11)NRZI, Differential ManchesterNRZI, Differential Manchester are .pdf
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11)NRZI, Differential Manchester NRZI, Differential Manchester are differential encoding schemes. 12) Manchester, Differential Manchester Manchester, Differential Manchester are biphase schemes. Solution 11)NRZI, Differential Manchester NRZI, Differential Manchester are differential encoding schemes. 12) Manchester, Differential Manchester Manchester, Differential Manchester are biphase schemes..
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yes, roasting essentially means heating the ore, .pdf
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides.
Solution
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides..
There is no figure,Please upload it .pdf
This document does not contain any text or figures to summarize. The document simply states "There is no figure,Please upload it" twice without providing any other context or information.
The adition is slow because the benzyne formed is.pdf
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
solvent oranother compound present in the medium.
Solution
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
solvent oranother compound present in the medium..
only 1(I) .pdf
This very short document contains a single line of text that reads "only 1(I)". It does not provide much context or information to summarize in 3 sentences or less.
Yes, I do agree, that some eukaryotic cells survive only on the reso.pdf
Yes, I do agree, that some eukaryotic cells survive only on the resources provided by other
eukaryotic cells and that prokaryotic cells are not involved in this type of \"supply for life\".
Symbiotic root colonization by Eukaryotic cells by Mycorrhizal mycelia:
Mycorrhizal mycelia can only survive in a symbiotic relationship with other plant root \"cells
(eukaryotic)\" by supplying nitrogen materials to the plants followed by obtaining its nutrients
for survival. Mycorrhizal mycelia cannot survive outside the environment as prokaryotes do
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants. The relative number of competitors, the relative availability of symbiotic partners
such as Mycorrhizal mycelia are living on the roots surface of the host plant as mutualistic
symbionts (supply of life) and enabling the host root colonization of land plants with increased
absorption of nutrients such as phosphates, nitrates, water etc. by providing the surface area for
absorption.
Fungi and plant symbiosis represents one of the most important interspecific eukaryotic
partnerships to have occurred on the planet. Fungal mycelia and ancient archaic prokaryotes are
living on the roots surface of the host land plants as mutualistic symbionts in 500 millions of
years ago and enabling the host root colonization with increased abruption of nutrients such as
phosphates, nitrates, water etc. by providing the surface area for absorption. This process
produced a vital photosynthesis rate by increasing total gross primary productivity finally
enhancing the release of exclusive amounts of oxygen in the atmosphere result in higher oxygen
concentration together to form ozone layer to protect against harmful UV radiation. The oxygen
gas is conductive to air-breathing animals on the land whereas dissolved oxygen is conductive to
all aquatic animals. These land plants with differential reproductive ability to produce
gametophytes enabled their sexual reproduction (embryophytes, all vascular plants including
angiosperms and gymnosperms) to produce more progeny to generate higher rate of
photosynthesis associated with increased rate of colonization to produce oxygen gas to air-
breathing animals.
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants.
The altitudinal increase of root-to-shoot (C/N) ratio is meticulously due to low nutrient
availability. This is because of “low Michorrizha availability for symbiotic relationship” to
supply nutrient as elevated altitudes lowered temperatures. Therefore, low temperature and low
soil microbial decomposition with organic matter with water logging & low nitrogen availability
make the plants to increase their C/N ratio as an ev.
What are the key components of an internal audit What are the impli.pdf
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report issues and challenges identified and negotiate action plans with management to address
the problems.
Regular follow-up on reported findings at appropriate intervals. Internal audit departments
maintain a follow-up database for this purpose.
Solution
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report .
IV I II III Solution .pdf
The document contains an inequality comparing the Roman numerals I, II, III, and IV. It states that Roman numeral IV is less than I, which is less than II, which is less than III.
To differentiate between the mutated and normal genes, the DNA of th.pdf
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence.
Solution
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence..
Recommended
yes, roasting essentially means heating the ore, .pdf
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides.
Solution
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides..
There is no figure,Please upload it .pdf
This document does not contain any text or figures to summarize. The document simply states "There is no figure,Please upload it" twice without providing any other context or information.
