a. Homo neanderthalensis, Homo heidelbergensis, and maybe Homo flore.pdf
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a. Homo neanderthalensis, Homo heidelbergensis, and maybe Homo floresiensis Homo heidelbergensis is an extinct, potentially distinct species of the genus Homo and may be the direct ancestor of H. neanderthalensis in Europe and H.floresiensis fits into the family tree of hominins which includes those species that evolved after the human lineage (of the genus Homo) split from the chimpanzees. Solution a. Homo neanderthalensis, Homo heidelbergensis, and maybe Homo floresiensis Homo heidelbergensis is an extinct, potentially distinct species of the genus Homo and may be the direct ancestor of H. neanderthalensis in Europe and H.floresiensis fits into the family tree of hominins which includes those species that evolved after the human lineage (of the genus Homo) split from the chimpanzees..
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molality = mole wt of solvent 0.11 = mole 21.pdf
molality = mole / wt of solvent 0.11 = mole / 217 mole = 0.11*217 = 23.87mole
Solution
molality = mole / wt of solvent 0.11 = mole / 217 mole = 0.11*217 = 23.87mole.
NaF would be having highest melting point because.pdf
NaF would be having highest melting point because melting point increases with
ionic character F is more electronegative and Na is more electro positive and hence NaF is more
ionic
Solution
NaF would be having highest melting point because melting point increases with
ionic character F is more electronegative and Na is more electro positive and hence NaF is more
ionic.
The four bases in DNA are Adenine, Guanine, Cytos.pdf
The four bases in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine
and Guanine are Purines (2 ring structure). Cytosine and Thymine are Pyrimidines (single ring
structure). Adenine pairs with Thymine through 2 hydrogen bonds. Guanine pairs with Cytosine
through 3 hydrogen bonds. Since all base pairs contain a Purine and a Pyrimidine, they have the
same molecular dimensions.
Solution
The four bases in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine
and Guanine are Purines (2 ring structure). Cytosine and Thymine are Pyrimidines (single ring
structure). Adenine pairs with Thymine through 2 hydrogen bonds. Guanine pairs with Cytosine
through 3 hydrogen bonds. Since all base pairs contain a Purine and a Pyrimidine, they have the
same molecular dimensions..
b) HNO2,NO2^+ as when H+ will leave then NO2- wil.pdf
b) HNO2,NO2^+ as when H+ will leave then NO2- will be left as conjucate base
and not NO2+
Solution
b) HNO2,NO2^+ as when H+ will leave then NO2- will be left as conjucate base
and not NO2+.
Number of moles of Calcium = 0.05Number of equivalents of calcium .pdf
Number of moles of Calcium = 0.05
Number of equivalents of calcium = Number of moles / Valency = 0.05 / 2 = 0.025
Normality = 0.025 / 1 = 0.025 Ea /L
Solution
Number of moles of Calcium = 0.05
Number of equivalents of calcium = Number of moles / Valency = 0.05 / 2 = 0.025
Normality = 0.025 / 1 = 0.025 Ea /L.
Recommended
molality = mole wt of solvent 0.11 = mole 21.pdf
molality = mole / wt of solvent 0.11 = mole / 217 mole = 0.11*217 = 23.87mole
Solution
molality = mole / wt of solvent 0.11 = mole / 217 mole = 0.11*217 = 23.87mole.
NaF would be having highest melting point because.pdf
NaF would be having highest melting point because melting point increases with
ionic character F is more electronegative and Na is more electro positive and hence NaF is more
ionic
Solution
NaF would be having highest melting point because melting point increases with
ionic character F is more electronegative and Na is more electro positive and hence NaF is more
ionic.
The four bases in DNA are Adenine, Guanine, Cytos.pdf
The four bases in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine
and Guanine are Purines (2 ring structure). Cytosine and Thymine are Pyrimidines (single ring
structure). Adenine pairs with Thymine through 2 hydrogen bonds. Guanine pairs with Cytosine
through 3 hydrogen bonds. Since all base pairs contain a Purine and a Pyrimidine, they have the
same molecular dimensions.
Solution
The four bases in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine
and Guanine are Purines (2 ring structure). Cytosine and Thymine are Pyrimidines (single ring
structure). Adenine pairs with Thymine through 2 hydrogen bonds. Guanine pairs with Cytosine
through 3 hydrogen bonds. Since all base pairs contain a Purine and a Pyrimidine, they have the
same molecular dimensions..
b) HNO2,NO2^+ as when H+ will leave then NO2- wil.pdf
b) HNO2,NO2^+ as when H+ will leave then NO2- will be left as conjucate base
and not NO2+
Solution
b) HNO2,NO2^+ as when H+ will leave then NO2- will be left as conjucate base
and not NO2+.
Number of moles of Calcium = 0.05Number of equivalents of calcium .pdf
Number of moles of Calcium = 0.05
Number of equivalents of calcium = Number of moles / Valency = 0.05 / 2 = 0.025
Normality = 0.025 / 1 = 0.025 Ea /L
Solution
Number of moles of Calcium = 0.05
Number of equivalents of calcium = Number of moles / Valency = 0.05 / 2 = 0.025
Normality = 0.025 / 1 = 0.025 Ea /L.
While doing capital investment, CFO analyze all the risk related to .pdf
While doing capital investment, CFO analyze all the risk related to project. There are several
methods that can be used by CFO for project evaluation, but the best method should consider the
project risks while calculating net present value of cashflow generated by the project.
If I were a CFO of an organization then I will consider risk adjusted discount rate because it
calculate NPV while considering the risk factors. There are many advantages of using these
method which are as follows:
1. It can be used for the calculation for different time period and one can use different discount
rate for risk factors for specific time period.
