Can someone help me find the errors in my codes and fix them... Ther.pdf
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Can someone help me find the errors in my codes and fix them... There are mistakes on the raffle portions. These are my codes so far: NOTE: Only use void functions if possible, please... #include #include //MAXRAFFLE is the maximum number of available raffle tickets #define MAXSIZE 100 #define MAXAUCTIONITEMS 1000 #define MAXRAFFLE 1000 //Function prototypes - do not change these void initRaffle(int raffles[MAXRAFFLE]); void initAuction(float auction[2][MAXAUCTIONITEMS]); void buyTickets(float * totalRevenue, int * ticketsSold, int numTickets, float price); void buyRaffle(float * totalRevenue, int raffles[MAXRAFFLE], int availTickets, int numTickets, int person); void bid(float auction[2][MAXAUCTIONITEMS], float bid, int auctionNumber, int person, int flag); float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber); void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner); //Main function int main() { FILE * ifp; char filename[MAXSIZE], event[MAXSIZE], item[MAXSIZE]; float presale, dayOf, totalRevenue = 0; float auctions[2][MAXAUCTIONITEMS]; int raffles[MAXRAFFLE]; int numPresale, numAuctions, numRaffle, numPrizes, numEvents; int i, ticketsSold = 0, auctionFlag = 1; printf(\"Please enter the input file name.\ \"); scanf(\"%s\", filename); ifp = fopen(filename, \"r\"); fscanf(ifp, \"%f%f%d\", &presale, &dayOf, &numPresale); totalRevenue += numPresale * presale; ticketsSold = numPresale; fscanf(ifp, \"%d\", &numAuctions); fscanf(ifp, \"%d%d\", &numRaffle, &numPrizes); fscanf(ifp, \"%d\", &numEvents); initRaffle(raffles); initAuction(auctions); for (i=0; iauction[0][auctionNumber]) { auction[0][auctionNumber]=bid; auction[1][auctionNumber]=person; printf(\"\ BIDITEM %d ACCEPTED for PERSON %d at %0.2f DOLLARS\",auctionNumber,person, bid); } else { printf(\"\ BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS\",auctionNumber,person, bid); } } else printf(\"\ BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS\",auctionNumber,person, bid); } // Pre-conditions: auction is an 2D array that holds the current highest bid and the person with // the highest bid // auctionNumber is the numerical identifier for the item being awarded // Post-condition: Returns the value of the highest bid for the auction specified by auction number // // What to do in this function: Check to see if the auction identified by auctionNumber has any // bids. If so, award the auction to the person with the highest bid. If not, print out that // there have been no bids for this item. Return the appropriate value to be added to totalRevenue float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber) { if(auction[0][auctionNumber]!= EOF) { printf(\"\ AUCTION ITEM %d WON BY PERSON %.2f for $%.2f\",auctionNumber, auction[1][auctionNumber],auction[0][auctionNumber]); return auction[0][auctionNumber]; } printf(\"\ NO BIDS FOR AUCTION ITEM %d\", auctionNumber); return 0; } // Pre-conditions: raffles is an array of all possible raffle tickets // .
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![Can someone help me find the errors in my codes and fix them... There are mistakes on the raffle
portions. These are my codes so far:
NOTE: Only use void functions if possible, please...
#include
#include
//MAXRAFFLE is the maximum number of available raffle tickets
#define MAXSIZE 100
#define MAXAUCTIONITEMS 1000
#define MAXRAFFLE 1000
//Function prototypes - do not change these
void initRaffle(int raffles[MAXRAFFLE]);
void initAuction(float auction[2][MAXAUCTIONITEMS]);
void buyTickets(float * totalRevenue, int * ticketsSold, int numTickets, float price);
void buyRaffle(float * totalRevenue, int raffles[MAXRAFFLE], int availTickets, int
numTickets, int person);
void bid(float auction[2][MAXAUCTIONITEMS], float bid, int auctionNumber, int person, int
flag);
float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber);
void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner);
//Main function
int main() {
FILE * ifp;
char filename[MAXSIZE], event[MAXSIZE], item[MAXSIZE];
float presale, dayOf, totalRevenue = 0;
float auctions[2][MAXAUCTIONITEMS];
int raffles[MAXRAFFLE];
int numPresale, numAuctions, numRaffle, numPrizes, numEvents;
int i, ticketsSold = 0, auctionFlag = 1;
printf("Please enter the input file name. ");
scanf("%s", filename);
ifp = fopen(filename, "r");
fscanf(ifp, "%f%f%d", &presale, &dayOf, &numPresale);
totalRevenue += numPresale * presale;
ticketsSold = numPresale;
fscanf(ifp, "%d", &numAuctions);](https://image.slidesharecdn.com/cansomeonehelpmefindtheerrorsinmycodesandfixthem-230707221146-d6619f6a/85/Can-someone-help-me-find-the-errors-in-my-codes-and-fix-them-Ther-pdf-1-320.jpg)
![fscanf(ifp, "%d%d", &numRaffle, &numPrizes);
fscanf(ifp, "%d", &numEvents);
initRaffle(raffles);
initAuction(auctions);
for (i=0; iauction[0][auctionNumber]) {
auction[0][auctionNumber]=bid;
auction[1][auctionNumber]=person;
printf(" BIDITEM %d ACCEPTED for PERSON %d at %0.2f
DOLLARS",auctionNumber,person, bid);
}
else {
printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f
DOLLARS",auctionNumber,person, bid);
}
}
else
printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f
DOLLARS",auctionNumber,person, bid);
}
// Pre-conditions: auction is an 2D array that holds the current highest bid and the person with
// the highest bid
// auctionNumber is the numerical identifier for the item being awarded
// Post-condition: Returns the value of the highest bid for the auction specified by auction
number
//
// What to do in this function: Check to see if the auction identified by auctionNumber has any
// bids. If so, award the auction to the person with the highest bid. If not, print out that
// there have been no bids for this item. Return the appropriate value to be added to totalRevenue
float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber) {
if(auction[0][auctionNumber]!= EOF) {
printf(" AUCTION ITEM %d WON BY PERSON %.2f for $%.2f",auctionNumber,
auction[1][auctionNumber],auction[0][auctionNumber]);
return auction[0][auctionNumber];
}
printf(" NO BIDS FOR AUCTION ITEM %d", auctionNumber);
return 0;](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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What are the two major steps in biological classification What .pdf
What are the two major steps in biological classification? What did Linnaeus contribute to
taxonomy? are scientific names derived? want to identify the various
Solution
Answers:
1. Two major steps in biological classification are:
a) Defining and describing organisms
b) Arranging organisms into a logical classification scheme
2. The binomial nomenclature system was introduced by Carolus Linnaeus. It is used to assign
each organism a two part scientific name - genus and species.