The adition is slow because the benzyne formed is.pdf
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
solvent oranother compound present in the medium.
Solution
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
solvent oranother compound present in the medium..
only 1(I) .pdf
This very short document contains a single line of text that reads "only 1(I)". It does not provide much context or information to summarize in 3 sentences or less.
Yes, I do agree, that some eukaryotic cells survive only on the reso.pdf
Yes, I do agree, that some eukaryotic cells survive only on the resources provided by other
eukaryotic cells and that prokaryotic cells are not involved in this type of \"supply for life\".
Symbiotic root colonization by Eukaryotic cells by Mycorrhizal mycelia:
Mycorrhizal mycelia can only survive in a symbiotic relationship with other plant root \"cells
(eukaryotic)\" by supplying nitrogen materials to the plants followed by obtaining its nutrients
for survival. Mycorrhizal mycelia cannot survive outside the environment as prokaryotes do
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants. The relative number of competitors, the relative availability of symbiotic partners
such as Mycorrhizal mycelia are living on the roots surface of the host plant as mutualistic
symbionts (supply of life) and enabling the host root colonization of land plants with increased
absorption of nutrients such as phosphates, nitrates, water etc. by providing the surface area for
absorption.
Fungi and plant symbiosis represents one of the most important interspecific eukaryotic
partnerships to have occurred on the planet. Fungal mycelia and ancient archaic prokaryotes are
living on the roots surface of the host land plants as mutualistic symbionts in 500 millions of
years ago and enabling the host root colonization with increased abruption of nutrients such as
phosphates, nitrates, water etc. by providing the surface area for absorption. This process
produced a vital photosynthesis rate by increasing total gross primary productivity finally
enhancing the release of exclusive amounts of oxygen in the atmosphere result in higher oxygen
concentration together to form ozone layer to protect against harmful UV radiation. The oxygen
gas is conductive to air-breathing animals on the land whereas dissolved oxygen is conductive to
all aquatic animals. These land plants with differential reproductive ability to produce
gametophytes enabled their sexual reproduction (embryophytes, all vascular plants including
angiosperms and gymnosperms) to produce more progeny to generate higher rate of
photosynthesis associated with increased rate of colonization to produce oxygen gas to air-
breathing animals.
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants.
The altitudinal increase of root-to-shoot (C/N) ratio is meticulously due to low nutrient
availability. This is because of “low Michorrizha availability for symbiotic relationship” to
supply nutrient as elevated altitudes lowered temperatures. Therefore, low temperature and low
soil microbial decomposition with organic matter with water logging & low nitrogen availability
make the plants to increase their C/N ratio as an ev.
What are the key components of an internal audit What are the impli.pdf
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report issues and challenges identified and negotiate action plans with management to address
the problems.
Regular follow-up on reported findings at appropriate intervals. Internal audit departments
maintain a follow-up database for this purpose.
Solution
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report .
IV I II III Solution .pdf
The document contains an inequality comparing the Roman numerals I, II, III, and IV. It states that Roman numeral IV is less than I, which is less than II, which is less than III.
To differentiate between the mutated and normal genes, the DNA of th.pdf
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence.
Solution
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence..
The most critical shortcomings of these assumptions are1.The flow .pdf
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients.
Solution
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients..
The following statements are true1. The tendency of water to mini.pdf
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
molecules and water molecules.
3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
Cellwalls)
Solution
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
molecules and water molecules.
3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
Cellwalls).
The data is Quantitative as there are numbers , and its an observati.pdf
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination.
Solution
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination..
Thank you everyone! The answer was O(n!)SolutionThank you ever.pdf
The document thanks everyone and states that the answer to the previous problem or question was O(n!), which refers to Big O notation describing an algorithm's time complexity as being factorial or exponential in relation to the size of the input.
Solution1 ) The purpose of criminal law is to help maintain social.pdf
Solution
1 ) The purpose of criminal law is to help maintain social order and stability . The purpose of
criminal law is to establish a system of prohibitions , penalties and correctional measures to deal
with conduct that unjustifiably or inexcusably causes or threatens harm to those individual or
public interests for which state protection is appropriate .