2. One can adjust the risk by incresing the time period.
3. While evaluating the risk for any project, it is the easiest one.
Risk adjusted discount rate method produces most accurate result at differing levels of project
risk and at the same time it is easiest to implement. One needs to calculate the risk factors before
using this method.
Solution
While doing capital investment, CFO analyze all the risk related to project. There are several
methods that can be used by CFO for project evaluation, but the best method should consider the
project risks while calculating net present value of cashflow generated by the project.
If I were a CFO of an organization then I will consider risk adjusted discount rate because it
calculate NPV while considering the risk factors. There are many advantages of using these
method which are as follows:
1. It can be used for the calculation for different time period and one can use different discount
rate for risk factors for specific time period.
2. One can adjust the risk by incresing the time period.
3. While evaluating the risk for any project, it is the easiest one.
Risk adjusted discount rate method produces most accurate result at differing levels of project
risk and at the same time it is easiest to implement. One needs to calculate the risk factors before
using this method..
What is Network security, or Securiing your data Through Network,N.pdf
What is Network security, or Securiing your data Through Network,
Network security consists of the policies and practices adopted to prevent and monitor
unauthorized access, misuse, modification, or denial of a computer network and network-
accessible resources. Network security involves the authorization of access to data in a network,
which is controlled by the network administrator.
Standards for Network Security are:-
The Network Security Standard provides measures to prevent, detect, and correct network
compromises. The standard is based on both new practices and best practices currently in use at
RIT.
The Cyber Security Technical team at World Level Decides the Standards necessary for the
Network to be secured.
Regulations:-
Cybersecurity regulation comprises directives from the Executive Branch and legislation from
Congress that safeguards information technology and computer systems. The purpose of
cybersecurity regulation is to force companies and organizations to protect their systems and
information from cyber-attacks. Cyber-attacks include viruses, worms, Trojan horses, phishing,
denial of service (DOS) attacks, unauthorized access (stealing intellectual property or
confidential information) and control system attacks.
Network Compilations:-
The term network compliance is a broad one, and can have many meanings. The word
compliance means “the state or act or conforming with or agreeing to do something, often in
response to legislation, rules or regulations or court order.” Complying with all the rules, laws
and orders in effect in the U.S. and other jurisdictions can be a tall order. Within the industry, the
term is usually used to refer to one of the following:
Laws:-
There are many laws in the field of Cyber Security or we can say Network Security, and are
increasing day by day, some of them are,
Difference Between Laws and Regulations:-
Laws are the products of written statutes, passed by either the U.S. Congress or state legislatures.
The legislatures create bills that, when passed by a vote, become law.
Regulations, on the other hand, are standards and rules adopted by administrative agencies that
govern how laws will be enforced. So an agency like the SEC can have its own regulations for
enforcing major securities laws.
Thank You
Solution
What is Network security, or Securiing your data Through Network,
Network security consists of the policies and practices adopted to prevent and monitor
unauthorized access, misuse, modification, or denial of a computer network and network-
accessible resources. Network security involves the authorization of access to data in a network,
which is controlled by the network administrator.
Standards for Network Security are:-
The Network Security Standard provides measures to prevent, detect, and correct network
compromises. The standard is based on both new practices and best practices currently in use at
RIT.
The Cyber Security Technical team at World Level Decides the Standards necessary for .
TrueClay it the important of soil for holding plant nutrients in t.pdf
True
Clay it the important of soil for holding plant nutrients in the soil. Clay particles have a negative
surface charge.
Solution
True
Clay it the important of soil for holding plant nutrients in the soil. Clay particles have a negative
surface charge..
The statement is FalseOnly single target target agents are sunject.pdf
The statement is False
Only single target target agents are sunjected to hetergeneity and clonal evolution, while the
combination therapies are used to minimize the heterogeneity among the poplulation
Solution
The statement is False
Only single target target agents are sunjected to hetergeneity and clonal evolution, while the
combination therapies are used to minimize the heterogeneity among the poplulation.
The reactants along with the oxidation states are Zn = 0 CuSO4.pdf
The reactants along with the oxidation states are:
Zn = 0
CuSO4: Cu = +2, S = +6, O = -2
The products along with the oxidation states are:
Cu = 0
ZnSO4: Zn = +2, S= +6, O=-2
Zn went from 0 to +2 i.e. it lost electrons so it was Oxidized
Cu went from +2 to 0 i.e. it gained electrons so it was Reduced.
Solution
The reactants along with the oxidation states are:
Zn = 0
CuSO4: Cu = +2, S = +6, O = -2
The products along with the oxidation states are:
Cu = 0
ZnSO4: Zn = +2, S= +6, O=-2
Zn went from 0 to +2 i.e. it lost electrons so it was Oxidized
Cu went from +2 to 0 i.e. it gained electrons so it was Reduced..
The scanner performs lexical analysis of a precise program (in our c.pdf
The scanner performs lexical analysis of a precise program (in our case, the straightforward
program). It reads the computer program as a sequence of characters and acknowledges \"larger\"
matter units known as tokens. for instance, if the supply programs contains the characters
VAR ics142: INTEGER; // variable declaration
the scanner would manufacture the tokens
VAR ID(ics142) COLON ID(INTEGER) punctuation
to be processed in later phases of the compiler. Note that the scanner discards white house and
comments between the tokens, i.e. they\'re \"filtered\" and not passed on to later phases. samples
of nontokens ar tabs, line feeds, carriage returns, etc.
How to use FLEX?