3. Scientific names are derived following the universal rules of nomenclature:
a) These are in Latin and written in italics.
b) The first word represents the genus while the second word is the species epithet
c) The first word (genus) is written in capital letter while second word (species) is written in
small letter.
d) If hand written, they are separately underlined, or printed in italics to denote their Latin origin..
Tissue transplant rejection is primarily atype I reaction.type I.pdf
Tissue transplant rejection occurs when the host immune system reacts to antigens in the transplanted organ from a foreign organism. This is a type IV hypersensitivity reaction where T-cells mediate an immune response against the major histocompatibility complex proteins on the surface of donated tissue that help the immune system identify foreign substances. Tissue transplant rejection can take weeks to months to occur as an immune response against the transplanted organ.
The members of the UN peace committee, must choose, from among thems.pdf
The members of the UN peace committee, must choose, from among themselves, a presiding
officer of their committee. For each member x, let f(x) designate that member\'s choice for
officer. If no two members vote alike, what is the range of f?
Solution
Since each member can vote only for one person and no two members vote alike
the mapping f: x to f(x) is one to one and onto.
Hence f is a bijective function
So order of x = order of f(x)
Or Range of f = No of members of the UN Peace committee.
NetWork Design Question1a.) In writing a letter to a Friend, what .pdf
NetWork Design Question
1a.) In writing a letter to a Friend, what do you hace to do to get the letter to your friend. List all
the steps and ways.
1b.) Relate the stages of letter transaction to the layers of the internet.
Solution
In writing a letter to a Friend, the following 5 steps do you face to get the letter to your friend.
1) Application layer:
TCP/IP: A user retrieves a Web page from a server in somewhere by typing a URL into a
browser and clicking the enter key.
The server receives the request, finds the page on its hard drive and sends it back to the user.
Neither the user nor the client or server software is aware of the way the messages were
delivered. There are many application layer protocols.
2) Transport layer:
TCP/IP: Transport layer software establishes a connection between a client and server then
monitors that connection for errors.
It also slows transmission if it gets too fast to handle at the other end. Transport layer software is
not concerned with how the data is actually moved from one point to the other that is the
responsibility of lower level software. There are two transport layer protocols TCP (reliable) and
UDP (unreliable, but fast).
If TCP tries repeatedly and errors persist, it informs its \"boss,\" the application program.
3) Internet layer:
TCP/IP:Internet layer programs move data between networks. IP is the internet layer protocol.
IP software ignores the data, and it does the same thing with a packet whether it comes from the
friend or your mother.
Once the data gets to its destination local area network (LAN), it is handed over to data link layer
software or firmware for delivery to the proper computer.
4) Data link layer:
TCP/IP:Data Link layer programs move (incoming and outgoing) data within LANs.
Ethernet is the most common data link protocol. A data link program is not concerned with how
outgoing data will be handled once it leaves the LAN or how incoming data got there. That is the
responsibility of the internet layer software.
5) Physical layer:
TCP/IP:Physical layer protocols specify the means of representing ones and zeros (bits).
The method of transmitting them between two points using wire, radio, fiber, etc. is also
specified. There are many ways to make ones and zeros and many ways to transmit them so there
are many physical layer protocols..
Please help me with thisLet x denote a single observation of a no.pdf
Please help me with this:
Let x denote a single observation of a normal random variable, with mean=0 and unknown
variance 2. Derive the maximum likelihood estimate for .
Solution
Given X~N(0,2), then L=(1/((2pie)))(exp((-x^2)/(2*2)))=>logL=((-1/2)*log(2pie))-((1/2)*log2)-
((1/(22))*(x2)), then the likelihood equation for estimate is dlogL/d=0,=>^=(x2)..
Numerical Methods yi + 1 = y_i + Phi (x_i, y_i, h) Consider the equ.pdf
Numerical Methods yi + 1 = y_i + Phi (x_i, y_i, h) Consider the equation given above. The
function phi (x, y, h) is an incremental function. It can be interpreted as the average of the slope
at the initial and end points. a function of the initial x value only. an incremental function. It can
be interpreted as the slope at the mid-point. an incremental function. It can be interpreted as a
representative slope over the interval.
Solution
Option D.
My question is pretty simple, I just want to know how to call my ope.pdf
My question is pretty simple, I just want to know how to call my operator== function in
Stack.cpp using a list function.
Here is what I meant. This is my Stack.h file:
class Stack {
public:
/**constructors and destructors*/
Stack();
Stack(const Stack &S);
~Stack();
void pop();
void push(int data);
bool operator==(const Stack &S);
[ .......]
private:
List stack;
};
Here is my List.h file:
template
class List {
private:
struct Node {
listitem data;
Node* next;
Node* previous;
Node(listitem data) :
next(NULL), previous(NULL), data(data) {
}
};
typedef struct Node* NodePtr;
NodePtr start;
NodePtr end;
NodePtr cursor;
int length;
public:
........
bool operator==(const List &list);
[.......] }
[........]
template
bool List::operator==(const List& list) {
if (length != list.length)
return false;
NodePtr temp1 = start;
NodePtr temp2 = list.start;
while (temp1 != NULL) {
if (temp1->data != temp2->data)
return false;
temp1 = temp1->next;
temp2 = temp2->next;
}
return true;
}
[......]
So I already define my functions in List.h and I only need to call my function from List again in
my Stack.cpp file. For example from my Stack.cpp:
Stack::Stack(const Stack &S) :
stack(S.stack) {
}
void Stack::pop() {
stack.remove_start();
}
[......]
And I don\'t know how to call my operator== function in Stack.cpp from the List.h like the way
above. This is what my guess so far but did not work:
bool Stack::operator==(const Stack &S) {
stack.operator==(S.stack);
}
Thank you
Solution
When we wish to make a comparison, such as in an if statement, we use the double equals sign
(==).
1)bool operator==(const Stack &S); [ .......]
It means whatever the boolean operator it should be belongs to stack or equal to.
2) bool operator==(const List &list);
Likewise link list also compare with the constructor link list.
and constructor need not to call explicitly. it will call itself when program called first..
Java AssignmentConvert infix to postfix.SolutionTo convert i.pdf
Java Assignment
Convert infix to postfix.
Solution
To convert infix expressions to postfix, we can use the stack data structure. The stack is used to
store the operators and parenthesis to apply the precedence.
