Present an overview of the gene organization in large-genome plants..pdf
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JAVA Information Security: Principles ad Practices. Write a program to help an analyst decrypt a simple substitution cipher. Your program should take the ciphertext as input, compute letter frequency counts, and display these for the analyst. The program should then allow the analyst to guess a key and display the results of the corresponding \"decryption\" with the putative key. BELOW IS WHAT I GOT SO FAR. ANALYST IS NOT GUESSING A KEY OR PUTATIVE KEY IS NOT COMING OUT. IF YOU HAVE BETTER PROGRAM, PLEASE IGNORE THE SOURCE BELOW. THANKS. import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.HashMap; import java.util.LinkedHashMap; import java.util.Map; publicclass SubstitutionCipher { Map letterFrequencyMap; Map key; String ciphertext; String plaintext; BufferedReader br; privatevoid getCiphertext() // get cipher text as input from analyst { System.out.println(\"----- Substitution Cipher ----\"); System.out.println(\" Enter the cipher text: \"); br = new BufferedReader(new InputStreamReader(System.in)); try { ciphertext = br.readLine(); } catch(Exception e) { System.err.println(\"Exception in cipher text reading \" + e ); e.printStackTrace(); } ciphertext = ciphertext.toUpperCase(); System.out.println(\"The ciphertext Entered is - \" + ciphertext); } privatevoid displayLetterFrequency() { letterFrequencyMap = new LinkedHashMap(); for (int i = 0; i < ciphertext.length(); i++) { Character ch = ciphertext.charAt(i); if (!( ch >= \'A\' && ch <=\'Z\')) continue; // check if letter is not in between A to Z Integer cnt = (Integer) letterFrequencyMap.get(ch); if(cnt == null) { letterFrequencyMap.put(ch,1); } else { letterFrequencyMap.put(ch,cnt + 1); } } System.out.println(letterFrequencyMap); } privatevoid keyGuess() { System.out.println(\"Hey Analyst , Guess a key from a to z\"); String putativeKey = null; try { putativeKey = br.readLine(); } catch(Exception e) { System.out.println(\"Excpetion in reading the putative key\"); e.printStackTrace(); } createKeyMap(putativeKey); } privatevoid createKeyMap(String putativeKey) { if(putativeKey.length() < 26) { System.out.println(\"putative key length is less then 26 letters. Invalid Entry. \"); return; } key = new HashMap(); putativeKey = putativeKey.toUpperCase(); for(int j = 0;j Solution import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.HashMap; import java.util.LinkedHashMap; import java.util.Map; class SubstitutionCipher { Map letterFrequencyMap; Map key; String ciphertext; String plaintext; BufferedReader br; private void getCiphertext() // get cipher text as input from analyst { System.out.println(\"----- Substitution Cipher ----\"); System.out.println(\" Enter the cipher text: \"); br = new BufferedReader(new InputStreamReader(System.in)); try { ciphertext = br.readLine(); } catch(Exception e) { System.err.println(\"Exception in cipher text reading \" + e ); e.printStackTrace(); } ciphertext = ciphertext.toUpperCase(); System.out.println(\.
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Execute the following code and identify the errors in the program. D.pdf
complete the proof Propost ion: If a positive intege r m is not the square of any integer, then Vm
is irrational.
Solution
Let nbe a positive integer such that there is no msuch that n=m2. Suppose nis rational. Then
there exists pand qwith no common factor (beside 1) such that
n=pq
Then
n=p2q2
However, nn is an positive integer and p and q have no common factors beside 1. So q=1.
This gives that
n=p2
Contradiction since it was assumed that nm2 for any m
c.
c
7down voteaccepted
Let nbe a positive integer such that there is no msuch that n=m2. Suppose nis rational. Then
there exists pand qwith no common factor (beside 1) such that
n=pq
Then
n=p2q2
However, nn is an positive integer and p and q have no common factors beside 1. So
q=1. This gives that
n=p2
Contradiction since it was assumed that nm2 for any m
c.
c.
Essay question The genomes of lots and lots of organisms (mostly ba.pdf
Complete the identity: cot(t) (tan (t) - sin (t) cos (t)) =? sin^2 (t) sin (t) tan (t) sec (t) csc (t) tan
(t) + 1
Solution
cot(t)(tan(t) -sin(t)cos(t))
=(cos(t)/sin(t))((sin(t)/cos(t)) -sin(t)cos(t))
=((cos(t)/sin(t))(sin(t)/cos(t)) - (cos(t)/sin(t))sin(t)cos(t))
=1- cos2(t)..................................................................................[sin2A+cos2A =1 is an identity ]
=sin2(t)
therefore cot(t)(tan(t) -sin(t)cos(t)) =sin2(t).
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ALSO CALLED JUNK DNA.
