchar(10),
addr varchar(10)
);
create table parts(
pid int primary key,
pname varchar2(10),
pdesc varchar(10)
);
create table sup_parts(
sid int references supplier(sid),
pid int references parts(pid),
cost int,
primary key(sid,pid)
);
insert into supplier values(101,'Ram','Pune');
insert into supplier values(102,'Sam','Mumbai');
insert into supplier values(103,'Gamsam','Satara');
insert into supplier values(104,'Balaji','Pune');
insert into supplier values(105,'Tata','Nashik');

insert into parts values(1,'Wheel','Car');
insert into parts values(2,'Handl','Bike');
insert into parts values(3,'Stand','Bike');
insert into parts values(4,'Breks','Bus');

insert into  sup_parts values(101,1,100);
insert into  sup_parts values(102,2,200);
insert into  sup_parts values(103,3,300);
insert into  sup_parts values(104,4,400);

select *from supplier;
select *from parts;
select *from sup_parts;

Que6

Q. 8)    Consider the following Entities and Relationships                [30 Marks]
Bill (billno, day, tableno, total)                   
Menu (dish_no, dish_desc, price)
Relation between Bill and Menu is Many to Many with quantity as descriptive attribute.

Constraint: Primary key, price should be > 0.
Create a Database in 3NF & write queries for following.
•    Display receipt which includes bill_no with Dish description, price, quantity and total amount of each menu.
•    Find total amount collected by hotel on date 08/01/2013
•    Count number of menus of billno 301.
•    Display menu details having   price between 100 and 500. 
•    Display total number of bills collected from each table on 01/12/2013.

Ans
drop table bill;
drop table menu;
drop table  bill_menu;

create table bill(
bno int primary key,
day1 date,
tableno int,
total int
);
create table menu(
dno int primary key,
ddesc varchar(10),
price int check(price>0)
);
create table bill_menu(
bno int references bill(bno),
dno int references menu(dno),
qnt int ,
primary key(bno,dno)
);
insert into bill values(300,'1-dec-2013',23,500);
insert into bill values(301,'8-jan-2013',24,100);
insert into bill values(302,'13-jeb-1013',26,300);
insert into bill values(303,'31-dec-2013',28,600);

insert into  menu values(11,'manturian',20);
insert into  menu values(12,'roti',20);
insert into  menu values(13,'panier',20);
insert into  menu values(14,'rice',20);
insert into  menu values(15,'plane-rice',20);

insert into bill_menu values(300,12,4);
insert into bill_menu values(300,15,2);
insert into bill_menu values(300,11,2);
insert into bill_menu values(301,11,1);
insert into bill_menu values(302,14,1);

select *from bill;
select *from menu;
select *from bill_menu;

************************************************
Q. 9)    Consider the following Entities and Relationships                [30 Marks]
Musician (mno, mname, addr, phno)
Album (title, copy_right_date, format)
Relation between Musicians and Album is One to Many.
Constraint: Primary key.
Create a Database in 3NF & write queries for following.
•    Display all albums composed by ‘A R Rehman’.
•    Display musician details who have composed Audio album.
•    Find all musicians who have composed maximum albums.
•    Display musician wise album details.
•    Display Musian details from 'Pune'

Ans.
create table musician(
mno int primary key,
mname varchar(20),
addr varchar(10),
phone int
);
create table album(
mnno int references musician(mno),
title varchar(30),
crdate date,
format varchar(10)
);
insert into musician values(1,'Shreya Ghoshal','mumbai',1245666);
insert into musician values(2,'Lata Mangeshkar','mumbai',1245666);
insert into musician values(3,'Asha Bhosle','mumbai',1245666);
insert into musician values(4,'A R Rehman','mumbai',1245666);

insert into album values(4,'jay ho','01-jan-2013','Audio');
insert into album values(2,'jana-gana','01-jan-2013','Audio');
insert into album values(2,'parichy','01-jan-2013','Audio');
insert into album values(3,'Ijaazat','01-jan-2013','Audio');
insert into album values(4,'vande matram','01-jan-2013','Audio');
insert into album values(4,'shivaji','01-jan-2013','Audio');
insert into album values(1,'Paheli','01-jan-2005','Audio');
select *from album;
select *from musician;

1]Display all albums composed by ‘A R Rehman’.
//select title from album where mnno in(select mnno from musician where mname='A R Rehman');
2]    Display musician details who have composed Audio album.
//select *from musician where musician.mnno=album.mno and album.format='Audio';
select *from musician where mno in(select mno from album where format='Audio');
3]    Find all musicians who have composed maximum albums.
//select mname from musician where mno in(select mnno,(mnno) from album group by mnno);
4]Display musician wise album details.
//
5]Display Musian details from 'Pune'
//select *from musician where addr='mumbai';