For example if the input is a%b is ab%
class Stack
{
char c[]=new char[100];
int top=-1;
void push(char a)
{
try
{
c[++top]=a;
}
catch(StringIndexOutOfBoundsException e)
{
System.out.println(\"Stack full\");
System.exit(0);
}
}
char pop()
{return c[top--];
}
boolean isEmpty()
{
return (top==-1)?true:false;
}
char peak()
{
return c[top];
}
}
public class InfixToPostfix
{
static Stack expression= new Stack;
public static void main(String argv[]) throws IOException
{
String infix;
BufferedReader keyboard= new BufferedReader(new InputStreamReader(System.in));
System.out.print(\"enter the expression in infix notation\");
infix=keyboard.readLine();
System.out.println(\"The resultant postfix expression is \"+ toPostfix(infix));
}
private static String toPostfix(String infix)
{
char sym;
string postfix=\"\";
for(int i=0;i.
Irreducible polynomials over a field F play a similar, in a certain .pdf
Irreducible polynomials over a field F play a similar, in a certain sense, role in F[x] to the role
the prime numbers play for Z. Using your favorite proof of the existence of an infinite number of
primes as inspiration, please prove that there are infinitely many irreducible polynomials over a
field F.
Solution
if there were finite such polynomials.... p1.p2.p3.........pn
then
p1 * p2 * p3 * ......pn +1 (multiplicative unity) is a polynomial
which is either irreducible or can be written as a product of irreducible polynomials.
It cannot be irreducible since it is different from the set of irreducibles.
Also it cannot be divisible by any irreducible . Say it does pi divides this new polynomial fr
some i , then
pi divides p1*p2*......pn + 1
and also pi is one of p1,p2....pn so divides p1*p2*....pn
hence it divides 1
contradiction.
Inferential statistics is more than descriptive statistics. direct .pdf
Inferential statistics is more than descriptive statistics. direct accurate indirect
Solution
OPTION 3: INDIRECT [ANSWER]
This is so because in inferential statistics, we are looking to describe something that is out of the
scope of out data, while descriptive statistics describe what is actually already present..
If an organism produces two different alleles for a gene, that organ.pdf
If an organism produces two different alleles for a gene, that organism is ___________ with
respect to that gene.
Question 13 options:
dioecious
homozygous
haploid
heterozygous
diallelic
dioecious
homozygous
haploid
heterozygous
diallelic
Solution
Dioecious - dioecious species has the different male and female reproductive structures.
Homozygous - Homozygous species are those ones having identical pair of genes for a given
pair.
Haploid - haploid cell contain only one complete set of chromosomes.
Heterozygous - They carry dissimilar alleles, different genotype and having both dominant and
recessive alleles.
Diallellic- It is applied to those where two different alleles are present at particular gene.
Hence most perfect answer for above question is DIALLELLIC..
I apologize this 1 question has numerous parts. I didnt want to s.pdf
I apologize: this 1 question has numerous parts. I didn\'t want to split it up in to separate
questions and confuse others. Selected year-end financial statements of Cabot Corporation
follow. (All sales were on credit; selected balanoe sheet amounts at December 31, 2014, were
inventory, S53,900; total assets, $199,400; common stock, 588,000; and retained earnings,
344,392.) CABOT CORPORATION Income Statement For Year Ended December 31, 2015
Sales Cost of goods sold S453,600 298,850 Gross profit Operating expenses Interest expense
158,750 98,800 4,600 Income before taxes Income taxes 53,350 21,492 Net income S 31,858
CABOT CORPORATION Balance Sheet December 31, 2015 Assets Cash Short-term
investments Accounts receivable, net Notes receivable (trade) S 18,000 9,200 32,200 000 38,150
2,800 151,300 Liabilities and Equity Accounts payable Accrued Income taxes payable Long-
term note payable, secured S 17,500 4,800 4,700 wages payable 5, Prepaid expenses Plant assets,
net by mortgage on plant assets Common stod Retained earnings 83,400 88,000 76,250 Total
assets S254,650 Total liabilities and equity S254,650 These are short-term notes receivable
arising from customer (trade) sales. Required: Compute the following: (1) current ratio, (2) acid-
test ratio, (3) days sales uncollected, (4) inventory turnover, (5) days\' sales in inventory, (8)
debt-to-equity ratio, (7) times interest earned, (8) profit margin ratio, (9) total asset turnover, (10)
return on total assets, and (11) return on common stockholders\' equity Do not round
intermediate calculations.) Choose Numerator: | Choose Denominator: | = Current Ratio Current
ratio 2015 to 1
Solution
Answer to Part 1.
Current Ratio = Current Assets / Current Liabilities
Current Assets = Cash + Short Term Investments + Accounts Receivable, net + Notes
Receivable (Trade) + Merchandise Inventory + Prepaid Expenses
Current Assets = $16,000 + $9,200 + $32,200 + $5,000 + $38,150 + $2,800
Current Assets = $103,350
Current Liabilities = Accounts Payable + Accrued Wages Payable + Income Taxes Payable
Current Liabilities = $17,500 + $4,800 + $4,700
Current Liabilities = $27,000
Current Ratio = 103,350 / 27,000
Current Ratio = 3.83 to 1
Answer to Part 2.
Acid-Test Ratio = Quick Assets / Current Liabilities
Quick Assets = Current Assets – Merchandise Inventory – Prepaid Expenses
Quick Assets = $103,350 - $38,150 - $2,800
Quick Assets = $62,400
Acid-Test Ratio = 62,400 /27,000
Acid-Test Ratio = 2.31 to 1
Answer to Part 3.
Days Collected Sales = Current Receivable / Net Sales * 365
Current Receivable = $32,200 + $5,000 = $37,200
Days Collected Sales = 37,200 / 453,600 * 100
Days Collected Sales = 29.93 days
Answer to Part 4.
Inventory Turnover Ratio = Cost of goods sold / Average Inventory
Average Inventory = (38,150 + 53,900) / 2
Average Inventory = $46,025
Inventory Turnover Ratio = 296,850 / 46,025
Inventory Turnover Ratio = 6.45 times
Answer to Part 5.
Days’ Sales in Inventory = Average Inventory / Cost of Goods Sol.
A sample survey by the Pew internet and American Life Project asked .pdf
A sample survey by the Pew internet and American Life Project asked a random sample of adults
whether they had a brodband internet connection at home, with the following results by type of
community:
Have broadband No broadband
Urban 320 276
Suburban 521 542
Rural 254 327
Test the hypotesis that there is a relationship between community type (urban, suburban, or rural)
and whether the
residents have broadband.
Solution
Set up null and alternate as
H0: There is no relation between community type and having broadband
Ha: There is relation between them
Two tailed chi square test
Prepare contingency table
chi-square = 11.7
degrees of freedom = 2
probability = 0.003
As p <0.01, reject null hypothesis.
There is relation between the two..