Non-coding DNA corresponds to the portions of an organism's genome that do not code for amino acids, the building blocks of proteins. Some non-coding DNA sequences are known to serve functional roles, such as in the regulation of gene expression, while other areas of non-coding DNA have no known function.Non-coding DNA is a type of DNA that does not code for proteins, but it could have functions related to the regulation of gene expression. The percentage of non-coding DNA content varies between species, but it's approximately 98% in humans.
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Non-coding DNA can also code for RNA molecules such as microRNA (which is used to blocked translation), ribosomal RNA (which are used to build ribosomes), and transfer RNA (which is used in the process of translation. DNA that has previously been shown to have no biological function are still deemed important due to their high degree of homology and their conservation during evolution. Furthermore, some of the functions of non-coding DNA are yet to be fully understood.
Non-Coding Regions of DNA
Non-coding regions of DNA are defined as regions that do not directly contribute to protein synthesis. Non-coding regions can fall into the form of introns within genes, satellite DNA, telomeres, and regulatory sequences such as promoters.
Non-coding DNA is a type of DNA that does not code for proteins, but it could have functions related to the regulation of gene expression. The percentage of non-coding DNA content varies between species, but it's approximately 98% in humans.
A significant portion of DNA was though to have no important biological function, and thus it was given the nickname "junk" DNA. A research study named "ENCODE," however, has revealed that around 80% of human junk DNA does seem to have relevant biochemical functions, due to its access to proteins such as transcription factors.
Non-coding DNA can also code for RNA molecules such as microRNA (which is used to blocked translation), ribosomal RNA (which are used to build ribosomes), and transfer RNA (which is used in the process of translation. DNA that has previously been shown to have no biological function are still deemed important due to their high degree of homology and their conservation during evolution. Furthermore, some of the functions of non-coding DNA are yet to be fully understood.
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The document discusses genome structure and genomics. It defines key terms like genome, repetitive DNA, and defines the major classes of repetitive DNA, which constitute around 45% of the human genome. It discusses how DNA denaturation and renaturation kinetics can be used to determine genome complexity and GC content. Cot analysis allows characterization of sequence complexity based on repetitiveness. Eukaryotic genomes have lower gene density and finding genes is more challenging compared to prokaryotes due to introns and other factors.
GENOME_STRUCTURE1.ppt
The document discusses various components that make up genomes, including genes, repetitive sequences, and different types of DNA. It describes the human genome in particular, noting it contains around 3 billion base pairs, with 3% coding for proteins. Around 40-50% is repetitive sequences from transposition. Genomics is defined as the study of genomes, including gene mapping and sequencing. Key components of genomes discussed include transposable elements like SINEs, LINEs, LTR retrotransposons, and other interspersed repeats. Comparative analysis of genome sequences can provide insights into gene number and function.
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1. Non-coding DNA makes up the majority of DNA in eukaryotic organisms but does not encode for proteins. It was once referred to as "junk DNA" but is now known to have important functions.
2. There are several types of non-coding DNA, including regulatory elements, introns, pseudogenes, repeats, transposons, telomeres, and various non-coding RNAs.
3. While the amount of non-coding DNA varies between species, it plays roles in chromosome structure, gene expression regulation, and protecting the genome from degradation.
Lecture 1.ppt
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- Prokaryotic genomes are usually circular and compact, with operons of related genes. Prokaryotic DNA is packaged via supercoiling without chromatin.
- Mitochondrial and chloroplast genomes are usually circular as well, encoding tRNAs, rRNAs and proteins. They resemble bacterial genomes but are much smaller due to gene loss.
- Organelle genomes are usually maternally inherited and range in size from 6kb to 2MB, encoding genes for respiration and photosynthesis respectively.
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3) The structure and function of mitochondrial DNA, including that it is maternally inherited and encodes 37 essential genes in humans.
4) The types and functions of RNA, including ribosomal RNA, messenger RNA, and
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The document discusses the history and techniques of forensic DNA analysis. It describes how early methods like RFLP used repeated DNA regions called VNTRs to differentiate individuals. More recent STR techniques use short tandem repeats that allow analysis of low-quantity and degraded DNA samples. The document provides an overview of DNA structure and function, describing the double helix structure and how DNA is organized into chromosomes, genes, and loci. It also summarizes sample collection sources for DNA analysis and the basic workflow from extraction to profiling and comparison.
Gene silencing and editing in plant PPT
Gene silencing and editing techniques allow for precise regulation of gene expression. RNA interference uses small RNA molecules like siRNA and miRNA to degrade target mRNA, decreasing protein production. Genome editing tools like CRISPR/Cas9 create targeted DNA breaks that are repaired, resulting in mutations like insertions or deletions that can knock out gene function. These techniques have applications in research including studying gene function and biotechnology to modify traits in crops.
