Q1) Show what part of SSL that protects against the following attack.pdf

•

0 likes•2 views

Q1) Show what part of SSL that protects against the following attacks :- a. Replay Attack: Earlier SSL handshake messages are replayed. b. Man-in-the-Middle Attack: An attacker interposes during key exchange, acting as the client to the server and as the server to the client. c. IP Spoofing: Uses forged IP addresses to fool a host into accepting bogus data. Q2) Describe the port forwarding operation. Using an example, describe the cases when it is very useful, and the cases when it is very dangerous. Solution The secure socket layer (SSL) is a computer networking protocol that manages server authentication, client authentication and encrypted communication between servers ant client . PROTACTION FROM REPLAY ATTACK- The SSL itself is protected againsts replay attack using Medium Access Control ( MAC ). By using MAC secret and the sequence number it ensures the communication integrity. ssl handshake protocol PROTECTION AGAINSTS MAN-IN-THE MIDDLE ATACK- SSL is based on public/ private key cryptography. This basically means that there is a key pair. The public key is used for encryption and secret key is used for decryption . A certificate is basically a public key with a label identifying the owner. So when your browser connects to an https server , the server will answer with its certificate. The browser checks if the certificate is valid- 1: The owner information needs to match the server name that the user requested. 2: The certificate needs to be signed by a trusted certification authority. if one of these condition is not met, the user is informed about the problem. PROTECTION AGAINSTS IP SPOOFING- An SSL certificate is specific to the particular domain for which it was created regardless of the IP address of the doimain. So if a malicious entity want to attempt to redirect trafiic for a certain website to their own malicious site through IP spoofing techniques. The fake site would not have a SSL certificate that was accurately registerd to the real website.This can be varified through a browser by viewing the certificate information and if the wrong address is shown, then the user will know that they are not on the site they think they are on.Secure Socket Layer Architecture ssl handshake protocolssl change cipher protocalssl alert protocolHTTPSSL record protocolTCPIP.

Education

Similar to Q1) Show what part of SSL that protects against the following attack.pdf

Secure socket layer

Introduction of an SSL Certificate

CheapSSLUSA

SSL/TLS Handshake

Arpit Agarwal

Certificate pinning in android applications

Arash Ramez

Secure sockets layer, ssl presentation

Amjad Bhutto

Introduction to Secure Sockets Layer

Nascenia IT

ch17.ppt

SomuPatil8

Certificates and Web of Trust

Yousof Alsatom

fengmei.ppt

ssuserec53e73

Ssl (Secure Sockets Layer)

Asad Ali

fengmei.ppt

ssuserec53e73

SSLtalk

Matthew Aylard

WEB SECURITY CRYPTOGRAPHY PPTeriu8t erhiut.ppt

SonukumarRawat

Transport Layer Security (TLS), and its now-deprecated predecessor, Secure Sockets Layer (SSL), are cryptographic protocols designed to provide communications security over a computer network. Several versions of the protocols find widespread use in applications such as web browsing, email, instant messaging, and voice over IP (VoIP). Websites can use TLS to secure all communications between their servers and web browsers.

Web Security and SSL - Secure Socket Layer

Akhil Nadh PC

Secure payment systems

SSL-image

Network Security

The last picks

SSL Secure socket layer

Ahmed Elnaggar

Ch17

Joe Christensen

Similar to Q1) Show what part of SSL that protects against the following attack.pdf (20)