8. On January 1, Wismer Corporation declared a $1.50 per share cash d.pdf
8. On January 1, Wismer Corporation declared a $1.50 per share cash dividend on its common
stock (18,000 shares) for stockholders on record as of January 15. Wismer paid the dividend on
January 31. Journalize the entries declaring the cash dividend and paying the dividend. (Record
debits first, then credits. Select the explanation on the last line of the journal entry table.)
Journalize the entry declaring the cash dividend. Date Accounts and Explanation Debit Credit
Jan. 1 Journalize the entry paying the dividend. Date Accounts and Explanation Debit Credit Jan.
31
Solution
1-Jan Cash dividends 27000 =18000*1.5 Dividends payable 27000 31-Jan
Dividends payable 27000 Cash 27000 2 Interest expense for 2019 =
59000*7%/12*5= 1721.
Graph one cycle of the following function. y = 5 sin(-2pi x + pi) St.pdf
Graph one cycle of the following function. y = 5 sin(-2pi x + pi) State the period, amplitude,
phase shift and vertical shift, period Amplitude phase shift vertical shift
Solution
y=-5sin(-2pi x + pi)
when x=1/2, y= -5sin(-pi+pi)=0
when x=3/2,y=-5sin(-2pi)=0
Hence the correct option is the last one.
explain why these three properties of water are importantmelting .pdf
explain why these three properties of water are important?
melting and freezing
cohesive and adhesive
An excellent solvent
Solution
Water melting and freezing properties are important because these properties are main colligative
properties useful to prepare polar biological & pharmaceutical products suitable to the internal
human body environment
The cohesive forces (cohesion) & adhesive forces of water molecules enable to acquire elastic
tendency with considerable surface area promote solubility. This tendency is referred as surface
tension. Surface tension of any fluid decreases with increasing temperature. This is due to
existence of cohesive intermolecular forces between molecules of the fluid is reduced and it is
based on the surrounding environment with adhesive intermolecular forces & viscosity of the
fluid, molecular thermal activity at the liquid interface
For example, water is a universal solvent and it has both polar because it has both positive charge
(H+ or H3O+) and negative charge (OH-). This polarity enables water to make substances to
dissolve in it as charged species. On the other hand, water molecules possess a higher attractive
power to each other so that these forces are referred as cohesive forces responsible for “surface
tension” phenomenon” and these attractions are predominantly due to hydrogen bonding. For
instance, 2 to 3 drops of water between the cover slip is imparted adhesive forces (another
example is capillary rise due to forces between glass and water in capillary tube) between the
two glass cover slips result in adhesion instead of cohesion that potentially differentiate
intermolecular forces between two different substances namely water and silicon. These
intermolecular forces render glass surface as hydrophilic. These intermolecular forces are
decreased as the temperature increases due to increase in molecular thermal activity of the fluid
molecules ate the interface of water molecules.
An excellent solvent
Water is a universal polar solvent and it has hydrogen bonding associated with highest heat
capacity and is referred as “specific heat capacity”. This specific heat is predominantly due to
intermolecular and intermolecular hydrogen bonding. In nature on land when heat is absorbed
result in breaking of hydrogen bonds specifically followed by production of heat. Normally
ocean water cool little bit slower when compared to the land water because of more heat
capacity. This sterile solvent has ability to dissolve a majority of biological products to prepare
pharmaceutical products finally used as diluents in saline, injectables etc.
Water properties and the biological importance of water:
1). Due of the presence of hydrogen bonds and strong intermolecular forces, water has high
specific heat index. Specific heat capacity of water is defined as; the amount of heat (energy in
joules) required for raise of temperature of water per unit mass by 10C.
Our body temperature is maintained higher than the surroundin.
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
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Which of the following is not a property of a normal curve A. Bell .pdf
Which of the following is not a property of a normal curve? A. Bell Shaped B. Symmetrical to
the mean C. Intersects the horizontal axis D. Contains two inflection points
Solution
It is
OPTION C: Intersects the horizontal axis.
****************
It does not intersect the horizontal axis; in fact, it is asymptotic to it; it draws closer and closer,
but it never touches the axis..
What tissue type has polarity and is avascular What tissue.pdf
What tissue type has polarity and is avascular
What tissue type has polarity and is avascular
Solution
The epithelial tissue do not receive vascular supply, so they are avascular. It is polar because it
contains apical, lateral and basal membrane domains.
The epithelial tissues form the sheet like outer covering of the body surfaces and provide
protection to the organs. One surface of this epithelial tissue is not bound to anything and
produces glandular secretions, also involved in absorption process.
The characteristics of the epithelial tissues include the following:
1). Cells undergo frequent cell division
2). Lack of blood vessels
3). Cells are tightly packed by cell junctions..
What organizations are involved in Internet2How is this developme.pdf
What organizations are involved in Internet2?
How is this development similar to the beginnings of our present Internet?
Does your school or employer participate in Internet2?
Solution
Answer:
Internet2 developing network capabilities . The organizations in involved in the internet2 are
mostly the US institutions of higher education , government agencies , and corporations.
Internet brought the world into new stage. Even we are seeing change in almost everything.
Bussinesses are changing to E-business , banking to e-banking all because of the internet.
Internet2 brought the higher capabilities of networking systems. Like before , we were using
Class A IP addresses . But now a days we are using the class C IP addresses due to increse of the
users.
Yes first when the internet 2 introduced, the US institutions of higher education participated with
full bhoom..
What is meant by modular software and why is it important How can t.pdf
What is meant by modular software and why is it important? How can the modularity of a
program code be quantifiably measured?
Solution
1) What is ment by modular software and why is it important?
Answer:-
How important is modularization of software projects:
Modularization is obviously important in software projects, but I want to know people\'s
opinions on how important and for what reasons it is important. I\'ve obviously got my own ideas
since I\'m asking this, but think of it like a \"common brainstorm\" of the reasons one should
modularize one\'s software projects..
Techopedia explains Modular Programming:
1) Modules in modular programming enforce logical boundaries between components and
improve maintainability. They are incorporated through interfaces. They are designed in such a
way as to minimize dependencies between different modules. Teams can develop modules
separately and do not require knowledge of all modules in the system.
2) Each and every modular application has a version number associated with it. This provides
developers flexibility in module maintenance. If any changes have to be applied to a module,
only the affected subroutines have to be changed. This makes the program easier to read and
understand.
3) Modular programming has a main module and many auxiliary modules. The main module is
compiled as an executable (EXE), which calls the auxiliary module functions.
2) How can the modularity of a program code be quantifiably measured?