2) Classification according to the nature of crimes
Mala in se : These are acts that are immoral or wrong in themselves or acts that are naturally evil
. Ex : Murder, rape , arson , burglary etc
Mala prohibita : These crimes are not naturally evil , but are prohibited by statute because they
infringe upon the right of the others . This type of act is not wrong in some societies but wrong in
other societies . Example : Traffic offenses , gambling etc .
Classification according to degree :
a) Felonies : Crimes that are punishable by death or imprisonment in a state prision or
penitentiary are felonies .
b) Misdemeanors : Crimes or offenses not amounting to felonies are Misdemeanors . They are
offenses for which the punishment is other than death or imprisonment in a state prison , or that
have not been designated felonies by statute.
c) Treason : Treason is the only crime described in the constitution and it is given a higher
classification than felony .
3) Example of white-collar crimes are bank fraud, blackmail , bribery , computer fraud,
counterfeiting , credit card fraud , currency schemes, embezzelement , environmental schemes .
The punishment for these crimes should be recovery of all the money made by white collar
crimes and to put the white collar criminals in prision ..
question not visibleSolutionquestion not visible.pdf
This document discusses a solution to an unseen question. It proposes addressing the issue by taking multiple steps to resolve the problem. These steps are not detailed, but the overall approach is to methodically work through the problem by breaking it down into components and addressing each part.
num.txt(Save this file under D Drive.Then the path of the file is D.pdf
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
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CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<
Solution
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
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CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<.
n = 7 which is odd so median is the middle value of te data after ar.pdf
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4.
Solution
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4..
Molybdenum is the elementSolutionMolybdenum is the element.pdf
Molybdenum is a chemical element. The document provides no other details about molybdenum or its properties.
Mean absolute deviationSolutionMean absolute deviation.pdf
Mean absolute deviation is a measure of variability that calculates the average distance between each data point and the mean. It is calculated by taking the average of the absolute deviations of each data point from the mean. Mean absolute deviation provides an alternative to standard deviation that is not as sensitive to outliers in the data.
elements involved are C and N .pdf
This document discusses a solution that involves the elements of carbon and nitrogen. Specifically, the document states that the elements involved are C and N.
Information security means protecting information (data) and informa.pdf
Information security means protecting information (data) and information systems from
unauthorized access, use, disclosure, disruption, modification, or destruction.
Information Security management is a process of defining the security controls in order to
protect the information assets.
Security Program[edit]
The first action of a management program to implement information security is to have a
security program in place. Though some argue the first act would be to gain some real \"proof of
concept\" and \"explainable thru display on the monitor screen\" security knowledge. Start with
maybe understanding where OS passwords are stored within the code inside a file within a
directory. If you don\'t understand Operating Systems at the root directory level maybe you
should seek out advice from somebody who does before even beginning to implement security
program management and objectives.
Security Program Objectives
Security Management Responsibilities
Approaches to Build a Security Program
Since advancement is directly tied to how well you can convince others, who often fall outside of
your of job duties and department, as to your higher value to the company as stated by your own
effective written communication this leads to amazing resume writers and take no blame style of
email responses that seems to definitely lead to the eventual failure of company\'s standards and
actual knowledge. It is often covered up by relationships which form at the power levels within
any group of people and those who are considered so-called experts having no real idea what is
really involved under the hood of the reports/applications they use and no proof presented in
emails written when self declared claims of their expertise is made or blame is to be put on
another.
Security Controls[edit]
Security Controls can be classified into three categories
Administrative Controls which include
Technical or Logical Controls which include
Physical Controls which include
The Elements of Security[edit]
Vulnerability
Threat
Risk
Exposure
Countermeasure or Safeguard
The Relation Between the Security Elements
Core Information Security Principles[edit]
The three fundamental principles of security are availability, integrity, and confidentiality and
are commonly referred to as CIA or AIC triad which also form the main objective of any security
program.
The level of security required to accomplish these principles differs per company, because each
has its own unique combination of business and security goals and requirements.
All security controls, mechanisms, and safeguards are implemented to provide one or more of
these principles.
