This is for a java program.
Payroll System Using Inheritance and
Polymorphism
1. Implement an interface called EmployeeInfo with the following
constant variables:
FACULTY_MONTHLY_SALARY = 6000.00
STAFF_MONTHLY_HOURS_WORKED = 160
2. Implement an abstract class Employee with the following requirements:
Attributes
last name (String)
first name (String)
ID number (String)
Sex - M or F
Birth date - Use the Calendar Java class to create a date object
Default argument constructor and argument constructors.
Public methods
toString - returning a string with the following format:
ID Employee number :_________
Employee name: __________
Birth date: _______
mutators and accessors
abstract method monthlyEarning that returns the monthly earning.
3. Implement a class called Staff extending from the class Employee with the following requirements:
Attribute
Hourly rate
Default argument and argument contructors
Public methods
get and set
The method monthlyEarning returns monthly salary (hourly rate times 160)
toString - returning a string with the following format:
ID Employee number :_________
Employee name: __________
Birth date: _______
Full Time
Monthly Salary: _________
Implelment a class Education with the following requirements:
Attributes
Degree (MS or PhD )
Major (Engineering, Chemistry, English, etc ... )
Research (number of researches)
Default argument and argument constructors.
Public methods
get and set
Implement a class Faculty extending from the class Employee with the following requirements:
Attributes
Level (Use enum Java)
"AS": assistant professor
"AO": associate professor
"FU": professor
Education object
Default argument and argument constructor
Public methods
mutators and accessors
The method monthlyEarning returns monthly salary based on the
faculty's level.
AS - faculty monthly salary
AO - 1.5 times faculty monthly salary
FU - 2.0 times faculty monthly salary
toString - returning a string with the following format:
ID Employee number :_________
Employee name: __________
Birth date: _______
XXXXX Professor where XXXXX can be Assistant, Associate or
Full
Monthly Salary: _________
Implement a class called Partime extending from the class Staff with the following requirements:
Attributes
Hours worked per week
Default argument and argument constructors
Public methods
mutators and accessors
The method monthlyEarning returns monthly salary . The monthly salary is equal to hourly rate times the hours worked in four weeks.
toString - returning a string with the following format:
ID Employee number :_________
Employee name: __________
Birth date: _______
Hours works per month: ______
Monthly Salary: _________
mplement a test driver program that creates a one-dimensional
array of class Employee to store the objects Staff, Faculty and
Partime.
Using polymorphism, display the following outputs:
a. Employee information using the method toString
Staff
Faculty
Part-time
b. Total monthly salary for all the part-time staff .
c. Total monthly salary for all employees.
d. Display all employee information descending by employee id using
interface Comparable
e. Display all employee information ascending by last name using
interface Comparer
f. Duplicate a faculty object using clone. Verify the
duplication.
Test Data
Staff
Last name: Allen
First name: Paita
ID: 123
Sex: M
Birth date: 2/23/59
Hourly rate: $50.00
Last name: Zapata
First Name: Steven
ID: 456
Sex: F
Birth date: 7/12/64
Hourly rate: $35.00
Last name:Rios
First name:Enrique
ID: 789
Sex: M
Birth date: 6/2/70
Hourly rate: $40.00
Faculty
Last name: Johnson
First name: Anne
ID: 243
Sex: F
Birth date: 4/27/62
Level: Full
Degree: Ph.D
Major: Engineering
Reseach: 3
Last name: Bouris
First name: William
ID: 791
Sex: F
Birth date: 3/14/75
Level: Associate
Degree: Ph.D
Major: English
Reseach: 1
Last name: Andrade
First name: Christopher
ID: 623
Sex: F
Birth date: 5/22/80
Level: Assistant
Degree: MS
Major: Physical Education
Research: 0
Part-time
Last name: Guzman
First name: Augusto
ID: 455
Sex: F
Birth date: 8/10/77
Hourly rate: $35.00
Hours worked per week: 30
Last name: Depirro
First name: Martin
ID: 678
Sex: F
Birth date: 9/15/87
Hourly rate: $30.00
Hours worked per week:15
Last name: Aldaco
First name: Marque
ID: 945
Sex: M
Birth date: 11/24/88
Hourly rate: $20.00
Hours worked per week: 35
In: Computer Science
A study compared three display panels used by air traffic controllers. Each display panel was tested for four different simulated emergency conditions. Twenty-four highly trained air traffic controllers were used in the study. Two controllers were randomly assigned to each display panel-emergency condition combination. The time (in seconds) required to stabilize the emergency condition was recorded. The following table gives the resulting data and the MINITAB output of a two-way ANOVA of the data.