Answer:-
A Software Metric is a standard of measure of a degree to which a software system or process
possesses some property. Even if a metric is not a measurement (metrics are functions, while
measurements are the numbers obtained by the application of metrics), often the two terms are
used as synonymous.
A Software development goal is obtaining objective, reproducible and quantifiable
measurements, which may have numerous valuable applications in schedule and budget
planning, cost estimation, quality assurance testing, software debugging, software performance
optimization, and optimal personnel task assignments..
What are the two major steps in biological classification What .pdf
What are the two major steps in biological classification? What did Linnaeus contribute to
taxonomy? are scientific names derived? want to identify the various
Solution
Answers:
1. Two major steps in biological classification are:
a) Defining and describing organisms
b) Arranging organisms into a logical classification scheme
2. The binomial nomenclature system was introduced by Carolus Linnaeus. It is used to assign
each organism a two part scientific name - genus and species.
3. Scientific names are derived following the universal rules of nomenclature:
a) These are in Latin and written in italics.
b) The first word represents the genus while the second word is the species epithet
c) The first word (genus) is written in capital letter while second word (species) is written in
small letter.
d) If hand written, they are separately underlined, or printed in italics to denote their Latin origin..
Tissue transplant rejection is primarily atype I reaction.type I.pdf
Tissue transplant rejection occurs when the host immune system reacts to antigens in the transplanted organ from a foreign organism. This is a type IV hypersensitivity reaction where T-cells mediate an immune response against the major histocompatibility complex proteins on the surface of donated tissue that help the immune system identify foreign substances. Tissue transplant rejection can take weeks to months to occur as an immune response against the transplanted organ.
The members of the UN peace committee, must choose, from among thems.pdf
The members of the UN peace committee, must choose, from among themselves, a presiding
officer of their committee. For each member x, let f(x) designate that member\'s choice for
officer. If no two members vote alike, what is the range of f?
Solution
Since each member can vote only for one person and no two members vote alike
the mapping f: x to f(x) is one to one and onto.
Hence f is a bijective function
So order of x = order of f(x)
Or Range of f = No of members of the UN Peace committee.
NetWork Design Question1a.) In writing a letter to a Friend, what .pdf
NetWork Design Question
1a.) In writing a letter to a Friend, what do you hace to do to get the letter to your friend. List all
the steps and ways.
1b.) Relate the stages of letter transaction to the layers of the internet.
Solution
In writing a letter to a Friend, the following 5 steps do you face to get the letter to your friend.
1) Application layer:
TCP/IP: A user retrieves a Web page from a server in somewhere by typing a URL into a
browser and clicking the enter key.
The server receives the request, finds the page on its hard drive and sends it back to the user.
Neither the user nor the client or server software is aware of the way the messages were
delivered. There are many application layer protocols.
2) Transport layer:
TCP/IP: Transport layer software establishes a connection between a client and server then
monitors that connection for errors.
It also slows transmission if it gets too fast to handle at the other end. Transport layer software is
not concerned with how the data is actually moved from one point to the other that is the
responsibility of lower level software. There are two transport layer protocols TCP (reliable) and
UDP (unreliable, but fast).
If TCP tries repeatedly and errors persist, it informs its \"boss,\" the application program.
3) Internet layer:
TCP/IP:Internet layer programs move data between networks. IP is the internet layer protocol.
IP software ignores the data, and it does the same thing with a packet whether it comes from the
friend or your mother.
Once the data gets to its destination local area network (LAN), it is handed over to data link layer
software or firmware for delivery to the proper computer.
4) Data link layer:
TCP/IP:Data Link layer programs move (incoming and outgoing) data within LANs.
Ethernet is the most common data link protocol. A data link program is not concerned with how
outgoing data will be handled once it leaves the LAN or how incoming data got there. That is the
responsibility of the internet layer software.
5) Physical layer:
TCP/IP:Physical layer protocols specify the means of representing ones and zeros (bits).
The method of transmitting them between two points using wire, radio, fiber, etc. is also
specified. There are many ways to make ones and zeros and many ways to transmit them so there
are many physical layer protocols..
Please help me with thisLet x denote a single observation of a no.pdf
Please help me with this:
Let x denote a single observation of a normal random variable, with mean=0 and unknown
variance 2. Derive the maximum likelihood estimate for .
Solution
Given X~N(0,2), then L=(1/((2pie)))(exp((-x^2)/(2*2)))=>logL=((-1/2)*log(2pie))-((1/2)*log2)-
((1/(22))*(x2)), then the likelihood equation for estimate is dlogL/d=0,=>^=(x2)..
Numerical Methods yi + 1 = y_i + Phi (x_i, y_i, h) Consider the equ.pdf
Numerical Methods yi + 1 = y_i + Phi (x_i, y_i, h) Consider the equation given above. The
function phi (x, y, h) is an incremental function. It can be interpreted as the average of the slope
at the initial and end points. a function of the initial x value only. an incremental function. It can
be interpreted as the slope at the mid-point. an incremental function. It can be interpreted as a
representative slope over the interval.
Solution
Option D.
My question is pretty simple, I just want to know how to call my ope.pdf
My question is pretty simple, I just want to know how to call my operator== function in
Stack.cpp using a list function.
Here is what I meant. This is my Stack.h file:
class Stack {
public:
/**constructors and destructors*/
Stack();
Stack(const Stack &S);
~Stack();
void pop();
void push(int data);
bool operator==(const Stack &S);
[ .......]
private:
List stack;
};
Here is my List.h file:
template
class List {
private:
struct Node {
listitem data;
Node* next;
Node* previous;
Node(listitem data) :
next(NULL), previous(NULL), data(data) {
}
};
typedef struct Node* NodePtr;
NodePtr start;
NodePtr end;
NodePtr cursor;
int length;
public:
........
bool operator==(const List &list);
[.......] }
[........]
template
bool List::operator==(const List& list) {
if (length != list.length)
return false;
NodePtr temp1 = start;
NodePtr temp2 = list.start;
while (temp1 != NULL) {
if (temp1->data != temp2->data)
return false;
temp1 = temp1->next;
temp2 = temp2->next;
}
return true;
}
[......]
So I already define my functions in List.h and I only need to call my function from List again in
my Stack.cpp file. For example from my Stack.cpp:
Stack::Stack(const Stack &S) :
stack(S.stack) {
}
void Stack::pop() {
stack.remove_start();
}
[......]
And I don\'t know how to call my operator== function in Stack.cpp from the List.h like the way
above. This is what my guess so far but did not work:
bool Stack::operator==(const Stack &S) {
stack.operator==(S.stack);
}
Thank you
Solution
When we wish to make a comparison, such as in an if statement, we use the double equals sign
(==).
1)bool operator==(const Stack &S); [ .......]