FLEX (Fast LEXical instrument generator) may be a tool for generating scanners. In office of
writing a scanner from scratch, you merely ought to establish the vocabulary of a precise
language (e.g. Simple), write a specification of patterns mistreatment regular expressions (e.g.
DIGIT [0-9]), and FLEX can construct a scanner for you. FLEX is mostly utilized in the way
represented here:
First, FLEX reads a specification of a scanner either from associate degree computer file *.lex, or
from commonplace input, and it generates as output a C supply file lex.yy.c. Then, lex.yy.c is
compiled associate degreed connected with the \"-lfl\" library to provide an workable a.out.
Finally, a.out analyzes its input stream and transforms it into a sequence of tokens.
Solution
The scanner performs lexical analysis of a precise program (in our case, the straightforward
program). It reads the computer program as a sequence of characters and acknowledges \"larger\"
matter units known as tokens. for instance, if the supply programs contains the characters
VAR ics142: INTEGER; // variable declaration
the scanner would manufacture the tokens
VAR ID(ics142) COLON ID(INTEGER) punctuation
to be processed in later phases of the compiler. Note that the scanner discards white house and
comments between the tokens, i.e. they\'re \"filtered\" and not passed on to later phases. samples
of nontokens ar tabs, line feeds, carriage returns, etc.
How to use FLEX?
FLEX (Fast LEXical instrument generator) may be a tool for generating scanners. In office of
writing a scanner from scratch, you merely ought to establish the vocabulary of a precise
language (e.g. Simple), write a specification of patterns mistreatment regular expressions (e.g.
DIGIT [0-9]), and FLEX can construct a scanner for you. FLEX is mostly utilized in the way
represented here:
First, FLEX reads a specification of a scanner either from associate degree computer file *.lex, or
from commonplace input, and it generates as output a C supply file lex.yy.c. Then, lex.yy.c is
compiled associate degreed connected with the \"-lfl\" library to provide an workable a.out.
Finally, a.out analyzes its input stream and transforms it into a sequence of tokens..
Solution Pseudoextinction refers to- D) a species evolving to t.pdf
Solution
:
Pseudoextinction refers to-
D) a species evolving to the point where it is no longer recongnized as a parental species.
In this type of extinction ancestral species is no longer exists and a new subspecies
evolved.Pseudoextinction leads to evolution of species.Currently present species of birds and
some animals are the result of evolution through pseudoextinction and their ancestral species not
exist now..
Share Holders Equity = Current Assets + Net Fixed Assets - Current.pdf
Share Holder\'s Equity = Current Assets + Net Fixed Assets - Current Liabilities - Long term
Debt
= $ 2210 +$ 9700 - $ 1370 - $ 4020
= $ 6,520
Working Capital = Current Assets - Current Liabilities
= $ 2,210 - $ 1,370
= $ 840
Solution
Share Holder\'s Equity = Current Assets + Net Fixed Assets - Current Liabilities - Long term
Debt
= $ 2210 +$ 9700 - $ 1370 - $ 4020
= $ 6,520
Working Capital = Current Assets - Current Liabilities
= $ 2,210 - $ 1,370
= $ 840.
The correct answer for m_1=2 is 2,3 for lSolutionThe corre.pdf
The correct answer for m_1=2 is 2,3 for l
Solution
The correct answer for m_1=2 is 2,3 for l.
solimportance for an organization to implement an enterprise system.pdf
sol:importance for an organization to implement an enterprise system
1)we can easily see into main processes of various deparrtments in an organization
2)there is a keen tracking of inter department activities
3)a single report will be given which will be analyzed for all the departments
4)there is no needed installing differet software for different departments ,just install same
software
for all the departments
5)by using this erp we can integrate the ecommerce integration ,so by using this we can web
based tracking
6)erp is an structured design so there is no need of implementing the whole software ,as it is
modular software
we can implement required piece of software
7)erp system are specificly used for international companies or globally dispered enterprise
organizations
Solution
sol:importance for an organization to implement an enterprise system
1)we can easily see into main processes of various deparrtments in an organization
2)there is a keen tracking of inter department activities
3)a single report will be given which will be analyzed for all the departments
4)there is no needed installing differet software for different departments ,just install same
software
for all the departments
5)by using this erp we can integrate the ecommerce integration ,so by using this we can web
based tracking
6)erp is an structured design so there is no need of implementing the whole software ,as it is
modular software
we can implement required piece of software
7)erp system are specificly used for international companies or globally dispered enterprise
organizations.
We use the complementation principle. Clearly the.pdf
We use the complementation principle. Clearly the 20 people can be seated in 20!
ways without any restrictions. So, we now want to substract the possibility where the
youngest(Y) and the oldest(O) are seated less than 3 seats appart. case1 - No seat in between -
Now we can treat (OY) or (YO) as just a single person. Hence # such seatings is 19! *2 (for OY
or YO ordering). Case 2 - 1 Seat in between. Now we observe that this duo can be seated in 18
ways - ie X^iOXYX^17-i ie some no of people on the left - can range from 0 to 17 no more. So,
seat them in 18 ways, permute OY,YO in 2 ways(this will multiply) and then permute the 18
people in 18! ways - 2*18*18! Case 2 - 2 in between, similar analysis gives us that the left guy
from O,Y can be seated in 17 ways - X^i O XX Y X^{16-i}. So i ranges from 0 to 16 ie 17
ways. ie 2* 17 * 18! So total possibilities = 20! - 2*18!(19 + 18 + 17) is the answer. The
previous answer has ignored more than 3 chairs possibility. Message me if you have any
doubts
Solution
We use the complementation principle. Clearly the 20 people can be seated in 20!