Secure socket layer

Introduction of an SSL Certificate

SSL/TLS Handshake

Certificate pinning in android applications

Secure sockets layer, ssl presentation

Introduction to Secure Sockets Layer

ch17.ppt

Certificates and Web of Trust

fengmei.ppt

Ssl (Secure Sockets Layer)

fengmei.ppt

SSLtalk

WEB SECURITY CRYPTOGRAPHY PPTeriu8t erhiut.ppt

Web Security and SSL - Secure Socket Layer

Secure payment systems

SSL-image

Network Security

The last picks

SSL Secure socket layer

Ch17

More from arishmarketing21

A series RL circuit includes a 9.05-V battery, a resistance of R = 0.755 , and an inductance of L = 2.97 H. What is the induced emf 1.33 s after the circuit has been closed? So, basically, I tried solving this equation using an RL circuit formula: I = emf/R(1-e^-t/(L/R)) and used V=IR to solve for the current, for which I plugged it back in to obtain the emf. I got an answer of about 31, which was wrong. If you could help me on this, it would make my day! Solution tau = L/R = 2.97/0.755 = 3.9337 V = V0*e^(-t/tau) V = 9.05*e^(-1.33/3.9337) V = 6.453 V.

A series RL circuit includes a 9.05-V battery, a resistance of R = 0.pdf

arishmarketing21

What is the dangling pointer? Explain with a proper example. Solution A dangling pointer is a pointer which nce pointed to some data, but because of memory deallocation, points to an invalid or null data. Example: Suppose we declare and allocate some memory to a pointer: char *temp = malloc(some_size); Then we de-allocate the memory allocated to the pointer temp by, free (temp); After execution of the above statement, temp becomes a dangling pointer..

What is the dangling pointer Explain with a proper example.Solut.pdf

arishmarketing21

Write a function in javascript that calculates the average element i.pdf

arishmarketing21

Which a not a likely location of a bacterial to be found? Atherosclerotic plaque Dialysis catheter Teeth Artificial heart value Solution Bacterial biofilm can easily found in teeth , in teeth bacteria staphylococcus aureus colonizes and make biofilms . Its called the pallised arrangement. Bacteria can\'t form a colony in blood or flowing paths of blood because of immune system. The dialysis catheter is a attractive place for bacteria but generally doctors used a sterile catheter..

Which a not a likely location of a bacterial to be found Atheroscle.pdf

arishmarketing21

What’s Love Got To Do With It? The Evolution of Human Mating By Meredith F. Small Reviewed By Michael S. Kimmel When Tina Turner first bitterly asked “What’s love got to do with it?” her voice was so drenched with pain and regret, the listener knew better than to believe her. Not so Meredith Small. A physical anthropologist whose main research has been on primate mating behavior, Small joins a coterie of such writers as Robert Ardrey, Desmond Morris and Lionel Tiger, who have sought to make evolutionary biology the foundation of studies of sexual behavior. In “What’s Love Got To Do With It,” Small argues that we’re genetically programmed to experience sexual pleasure; hard-wired to search for young, attractive mates; driven to copulate “Way down deep, underneath all the love and lust, underlying the attachment and intimacy, is DNA pushing us along” Like those male writers, she claims that our genetic predisposition toward reproductive success is the driving force of our sex-seeking behavior. And like them, she assumes that sex is a constant, universal and relatively unchanging experience. But unlike them, Small places no value on judgments on the strategies that individuals may adopt to achieve that reproductive success. Women, she argues, are as sexual as men; homosexuality is as “normal” as heterosexuality. Small exposes many of her colleagues’ conclusions as based more on myth than on evolution. Take, for example, the familiar canard that men are “naturally” predisposed toward promiscuity, while women are “naturally” monogamous. Males’ prodigious sperm production, the argument goes, means that it is in men’s interest to fertilize as many women as possible. But women produce only one egg at a time, and gestation and child care take an extraordinarily long time, which means that it is in their interest to mate with only one man for life. Not only do such arguments assume what they are trying to explain, but their assumptions are wrong to begin with. Sperm and eggs do not have little personalities, complete with ambitions and motivations. What’s more, the evidence actually tilts the other way. Female humans are among the only mammals who conceal estrus; that is, they do not go into “heat” when they are sexually receptive. They are potentially sexually receptive all through their cycle. What’s more, a woman has a clitoris, which “exists only to receive and give off sexual pleasure.” What could be the evolutionary function of that? Small argues that it is just as likely that women may be designed to maximize their reproductive chances by promiscuity. If they have sex with a large number of male partners, they can ensure that their offspring will have a significant number of men who believe that they are the father, and who will, therefore, protect the baby and provide it with food. “The ability to move from male to male certainly improves a female’s chance of finding the best genes for her future offspring,” Small writes. Males, by contrast, wo.