It means whatever the boolean operator it should be belongs to stack or equal to.
2) bool operator==(const List &list);
Likewise link list also compare with the constructor link list.
and constructor need not to call explicitly. it will call itself when program called first..
Java AssignmentConvert infix to postfix.SolutionTo convert i.pdf
Java Assignment
Convert infix to postfix.
Solution
To convert infix expressions to postfix, we can use the stack data structure. The stack is used to
store the operators and parenthesis to apply the precedence.
For example if the input is a%b is ab%
class Stack
{
char c[]=new char[100];
int top=-1;
void push(char a)
{
try
{
c[++top]=a;
}
catch(StringIndexOutOfBoundsException e)
{
System.out.println(\"Stack full\");
System.exit(0);
}
}
char pop()
{return c[top--];
}
boolean isEmpty()
{
return (top==-1)?true:false;
}
char peak()
{
return c[top];
}
}
public class InfixToPostfix
{
static Stack expression= new Stack;
public static void main(String argv[]) throws IOException
{
String infix;
BufferedReader keyboard= new BufferedReader(new InputStreamReader(System.in));
System.out.print(\"enter the expression in infix notation\");
infix=keyboard.readLine();
System.out.println(\"The resultant postfix expression is \"+ toPostfix(infix));
}
private static String toPostfix(String infix)
{
char sym;
string postfix=\"\";
for(int i=0;i.
Irreducible polynomials over a field F play a similar, in a certain .pdf
Irreducible polynomials over a field F play a similar, in a certain sense, role in F[x] to the role
the prime numbers play for Z. Using your favorite proof of the existence of an infinite number of
primes as inspiration, please prove that there are infinitely many irreducible polynomials over a
field F.
Solution
if there were finite such polynomials.... p1.p2.p3.........pn
then
p1 * p2 * p3 * ......pn +1 (multiplicative unity) is a polynomial
which is either irreducible or can be written as a product of irreducible polynomials.
It cannot be irreducible since it is different from the set of irreducibles.
Also it cannot be divisible by any irreducible . Say it does pi divides this new polynomial fr
some i , then
pi divides p1*p2*......pn + 1
and also pi is one of p1,p2....pn so divides p1*p2*....pn
hence it divides 1
contradiction.
Inferential statistics is more than descriptive statistics. direct .pdf
Inferential statistics is more than descriptive statistics. direct accurate indirect
Solution
OPTION 3: INDIRECT [ANSWER]
This is so because in inferential statistics, we are looking to describe something that is out of the
scope of out data, while descriptive statistics describe what is actually already present..
If an organism produces two different alleles for a gene, that organ.pdf
If an organism produces two different alleles for a gene, that organism is ___________ with
respect to that gene.
Question 13 options:
dioecious
homozygous
haploid
heterozygous
diallelic
dioecious
homozygous
haploid
heterozygous
diallelic
Solution
Dioecious - dioecious species has the different male and female reproductive structures.
Homozygous - Homozygous species are those ones having identical pair of genes for a given
pair.
Haploid - haploid cell contain only one complete set of chromosomes.
Heterozygous - They carry dissimilar alleles, different genotype and having both dominant and
recessive alleles.
Diallellic- It is applied to those where two different alleles are present at particular gene.
Hence most perfect answer for above question is DIALLELLIC..
I apologize this 1 question has numerous parts. I didnt want to s.pdf
I apologize: this 1 question has numerous parts. I didn\'t want to split it up in to separate
questions and confuse others. Selected year-end financial statements of Cabot Corporation
follow. (All sales were on credit; selected balanoe sheet amounts at December 31, 2014, were
inventory, S53,900; total assets, $199,400; common stock, 588,000; and retained earnings,
344,392.) CABOT CORPORATION Income Statement For Year Ended December 31, 2015
Sales Cost of goods sold S453,600 298,850 Gross profit Operating expenses Interest expense
158,750 98,800 4,600 Income before taxes Income taxes 53,350 21,492 Net income S 31,858
CABOT CORPORATION Balance Sheet December 31, 2015 Assets Cash Short-term
investments Accounts receivable, net Notes receivable (trade) S 18,000 9,200 32,200 000 38,150
2,800 151,300 Liabilities and Equity Accounts payable Accrued Income taxes payable Long-
term note payable, secured S 17,500 4,800 4,700 wages payable 5, Prepaid expenses Plant assets,
net by mortgage on plant assets Common stod Retained earnings 83,400 88,000 76,250 Total
assets S254,650 Total liabilities and equity S254,650 These are short-term notes receivable
arising from customer (trade) sales. Required: Compute the following: (1) current ratio, (2) acid-
test ratio, (3) days sales uncollected, (4) inventory turnover, (5) days\' sales in inventory, (8)
debt-to-equity ratio, (7) times interest earned, (8) profit margin ratio, (9) total asset turnover, (10)
return on total assets, and (11) return on common stockholders\' equity Do not round
intermediate calculations.) Choose Numerator: | Choose Denominator: | = Current Ratio Current
ratio 2015 to 1
Solution
Answer to Part 1.
Current Ratio = Current Assets / Current Liabilities
Current Assets = Cash + Short Term Investments + Accounts Receivable, net + Notes
Receivable (Trade) + Merchandise Inventory + Prepaid Expenses
Current Assets = $16,000 + $9,200 + $32,200 + $5,000 + $38,150 + $2,800
Current Assets = $103,350
Current Liabilities = Accounts Payable + Accrued Wages Payable + Income Taxes Payable
Current Liabilities = $17,500 + $4,800 + $4,700
Current Liabilities = $27,000
Current Ratio = 103,350 / 27,000
Current Ratio = 3.83 to 1
Answer to Part 2.
Acid-Test Ratio = Quick Assets / Current Liabilities
Quick Assets = Current Assets – Merchandise Inventory – Prepaid Expenses
Quick Assets = $103,350 - $38,150 - $2,800
Quick Assets = $62,400
Acid-Test Ratio = 62,400 /27,000
Acid-Test Ratio = 2.31 to 1
Answer to Part 3.
Days Collected Sales = Current Receivable / Net Sales * 365
Current Receivable = $32,200 + $5,000 = $37,200
Days Collected Sales = 37,200 / 453,600 * 100
Days Collected Sales = 29.93 days
Answer to Part 4.
Inventory Turnover Ratio = Cost of goods sold / Average Inventory
Average Inventory = (38,150 + 53,900) / 2
Average Inventory = $46,025
Inventory Turnover Ratio = 296,850 / 46,025
Inventory Turnover Ratio = 6.45 times
Answer to Part 5.
Days’ Sales in Inventory = Average Inventory / Cost of Goods Sol.