ways without any restrictions. So, we now want to substract the possibility where the
youngest(Y) and the oldest(O) are seated less than 3 seats appart. case1 - No seat in between -
Now we can treat (OY) or (YO) as just a single person. Hence # such seatings is 19! *2 (for OY
or YO ordering). Case 2 - 1 Seat in between. Now we observe that this duo can be seated in 18
ways - ie X^iOXYX^17-i ie some no of people on the left - can range from 0 to 17 no more. So,
seat them in 18 ways, permute OY,YO in 2 ways(this will multiply) and then permute the 18
people in 18! ways - 2*18*18! Case 2 - 2 in between, similar analysis gives us that the left guy
from O,Y can be seated in 17 ways - X^i O XX Y X^{16-i}. So i ranges from 0 to 16 ie 17
ways. ie 2* 17 * 18! So total possibilities = 20! - 2*18!(19 + 18 + 17) is the answer. The
previous answer has ignored more than 3 chairs possibility. Message me if you have any
doubts.
right idea but numbers are wrong all over the placeSolution.pdf
right idea but numbers are wrong all over the place
Solution
right idea but numbers are wrong all over the place.
Please follow the data and description EBNF In the stream of c.pdf
Please follow the data and description :
EBNF :
In the stream of computer science, Extended Backus–Naur Form (EBNF) is a family of
metasyntax notations, any of which can be used to express a context-free grammar. They are said
to be the extensions of the basic Backus–Naur Form (BNF) metasyntax notation. The earliest
EBNF was originally developed by Niklaus Wirth incorporating some of the concepts from
Wirth syntax notation. However, many variants of EBNF are in use. The International
Organization for Standardization has adopted an EBNF standard (ISO/IEC 14977).
BNF :
BNF is the abbrevation of Backus Normal Form or Backus–Naur Form which is one of the two
main notation techniques for context-free grammars, often used to describe the syntax of
languages used in computing, such as computer programming languages, document formats,
instruction sets and communication protocols. They are applied wherever exact descriptions of
languages are needed, for instance, in official language specifications, in manuals, and in
textbooks on programming language theory.
EBNF To BNF :
Given,
-> {b}
-> a[b]
a) -> {b} =
Here S is a sequence of A that is optionally followed by a sequence of b and S together.
->
| b ;
b) -> a[b] =
Here A is composed of \'a\', that is then followed by an optional \'b\', followed by another A
combinedly.
-> a
| a b ;
Hope this is helpful.
Solution
Please follow the data and description :
EBNF :
In the stream of computer science, Extended Backus–Naur Form (EBNF) is a family of
metasyntax notations, any of which can be used to express a context-free grammar. They are said
to be the extensions of the basic Backus–Naur Form (BNF) metasyntax notation. The earliest
EBNF was originally developed by Niklaus Wirth incorporating some of the concepts from
Wirth syntax notation. However, many variants of EBNF are in use. The International
Organization for Standardization has adopted an EBNF standard (ISO/IEC 14977).
BNF :
BNF is the abbrevation of Backus Normal Form or Backus–Naur Form which is one of the two
main notation techniques for context-free grammars, often used to describe the syntax of
languages used in computing, such as computer programming languages, document formats,
instruction sets and communication protocols. They are applied wherever exact descriptions of
languages are needed, for instance, in official language specifications, in manuals, and in
textbooks on programming language theory.
EBNF To BNF :
Given,
-> {b}
-> a[b]
a) -> {b} =
Here S is a sequence of A that is optionally followed by a sequence of b and S together.
->
| b ;
b) -> a[b] =
Here A is composed of \'a\', that is then followed by an optional \'b\', followed by another A
combinedly.
-> a
| a b ;
Hope this is helpful..
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
The sample proportion falls between 0.4 and 0.6Solution The s.pdf
The sample proportion falls between 0.4 and 0.6
Solution
The sample proportion falls between 0.4 and 0.6.
namespace MonthNamesGUI{public partial class Form1 Form{en.pdf
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
int monthNumber;
int.TryParse(inputTextBox.Text, out monthNumber);
if (monthNumber < 1 || monthNumber > 12)
{
MessageBox.Show(\"Please enter an integer between 1 and 12\");
inputTextBox.Focus();
return;
}
string monthName = ((Month)monthNumber).ToString();
MessageBox.Show(\"You entered the month number for \" + monthName);
inputTextBox.Clear();
inputTextBox.Focus();
}
}
}
Solution
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
int monthNumber;
int.TryParse(inputTextBox.Text, out monthNumber);
if (monthNumber < 1 || monthNumber > 12)
{
MessageBox.Show(\"Please enter an integer between 1 and 12\");
inputTextBox.Focus();
return;
}
string monthName = ((Month)monthNumber).ToString();
MessageBox.Show(\"You entered the month number for \" + monthName);
inputTextBox.Clear();
inputTextBox.Focus();
}
}
}.
+0.35, as the scatter is not showing any significant relationship to.pdf
+0.35, as the scatter is not showing any significant relationship to generate higher values of other
options
Solution
+0.35, as the scatter is not showing any significant relationship to generate higher values of other
options.
Here ABE is one planeBCD is another perpendicular plane from the g.pdf
Here ABE is one plane
BCD is another perpendicular plane from the given details.
AE lies on ABE plane and CE lies on BCD plane.
Hence , perpendiculat to each other.
Solution
Here ABE is one plane
BCD is another perpendicular plane from the given details.
AE lies on ABE plane and CE lies on BCD plane.
Hence , perpendiculat to each other..