| Emergency Condition | ||||
| Display Panel | 1 | 2 | 3 | 4 |
| A | 20 | 26 | 33 | 11 |
| 20 | 25 | 35 | 11 | |
| B | 15 | 20 | 29 | 12 |
| 10 | 19 | 30 | 9 | |
| C | 22 | 30 | 35 | 10 |
| 23 | 29 | 36 | 16 | |
| Two-way ANOVA: Time versus Panel, Condition | |||||
| Source | DF | SS | MS | F | P |
| Panel | 2 | 280.583 | 140.292 | 43.73 | .0000 |
| Condition | 3 | 1,427.46 | 475.819 | 148.31 | .0000 |
| Interaction | 6 | 20.42 | 3.403 | 1.06 | .4361 |
| Error | 12 | 38.50 | 3.208 | ||
| Total | 23 | 1,766.96 | |||
| Tabulated statistics: Panel, Condition | |||||
| Rows: | Panel | Columns: | Condition | ||
| 1 | 2 | 3 | 4 | All | |
| A | 17.00 | 25.50 | 34.00 | 11.50 | 22.00 |
| B | 12.50 | 19.50 | 29.50 | 9.50 | 17.75 |
| C | 22.50 | 29.50 | 35.50 | 17.00 | 26.13 |
| All | 17.67 | 24.83 | 33.00 | 12.67 | 21.96 |
Figure 12.12
(a) Interpret the interaction plot in the above table. Then test for interaction with α = .05.
| Panel B requires (Click to select) more time less time to stabilize the emergency condition. | |
| F(int)= 1.06, p-value= .436;; (Click to select) cannot can reject H0, no interaction exists. |
(b) Test the significance of display panel effects with α = .05.
F = 43.73, p-value = .0000; (Click to select) do not reject reject H0
(c) Test the significance of emergency condition effects with α = .05.
F = 148.31, p-value = .0000; (Click to select) do not reject reject H0
(d) Make pairwise comparisons of display panels A, B , and C by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
| Tukey q.05 = , MSE = 3.208 | |
| uA – uB: | [ , ] |
| uA – uC: | [ , ] |
| uB – uC: | [ , ] |
(e) Make pairwise comparisons of emergency conditions 1, 2, 3, and 4 by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
| u1 – u2: | [ , ] | |
| u1 – u3: | [ , ] | |
| u1 – u4: | [ , ] | |
| u2 – u3: | [ , ] | |
| u2 – u4: | [ , ] | |
| u3 – u4: | [ , ] | |
(f) Which display panel minimizes the time required to stabilize an emergency condition? Does your answer depend on the emergency condition? Why?
|
(Click to select) Panel
A Panel C Panel
B minimizes the time required to stabilize an emergency
condition. (Click to select) No Yes , there is (Click to select) no some interaction. |
(g) Calculate a 95 percent (individual) confidence interval for the mean time required to stabilize emergency condition 4 using display panel B. (Round your answers to 2 decimal places.)
Confidence interval [ , ]
In: Statistics and Probability
A study compared three display panels used by air traffic controllers. Each display panel was tested for four different simulated emergency conditions. Twenty-four highly trained air traffic controllers were used in the study. Two controllers were randomly assigned to each display panel-emergency condition combination. The time (in seconds) required to stabilize the emergency condition was recorded. The following table gives the resulting data and the MINITAB output of a two-way ANOVA of the data.
| Emergency Condition | ||||
| Display Panel | 1 | 2 | 3 | 4 |
| A | 19 | 25 | 34 | 12 |
| 19 | 26 | 37 | 12 | |
| B | 16 | 20 | 28 | 14 |
| 11 | 19 | 28 | 7 | |
| C | 22 | 30 | 33 | 10 |
| 25 | 30 | 39 | 16 | |
| Two-way ANOVA: Time versus Panel, Condition | |||||
| Source | DF | SS | MS | F | P |
| Panel | 2 | 361.750 | 180.875 | 29.73 | .0000 |
| Condition | 3 | 1,381.50 | 460.500 | 75.70 | .0000 |
| Interaction | 6 | 28.25 | 4.708 | .77 | .6052 |
| Error | 12 | 73.00 | 6.083 | ||
| Total | 23 | 1,844.50 | |||
| Tabulated statistics: Panel, Condition | |||||
| Rows: | Panel | Columns: | Condition | ||
| 1 | 2 | 3 | 4 | All | |
| A | 16.00 | 25.50 | 35.50 | 13.00 | 22.50 |
| B | 13.50 | 19.50 | 28.00 | 8.50 | 17.38 |
| C | 23.50 | 30.00 | 36.00 | 18.00 | 26.88 |
| All | 17.67 | 25.00 | 33.17 | 13.17 | 22.25 |
Figure 12.12
(a) Interpret the interaction plot in the above table. Then test for interaction with α = .05.
| Panel B requires (Click to select) less time more time to stabilize the emergency condition. | |
| F(int)= .77, p-value= .605;; (Click to select) can cannot reject H0, no interaction exists. |
(b) Test the significance of display panel effects with α = .05.