A sample survey by the Pew internet and American Life Project asked .pdf
A sample survey by the Pew internet and American Life Project asked a random sample of adults
whether they had a brodband internet connection at home, with the following results by type of
community:
Have broadband No broadband
Urban 320 276
Suburban 521 542
Rural 254 327
Test the hypotesis that there is a relationship between community type (urban, suburban, or rural)
and whether the
residents have broadband.
Solution
Set up null and alternate as
H0: There is no relation between community type and having broadband
Ha: There is relation between them
Two tailed chi square test
Prepare contingency table
chi-square = 11.7
degrees of freedom = 2
probability = 0.003
As p <0.01, reject null hypothesis.
There is relation between the two..
8. On January 1, Wismer Corporation declared a $1.50 per share cash d.pdf
8. On January 1, Wismer Corporation declared a $1.50 per share cash dividend on its common
stock (18,000 shares) for stockholders on record as of January 15. Wismer paid the dividend on
January 31. Journalize the entries declaring the cash dividend and paying the dividend. (Record
debits first, then credits. Select the explanation on the last line of the journal entry table.)
Journalize the entry declaring the cash dividend. Date Accounts and Explanation Debit Credit
Jan. 1 Journalize the entry paying the dividend. Date Accounts and Explanation Debit Credit Jan.
31
Solution
1-Jan Cash dividends 27000 =18000*1.5 Dividends payable 27000 31-Jan
Dividends payable 27000 Cash 27000 2 Interest expense for 2019 =
59000*7%/12*5= 1721.
Graph one cycle of the following function. y = 5 sin(-2pi x + pi) St.pdf
Graph one cycle of the following function. y = 5 sin(-2pi x + pi) State the period, amplitude,
phase shift and vertical shift, period Amplitude phase shift vertical shift
Solution
y=-5sin(-2pi x + pi)
when x=1/2, y= -5sin(-pi+pi)=0
when x=3/2,y=-5sin(-2pi)=0
Hence the correct option is the last one.
explain why these three properties of water are importantmelting .pdf
explain why these three properties of water are important?
melting and freezing
cohesive and adhesive
An excellent solvent
Solution
Water melting and freezing properties are important because these properties are main colligative
properties useful to prepare polar biological & pharmaceutical products suitable to the internal
human body environment
The cohesive forces (cohesion) & adhesive forces of water molecules enable to acquire elastic
tendency with considerable surface area promote solubility. This tendency is referred as surface
tension. Surface tension of any fluid decreases with increasing temperature. This is due to
existence of cohesive intermolecular forces between molecules of the fluid is reduced and it is
based on the surrounding environment with adhesive intermolecular forces & viscosity of the
fluid, molecular thermal activity at the liquid interface
For example, water is a universal solvent and it has both polar because it has both positive charge
(H+ or H3O+) and negative charge (OH-). This polarity enables water to make substances to
dissolve in it as charged species. On the other hand, water molecules possess a higher attractive
power to each other so that these forces are referred as cohesive forces responsible for “surface
tension” phenomenon” and these attractions are predominantly due to hydrogen bonding. For
instance, 2 to 3 drops of water between the cover slip is imparted adhesive forces (another
example is capillary rise due to forces between glass and water in capillary tube) between the
two glass cover slips result in adhesion instead of cohesion that potentially differentiate
intermolecular forces between two different substances namely water and silicon. These
intermolecular forces render glass surface as hydrophilic. These intermolecular forces are
decreased as the temperature increases due to increase in molecular thermal activity of the fluid
molecules ate the interface of water molecules.
An excellent solvent
Water is a universal polar solvent and it has hydrogen bonding associated with highest heat
capacity and is referred as “specific heat capacity”. This specific heat is predominantly due to
intermolecular and intermolecular hydrogen bonding. In nature on land when heat is absorbed
result in breaking of hydrogen bonds specifically followed by production of heat. Normally
ocean water cool little bit slower when compared to the land water because of more heat
capacity. This sterile solvent has ability to dissolve a majority of biological products to prepare
pharmaceutical products finally used as diluents in saline, injectables etc.
Water properties and the biological importance of water:
1). Due of the presence of hydrogen bonds and strong intermolecular forces, water has high
specific heat index. Specific heat capacity of water is defined as; the amount of heat (energy in
joules) required for raise of temperature of water per unit mass by 10C.
Our body temperature is maintained higher than the surroundin.
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- 1. Can someone help me find the errors in my codes and fix them... There are mistakes on the raffle portions. These are my codes so far: NOTE: Only use void functions if possible, please... #include #include //MAXRAFFLE is the maximum number of available raffle tickets #define MAXSIZE 100 #define MAXAUCTIONITEMS 1000 #define MAXRAFFLE 1000 //Function prototypes - do not change these void initRaffle(int raffles[MAXRAFFLE]); void initAuction(float auction[2][MAXAUCTIONITEMS]); void buyTickets(float * totalRevenue, int * ticketsSold, int numTickets, float price); void buyRaffle(float * totalRevenue, int raffles[MAXRAFFLE], int availTickets, int numTickets, int person); void bid(float auction[2][MAXAUCTIONITEMS], float bid, int auctionNumber, int person, int flag); float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber); void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner); //Main function int main() { FILE * ifp; char filename[MAXSIZE], event[MAXSIZE], item[MAXSIZE]; float presale, dayOf, totalRevenue = 0; float auctions[2][MAXAUCTIONITEMS]; int raffles[MAXRAFFLE]; int numPresale, numAuctions, numRaffle, numPrizes, numEvents; int i, ticketsSold = 0, auctionFlag = 1; printf("Please enter the input file name. "); scanf("%s", filename); ifp = fopen(filename, "r"); fscanf(ifp, "%f%f%d", &presale, &dayOf, &numPresale); totalRevenue += numPresale * presale; ticketsSold = numPresale; fscanf(ifp, "%d", &numAuctions);