For governments that implement GASB Statement No. 34, the fund finan.pdf
For governments that implement GASB Statement No. 34, the fund financial statements will
accompany Management\'s Discussion and Analysis and the government-wide statements. The
fund financial statements are used to report additional detailed information about the primary
government not found in the government-wide statements and focus on major funds of the
primary government.
The GWFS are reported using the economic resources measurement focus and the full accrual
basis of accounting. GWFS report information about the government as a whole without
displaying individual funds or fund types. Distinctions should be made in the GWFS between the
primary government and its discretely presented component units, as well as between
governmental activities and business-type activities. Fiduciary activities are not included in the
GWFS.
GWFS include:
The Government-wide Financial Statements and Fund Financial Statements have a different
measurement focus and basis of accounting. To help the user understand and assess the
relationship between fund-based and government-wide financial statements, local governments
will be required to present a summary reconciliationthat will show the interplay between the two
types of statements.
Balance sheet - The basic statement of position for governmental funds reports the difference
between assets and liabilities as fund balance. Fund balance is divided into reserved and
unreserved components. This is the same statement that is contained in the AUD and ST3.
Statement of revenues, expenditures and changes in fund balance-The basic statement of
activities for governmental funds reports revenues by source and expenditures by function just as
they are currently reported. Since long-term liabilities and capital assets have been incorporated
into the government-wide statements the accounts groups will no longer be reported in the
statements. A revised supplemental schedule will be included in the AUD and ST3 to continue to
accumulate this information. Statement of net assets- The basic statement of position for
proprietary funds reports the difference between assets and liabilities as net assets. Net assets are
divided into invested in capital assets, restricted and unrestricted.Statement of revenues,
expenses and changes in fund net assets- The basic statement of activities for proprietary funds
distinguishes operating and non-operating revenues and expenses.Statement of cash flows - This
statement will now be required to be prepared on the direct method where operating cash flows
will be reported by source. This is not a change for local governments since OSC has already
required this method.
Solution
For governments that implement GASB Statement No. 34, the fund financial statements will
accompany Management\'s Discussion and Analysis and the government-wide statements. The
fund financial statements are used to report additional detailed information about the primary
government not found in the government-wide statement.
Ethical theories are often divided into three types -1) Consequent.pdf
Ethical theories are often divided into three types -
1) Consequentiolist Theories -
2) Non - Consequentialist Approach -
3) Agent - Centered Theories -
Solution
Ethical theories are often divided into three types -
1) Consequentiolist Theories -
2) Non - Consequentialist Approach -
3) Agent - Centered Theories -.
More Related Content
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While doing capital investment, CFO analyze all the risk related to .pdf
While doing capital investment, CFO analyze all the risk related to project. There are several
methods that can be used by CFO for project evaluation, but the best method should consider the
project risks while calculating net present value of cashflow generated by the project.
If I were a CFO of an organization then I will consider risk adjusted discount rate because it
calculate NPV while considering the risk factors. There are many advantages of using these
method which are as follows:
1. It can be used for the calculation for different time period and one can use different discount
rate for risk factors for specific time period.
2. One can adjust the risk by incresing the time period.
3. While evaluating the risk for any project, it is the easiest one.
Risk adjusted discount rate method produces most accurate result at differing levels of project
risk and at the same time it is easiest to implement. One needs to calculate the risk factors before
using this method.
Solution
While doing capital investment, CFO analyze all the risk related to project. There are several
methods that can be used by CFO for project evaluation, but the best method should consider the
project risks while calculating net present value of cashflow generated by the project.
If I were a CFO of an organization then I will consider risk adjusted discount rate because it
calculate NPV while considering the risk factors. There are many advantages of using these
method which are as follows:
1. It can be used for the calculation for different time period and one can use different discount
rate for risk factors for specific time period.
2. One can adjust the risk by incresing the time period.
3. While evaluating the risk for any project, it is the easiest one.
Risk adjusted discount rate method produces most accurate result at differing levels of project
risk and at the same time it is easiest to implement. One needs to calculate the risk factors before
using this method..
What is Network security, or Securiing your data Through Network,N.pdf
What is Network security, or Securiing your data Through Network,
Network security consists of the policies and practices adopted to prevent and monitor
unauthorized access, misuse, modification, or denial of a computer network and network-
accessible resources. Network security involves the authorization of access to data in a network,
which is controlled by the network administrator.
Standards for Network Security are:-
The Network Security Standard provides measures to prevent, detect, and correct network
compromises. The standard is based on both new practices and best practices currently in use at
RIT.
The Cyber Security Technical team at World Level Decides the Standards necessary for the
Network to be secured.
Regulations:-
Cybersecurity regulation comprises directives from the Executive Branch and legislation from
Congress that safeguards information technology and computer systems. The purpose of
cybersecurity regulation is to force companies and organizations to protect their systems and
information from cyber-attacks. Cyber-attacks include viruses, worms, Trojan horses, phishing,
denial of service (DOS) attacks, unauthorized access (stealing intellectual property or
confidential information) and control system attacks.
Network Compilations:-
The term network compliance is a broad one, and can have many meanings. The word
compliance means “the state or act or conforming with or agreeing to do something, often in
response to legislation, rules or regulations or court order.” Complying with all the rules, laws
and orders in effect in the U.S. and other jurisdictions can be a tall order. Within the industry, the
term is usually used to refer to one of the following:
Laws:-
There are many laws in the field of Cyber Security or we can say Network Security, and are
increasing day by day, some of them are,
Difference Between Laws and Regulations:-
Laws are the products of written statutes, passed by either the U.S. Congress or state legislatures.
The legislatures create bills that, when passed by a vote, become law.