1 human genome
The human genome consists of DNA sequences within 23 chromosome pairs in cell nuclei and mitochondrial DNA that contain the complete set of genetic information for humans. It includes both protein-coding genes and noncoding DNA. While differences exist among individuals, they are smaller than differences between humans and other species like chimpanzees. Noncoding DNA performs important biological functions and can contain disease-causing genetic variants. Various components of genomes like introns, pseudogenes, telomeres and repeat sequences are described. Protein coding sequences contain the information to produce polypeptides. Genetic disorders can be caused by variations in single genes or combinations of genes and environmental factors.
Genome analysis
This document discusses genome analysis and sequencing. It provides background on identifying genes and studying disease processes through genome sequencing. It also describes goals of identifying gene function through experiments and challenges like gene prediction and repetitive sequences. Specific projects aimed at tracking human genetic variations and the first bacterial genome sequencing are summarized. Criteria for selecting early genomes to sequence are outlined. Key differences between prokaryotic and eukaryotic genomes are noted, including the presence of chromosomes, repeats, introns and heterochromatin/euchromatin. Different types of repetitive sequences like satellites, minisatellites and microsatellites are defined. Transposable elements in eukaryotes are also briefly introduced.
Genomics
Genomics is the field of studying genomes through techniques like DNA sequencing and bioinformatics. It aims to understand genome content, organization, function, and evolution. Genomics has two main areas - structural genomics determines genome sequence and organization, while functional genomics studies gene function. A third area, comparative genomics, compares genomes across species. Genomics research has contributed to human health, agriculture and other fields by providing gene sequences. Comparing genome sequences is also improving understanding of evolution and life's history.
GENOME ORGANISATION IN EUKARYOTES
The document discusses genome organization in eukaryotes. It begins by defining the genome as an organism's entire hereditary information, encoded in DNA or RNA. In eukaryotes, DNA is associated with histone proteins to form chromatin fibers, which condense into chromosomes. The document then discusses various levels of chromatin organization, from DNA wrapping around nucleosomes to form beads on a string, to higher-order folding forming metaphase chromosomes. Chromatin exists in two types - loosely packed euchromatin and tightly packed heterochromatin. Overall, the document provides an overview of eukaryotic genome and chromatin organization from nucleosomes to chromosomes.
Content of the genome
basic of the genome content,types of genomes,its size, factors altering the genome size, Genome complexity and c-value paradox
Similar to Present an overview of the gene organization in large-genome plants..pdf (20)
Genome organisation in eukaryotes...........!!!!!!!!!!!
Genome organisation in eukaryotes...........!!!!!!!!!!!
More from ezzi552
Execute the following code and identify the errors in the program. D.pdf
complete the proof Propost ion: If a positive intege r m is not the square of any integer, then Vm
is irrational.
Solution
Let nbe a positive integer such that there is no msuch that n=m2. Suppose nis rational. Then
there exists pand qwith no common factor (beside 1) such that
n=pq
Then
n=p2q2
However, nn is an positive integer and p and q have no common factors beside 1. So q=1.
This gives that
n=p2
Contradiction since it was assumed that nm2 for any m
c.
c
7down voteaccepted
Let nbe a positive integer such that there is no msuch that n=m2. Suppose nis rational. Then
there exists pand qwith no common factor (beside 1) such that
n=pq
Then
n=p2q2
However, nn is an positive integer and p and q have no common factors beside 1. So
q=1. This gives that
n=p2
Contradiction since it was assumed that nm2 for any m
c.
c.
Essay question The genomes of lots and lots of organisms (mostly ba.pdf
Complete the identity: cot(t) (tan (t) - sin (t) cos (t)) =? sin^2 (t) sin (t) tan (t) sec (t) csc (t) tan
(t) + 1
Solution
cot(t)(tan(t) -sin(t)cos(t))
=(cos(t)/sin(t))((sin(t)/cos(t)) -sin(t)cos(t))
=((cos(t)/sin(t))(sin(t)/cos(t)) - (cos(t)/sin(t))sin(t)cos(t))
=1- cos2(t)..................................................................................[sin2A+cos2A =1 is an identity ]
=sin2(t)
therefore cot(t)(tan(t) -sin(t)cos(t)) =sin2(t).
Consider the language L = { anb2n n 0 }.Give an implementation.pdf
C++ LANGUAGE!!!!
Write a program to calculate stock transaction. The program will ask the user to enter the follow
information:
The name of the share purchased
The number of shares purchased
The price per share purchased
the price per share sold
In this question, we assume the user sold all the shares purchased. The user will pay the
stockbroker a commission that amounted to 2% of the amount he/she paid for the stock when the
user purchased the stock. The user will pay the stockbroker another commission that also
amounted to 2% of the amount he/she paid for the stock when the user sold the stock.