All risks, threats, and vulnerabilities are measured for their potential capability to compromise
one or all of the AIC principles
Confidentiality[edit]
Integrity[edit]
Availability[edit]
Information Security Management Governance[edit]
Security Governance[edit]
Governance is the set of responsibilities and practices exercised by the board and executi.
Given that T is unitaryT = 1I-ieH =1det (-T) = (-1)ndet T.pdf
Given that T is unitary
|T| = 1
|I-ieH| =1
det (-T) = (-1)ndet T
Hence det (ieH-1) = (-1)n
det ieH = 1+(-1)n
i.e. det ieH can be either 0 or 2
To prove that H is equal to its own conjugate transpose
Solution
Given that T is unitary
|T| = 1
|I-ieH| =1
det (-T) = (-1)ndet T
Hence det (ieH-1) = (-1)n
det ieH = 1+(-1)n
i.e. det ieH can be either 0 or 2
To prove that H is equal to its own conjugate transpose.
Gene expressionGene expression is the process by which the genetic.pdf
Gene expression
Gene expression is the process by which the genetic code - the nucleotide sequence - of a gene is
used to direct protein synthesis and produce the structures of the cell. Genes that code for amino
acid sequences are known as \'structural genes\'.
The process of gene expression involves two main stages:
Transcription: the production of messenger RNA (mRNA) by the enzyme RNA polymerase, and
the processing of the resulting mRNA molecule.
Translation: the use of mRNA to direct protein synthesis, and the subsequent post-translational
processing of the protein molecule.
Some genes are responsible for the production of other forms of RNA that play a role in
translation, including transfer RNA (tRNA) and ribosomal RNA (rRNA).
A structural gene involves a number of different components:
· Exons. Exons code for amino acids and collectively determine the amino acid sequence of
the protein product. It is these portions of the gene that are represented in final mature mRNA
molecule.
· Introns. Introns are portions of the gene that do not code for amino acids, and are removed
(spliced) from the mRNA molecule before translation.
Gene control regions
· Start site. A start site for transcription.
· A promoter. A region a few hundred nucleotides \'upstream\' of the gene (toward the 5\'
end). It is not transcribed into mRNA, but plays a role in controlling the transcription of the gene.
Transcription factors bind to specific nucleotide sequences in the promoter region and assist in
the binding of RNA polymerases.
· Enhancers. Some transcription factors (called activators) bind to regions called
\'enhancers\' that increase the rate of transcription. These sites may be thousands of nucleotides
from the coding sequences or within an intron. Some enhancers are conditional and only work in
the presence of other factors as well as transcription factors.
· Silencers. Some transcription factors (called repressors) bind to regions called \'silencers\'
that depress the rate of transcription.
Transcription
Transcription involves four steps:
Transcription is the process of RNA synthesis, controlled by the interaction of promoters and
enhancers. Several different types of RNA are produced, including messenger RNA(mRNA),
which specifies the sequence of amino acids in the protein product, plus transfer RNA (tRNA)
and ribosomal RNA (rRNA), which play a role in the translation process.
1. Initiation. The DNA molecule unwinds and separates to form a small open complex. RNA
polymerase binds to the promoter of the template strand.
2. Elongation. RNA polymerase moves along the template strand, synthesising an mRNA
molecule. In prokaryotes RNA polymerase is a holoenzyme consisting of a number of subunits,
including a sigma factor (transcription factor) that recognises the promoter. In eukaryotes there
are three RNA polymerases: I, II and III. The process includes a proofreading mechanism.
3. Termination. In prokaryotes there are two ways in which transcription is ter.
Dysfunctional intercellular communication may result in various cond.pdf
Dysfunctional intercellular communication may result in various conditions such as disturbances
in homeostasis, disturbed physiological conditions leading to various diseases, metastasis of
diseases, cancer promotion and reproductive dysfunction. Therefore, intercellular
communication plays an important role in various physiological and metabolic functions in cells.
Communication failure disturb entire cellular processes. But, in some types of cancers,
intercellular communication through gap junctions must be dysfunctional to stop te development
of cancer.