- 2. fscanf(ifp, "%d%d", &numRaffle, &numPrizes); fscanf(ifp, "%d", &numEvents); initRaffle(raffles); initAuction(auctions); for (i=0; iauction[0][auctionNumber]) { auction[0][auctionNumber]=bid; auction[1][auctionNumber]=person; printf(" BIDITEM %d ACCEPTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } else { printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } } else printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } // Pre-conditions: auction is an 2D array that holds the current highest bid and the person with // the highest bid // auctionNumber is the numerical identifier for the item being awarded // Post-condition: Returns the value of the highest bid for the auction specified by auction number // // What to do in this function: Check to see if the auction identified by auctionNumber has any // bids. If so, award the auction to the person with the highest bid. If not, print out that // there have been no bids for this item. Return the appropriate value to be added to totalRevenue float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber) { if(auction[0][auctionNumber]!= EOF) { printf(" AUCTION ITEM %d WON BY PERSON %.2f for $%.2f",auctionNumber, auction[1][auctionNumber],auction[0][auctionNumber]); return auction[0][auctionNumber]; } printf(" NO BIDS FOR AUCTION ITEM %d", auctionNumber); return 0;
- 3. } // Pre-conditions: raffles is an array of all possible raffle tickets // winner is the winning ticket number // raffleNumber is the current raffle being held // // Post-condition: prints out the numerical identifier of the person who // holds the winning ticket number void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner) { while (raffles[winner] != EOF) { printf (" RAFFLE %d WON BY PERSON %d", raffleNumber,raffles[winner]); break; } } INPUT: 15 25 200 3 100 4 27 BUY TICKET 8 BUY RAFFLE 30 22 BIDITEM 0 2 5 BIDITEM 0 3 20 BUY TICKET 10 BUY RAFFLE 40 209 BUY TICKET 1 BIDITEM 1 17 50 BIDITEM 2 5 30 BUY RAFFLE 26 123 BUY TICKET 3 BIDITEM 1 16 65 BUY RAFFLE 19 211 BIDITEM 1 17 70 BIDITEM 2 200 45 BIDITEM 2 201 54 BUY RAFFLE 1 27 CLOSEAUCTION
- 4. BIDITEM 1 100 10000 AWARD RAFFLE 0 13 AWARD RAFFLE 1 98 AWARD RAFFLE 2 30 AWARD RAFFLE 3 85 AWARD AUCTION 0 AWARD AUCTION 1 AWARD AUCTION 2 TOTAL REVENUE CORRECT OUTPUT: SOLD TICKETS 200 - 207 RAFFLE TICKETS 0 - 29 given to PERSON 22 BIDITEM 0 ACCEPTED for PERSON 2 at 5.00 DOLLARS BIDITEM 0 ACCEPTED for PERSON 3 at 20.00 DOLLARS SOLD TICKETS 208 - 217 RAFFLE TICKETS 30 - 69 given to PERSON 209 SOLD TICKETS 218 - 218 BIDITEM 1 ACCEPTED for PERSON 17 at 50.00 DOLLARS BIDITEM 2 ACCEPTED for PERSON 5 at 30.00 DOLLARS RAFFLE TICKETS 70 - 95 given to PERSON 123 SOLD TICKETS 219 - 221 BIDITEM 1 ACCEPTED for PERSON 16 at 65.00 DOLLARS RAFFLE TICKETS 96 - 99 given to PERSON 211 BIDITEM 1 ACCEPTED for PERSON 17 at 70.00 DOLLARS BIDITEM 2 ACCEPTED for PERSON 200 at 45.00 DOLLARS BIDITEM 2 ACCEPTED for PERSON 201 at 54.00 DOLLARS NO RAFFLE TICKETS given to PERSON 27 CLOSEAUCTION BIDITEM 1 REJECTED for PERSON 100 at 10000.00 DOLLARS RAFFLE 0 WON BY PERSON 22 RAFFLE 1 WON BY PERSON 211 RAFFLE 2 WON BY PERSON 209 RAFFLE 3 WON BY PERSON 123 AUCTION ITEM 0 WON BY PERSON 3 for $20.00 AUCTION ITEM 1 WON BY PERSON 17 for $70.00 AUCTION ITEM 2 WON BY PERSON 201 for $54.00
- 5. TOTALREVENUE is $3894.00 Solution #include #include //MAXRAFFLE is the maximum number of available raffle tickets #define MAXSIZE 100 #define MAXAUCTIONITEMS 1000 #define MAXRAFFLE 1000 //Function prototypes - do not change these void initRaffle(int raffles[MAXRAFFLE]); void initAuction(float auction[2][MAXAUCTIONITEMS]); void buyTickets(float * totalRevenue, int * ticketsSold, int numTickets, float price); void buyRaffle(float * totalRevenue, int raffles[MAXRAFFLE], int availTickets, int numTickets, int person); void bid(float auction[2][MAXAUCTIONITEMS], float bid, int auctionNumber, int person, int flag); float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber); void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner); //Main function int main() { FILE * ifp; char filename[MAXSIZE], event[MAXSIZE], item[MAXSIZE]; float presale, dayOf, totalRevenue = 0; float auctions[2][MAXAUCTIONITEMS]; int raffles[MAXRAFFLE]; int numPresale, numAuctions, numRaffle, numPrizes, numEvents; int i, ticketsSold = 0, auctionFlag = 1; printf("Please enter the input file name. "); scanf("%s", filename);
- 6. ifp = fopen(filename, "r"); fscanf(ifp, "%f%f%d", &presale, &dayOf, &numPresale); totalRevenue += numPresale * presale; ticketsSold = numPresale; fscanf(ifp, "%d", &numAuctions); fscanf(ifp, "%d%d", &numRaffle, &numPrizes); fscanf(ifp, "%d", &numEvents); initRaffle(raffles); initAuction(auctions); for (i=0; iauction[0][auctionNumber]) { auction[0][auctionNumber]=bid; auction[1][auctionNumber]=person; printf(" BIDITEM %d ACCEPTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } else { printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } } else printf(" BIDITEM %d REJECTED for PERSON %d at %0.2f DOLLARS",auctionNumber,person, bid); } // Pre-conditions: auction is an 2D array that holds the current highest bid and the person with // the highest bid // auctionNumber is the numerical identifier for the item being awarded // Post-condition: Returns the value of the highest bid for the auction specified by auction number
- 7. // // What to do in this function: Check to see if the auction identified by auctionNumber has any // bids. If so, award the auction to the person with the highest bid. If not, print out that // there have been no bids for this item. Return the appropriate value to be added to totalRevenue float awardAuction(float auction[2][MAXAUCTIONITEMS], int auctionNumber) { if(auction[0][auctionNumber]!= EOF) { printf(" AUCTION ITEM %d WON BY PERSON %.2f for $%.2f",auctionNumber, auction[1][auctionNumber],auction[0][auctionNumber]); return auction[0][auctionNumber]; } printf(" NO BIDS FOR AUCTION ITEM %d", auctionNumber); return 0; } // Pre-conditions: raffles is an array of all possible raffle tickets // winner is the winning ticket number // raffleNumber is the current raffle being held // // Post-condition: prints out the numerical identifier of the person who // holds the winning ticket number void awardRaffle(int raffles[MAXRAFFLE], int raffleNumber, int winner) { while (raffles[winner] != EOF) { printf (" RAFFLE %d WON BY PERSON %d", raffleNumber,raffles[winner]); break; } }