Then the program will calculate the following:
the amount of money paid for the stock for the stock purchase
the amount of commission paid to the broker for stock purchase
the total amount of money paid for the stock purchase transaction
the amount of money sold for the stock
the amount of commission paid to the broker for stock sale
the total amount of money the user got for stock sale after paying the commission
the net profit after selling the stock and paying two commissions to the broker
Solution
// C++ code
#include
#include
#include
#include
#include
#include
#include // std::setprecision
using namespace std;
int main()
{
double brokersCommission = 0.02; // Broker commission rate
string name;
double number, pricePurchased, priceSold;
cout << \"Enter name of the share purchased: \";
getline(cin , name);
cout << \"Enter number of shares purchased: \";
cin >> number;
cout << \"Enter price per share purchased: \";
cin >> pricePurchased;
cout << \"Enter price per share sold: \";
cin >> priceSold;
double purchaseAmount = number * pricePurchased;
double purchaseCommission = purchaseAmount * brokersCommission ;
double totalAmountPaid = purchaseAmount + purchaseCommission;
double stockSaleAmount = number * priceSold;
double sellingCommission = (number * priceSold) * brokersCommission;
double totalAmountRecieved = stockSaleAmount - sellingCommission;
double netProfitLoss = totalAmountRecieved - totalAmountPaid;
cout << \"\ The amount of money paid for the stock for the stock purchase: \" <<
purchaseAmount << endl;
cout << \"The amount of commission paid to the broker for stock purchase: \" <<
purchaseCommission << endl;
cout << \"The total amount of money paid for the stock purchase transaction: \" <<
totalAmountPaid << endl;
cout << \"The amount of money sold for the stock: \" << stockSaleAmount << endl;
cout << \"The amount of commission paid to the broker for stock sale: \" << sellingCommission
<< endl;
cout << \"The total amount of money you got for stock sale after paying the commission: \" <<
totalAmountRecieved << endl;
cout << \"The net profit after selling the stock and paying two commissions to the broker: \" <<
netProfitLoss << endl;
return 0;
}
/*
output:
Enter name of the share purchased: Nifty
Enter number of shares purchased: 1000
Enter price per share purchased: 32.87
Enter price per share sold: 33.92
The amount of money paid for the sto.
Case Study.You are the Chair of the Department of Surgery at a lar.pdf
Assume that the City of Coyote has already produced its financial statements for December 31,
2015, and the year then ended. The city\'s General Fund was only for education and parks. Its
Capital Projects Funds worked with each of these functions at times during the current year. The
city also had established an Enterprise Fund to account for its art museum The government-wide
financial statements indicated the following figures Education reported net expenses of $748,000
Parks reported net expenses of $151,000 Art museum reported net revenues of $54,000 General
government revenues for the year were $1,053,000 with an overall increase in the city\'s net
position of $208,000 The fund financial statements indicated the following for the entire year:
The General Fund reported a $35,250 increase in its fund balance The Capital Projects Fund
reported a $54,500 increase in its fund balance The Enterprise Fund reported a $62,000 increase
in its net position The CPA firm of Abernethy and Chapman has been asked to review several
transactions that occurred during 2015 and indicate how to correct any erroneous reporting and
the impact of each error. View each of the following situations as independent. An art display set
up for the City of Coyote was recorded within the General Fund and generated revenues of
$13,400 but had expenditures of $55,700 ($20,000 in expenses and $35,700 to buy land for the
display) The CPA firm has determined that this program should have been recorded as an
Enterprise Fund activity because it was offered in association with the art museum. a. Based on
the information provided above, what was the correct change in the fund balance for the General
Fund for 2015? Correct change b. What was the correct overall change in the city\'s net position
on the government-wide financial statements? Correct change c. What was the correct change in
the net position of the Enterprise Fund on the fund financial statements? Correct change
Solution
The city has already produced its financial statements for December 31, 2015.
The city’s general fund consists of : i.) Education and ii.) Parks
The Enterprise fund consisted of : i.) Art Museum.
As per the given information, there was an art display set up for the city of Coyote which was
recorded with General Fund but it should have been recorded in Enterprise Fund.
The Art revenue $13,400 and expenses recorded $55700 which nets to loss of
$(13400-55700)= $42300.
Therefore,
Fund Calculation:
Particulars
General Fund
Capital Project Fund
Enterprise Fund
Increase/Decrease in position recorded
$ 35250 (increase)
$ 54500 (increase)
$ 62000 (increase)
Add: Loss from Art display wrongly included in General Fund
+ $42300
-
-
Less: Loss from Art display not included in Enterprise Fund
-
-
- $(42300)
Net increase/decrease in position
$77550
$54500
$19700
Ans a) The correct change in the net position in fund balance for General Fund for 2015 was net
increase of $77550.
Ans b) The correct overall change in the.
An Unsorted Type ADT is to be extended by the addition of function S.pdf
A wind generator is 212 feet tall with 116 ft blades, attached at the top of the tower. A bug lands
on the tip of one of the blades at the moment that it is passing through the horizontal, going
down. if the blades are spinning at the constant rate so that they complete one revolution every 6
seconds. Give the function h(t) which describes the height of the bug t seconds after landing.