Regulations, on the other hand, are standards and rules adopted by administrative agencies that
govern how laws will be enforced. So an agency like the SEC can have its own regulations for
enforcing major securities laws.
Thank You
Solution
What is Network security, or Securiing your data Through Network,
Network security consists of the policies and practices adopted to prevent and monitor
unauthorized access, misuse, modification, or denial of a computer network and network-
accessible resources. Network security involves the authorization of access to data in a network,
which is controlled by the network administrator.
Standards for Network Security are:-
The Network Security Standard provides measures to prevent, detect, and correct network
compromises. The standard is based on both new practices and best practices currently in use at
RIT.
The Cyber Security Technical team at World Level Decides the Standards necessary for .
TrueClay it the important of soil for holding plant nutrients in t.pdf
True
Clay it the important of soil for holding plant nutrients in the soil. Clay particles have a negative
surface charge.
Solution
True
Clay it the important of soil for holding plant nutrients in the soil. Clay particles have a negative
surface charge..
The statement is FalseOnly single target target agents are sunject.pdf
The statement is False
Only single target target agents are sunjected to hetergeneity and clonal evolution, while the
combination therapies are used to minimize the heterogeneity among the poplulation
Solution
The statement is False
Only single target target agents are sunjected to hetergeneity and clonal evolution, while the
combination therapies are used to minimize the heterogeneity among the poplulation.
The reactants along with the oxidation states are Zn = 0 CuSO4.pdf
The reactants along with the oxidation states are:
Zn = 0
CuSO4: Cu = +2, S = +6, O = -2
The products along with the oxidation states are:
Cu = 0
ZnSO4: Zn = +2, S= +6, O=-2
Zn went from 0 to +2 i.e. it lost electrons so it was Oxidized
Cu went from +2 to 0 i.e. it gained electrons so it was Reduced.
Solution
The reactants along with the oxidation states are:
Zn = 0
CuSO4: Cu = +2, S = +6, O = -2
The products along with the oxidation states are:
Cu = 0
ZnSO4: Zn = +2, S= +6, O=-2
Zn went from 0 to +2 i.e. it lost electrons so it was Oxidized
Cu went from +2 to 0 i.e. it gained electrons so it was Reduced..
The scanner performs lexical analysis of a precise program (in our c.pdf
The scanner performs lexical analysis of a precise program (in our case, the straightforward
program). It reads the computer program as a sequence of characters and acknowledges \"larger\"
matter units known as tokens. for instance, if the supply programs contains the characters
VAR ics142: INTEGER; // variable declaration
the scanner would manufacture the tokens
VAR ID(ics142) COLON ID(INTEGER) punctuation
to be processed in later phases of the compiler. Note that the scanner discards white house and
comments between the tokens, i.e. they\'re \"filtered\" and not passed on to later phases. samples
of nontokens ar tabs, line feeds, carriage returns, etc.
How to use FLEX?
FLEX (Fast LEXical instrument generator) may be a tool for generating scanners. In office of
writing a scanner from scratch, you merely ought to establish the vocabulary of a precise
language (e.g. Simple), write a specification of patterns mistreatment regular expressions (e.g.
DIGIT [0-9]), and FLEX can construct a scanner for you. FLEX is mostly utilized in the way
represented here:
First, FLEX reads a specification of a scanner either from associate degree computer file *.lex, or
from commonplace input, and it generates as output a C supply file lex.yy.c. Then, lex.yy.c is
compiled associate degreed connected with the \"-lfl\" library to provide an workable a.out.
Finally, a.out analyzes its input stream and transforms it into a sequence of tokens.
Solution
The scanner performs lexical analysis of a precise program (in our case, the straightforward
program). It reads the computer program as a sequence of characters and acknowledges \"larger\"
matter units known as tokens. for instance, if the supply programs contains the characters
VAR ics142: INTEGER; // variable declaration
the scanner would manufacture the tokens
VAR ID(ics142) COLON ID(INTEGER) punctuation
to be processed in later phases of the compiler. Note that the scanner discards white house and
comments between the tokens, i.e. they\'re \"filtered\" and not passed on to later phases. samples
of nontokens ar tabs, line feeds, carriage returns, etc.
How to use FLEX?
FLEX (Fast LEXical instrument generator) may be a tool for generating scanners. In office of
writing a scanner from scratch, you merely ought to establish the vocabulary of a precise
language (e.g. Simple), write a specification of patterns mistreatment regular expressions (e.g.
DIGIT [0-9]), and FLEX can construct a scanner for you. FLEX is mostly utilized in the way
represented here:
First, FLEX reads a specification of a scanner either from associate degree computer file *.lex, or
from commonplace input, and it generates as output a C supply file lex.yy.c. Then, lex.yy.c is
compiled associate degreed connected with the \"-lfl\" library to provide an workable a.out.
Finally, a.out analyzes its input stream and transforms it into a sequence of tokens..
Solution Pseudoextinction refers to- D) a species evolving to t.pdf
Solution
:
Pseudoextinction refers to-
D) a species evolving to the point where it is no longer recongnized as a parental species.
In this type of extinction ancestral species is no longer exists and a new subspecies
evolved.Pseudoextinction leads to evolution of species.Currently present species of birds and
some animals are the result of evolution through pseudoextinction and their ancestral species not
exist now..