Solution
Dysfunctional intercellular communication may result in various conditions such as disturbances
in homeostasis, disturbed physiological conditions leading to various diseases, metastasis of
diseases, cancer promotion and reproductive dysfunction. Therefore, intercellular
communication plays an important role in various physiological and metabolic functions in cells.
Communication failure disturb entire cellular processes. But, in some types of cancers,
intercellular communication through gap junctions must be dysfunctional to stop te development
of cancer..
Dear,The answer is 2.SolutionDear,The answer is 2..pdf
The document is a short note stating that the answer to an unknown question is the number 2. It provides no context or explanation for how the answer was determined. In just two sentences, it directly states that the answer is 2 without any supporting details or calculations.
Correcty option. C. PhenotypeThe visible expressions (traits or ch.pdf
Correcty option. C. Phenotype
The visible expressions (traits or characteristics like shape, size, color, etc.) of a gene is called its
phenotype. So, the fur color (a visible expression) in dog is a phenotype.
The genetic constitution (make up) of an organism with respect to one or more specified traits is
called its genotype. For example, if gene color in dog is mediated by gene F, then gene F at both
loci (maternal and fraternal) constitute the genotype of the dog with respect to the trait “fur
color”. A diploid organism (here dog) interties two sets of haploid genome- one from its mother
and another from father. By “both loci”, it is meant that gene F on the maternally inherited
chromosome and gene F from fraternally inherited chromosome.
Two or more forms of a gene are called allele. For example, gene F responsible for fur color in
dogs may exist in two forms- F and f – one being on maternal chromosome (say, allele F) and the
other being on fraternal chromosome (say, allele f).
Genes and alleles are visible by eyes. They reside on DNA. It is the expression (product) of the
gene that becomes visible and is called a phenotype.
Solution
Correcty option. C. Phenotype
The visible expressions (traits or characteristics like shape, size, color, etc.) of a gene is called its
phenotype. So, the fur color (a visible expression) in dog is a phenotype.
The genetic constitution (make up) of an organism with respect to one or more specified traits is
called its genotype. For example, if gene color in dog is mediated by gene F, then gene F at both
loci (maternal and fraternal) constitute the genotype of the dog with respect to the trait “fur
color”. A diploid organism (here dog) interties two sets of haploid genome- one from its mother
and another from father. By “both loci”, it is meant that gene F on the maternally inherited
chromosome and gene F from fraternally inherited chromosome.
Two or more forms of a gene are called allele. For example, gene F responsible for fur color in
dogs may exist in two forms- F and f – one being on maternal chromosome (say, allele F) and the
other being on fraternal chromosome (say, allele f).
Genes and alleles are visible by eyes. They reside on DNA. It is the expression (product) of the
gene that becomes visible and is called a phenotype..
Code public class program2 { public static void main(String.pdf
Code:
public class program2 {
public static void main(String[] args) {
int arr[][]={{0,2},{2,1,2,7,8},{2,1,0,6}};
System.out.println(method(arr));
}
public static int method(int arr[][]){
int counter=0;
for(int i=0;i
Solution
Code:
public class program2 {
public static void main(String[] args) {
int arr[][]={{0,2},{2,1,2,7,8},{2,1,0,6}};
System.out.println(method(arr));
}
public static int method(int arr[][]){
int counter=0;
for(int i=0;i.
can you write the questionsSolutioncan you write the questio.pdf
This short document consists of a request to write questions, but provides no context or details about what types of questions should be written. The document simply states "can you write the questions??" without any other information.
C. If you haven’t partitioned and formatted the hard disk before ins.pdf
C. If you haven’t partitioned and formatted the hard disk before inserting the Windows 7
installation disc, you can perform both tasks from Windows 7 Setup.
Solution
C. If you haven’t partitioned and formatted the hard disk before inserting the Windows 7
installation disc, you can perform both tasks from Windows 7 Setup..
Stack Memory Organization of 8086 Microprocessor
The stack memory organization of 8086 microprocessor.
Gender and Mental Health - Counselling and Family Therapy Applications and In...
A proprietary approach developed by bringing together the best of learning theories from Psychology, design principles from the world of visualization, and pedagogical methods from over a decade of training experience, that enables you to: Learn better, faster!