Solution
comparing with h(t) =Asin(Bt) +D
wind generator is 212 feet tall => D=212
with 116 ft blades =>A=116
the blades are spinning at the constant rate so that they complete one revolution every 6 seconds
=>2/B =6
=>B=/3
bug lands on the tip of one of the blades at the moment that it is passing through the horizontal,
going down
=>A is negative
function h(t) which describes the height of the bug t seconds after landing. is h(t) =-
(116sin((/3)t))+212.
alue 0.83 points M7-6 Calculating Cost of Goods Available for Sale, E.pdf
A surfer is riding a wave with an amplitude of 20 feet and a period of 40 feet. If the wave
completes a period in 10 seconds, what is the speed of the surfer? Select one: a. 400 feet per
second b. 40 feet per second c. 4 feet per second d. 1414 feet per second
Solution
Amplitude = 20 feet
Period = 40 feet
The wave completes a period in 10 seconds.
Speed of the surfer =Distance/Time =Period length /Time
= 40 /10 = 4 feet per second
Option C.
1)Using general mass-media (such as news sites) identify a recent co.pdf
1)Using general mass-media (such as news sites) identify a recent computer security incident.
Discuss the incident in few sentences. Next, find at least three articles on computer security sites
that
discuss the technology behind the attack / prevention / reaction etc. for the incident. Remember
to
provide the links also
Solution
Cyberterrorism is also clearly an emerging threat. Terrorist groups are increasingly computer
savvy, and some probably are acquiring the ability to use cyber attacks to inflict isolated and
brief disruptions of US infrastructure. Due to the prevalence of publicly available hacker tools,
many of these groups probably already have the capability to launch denial-of-service and other
nuisance attacks against Internet-connected systems. As terrorists become more computer savvy,
their attack options will only increase.” (War on Terrorism, 2003)
This is what Robert Mueller, FBI Director, testified on 11 February 2003 before the US Senate
on a hearing about War On Terrorism against Al-Qaeda and other terrorist organizations. The US
and global media organizations picked up this testimony and begun speculating on the possibility
of a large-scale Cyberterrorist attack. So far, such an attack has not materialised. At the same
time a similar term, Cybercrime, is used to describe criminal activities on the Internet such as
identity theft, copyright infringement and bank fraud, but many times these two terms
(Cybercrime and Cyberterrorism) end up been used inter-changeably and their meaning,
especially to the public, becomes blurred and unclear. Governments, policy networks and the
media around the globe have engaged in an effort to build defences against Cyberattacks, bring
new regulations in effect while maintaining an almost mythological atmosphere over the threats
and risks of potential Cybercrime and Cyberterrorist attacks.
This essay will explore the concepts of Cybercrime and Cyberterrorism, how such acts are
portrayed in the media and the government policies that come in effect. The focus will be on how
all these processes affect people’s psychology in regards to induced feelings of anxiety and fear
and not the illegal activities themselves. In essence the essay aims to investigate Furedi’s (2002,
p. 34) statement that “what we already fear can now thrive in the new space provided by the
Internet”..
1. The notion that two network exist in the brain, one for emotional.pdf
1. The notion that two network exist in the brain, one for emotional responses and one for
analytical reasoning, describes:
A. formal operational thought
B. the dual process model
C. intuitive thought
D. hypothetical thought
2. Secondary education refers to:
A. after-school programs
B. college
C. vocational educationa programs, where one learns a job skill
D. grades 7 through 12
Solution
1. The two network exists in the brain describes the dual process model. There are two system of
reasoning. System 1 and System 2. system 1 is for unconscious reasoning and system 2 is for
conscious reasoning.
2. Secondary education refers to the grades 7 through 12.
Which of the following is a bank liabilityA. Reserve deposits at .pdf
The net revenue for a certain car company (in billions of dollars)between the years 2004 abd
2012 is approximated by R(X)=-0.717x^3+23.8x^2-157.6x+423.2, where x=4 correspoinds to
the year 2004. Complete parts (a) and (b)
(a) what is the revenue in 2009 (in billions)
(b) What is the revenue in 2010
Solution
The revenue is given by the formula:
R(x) = -0.717 x3 + 23.8x2 - 157.6x + 423.2 where x = 4 for year 2004
So,
for year 2009, x = 9 and x = 10 would correspond to year 2010
So,
revenue in 2009 =
R(9) = 409.907 billion dollars (just put x = 9 in the above equation of R(x) )
R (10) = 510.2 billion dollars
Hope this helps,.
What was Eisenhower’s reinsurance plan What was Eisenhower’s r.pdf
The Food and Drug Administration (FDA) produces a quarterly report called Total Diet Study.