Share Holders Equity = Current Assets + Net Fixed Assets - Current.pdf
Share Holder\'s Equity = Current Assets + Net Fixed Assets - Current Liabilities - Long term
Debt
= $ 2210 +$ 9700 - $ 1370 - $ 4020
= $ 6,520
Working Capital = Current Assets - Current Liabilities
= $ 2,210 - $ 1,370
= $ 840
Solution
Share Holder\'s Equity = Current Assets + Net Fixed Assets - Current Liabilities - Long term
Debt
= $ 2210 +$ 9700 - $ 1370 - $ 4020
= $ 6,520
Working Capital = Current Assets - Current Liabilities
= $ 2,210 - $ 1,370
= $ 840.
The correct answer for m_1=2 is 2,3 for lSolutionThe corre.pdf
The correct answer for m_1=2 is 2,3 for l
Solution
The correct answer for m_1=2 is 2,3 for l.
solimportance for an organization to implement an enterprise system.pdf
sol:importance for an organization to implement an enterprise system
1)we can easily see into main processes of various deparrtments in an organization
2)there is a keen tracking of inter department activities
3)a single report will be given which will be analyzed for all the departments
4)there is no needed installing differet software for different departments ,just install same
software
for all the departments
5)by using this erp we can integrate the ecommerce integration ,so by using this we can web
based tracking
6)erp is an structured design so there is no need of implementing the whole software ,as it is
modular software
we can implement required piece of software
7)erp system are specificly used for international companies or globally dispered enterprise
organizations
Solution
sol:importance for an organization to implement an enterprise system
1)we can easily see into main processes of various deparrtments in an organization
2)there is a keen tracking of inter department activities
3)a single report will be given which will be analyzed for all the departments
4)there is no needed installing differet software for different departments ,just install same
software
for all the departments
5)by using this erp we can integrate the ecommerce integration ,so by using this we can web
based tracking
6)erp is an structured design so there is no need of implementing the whole software ,as it is
modular software
we can implement required piece of software
7)erp system are specificly used for international companies or globally dispered enterprise
organizations.
We use the complementation principle. Clearly the.pdf
We use the complementation principle. Clearly the 20 people can be seated in 20!
ways without any restrictions. So, we now want to substract the possibility where the
youngest(Y) and the oldest(O) are seated less than 3 seats appart. case1 - No seat in between -
Now we can treat (OY) or (YO) as just a single person. Hence # such seatings is 19! *2 (for OY
or YO ordering). Case 2 - 1 Seat in between. Now we observe that this duo can be seated in 18
ways - ie X^iOXYX^17-i ie some no of people on the left - can range from 0 to 17 no more. So,
seat them in 18 ways, permute OY,YO in 2 ways(this will multiply) and then permute the 18
people in 18! ways - 2*18*18! Case 2 - 2 in between, similar analysis gives us that the left guy
from O,Y can be seated in 17 ways - X^i O XX Y X^{16-i}. So i ranges from 0 to 16 ie 17
ways. ie 2* 17 * 18! So total possibilities = 20! - 2*18!(19 + 18 + 17) is the answer. The
previous answer has ignored more than 3 chairs possibility. Message me if you have any
doubts
Solution
We use the complementation principle. Clearly the 20 people can be seated in 20!
ways without any restrictions. So, we now want to substract the possibility where the
youngest(Y) and the oldest(O) are seated less than 3 seats appart. case1 - No seat in between -
Now we can treat (OY) or (YO) as just a single person. Hence # such seatings is 19! *2 (for OY
or YO ordering). Case 2 - 1 Seat in between. Now we observe that this duo can be seated in 18
ways - ie X^iOXYX^17-i ie some no of people on the left - can range from 0 to 17 no more. So,
seat them in 18 ways, permute OY,YO in 2 ways(this will multiply) and then permute the 18
people in 18! ways - 2*18*18! Case 2 - 2 in between, similar analysis gives us that the left guy
from O,Y can be seated in 17 ways - X^i O XX Y X^{16-i}. So i ranges from 0 to 16 ie 17
ways. ie 2* 17 * 18! So total possibilities = 20! - 2*18!(19 + 18 + 17) is the answer. The
previous answer has ignored more than 3 chairs possibility. Message me if you have any
doubts.
right idea but numbers are wrong all over the placeSolution.pdf
right idea but numbers are wrong all over the place
Solution
right idea but numbers are wrong all over the place.
Please follow the data and description EBNF In the stream of c.pdf
Please follow the data and description :
EBNF :
In the stream of computer science, Extended Backus–Naur Form (EBNF) is a family of
metasyntax notations, any of which can be used to express a context-free grammar. They are said
to be the extensions of the basic Backus–Naur Form (BNF) metasyntax notation. The earliest
EBNF was originally developed by Niklaus Wirth incorporating some of the concepts from
Wirth syntax notation. However, many variants of EBNF are in use. The International
Organization for Standardization has adopted an EBNF standard (ISO/IEC 14977).
BNF :
BNF is the abbrevation of Backus Normal Form or Backus–Naur Form which is one of the two
main notation techniques for context-free grammars, often used to describe the syntax of
languages used in computing, such as computer programming languages, document formats,
instruction sets and communication protocols. They are applied wherever exact descriptions of
languages are needed, for instance, in official language specifications, in manuals, and in
textbooks on programming language theory.
EBNF To BNF :
Given,
-> {b}
-> a[b]
a) -> {b} =
Here S is a sequence of A that is optionally followed by a sequence of b and S together.
->
| b ;
b) -> a[b] =
Here A is composed of \'a\', that is then followed by an optional \'b\', followed by another A
combinedly.
-> a
| a b ;
Hope this is helpful.