What’s Love Got To Do With ItThe Evolution of Human MatingB.pdf

arishmarketing21

What is the Surface characterization techniques of Fourier-transform infrared spectroscopy (FTIR) and Optical Imaging and Spectroscopy (microscopy, TIRF) explain what are they measuring, their uniqueness, working principle briefly Solution Fourier transform infrared spectroscopy (FTIR) is a powerful analysis tool for characterizing and identifying organic molecules. It is the spectroscopic technique that is the most widely used for determining the characteristics of new membranes. In attenuated total reflectance mode, this type of spectroscopy enables functional groups present over a depth of about 1 m to be identified. During ATR analysis, the sample is kept in contact with a crystal allowing total internal reflection. An infrared ray arrives at the crystal where the material under study has been placed. The internal reflection of the ray in the crystal gives rise to an evanescent wave which, at each reflection, continues beyond the surface of the crystal and penetrates the sample over about 1 m. The penetration depth depends on the wavelength, the angle of incidence of the beam on the crystal, and the nature of the crystal. Spectra are thus obtained (curves of absorbance vs. wavelength) that have absorption peaks characteristic of the functions present at the membrane surface. FTIR-ATR is a sensitive, nondestructive method that can be used qualitatively and quantitatively. However, it requires prior drying of the membrane sample. One of the applications of FTIR-ATR is the characterization of modified surfaces. In this case, the spectra show bands characteristic of the basic membrane with, in most cases, bands characteristic of the new functional groups related to the modification. FTIR-ATR also enables the efficiency of membrane cleaning to be assessed . This method can also be used to analyze the adsorption of macromolecules at the membrane surface and to check whether the conformation of the adsorbed compounds (e.g., proteins) has been modified by comparing the spectra of the adsorbed product with those of the same product in solution . When the deposits are very small, however, it is difficult to determine the presence of fouling agents; this necessitates the use of elaborate data-processing methods that eliminate the contributions of the membrane and water from the raw spectra. UNIQUENESS: There are three principal advantages for an FT spectrometer compared to a scanning (dispersive) spectrometer. TIRF Total internal reflection fluorescence (TIRF) microscopy (TIRFM) is an elegant optical technique that provides for the excitation of fluorophores in an extremely thin axial region (‘optical section’). The method is based on the principle that when excitation light is totally internally reflected in a transparent solid (e.g., coverglass) at its interface with liquid an electromagnetic field, called the evanescent wave, is generated in the liquid at the solid-liquid interface and is the same frequency as the excitation light. Since the inte.

What is the Surface characterization techniques of Fourier-transform.pdf

arishmarketing21

What is the running time complexity and space complexity of the following procedure (n = R - L + 1). Explain your answer. def foo(V, L, R): if L==R: return V[L] mid = (L+R)//2 m1 = foo(V, L, mid) m2 = foo(V, mid+1, R) if m1==m2: return ml count1, count2 = 0, 0 for i in range(L, R+1): if V[i] == ml: count1 += 1 if V[i] == m2: count2 += 1 if count 1 > (R-L+D/2: return ml if count2 > (R-L+D/2: return m2 return None Solution We are removing half the data at each turn so time complexity is O(log n). We divide it into 2 till we find 1. Thus, 1= N/2x 2x = N Thus x = log2 n.