The FDA\'s report covers more than 200 food items, each of which is analyzed for dangerous
chemical compounds. One Total Diet Study reported that no pesticides at all were found in 65%
of the domestically produced food samples. Consider a random sample of 800 food items
analyzed for the presence of pesticides. Compute Mu and Sigma for the random variable x, the
number of food items with no trace of pesticide. Is it likely that you would observe fewer then
half of food items without any traces of pesticide Why or Why not
Solution
a)
the expected value or miu will be = n*p for a binomial distribution and why is a binomial
distribution? because we have a probability of success that in this case is 65% (no pesticides)
miu = 0.65 * 800 = 520
sigma = srqt (n*p*q) = srqt ( 0.65*0.35*520) = 10.88
b)
No it is not likely because our probability of no pesticide is 65%
that means that more than half of food items are without any pesticide.
What does the following program do What’s its time complexity Just.pdf
The aqueous reaction between phenol (C6H5OH) and ammonia shown below, has initial
concentrations of
0.200 M C6H5OH(aq) and 0.120 M NH3(aq). At equilibrium, the C6H5O–(aq) concentration
is 0.050 M.
Calculate the equilibrium constant Kc for the reaction.
C6H5OH (aq) + NH3(aq) --> C6H5O– + NH4+(aq)
Answer is 0.24 - need steps - thanks!
Solution
Assuming it to be balanced reaction
At equilibrium
C6H5OH = 0.2 - x
NH3 = 0.12 - x
C6H5O- = 0.05 = x
and NH4 + = x = 0.05
Hence Kc = (0.05 * 0.05)/(0.2 - 0.05)(.12 - 0.05) = 0.238.
What is not true about the membranes of prokaryotic organism a. The.pdf
The correlation coefficient of two input sequences is given by r = n sigma x_i y_i - sigma x_i
sigma y_i/Squareroot [n sigma x_i^2 - (sigma x_i)^2][n sigma y_i^2 - (sigma y_i)^2] Adapt
Example 4.11 to compute the correlation coefficient from a sequence of pairs of input values
from an input file.
Solution
#include
using namespace std;
//Class for standard deviation
class stdDev
{
private:
int max;
double value[100];
double mean;
public:
double CalMean()
{
double sum = 0;
for(int i = 0; i < max; i++)
sum += value[i];
return (sum / max);
}
double CalVariane()
{
mean = CalMean();
double temp = 0;
for(int i = 0; i < max; i++)
{
temp += (value[i] - mean) * (value[i] - mean) ;
}
return temp / max;
}
double CalSampleVar()
{
mean = CalMean(); //get mean
double temp = 0;
for(int i = 0; i < max; i++)
{
temp += (value[i] - mean) * (value[i] - mean) ; //apply formula
}
return temp / (max - 1); //returns sample variance
}
int SetValues(double *p, int count)
{
if(count > 100)
return -1;
max = count;
for(int i = 0; i < count; i++)
value[i] = p[i];
return 0;
}
double CalStdDev()//returns Standard deviation
{
return sqrt(CalVariane());
}
double SampleStdDev()//returns sample deviation
{
return sqrt(CalSampleVar());
}
};
class Calculate
{
private:
double XTerms[100];
double YTerms[100];
int max;
stdDev x; // Standard deviation for x
stdDev y;// Standard deviation for y
public:
//set x and y series values to class members
void SetValues(double *xval, double *yval, int count)
{
for(int i = 0; i < count; i++)
{
XTerms[i] = xval[i];
YTerms[i] = yval[i];
}
x.SetValues(xval, count);
y.SetValues(yval, count);
max = count;
}
double Calculate_Covariance()
{
double xmean = x.CalMean();//Get X meam
double ymean = y.CalMean();//get y mean
double total = 0;
for(int i = 0; i < max; i++)
{
total += (XTerms[i] - xmean) * (YTerms[i] - ymean);
}
return total / max;//return Covariance
}
double Calculate_Correlation()//Find correlation
{
double cov = Calculate_Covariance();
double correlation = cov / (x.CalStdDev() * y.CalStdDev());
return correlation;
}
};
int main()
{
Calculate calc; //create object of class
double xarr[] = { 0, 2, 4, 6, 8 }; // x series
double yarr[] = { 6, 13, 15, 16, 20 };// y series
calc.SetValues(xarr,yarr,sizeof(xarr) / sizeof(xarr[0]));//set series values
cout<<\"X series ={ 0, 2, 4, 6, 8 }\ Y series={ 6, 13, 15, 16, 20 }\ \";
cout<<\"Correlation \"<< calc.Calculate_Correlation();
}
=====================================================================
========
Output:
akshay@akshay-Inspiron-3537:~/Chegg$ g++ coeffici.cpp
akshay@akshay-Inspiron-3537:~/Chegg$ ./a.out
X series ={ 0, 2, 4, 6, 8 }
Y series={ 6, 13, 15, 16, 20 }
Correlation 0.952157.
What behavior characteristics are associated with each of the four s.pdf
The answer has to be original.
For this week’s discussion, complete the following two scenario questions below in detail. Please
discuss thoroughly and substantivelyin your post.
1.)You are a digital forensic examiner and have been asked to examine a hard drive for potential
evidence. Give examples of how the hard drive (or the data on it) could be used as (or lead to the
presentation of) all four types of evidence in court; testimonial, real, documentary, and
demonstrative. If you do not believe one or more of the types of evidence would be included,
explain why not.