F = 29.73, p-value = .0000; (Click to select) do not reject reject H0
(c) Test the significance of emergency condition effects with α = .05.
F = 75.70, p-value = .0000; (Click to select) do not reject reject H0
(d) Make pairwise comparisons of display panels A, B , and C by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
| Tukey q.05 = , MSE = 6.083 | |
| uA – uB: | [ , ] |
| uA – uC: | [ , ] |
| uB – uC: | [ , ] |
(e) Make pairwise comparisons of emergency conditions 1, 2, 3, and 4 by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
| u1 – u2: | [ , ] | |
| u1 – u3: | [ , ] | |
| u1 – u4: | [ , ] | |
| u2 – u3: | [ , ] | |
| u2 – u4: | [ , ] | |
| u3 – u4: | [ , ] | |
(f) Which display panel minimizes the time required to stabilize an emergency condition? Does your answer depend on the emergency condition? Why?
|
(Click to select) Panel
C Panel B Panel
A minimizes the time required to stabilize an emergency
condition. (Click to select) Yes No , there is (Click to select) no some interaction. |
(g) Calculate a 95 percent (individual) confidence interval for the mean time required to stabilize emergency condition 4 using display panel B. (Round your answers to 2 decimal places.)
Confidence interval [ , ]
In: Statistics and Probability
Programming language is python 3
For this project, you will import the json module.
Write a class named NeighborhoodPets that has methods for adding a pet, deleting a pet, searching for the owner of a pet, saving data to a JSON file, loading data from a JSON file, and getting a set of all pet species. It will only be loading JSON files that it has previously created, so the internal organization of the data is up to you.
For example, your class could be used like this:
np = NeighborhoodPets()
np.add_pet("Fluffy", "gila monster", "Oksana")
np.add_pet("Tiny", "stegasaurus", "Rachel")
np.add_pet("Spot", "zebra", "Farrokh")
np.save_as_json("pets.json")
np.delete_pet("Tiny")
spot_owner = np.get_owner("Spot")
np.read_json("other_pets.json") # where other_pets.json is a file it saved in some previous session
species_set = np.get_all_species()
If you implement a Pet class (which is a natural option, but not required), then when you save it, you'll want to translate the information into one of the built-in object types the json module recognizes, and translate it back the other way when you read it.
The file must be named: NeighborhoodPets.py
In: Computer Science
In Casino Las Vegas, the accounts payable clerk Mr. Alexi Lalas discovered a flaw in the accounting system (accounting software used for general ledger), the accounts payable clerk discovered that he was able to change the names, address and phone numbers of the Casino’s vendors in the computer system to his name or to any other name, there were no evident access controls to the suppliers ( vendors) master file. He would create false invoices and then create a check for the same false invoice. The name on the check would be changed to the name of the employee or any other beneficiary not related to the transaction. After the check was printed, the name in the system could then be changed back to the appropriate vendor. The check register (listing that usually shows listing of checks by date, name and amount) is used by other employees in other departments for many reasons. The fraudulent employee had authorization to sign checks under $1,000. By writing small checks over a period of time, he was able to defraud the company for more than $100,000 over the years of his employment. This fraud was caught by accident. An employee of another department was looking through the vendor list and the check register on her computer after the fraudulent employee had changed the vendor name to his name. A few entries later, the vendor name changed again to the true name of the vendor. She wondered how this could occur and asked her supervisor for an explanation. Soon after, the company’s management hired an independent fraud investigator, after further investigation and inquiries, the fraudulent employee was caught and an admission was received to proceed with the next stage. Check Register Image Inspect the check register and list your notifications with your professional findings and analysis What kind of fraud being committed in this case?, explain how such scheme is possible How could this fraud have been prevented? What actions should be taken against the fraudster(s) and why?
In: Finance
'''
Problem 2: Functions that give answers
Define and complete the functions described below.
The functions are called in the code at the very bottom. So you
should be
able simply to run the script to test that your functions work as
expected.
'''
'''
* function name: get_name
* parameters: none
* returns: string
* operation:
Here, I just want you to return YOUR name. The expected output
below assumes
that your name is Paul. Of course, replace this with the real
article.
* expected output:
# JUST RETURNS THE NAME...TO VIEW IT YOU CAN PRINT IT AS
BELOW
>>> print(get_name())
Paul
'''
'''
* function name: get_full_name
* parameters: fname (string)
lname (string)
first_last (boolean)
* returns: string
* operation:
Return (again, NOT print) the full name based on the first and last
names
passed in as arguments. The first_last argument will be True if you
should
return the name as <fname lname> and False if you shoudl
return the name
as <lname, fname>.