What is the running time complexity and space complexity of the follo.pdf

arishmarketing21

A species has a diploid number of chromosomes of 6. If a cell from an individual of this species undergoes meiosis, what is the probability that a gamete made from this diploid cell will inherit only maternal copies of chromosomes (ie, that the gamete will not inherit any paternal chromosomes) a. 0 b. 1/4 c. 1/3 d. 1/8 Solution b)1/4 1 diploid produces =4 haploid childs =>6 diploid =>6*4=24 haploid childs ratio: diploid/haploid=6/24=1/4.

A species has a diploid number of chromosomes of 6. If a cell from a.pdf

arishmarketing21

What are the security requirements and challenges of Grid and Cloud Computing? Solution security issues: A. Policies: 1) Inside Threats: good supervision should be done for having trusted employees . 2) Access Control:Digital signature can be implemented for access control. 3) System Portability: The problem of vendor lock-in should be handled . B. Software Security: 1) Virtualization technology: up to date version of virtualization product should be installed for the security reasons. 2) Host Operating System:should be up to date and secure from hackers 3) Guest Operating System: should be up to date and secure from hackers 4) Data Encryption:should be done on all the data for its safety. C)Physical security: 1) Backup:Either a backup plan should be provided automatically for each customer, or they can use the plans provided elsewhere in the cloud. 2) Server Location:It should be at appropriate place. Room should have adequate space and isolated. A Cooling System and Fire Suppression System should be installed there. 3) Firewall: Cloud Computing service providers should provide a complete firewall solution to their clients. Challenges: A)No clear standard:Grid computing uses various standards, but all grids are not use same standards. Example all grid operating system such as Linux, Apache,and My SQL are using WSRF, WWW, SOAP and XML standards. B)Distributed computing Vs Grid computing: Grid computing involves dynamic virtual organisation, resource sharing and peer to peer computing. The Grid intends to make access to computing power, scientific data repositories and experimental facilities as easy as the Web makes access to information. Same all facilities provide the grid computing. So it is a challenge for grid computing. C)Lack of grid enabled software: The software, which are enabled the grid computing are less, It has limited software on Grid. Much software has not copyright issues and source code of licence. It is need for more company developing grid-enabled version, need more developers on grid development and need to develop open source software. D)Sharing Resources between Various types of Services:Grid used for sharing resource from various sites and grid hosts. It handles a massive amount of data as a grid platform. A lot of sites and multiple servers gathered there it is so complex infrastructure. It provides difficulty for hardware resource sharing within virtual organisation..

What are the security requirements and challenges of Grid and Cloud .pdf

arishmarketing21

Using the man command, determine which ls command option (flag) will.pdf

arishmarketing21

There a six seats in a bar. Your friend took the second seat from th.pdf

arishmarketing21

The basic economic problem is that we only have so many resources, b.pdf

arishmarketing21

The organization of interrupted genes is often conserved between species. What does this mean and what is an example of this conservation (diagram/explain)? Solution First we have know about interrupted genes is often conserved between species,the intricate cellular machinery that converts genetic information into functional protein and RNA molecules and the many ways in which gene expression is regulated by the cell,we discuss some of the ways that genes and genomes have evolved over time to produce the vast diversity of modern- day life forms on our planet uncovering an astonishing wealth of information about the family relationships among organisms and evolutionary mechanisms,but the great revelation of the past 20 years has been the discovery that the actual nucleotide sequences of many genes are sufficiently well conserved that homologous genes. The recognition of sequence homology has become a major tool for inferring gene and protein function thus it is often possible to predict the function of a gene in humans for which no biochemical or genetic information is available simply by comparing its sequence to that of an intensively studied gene in another organism. Gene sequences are often far more tightly conserved than is overall genome structure as discussed features of genome organization such as genome size,cells do not have specialized mechanisms for creating changes in the structures of their genomes evolution depends instead on accidents and mistakes. Introns can be detected by the presence of additional regions when genes are compared with their RNA products by restriction mapping or electron microscopy The positions of introns are usually conserved when homologous genes are compared between different organisms but the lengths of the corresponding introns may vary greatly..