2.) You have been asked to assist a law enforcement team serving a search warrant related to a
child pornography investigation. You are the digital forensic expert for the team, and, as such,
have been assigned the task of identifying and collecting the digital evidence at the search
location. Answer the following questions about your assignment.
What steps should you take before the search to serve the search warrant?
What types of evidence should you be on the alert for, when searching the residence?
What types of items would you seize?
Solution.
What are someways to better understand accounting There is a lot of.pdf
Take the problem 4284 divided by 14 using the standard division algorithm. the answer is 306.
(a) when we write the 3 in the quoitent, and then mulitply 3 by 14 to get 42, what does the 42
represent?
(b) explain the mathmatical reason why you have to \" move to the right\" as you do the
algorithm.
(c) is may be confusing for children to understand why the answer is 306 instead of 36. explain
clearly why a 0 must be put into the answer.
Solution
a) it is actually 4200. Not 42. It is the way of writing. Because we actually subtract 4200 from
4284.
b) we move to right because we are going towards units place. Like 1000x+100y+10z+a this is
the format.
C)we must understand we are taking all places one by one. Here hundred\'s place is 3.so 300.
Tens place is 0. So 0. Units place is 6. Therfore 300+0+6=306.
Based on the elements from the Ruby Payne readings create a resource.pdf
alue 0.83 points M7-6 Calculating Cost of Goods Available for Sale, Ending Inventory, Sales,
Cost of Goods Sold, and Gross Profit under Periodic FIFO, LIFO, and Weighted Average Cost
[LO 7-3 The following are the transactions for the month of July July 1 Beginning Inventory 49
July 25 Sold July 31 Ending Inventory194 Units Unit Cost Selling Price 245 11 100) S10 $14
Solution
Answers
FIFO (Periodic)
Units
Cost per unit
Total
Beginning Inventory
49
$ 10.00
$ 490.00
Purchases:
Jul-13
245
$ 11.00
$ 2,695.00
Goods Available for Sale
$ 3,185.00
Cost of Goods Sold:
Beginning Inventory
49
$ 10.00
$ 490.00
Jul-13
51
$ 11.00
$ 561.00
Total Cost of Goods Sold
$ 1,051.00
Ending Inventory
$ 2,134.00
Sales
$ 1,400.00
Cost of Goods Sold
$ 1,051.00
Gross Profit
$ 349.00
LIFO (Periodic)
Units
Cost per unit
Total
Beginning Inventory
49
$ 10.00
$ 490.00
Purchases:
Jul-13
245
$ 11.00
$ 2,695.00
Goods Available for Sale
$ 3,185.00
Cost of Goods Sold:
Jul-13
100
$ 11.00
$ 1,100.00
$ -
Total Cost of Goods Sold
$ 1,100.00
Ending Inventory
$ 2,085.00
Sales
$ 1,400.00
Cost of Goods Sold
$ 1,100.00
Gross Profit
$ 300.00
Weighted Average cost per unit:
Goods Available for sale = $ 3,185
Beginning + Purchased units = 49+ 245 = 294 units
Weighted Average cost per unit = 3185 / 294 = $ 10.83 per unit
Weighted Average (Periodic)
Units
Cost per unit
Total
Beginning Inventory
49
$ 10.00
$ 490.00
Purchases
245
$ 11.00
$ 2,695.00
Goods Available for Sale
$ 3,185.00
Cost of Goods Sold:
100
$ 10.83
$ 1,083.00
Ending Inventory
$ 2,102.00
Sales
$ 1,400.00
Cost of Goods Sold
$ 1,083.00
Gross Profit
$ 317.00
FIFO (Periodic)
Units
Cost per unit
Total
Beginning Inventory
49
$ 10.00
$ 490.00
Purchases:
Jul-13
245
$ 11.00
$ 2,695.00
Goods Available for Sale
$ 3,185.00
Cost of Goods Sold:
Beginning Inventory
49
$ 10.00
$ 490.00
Jul-13
51
$ 11.00
$ 561.00
Total Cost of Goods Sold
$ 1,051.00
Ending Inventory
$ 2,134.00
Sales
$ 1,400.00
Cost of Goods Sold
$ 1,051.00
Gross Profit
$ 349.00.
The answer has to be original.For this week’s discussion, complete.pdf
Problem 12-5A Bramble Corp.\'s income statement contained the condensed information below.
Bramble Corp. Income Statement For the Year Ended December 31, 2017 $3,171,900 Service
revenue Operating expenses, excluding depreciation 2,007,780 Depreciation expense Loss on
disposal of plant assets Income before income taxes Income tax expense Net income 179,850
52,320 2,239,950 931,950 183,120 $748,830 Bramble Corp.\'s balance sheet contained the
compartive data at December 31 2017 2016 Accounts receivable Accounts payable Income taxes
payable $228,900 134,070 42,510 $196,200 104,640 22,890 Accounts payable pertain to
operating expenses. Prepare the operating activities section of the statement of cash flows using
the indirect parenthesis e.g. (15,000).)
