CSC316 TUI SQL Statements: Employee table This assignment consists of 2 parts: Part 1: Answer questions Write Oracle SQL statements to create a table cal

CSC316 TUI SQL Statements: Employee table This assignment consists of 2 parts:

Part 1: Answer questions

Write Oracle SQL statements to create a table called Employee with the following attributes and associated data types.
Write Oracle SQL statements to insert new records to the Employee table. The program should do the following:
Check whether data has been loaded to the table properly.

Employee

EmployeeID

CHAR(9)

LName

CHAR(15)

FName

CHAR(15)

PhoneNo

CHAR(10)

— Trident Bookstore —

Please enter the Employee ID: Jimmy2345

Please enter employee First Name: Jimmy

Please enter employee Last Name: William

Please enter the phone number of the employee: 714-456-3456

Note: Add at least 2 records of the data

Oracle live SQL does not support some of the commands such as ACCEPT. It is helpful to learn these commands though.

Part 2: Write Essay

Database Queries and Reports

You are to finish the project by applying what you have learned to generate queries (ask questions) of the database.

For the same bookstore database, write SQL statements to provide the following information for the output specified in the SLP:

Employee information (e.g., employee name, address, phone number, hiring date, salary, etc.)

Information about the books and CDs in the bookstore (e.g., title, author, price, etc.)

Sales of CDs and books, searchable by date

Current inventory status of books and CDs

Any other output you think appropriate

Write a 2- to 5-page paper that includes your SQL statements and screenshots of the results tables in the DBMS (database management system). use the included powerpoint presentation and word document to supplement the tables in the powerpoint for part 2 Forecasting Dataset
Spring 2019
Month/Yr.
June 2016
Jan. 2017
Jan. 2018
Jan. 2019
PERIOD PRICE
1
6.1
2
5.75
3
5.7
4
5.7
5
5.6
6
5.6
7
5.6
8
6.3
9
6.4
10
6.2
11
5.9
12
5.9
13
5.7
14
5.75
15
5.75
16
5.8
17
5.7
18
5.8
19
5.7
20
5.8
21
5.8
22
5.75
23
5.7
24
5.55
25
5.6
26
5.65
27
5.7
28
5.75
29
5.8
30
5.3
31
5.4
32
5.7
AIP
5.8
6
6.3
5.7
5.85
5.8
5.75
5.85
5.65
6
6.1
6
6.1
6.2
6.1
6.1
6.2
6.3
6.1
5.75
5.75
5.65
5.9
5.65
6.1
6.25
5.65
5.75
5.85
6.25
6.3
6.4
DIFF
-0.3
0.25
0.6
0
0.25
0.2
0.15
-0.45
-0.75
-0.2
0.2
0.1
0.4
0.45
0.35
0.3
0.5
0.5
0.4
-0.05
-0.05
-0.1
0.2
0.1
0.5
0.6
-0.05
0
0.05
0.95
0.9
0.7
Feb. 2019
Mar-19
Apr-19
May-19
33
34
35
36
5.9
6.5
0.6
ADV
5.3
6.75
7.25
7.3
7.2
6.5
6.75
6.89
5.8
5.5
6.5
6.25
7
6.9
6.8
6.8
7.1
7
6.8
6.5
8.1
7.7
7.3
7.5
8.1
8.3
8.7
9.2
8.4
8.8
9.5
9.3
DEMAND
14.4
15.3
16.5
16.1
16
15.5
15.2
13.9
13.3
13.12
13.8
14.8
15.3
16.3
17.5
17.4
17.1
16.8
16.5
16
15.2
15.3
15.9
16.2
17.5
18.4
19.4
19.1
18.7
18.2
18.4
17.5
9.1
17.1
GRAPHS
Price vs. Periods
8
4
Price
Price
6
y = -0.0089x + 5.9193
R² = 0.1474
2
0
0
5
10
15
20
25
30
35
Periods
AIP vs. Periods
y = -4E-07×5 + 5E-05×4 – 0.0022×3 + 0.0367×2 – 0.2178x +
6.2272
R² = 0.4507
7
AIP
6.5
6
5.5
0
5
10
15
20
25
30
35
Periods
DIFF vs. Periods
1.5
y = -5E-07×6 + 5E-05×5 – 0.0019×4 + 0.0348×3 – 0.3009×2 +
1.0748x – 0.9937
R² = 0.5042
1
DIFF
0.5
0
-0.5
-1
10
0
5
10
15
20
25
30
35
Periods
ADV vs. Periods
6
4
y = 0.0916x + 5.8024
ADV
ADV
8
y = 0.0916x + 5.8024
R² = 0.663
ADV
ADV
4
2
0
0
5
10
15
20
25
30
35
Periods
Demand vs. Periods
y = 0.1173x + 14.3
R² = 0.4775
25
Demand
Demand
20
15
10
5
0
0
5
10
15
20
Periods
25
30
35
QUESTION 2
GRAPHS
ANALYSIS
Price vs. Periods
y = -7E-07×5 + 7E-05×4 – 0.0022×3 + 0.0299×2 – 0.1461x +
6.0139
R² = 0.2598
Price
6.5
6
5.5
5
0
5
10
15
20
25
30
35
Studying the first paragraph it seems that there is no ev
the price has changed considerably during the periods u
Visually, we would say that the price is almost constant d
but once we calculate the R-squared we see that the li
poor (R-squared < 0.5). Trying to obtain a better fit w describe the data with a polinomical equation of 5th de we still got a r-squared < 0.5 (poor fit) Periods In the other hand, the average industry price has chan time from 5.6 to almost 6.5. Visually there is not a defin However, if we fit the data to a polinomical of 5th degr almost a good fit, obtaining an R-squared of 0.4 Also, the price difference has varied duing time from alm Visually there is not a define pattern that describes the results. However, if we fit the data to a polinomical of 5 we get almost a good fit, obtaining an R-squared of 12 10 ADV vs. Periods ADV 8 6 4 y = 0.0046x2 - 0.066x + 6.7221 Something different happens in the case of ADV. Even behaviour is not totally linear, the linear fit is good, obta squared > 0.5 (0.63). However, if we want to be even m
a polinomical fit (2 degrees) is even better (R-squared
ADV
y = 0.0046×2 – 0.066x + 6.7221
R² = 0.7858
4
Something different happens in the case of ADV. Even
behaviour is not totally linear, the linear fit is good, obta
squared > 0.5 (0.63). However, if we want to be even m
a polinomical fit (2 degrees) is even better (R-squared
2
0
0
5
10
15
20
25
30
35
Periods
Demand vs. Periods
Demand
4
3
2
25y = -3E-05x + 0.0015x – 0.0134x – 0.0378x + 15.501
R² = 0.5723
20
Finally, let’s talk about demand. A linear fit means an R0.4775 (relatively poor, rsquared
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