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More from ANSAPPARELS
The most critical shortcomings of these assumptions are1.The flow .pdf
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients.
Solution
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients..
The following statements are true1. The tendency of water to mini.pdf
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
molecules and water molecules.
3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
Cellwalls)
Solution
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
molecules and water molecules.
3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
Cellwalls).
The data is Quantitative as there are numbers , and its an observati.pdf
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination.
Solution
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination..
Thank you everyone! The answer was O(n!)SolutionThank you ever.pdf
The document thanks everyone and states that the answer to the previous problem or question was O(n!), which refers to Big O notation describing an algorithm's time complexity as being factorial or exponential in relation to the size of the input.
Solution1 ) The purpose of criminal law is to help maintain social.pdf
Solution
1 ) The purpose of criminal law is to help maintain social order and stability . The purpose of
criminal law is to establish a system of prohibitions , penalties and correctional measures to deal
with conduct that unjustifiably or inexcusably causes or threatens harm to those individual or
public interests for which state protection is appropriate .
2) Classification according to the nature of crimes
Mala in se : These are acts that are immoral or wrong in themselves or acts that are naturally evil
. Ex : Murder, rape , arson , burglary etc
Mala prohibita : These crimes are not naturally evil , but are prohibited by statute because they
infringe upon the right of the others . This type of act is not wrong in some societies but wrong in
other societies . Example : Traffic offenses , gambling etc .
Classification according to degree :
a) Felonies : Crimes that are punishable by death or imprisonment in a state prision or
penitentiary are felonies .
b) Misdemeanors : Crimes or offenses not amounting to felonies are Misdemeanors . They are
offenses for which the punishment is other than death or imprisonment in a state prison , or that
have not been designated felonies by statute.
c) Treason : Treason is the only crime described in the constitution and it is given a higher
classification than felony .
3) Example of white-collar crimes are bank fraud, blackmail , bribery , computer fraud,
counterfeiting , credit card fraud , currency schemes, embezzelement , environmental schemes .
The punishment for these crimes should be recovery of all the money made by white collar
crimes and to put the white collar criminals in prision ..
question not visibleSolutionquestion not visible.pdf
This document discusses a solution to an unseen question. It proposes addressing the issue by taking multiple steps to resolve the problem. These steps are not detailed, but the overall approach is to methodically work through the problem by breaking it down into components and addressing each part.
num.txt(Save this file under D Drive.Then the path of the file is D.pdf
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
1
2
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4
5
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9
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13
14
15
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17
18
19
20
CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<
Solution
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<.
n = 7 which is odd so median is the middle value of te data after ar.pdf
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4.
Solution
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4..
Molybdenum is the elementSolutionMolybdenum is the element.pdf
Molybdenum is a chemical element. The document provides no other details about molybdenum or its properties.
Mean absolute deviationSolutionMean absolute deviation.pdf
Mean absolute deviation is a measure of variability that calculates the average distance between each data point and the mean. It is calculated by taking the average of the absolute deviations of each data point from the mean. Mean absolute deviation provides an alternative to standard deviation that is not as sensitive to outliers in the data.
elements involved are C and N .pdf
This document discusses a solution that involves the elements of carbon and nitrogen. Specifically, the document states that the elements involved are C and N.
Information security means protecting information (data) and informa.pdf
Information security means protecting information (data) and information systems from
unauthorized access, use, disclosure, disruption, modification, or destruction.
Information Security management is a process of defining the security controls in order to
protect the information assets.
Security Program[edit]
The first action of a management program to implement information security is to have a
security program in place. Though some argue the first act would be to gain some real \"proof of
concept\" and \"explainable thru display on the monitor screen\" security knowledge. Start with
maybe understanding where OS passwords are stored within the code inside a file within a
directory. If you don\'t understand Operating Systems at the root directory level maybe you
should seek out advice from somebody who does before even beginning to implement security
program management and objectives.