Solution
Please follow the data and description :
EBNF :
In the stream of computer science, Extended Backus–Naur Form (EBNF) is a family of
metasyntax notations, any of which can be used to express a context-free grammar. They are said
to be the extensions of the basic Backus–Naur Form (BNF) metasyntax notation. The earliest
EBNF was originally developed by Niklaus Wirth incorporating some of the concepts from
Wirth syntax notation. However, many variants of EBNF are in use. The International
Organization for Standardization has adopted an EBNF standard (ISO/IEC 14977).
BNF :
BNF is the abbrevation of Backus Normal Form or Backus–Naur Form which is one of the two
main notation techniques for context-free grammars, often used to describe the syntax of
languages used in computing, such as computer programming languages, document formats,
instruction sets and communication protocols. They are applied wherever exact descriptions of
languages are needed, for instance, in official language specifications, in manuals, and in
textbooks on programming language theory.
EBNF To BNF :
Given,
-> {b}
-> a[b]
a) -> {b} =
Here S is a sequence of A that is optionally followed by a sequence of b and S together.
->
| b ;
b) -> a[b] =
Here A is composed of \'a\', that is then followed by an optional \'b\', followed by another A
combinedly.
-> a
| a b ;
Hope this is helpful..
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
The sample proportion falls between 0.4 and 0.6Solution The s.pdf
The sample proportion falls between 0.4 and 0.6
Solution
The sample proportion falls between 0.4 and 0.6.
namespace MonthNamesGUI{public partial class Form1 Form{en.pdf
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
int monthNumber;
int.TryParse(inputTextBox.Text, out monthNumber);
if (monthNumber < 1 || monthNumber > 12)
{
MessageBox.Show(\"Please enter an integer between 1 and 12\");
inputTextBox.Focus();
return;
}
string monthName = ((Month)monthNumber).ToString();
MessageBox.Show(\"You entered the month number for \" + monthName);
inputTextBox.Clear();
inputTextBox.Focus();
}
}
}
Solution
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
namespace MonthNamesGUI
{
public partial class Form1 : Form
{
enum Month
{
January = 1,
February,
March,
April,
May,
June,
July,
August,
September,
October,
November,
December
};
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
int monthNumber;
int.TryParse(inputTextBox.Text, out monthNumber);
if (monthNumber < 1 || monthNumber > 12)
{
MessageBox.Show(\"Please enter an integer between 1 and 12\");
inputTextBox.Focus();
return;
}
string monthName = ((Month)monthNumber).ToString();
MessageBox.Show(\"You entered the month number for \" + monthName);
inputTextBox.Clear();
inputTextBox.Focus();
}
}
}.
+0.35, as the scatter is not showing any significant relationship to.pdf
+0.35, as the scatter is not showing any significant relationship to generate higher values of other
options
Solution
+0.35, as the scatter is not showing any significant relationship to generate higher values of other
options.
Here ABE is one planeBCD is another perpendicular plane from the g.pdf
Here ABE is one plane
BCD is another perpendicular plane from the given details.
AE lies on ABE plane and CE lies on BCD plane.
Hence , perpendiculat to each other.
Solution
Here ABE is one plane
BCD is another perpendicular plane from the given details.
AE lies on ABE plane and CE lies on BCD plane.
Hence , perpendiculat to each other..
For governments that implement GASB Statement No. 34, the fund finan.pdf
For governments that implement GASB Statement No. 34, the fund financial statements will
accompany Management\'s Discussion and Analysis and the government-wide statements. The
fund financial statements are used to report additional detailed information about the primary
government not found in the government-wide statements and focus on major funds of the
primary government.
The GWFS are reported using the economic resources measurement focus and the full accrual
basis of accounting. GWFS report information about the government as a whole without
displaying individual funds or fund types. Distinctions should be made in the GWFS between the
primary government and its discretely presented component units, as well as between
governmental activities and business-type activities. Fiduciary activities are not included in the
GWFS.
GWFS include:
The Government-wide Financial Statements and Fund Financial Statements have a different
measurement focus and basis of accounting. To help the user understand and assess the
relationship between fund-based and government-wide financial statements, local governments
will be required to present a summary reconciliationthat will show the interplay between the two
types of statements.
Balance sheet - The basic statement of position for governmental funds reports the difference
between assets and liabilities as fund balance. Fund balance is divided into reserved and
unreserved components. This is the same statement that is contained in the AUD and ST3.
Statement of revenues, expenditures and changes in fund balance-The basic statement of
activities for governmental funds reports revenues by source and expenditures by function just as
they are currently reported. Since long-term liabilities and capital assets have been incorporated
into the government-wide statements the accounts groups will no longer be reported in the
statements. A revised supplemental schedule will be included in the AUD and ST3 to continue to
accumulate this information. Statement of net assets- The basic statement of position for
proprietary funds reports the difference between assets and liabilities as net assets. Net assets are
divided into invested in capital assets, restricted and unrestricted.Statement of revenues,
expenses and changes in fund net assets- The basic statement of activities for proprietary funds
distinguishes operating and non-operating revenues and expenses.Statement of cash flows - This
statement will now be required to be prepared on the direct method where operating cash flows
will be reported by source. This is not a change for local governments since OSC has already
required this method.
Solution
For governments that implement GASB Statement No. 34, the fund financial statements will
accompany Management\'s Discussion and Analysis and the government-wide statements. The
fund financial statements are used to report additional detailed information about the primary
government not found in the government-wide statement.
Ethical theories are often divided into three types -1) Consequent.pdf
Ethical theories are often divided into three types -
1) Consequentiolist Theories -
2) Non - Consequentialist Approach -
3) Agent - Centered Theories -
Solution
Ethical theories are often divided into three types -
1) Consequentiolist Theories -
2) Non - Consequentialist Approach -
3) Agent - Centered Theories -.
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