* expected output:
# AGAIN JUST RETURNS THE NAME...TO VIEW IT YOU CAN PRINT IT AS
BELOW
>>> print(get_full_name("Paul","York",True))
Paul York
>>> print(get_full_name("Paul","York",False))
York, Paul
'''
'''
* function name: get_circle_area
* parameters: radius (float)
* returns: float
* operation:
Return the area of a circle with the given radius. Use 3.14 as Pi.
And Google if for
some reason you've forgotten how to get the area of a circle.
* expected output:
# YET AGAIN JUST RETURNS THE VALUE
>>> print(get_circle_area(5.0))
78.5
>>> print(get_circle_area(2.5))
19.625
'''
# FUNCTIONS ARE CALLED BELOW HERE...NO NEED TO TOUCH
ANYTHING
# UNLESS YOU WANT TO COMMENT SOMETHING OUT TO TEST THINGS
# ALONG THE WAY...
print(get_name())
print(type(get_name())) # >>> <class 'str'>
print(get_full_name("Darth","Vader",True))
print(get_full_name("Luke","Skywalker",False))
print(type(get_full_name("Han","Solo",False))) # >>>
<class 'str'>
print(get_circle_area(5.0)) # 78.5
print(get_circle_area(2.5)) # 19.625
print(get_circle_area(12.25)) # 471.19625
In: Computer Science
|
Project Instructions
Illini Company, Inc. Balance Sheet as of 12/31/20X0 Assets Current Assets: Cash 1,500,000 Accounts receivable, net 18,000 Inventory 50,000 Total current assets 1,568,000 Equipment 90,000 Goodwill 20,000 Total assets 1,678,000 Liabilities and shareholders' equity Shareholders' equity: Common stock, 20,000 shares outstanding, $1 par 20,000 Additional paid-in capital 280,000 Retained earnings 1,378,000 Total shareholders' equity 1,678,000 Total liabilities and shareholders' equity 1,678,000 Note that all additional paid-in capital (APIC) sub accounts (e.g., APIC-options and APIC-treasury stock), if any, are tracked in the “Additional paid-in capital” account on the Balance Sheet. Project 1.1 Topic: Accrual Accounting
Project 1.1 Balance Sheet
Enter the correct journal entry for part [B]. [C]. [D]. [E]. [F]. [G]. [H]. [I]. [J]. [K]. [L]. [M]. [N]. [O]. [P].[Q].[R].[S].[T]. [U]. [V] |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
In: Finance
In an experiment to determine the effect of ambient temperature on the emissions of oxides of nitrogen of diesel trucks, ten trucks were run at temperatures of 40°F and 80°F. The emissions, in parts per billion, are presented in the following table.
|
Truck |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
40° |
834.7 |
753.2 |
855.7 |
901.2 |
785.4 |
862.9 |
882.7 |
740.3 |
748.0 |
848.6 |
|
80° |
815.2 |
765.2 |
842.6 |
797.1 |
764.3 |
819.5 |
783.6 |
694.5 |
772.9 |
794.7 |
Test the claim that there is a significant difference in mean emissions between the two temperatures. Use α = 0.05
In: Statistics and Probability
An outcome in the cancer experiment is measured by a test for the presence of cells that trigger an immune response in the body and so may help fight cancer. Here are the data of 11 subjects: counts of active cells per 100,000 cells before and after infusion of thee modified cells. The difference ( after minus before) is the response variable.
|
Before |
14 |
0 |
1 |
0 |
0 |
0 |
0 |
20 |
1 |
6 |
0 |
|
After |
41 |
7 |
1 |
215 |
20 |
700 |
13 |
530 |
35 |
92 |
108 |
|
Difference |
27 |
7 |
0 |
215 |
20 |
700 |
13 |
510 |
34 |
86 |
108 |
a). Do the data give an evidence that the count of active cells is higher after treatment ?
b). Construct a 95% confidence interval for the data ?
In: Statistics and Probability
Consider a binomial experiment with
n = 10
and
p = 0.40.
(a)
Compute
f(0).
(Round your answer to four decimal places.)
f(0) =
(b)
Compute
f(2).
(Round your answer to four decimal places.)
f(2) =
(c)
Compute
P(x ≤ 2).
(Round your answer to four decimal places.)
P(x ≤ 2) =
(d)
Compute
P(x ≥ 1).
(Round your answer to four decimal places.)
P(x ≥ 1) =
(e)
Compute
E(x).
E(x) =
(f)
Compute
Var(x)
and σ. (Round your answer for σ to two decimal places.)
Var(x)
=σ=
In: Statistics and Probability