Que 5

Q.6)    Consider the following Entities and Relationships                [30 Marks]
Property (pno, desc, area, rate)
Owner (owner_name, addr, phno)
Relation between owner and Property is One to Many.
Constraint: Primary key, rate should be > 0
Create a Database in 3NF & write queries for following.
•    Display area wise property details.
•    Display property owned by 'Mr.Patil' having minimum rate.
•    Display all properties with owner name that having highest rate of  properties located in Chinchwad area.
•    Display owner wise property detail.
•    Display owner name having maximum no. of properties.
Ans.
create table owner1(
oname varchar2(20) primary key,
addr varchar2(10),
phone int
);
create table property1(
oname varchar2(20) references owner1(oname),
pno int primary key,
desc1 varchar2(10),
area varchar2(10),
rate int
);
insert into owner1 values('Mukesh Ambani','Gujarat',123496867);
insert into owner1 values('Dilip Shanghvi','Gujarat',123496862);
insert into owner1 values('Lakshmi Mittal','Parsi',123496863);
insert into owner1 values('Azim Premji','mumbai',123496865);
insert into owner1 values('Shiv Nadar','mumbai',123496866);
insert into owner1 values('S P Hinduja','mumbai',123496868);
insert into owner1 values('patil','mumbai',123496868);

insert into  property1 values('Mukesh Ambani',111,'1 plot','deccan',1000);
insert into  property1 values('Dilip Shanghvi',222,'2 plot','baramati',20000);
insert into  property1 values('Lakshmi Mittal',333,'3 plot','faltan',3000);
insert into  property1 values('Azim Premji',444,'plot','chinchwad',40000);
insert into  property1 values('S P Hinduja',555,'banglow','chinchwad',6000);
insert into  property1 values('patil',666,'banglow','pune',6000);
insert into  property1 values('patil',888,'plot','pune',9000);

select *from owner1;
select *from property1;

1]Display area wise property details.
select

2]Display property owned by 'Mr.Patil' having minimum rate.

select oname,min(rate) from property1 group by oname having oname='patil';

3]Display all properties with owner name that having highest rate of  properties located in Chinchwad area.

select oname from property1 where area =

(select max(rate) from  property1  group by area having area='chinchwad');
4]Display owner wise property detail.


5]Display owner name having maximum no. of properties.
//////select oname from property1 where    ;

*********************************************
Q. 7)    Consider the following Entities and Relationships                [30 Marks]
Employee (emp_no, name, skill, payrate)
Position (posting_no, skill)
Relation between Employee and Position is Many to Many with day and shift as descriptive attribute.   
      
    Constraint: Primary key, payrate should be > 0.
       Create a Database in 3NF & write queries for following.
•    Find the names and rate of pay all employees who allocated a duty.
•    Give employee number who are working at posting_no. 201, but don’t have the skills of waiter.
•    Display a list of names of employees who have skill of chef and who has assigned a duty.
•    Display emp_no and dates for all employees who are working on Tuesday and at least one other day.
•    Display shiftwise employee details

Ans
drop table emp_position;
drop table emp;
drop table position;

Create table emp(
eno int primary key,
ename varchar2(20),
skill varchar2(10),
payrate int check(payrate>0)
);
create table position(
pno int primary key,
skill varchar2(10)
);
create table emp_position(
eno int references emp(eno),
pno int references position(pno),
day1 varchar(7),
shift char(10) check(shift in('morning','evening')),
primary key(eno,pno)
);
insert into emp values(101,'sachin','watchman',2000);
insert into emp values(102,'pravin','manager',3000);
insert into emp values(103,'prasad','waiter',4000);
insert into emp values(104,'kiran','waiter',5000);
insert into emp values(105,'kartik','chef',7000);

insert into position values(1000,'manager');
insert into position values(1001,'waiter');
insert into position values(1002,'chef');
insert into position values(1003,'watchman');
insert into position values(1004,'account');

insert into emp_position values(101,1003,'Monday','morning');
insert into emp_position values(102,1000,'Tuesday','evening');
insert into emp_position values(103,1001,'Tuesday','morning');
insert into emp_position values(104,1001,'Monday','evening');
insert into emp_position values(105,1002,'Monday','morning');

select *from emp;
select *from position;
select *from emp_position;

1]Find the names and rate of pay all employees who allocated a duty.
select name,rate  from
2]Give employee number who are working at posting_no. 201, but don’t have the skills of waiter.

3]Display a list of names of employees who have skill of chef and who has assigned a duty.

4]Display emp_no and dates for all employees who are working on Tuesday and at least one other day.

5]Display shiftwise employee details

Que3

Q.3]Consider the following Entities and Relationships                [30 Marks]
Project (pno, pname, start_date, budget, status)
Department (dno, dname, HOD)
Relation between Project and Department is Many to One
          Constraint: Primary key.
                      Project Status Constraints: C – completed,
                      P-Progressive, I-Incomplete 
         Create a Database in 3NF & write queries for following.
•    List the project name and department details worked in projects that are ‘Complete’.
•    Display total budget of each department.
•    Display incomplete project of each department
•    Find the names of departments that have total budget greater than 50000   .
•    Display all project working under 'Mr.Desai'.

Ans.

drop table project cascade constraints;
drop table department2 cascade constraints;
create table project(
dno int references department2(dno),
pno int primary key,
pname varchar2(20),
sdate date,
budget int,
status char(1) check(status in ('y','n'))
);

create table department2(
dno int primary key,
dname varchar2(20),
hod varchar2(20)
);

insert into department2 values(101,'computer','james');
insert into department2 values(102,'networking','bill');
insert into department2 values(103,'database','mark');
insert into department2 values(104,'static','steve');

insert into project values(101,11,'image-processing','1-jan-2017',3000,'y');
insert into project values(101,12,'driverless car','1-jan-2017',3000,'y');
insert into project values(103,13,'wather mark','1-jan-2017',3000,'y');
insert into project values(104,14,'mobile jammer','1-jan-2017',3000,'y');
insert into project values(104,15,'mobile jammer','1-jan-2017',3000,'y');
insert into project values(103,16,'airtel shop','1-jan-2017',3000,'n');
insert into project values(103,17,'mobile database','1-jan-2017',3000,'n');
insert into project values(102,18,'router','1-jan-2017',3000,'n');

select *from project;
select *from department2;