The organization of interrupted genes is often conserved between spe.pdf

arishmarketing21

The daisy has which inflorescence morphology type? campanulte tubular cruciform rotate ligulate The hibiscus has which inflorescence type? campanulte tubular cruciform rotate ligulate The ixora has which inflorescence type? campanulte tubular cruciform rotate ligulate Solution ANSWER: 16)THE DAISY HAS WHICH INFLORESCENCE MORPHOLOGY TYPE:5)LIGULATE. 17)THE HIBISCUS WHICH INFLORESCENCE TYPE:2)TUBULAR. 18)THE IXORA HAS WHICH INFLORESCENC E TYPE:2)TUBULAR..

The daisy has which inflorescence morphology type campanulte tubul.pdf

arishmarketing21

Suppose that CaO is present as an impurity to Li2O. The Ca2+ ion sub.pdf

arishmarketing21

Resistance A primitive adaptive immune Zone of inhibition The ability of a microbe to grow Bacterial lawn Bacteriophage or Enrichment Dense, solid growth across the surface of a petri dish. A lawn of bacteria is prepared and filter disks containing antimicrobial agents are applied A molecule that specifically kills or inhibits bacterial cells Mixture of a bacterial host culture, viral sample, and bacterial media A primitive adaptive immune response in bacteria; that is, bacteria can \"learn\" to become resistant General term to describe a molecule or substance that kills or inhibits microbes The ability of a microbe to grow in the presence of an otherwise toxic molecule, such as an antimic Dramatic change in population-level genetics, in response to a selective pressure A kind of virus that preferentially and exclusively infects bacterial cells A circular clearing within a lawn of growth; surrounding a disk or colony containing an antimicrobial Solution The basic principle of Enrichment is that of selection. Enrichment culture provides favourable growth conditions for the organism of intrest and unfavourable conditions for the competing organisms. This technique can be used in many ways. for example, if we want to culture a thermophile bacteria, incubate the sample at high temperature i.e., at above 55 degrees centigrade because at that temperature, only the thermophilic bacteria can grow while this temperature hampers the growth of other bacteria in the sample. Likewise, enrichment culture may also contains antimicrobial agents to which the selective organism is resistant hence the required organism\'s growth is facilitated hampering the growth of other organisms which are not resistant to the antimicrobial agent. Enrichment culture may also contains pesticides or N2 (used for selective growth of nitrogen fixing bacteria) depending upon the type of microorganism selected. Hence the Answer is a lawn of bacteria is prepared and filter discs containing antimicrobial agents are applied..