Solution
Prepare operating activities indirect method :Net income748830Adjustment to reconcile net
incomeDepreciation expense179850Loss on disposal of plant assets52320Increase account
receivable-32700Increase account payable29430Increase income tax payable19620248520Net
cash flow from operating activities997350.
Assume you have a scanner object (called input).Declare an integer.pdf
abestion Tu 10 points Save Acower p Company purchased a 90% interest in Company on an ay
2 2013 it accounts for its investment n s company ng the cost the d P Co b Company because s
Company was its primary supplier of merchandise for resale. During 2011. P Company bought
merchandise from S Company. The selling price to P Company was S300.000, S Company uses
a 25% markup on cost. At the end of 2013, P Company still had in its books 2S percent of the
inventory purchased from S Company. In 2014, the intercompany sia profit in ending inventory
at the end of 2014? OA 544,800 les totalled $280.000 with 20 percent left in inventory at the end
of the year. What is the unrealized O B. $10.400 OC.$11.200 O D.$56.000
Solution
ANSWER IS C($ 11200)
DETAIL WORKING IS GIVEN AS UNDER:
INTER COMPANY SALE S= $ 280000
INVENTORY LEFT AT THE END =$280000X20%=$56000
UNREALISED PROFIT = $56000X25/125 = $11200
_S COMPANY USES A 25% MARKUP ON COST:
AS SUCH IF COSY TO S COMPANY IS $100
THAN BY ADDING 25% ON COST(I.E.,$25) SELLING PRICE OF S COMPANY
TO P COMPANY WILL BE $125
THAN UNREALISED PROFIT WILL BE 25/125.
6. Establishing priorities is an issue that local governments strugg.pdf
6. Below is the trial balance for Logan, Inc. for the year ended December 31, 201X Prepare a
multiple-step income statement in good form (proper format). (Check: Net income is greater than
$40,000. (6 points)
Solution
PARTICULARS Sub Amount Amount Sales $ 320,000.00 Sales Discount $ (6,000.00)
Sales Returns $ (16,500.00) Net Sales $ 297,500.00 Cost of Goods Sold $ 180,000.00
Gross Profit $ 117,500.00 Operating Expenses Selling Expenses $ 42,000.00 General
Expenses $ 19,800.00 $ 61,800.00 Non Operating Expenses Interest Cost $ 600.00
Net Profit $ 55,100.00.
Suppose that A and B are attributes of a certain relational table. G.pdf
please help with this question. 3. Write a counterexample for each property the relation does not
possess. (1,1), (1,2), (2,1), (2,2) (b) apy z
Solution
(a) Here let us consider the three properties. They are:
Reflexive: For every element 1 in A, (1,1) is in R.
Symmetric: If an ordered pair (1,2) in R, (2,1) is in R.
Transitive: If ordered pairs (1,2) and (2,1) in R, (1,1) in is R.
Here R is reflexive and symmetric and we have to prove that it is not transitive..
RHCE is least associated with which Rh group (D--, R1r, Ror, Rzy) .pdf
No hand soving at all! Please only do them as MATLAB/WOLFRAM\'s MATHMATICA
CODE/SCRIPT. NO SELF CALCULATIONS
MATLAB/WOLFRAM\'s MATHMATICA CODE/SCRIPT ONLY
MATLAB/WOLFRAM\'s MATHMATICA CODE/SCRIPT ONLY PLEASE
Solution
function fn=f(x)
if x<0 && x>-50
fn=x+50;
elseif x>0 && x<50
fn=x-50;
else
fn=0;
end
end
function cs=cs(n)
T=100;
if n~=0
fun=@(t)f(t).*cos((2.*pi.*n.*t)./T);
cs=integral(fun,-50,50);
cs=(2./T).*cs;
else
fun=@(t) f(t);
cs=integral(fun,-50,50);
cs=cs./T;
end
end
function sn=sn(n)
T=100;
fun=@(t) f(t).*sin((2.*pi.*n.*t)./T);
sn=integral(fun,-50,50);
sn=(2./T).*sn;
end
T=100;
disp(\'first five cos terms\');
for m=0:4
cs(m);
end
disp(\'first five sin terms\');
for n=1:5
sn(n)
end
A0=cs(0);
N=10;
A=zeros(1,N);
B=A;
for k=1:N
A(k)=cs(k);
B(k)=sn(k);
end
t=-50:0.1:50;
fn_orginal=f(t);
fun_trn=zeros(1,size(t));
for n=1:N
fun_trn=fun_trn+A(k).*cos((2.*pi.*n.*t)./T)+B(k).*sin((2.*pi.*n.*t)./T);
end
fun_trn=fun_trn+A0;
plot(t,fun_trn).
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