Resistance A primitive adaptive immune Zone of inhibition The ability.pdf

arishmarketing21

Refer to my progress on this assignment below In this problem you will make it “more” object-oriented in the following ways: -You will change its name to SortedList -You will change the constructor that takes no arguments to be more traditional and initialize the member fields to dummy values. -You will add a constructor that takes in an initialized array and a size -You will add an insert function that adds a value to the list and maintains its sorted-ness -You will add a quicksort function check the below code. -You will make updates as necessary to the main function so that it still runs and tests your code. The code below seems to sort correctly but it skips over some items in the array and I am not sure what is happening. Please help! Thanks! import java.util.Scanner; class SortedList { private static int array[]; private static int n; public SortedList() { Scanner in = new Scanner(System.in); System.out.println(\"Enter number of elements\"); n = in.nextInt(); array = new int[n]; if(n == 0) { System.out.print(\"Since no arguments array set to: \"); } else { System.out.print(\"Creating array size \" + n + \": \"); } for(int i = 0; i < n; i++) { array[i] = 0; } //System.out.println(\"Enter \" + n + \" integers in ascending order\"); /*for (c = 0; c < n; c++) array[c] = in.nextInt();*/ for(int i = 0; i < n; i++) { System.out.print(array[i] + \" \"); } System.out.print(\"\ \"); } public SortedList(int a[], int size) { array = a; n = size; } public int binsearch(int search) { int first, last, middle; first = 0; last = n - 1; middle = (first + last)/2; while( first <= last ) { if ( array[middle] < search ) first = middle + 1; else if ( array[middle] == search ) { //System.out.println(search + \" found at location \" + (middle + 1) + \".\"); return middle+1;//+1 for the non-CS people who don\'t start counting at zero. } else last = middle - 1; middle = (first + last)/2; } return -1; //System.out.println(search + \" is not present in the list.\ \"); } public static int partition(int input[], int p, int r) { int pivot = input[r]; while(p < r) { while(input[p] < pivot) { p++; } while(input[r] > pivot) { r--; } if(input[p] == input[r]) { p++; } else if(p < r) { int tmp = input[p]; input[p] = input[r]; input[r] = tmp; } } return r; } public static void quicksort(int input[], int p, int r) { if(p < r) { int j = partition(input, p, r); quicksort(input, p, j-1); quicksort(input, j+1, r); } } public static void insert(int value, int cell) { array[cell] = value; quicksort(array, 0, n-1); } public static void main(String args[]) { int c; Scanner in = new Scanner(System.in); SortedList b = new SortedList(); System.out.println(\"Input numbers\"); for(int i = 0; i < n; i++) { c = in.nextInt(); in.nextLine(); insert(c, i); } in.close(); for(int j = 0; j < n; j++) { System.out.print(array[j] + \", \"); } } } Solution //the whole program that you wrote is correct except one line , when you call the quicksort everytime you insert a //value dont pass n pass the i.

Refer to my progress on this assignment belowIn this problem you w.pdf

arishmarketing21

public class Patient extends Person { //=========== Properties =============== private int pId; private String address; //=========== Constructors ============= public Patient(){ super();//set 4 properties to blank pId = 0; address = \"\"; } public Patient(int p,String f,String l,String e, int pi,String a){ super(p,f,l,e);//pass 4 properties to super class pId = pi; address = a; } //=========== Behavior ================= public void setPId( int pi){pId = pi;} public int getPId(){ return pId;} public void setAddress(String a){address = a;} public String getAddress(){ return address;} public void display(){ super.display();//calls display in the super class to display 4 properties System.out.println(\"Patient Id = A\"+getPId()); System.out.println(\"Address = \"+getAddress()); }//end display() public static void main(String args[]){ //Object 1 Patient pa1; pa1 = new Patient();//calls no arg constructor pa1.setPId(900); pa1.setAddress(\"Marietta\"); pa1.display(); System.out.println(\"\ \");//space between to objects //Object 2 Patient pa2; pa2 = new Patient(1234,\"Jimmy\",\"Hawkins\",\"jhawkins@yahoo.com\",901,\"Acworth\"); //calls multi argument constructor pa2.display(); }//end main }//end class Let’s make it so that we can look up and find a Patient in the “Patients.txt” file. The Patients are organized by Patients Code. So we should be able to look in the File for Patient “A900”, and it should give us back all the data about that Patient, like, Patinet Id, etc. So we will need to read from the “Patients.txt” file and select the Patient Id “A900”. The File is delimited by “:”(colons). Take a look at the file. use FileInputStream and Buffered reader to read from the file. Code for testing ‘Select’ that goes in main: Patient p1 = new Patient(); p1.select(“A900”); p1.display(); Let’s also make it so that we can add a new Patient to the “Patients.txt” file.use PrintStream and Buffered writer to write to the file. We should be able to append a new line to the “Patients.txt” file with all the data for a new Patient, like PatientId,etc. Code for testing ‘Insert’ that goes in main: Patient p1 = new Patient(); p1.insert(“A900\",\"1234\",\"Jimmy:Hawkins\",\"Marietta\",\"jhawk@yahoo.com\",\"Cigna\"); //now go look in file to see if new line was added Solution import java.util.*; public class Patient { private int patientId; private String patientName; private String patientAddress; private String patientPhone; private Date patientDOB; public Patient(int patientId, String patientName, String patientAddress, String patientPhone, Date patientDOB) { // initialise instance variables this.patientId = patientId; this.patientName = patientName; this.patientAddress = patientAddress; this.patientPhone = patientPhone; this.patientDOB = patientDOB; } public void setId (int patientId) { this.patientId = patientId; } public void setName (String patientName){ this.patientName = patientName; } public void setAddress (String patientAddress){ this.patientAddress = patie.

public class Patient extends Person {=========== Properties ====.pdf

arishmarketing21

8. A human T lymphocyte is infected by a HIV. The viral genome present inside the cell can be any of the following molecules except _________. A. ssDNA B. dsDNA C. ssRNA D. dsRNA Solution dsRNA Retroviruses have a single stranded RNA genome. Two copies of the sense strand of the genome are present with the viral particle. When they infect a T cell, the single stranded RNA is converted into dsDNA copy by the reverse transcriptase. Replication and transcription occur from this dsDNA intermediate, the provirus, whihc is integrated into the host cell genome by a viral integrase enzye,..

8. A human T lymphocyte is infected by a HIV. The viral genome prese.pdf

arishmarketing21

please help with java questions JAVA CODE please check my code and help my questions CLASS ADDRESS public class Address{ private int streetNumber; private String streetName; private String city; private String state; private String country; // streetNumber // streetName // city // state // country public Address(int streetNumber, String streetName, String city, String state, String country) { this.streetNumber = streetNumber; this.streetName = streetName; this.city = city; this.state = state; this.country = country; } //return the streetNumber public int getStreetNumber() { return streetNumber; } // set street number public void setStreetNumber(int streetNumber) { this.streetNumber = streetNumber; } //return the streetName public String getStreetName() { return streetName; } // set street name public void setStreetName(String streetName) { this.streetName = streetName; } //return the city public String getCity() { return city; } //set the city public void setCity(String city) { this.city = city; } //return the state public String getState() { return state; } //the state to set public void setState(String state) { this.state = state; } // return the country public String getCountry() { return country; } // tset country public void setCountry(String country) { this.country = country; } // see java.lang.Object#toString() @Override public String toString() { return \" [streetNumber=\" + streetNumber + \", streetName=\" + streetName + \", city=\" + city + \", state=\" + state + \", country=\" + country + \"]\"; } } //////////////////////////////////////////////////////////////////////////////////////// CLASS PERSON public class Person extends Address { private String name; public Person (int streetNumber, String streetName, String city, String state, String country, String name) { super(streetNumber, streetName, city, state, country); this.name = name; } //retuurn name public String getName() { return name; } // set name public void setName(String name) { this.name = name; } //see java.lang.Object#toString() @Override public String toString() { return \"Person [name=\" + name + \", getStreetNumber()=\" + getStreetNumber() + \", getStreetName()=\" + getStreetName() + \", getCity()=\" + getCity() + \", getState()=\" + getState() + \", getCountry()=\" + getCountry() + \", toString()=\" + super.toString() + \", getClass()=\" + getClass() + \", hashCode()=\" + hashCode() + \"]\"; } } ///////////////////////////////////////////////////////////////// CLASS STUDENT import java.util.ArrayList; import java.util.List; public class Student extends Person { private String CIN; List courses; public Student(int streetNumber, String streetName, String city, String state, String country, String name, String CIN) { super(streetNumber, streetName, city, state, country, name); // Auto-generated constructor stub this.CIN = CIN; courses = new ArrayList(); } // reutrn cin public String getCIN() { return CIN; } // set cin public void setCIN(String cIN) { CIN = cIN; } pub.

please help with java questionsJAVA CODEplease check my code and.pdf

arishmarketing21

More from arishmarketing21 (20)