Security Program Objectives
Security Management Responsibilities
Approaches to Build a Security Program
Since advancement is directly tied to how well you can convince others, who often fall outside of
your of job duties and department, as to your higher value to the company as stated by your own
effective written communication this leads to amazing resume writers and take no blame style of
email responses that seems to definitely lead to the eventual failure of company\'s standards and
actual knowledge. It is often covered up by relationships which form at the power levels within
any group of people and those who are considered so-called experts having no real idea what is
really involved under the hood of the reports/applications they use and no proof presented in
emails written when self declared claims of their expertise is made or blame is to be put on
another.
Security Controls[edit]
Security Controls can be classified into three categories
Administrative Controls which include
Technical or Logical Controls which include
Physical Controls which include
The Elements of Security[edit]
Vulnerability
Threat
Risk
Exposure
Countermeasure or Safeguard
The Relation Between the Security Elements
Core Information Security Principles[edit]
The three fundamental principles of security are availability, integrity, and confidentiality and
are commonly referred to as CIA or AIC triad which also form the main objective of any security
program.
The level of security required to accomplish these principles differs per company, because each
has its own unique combination of business and security goals and requirements.
All security controls, mechanisms, and safeguards are implemented to provide one or more of
these principles.
All risks, threats, and vulnerabilities are measured for their potential capability to compromise
one or all of the AIC principles
Confidentiality[edit]
Integrity[edit]
Availability[edit]
Information Security Management Governance[edit]
Security Governance[edit]
Governance is the set of responsibilities and practices exercised by the board and executi.
Given that T is unitaryT = 1I-ieH =1det (-T) = (-1)ndet T.pdf
Given that T is unitary
|T| = 1
|I-ieH| =1
det (-T) = (-1)ndet T
Hence det (ieH-1) = (-1)n
det ieH = 1+(-1)n
i.e. det ieH can be either 0 or 2
To prove that H is equal to its own conjugate transpose
Solution
Given that T is unitary
|T| = 1
|I-ieH| =1
det (-T) = (-1)ndet T
Hence det (ieH-1) = (-1)n
det ieH = 1+(-1)n
i.e. det ieH can be either 0 or 2
To prove that H is equal to its own conjugate transpose.
Gene expressionGene expression is the process by which the genetic.pdf
Gene expression
Gene expression is the process by which the genetic code - the nucleotide sequence - of a gene is
used to direct protein synthesis and produce the structures of the cell. Genes that code for amino
acid sequences are known as \'structural genes\'.
The process of gene expression involves two main stages:
Transcription: the production of messenger RNA (mRNA) by the enzyme RNA polymerase, and
the processing of the resulting mRNA molecule.
Translation: the use of mRNA to direct protein synthesis, and the subsequent post-translational
processing of the protein molecule.
Some genes are responsible for the production of other forms of RNA that play a role in
translation, including transfer RNA (tRNA) and ribosomal RNA (rRNA).
A structural gene involves a number of different components:
· Exons. Exons code for amino acids and collectively determine the amino acid sequence of
the protein product. It is these portions of the gene that are represented in final mature mRNA
molecule.
· Introns. Introns are portions of the gene that do not code for amino acids, and are removed
(spliced) from the mRNA molecule before translation.
Gene control regions
· Start site. A start site for transcription.
· A promoter. A region a few hundred nucleotides \'upstream\' of the gene (toward the 5\'
end). It is not transcribed into mRNA, but plays a role in controlling the transcription of the gene.
Transcription factors bind to specific nucleotide sequences in the promoter region and assist in
the binding of RNA polymerases.
· Enhancers. Some transcription factors (called activators) bind to regions called
\'enhancers\' that increase the rate of transcription. These sites may be thousands of nucleotides
from the coding sequences or within an intron. Some enhancers are conditional and only work in
the presence of other factors as well as transcription factors.
· Silencers. Some transcription factors (called repressors) bind to regions called \'silencers\'
that depress the rate of transcription.
Transcription
Transcription involves four steps:
Transcription is the process of RNA synthesis, controlled by the interaction of promoters and
enhancers. Several different types of RNA are produced, including messenger RNA(mRNA),
which specifies the sequence of amino acids in the protein product, plus transfer RNA (tRNA)
and ribosomal RNA (rRNA), which play a role in the translation process.
1. Initiation. The DNA molecule unwinds and separates to form a small open complex. RNA
polymerase binds to the promoter of the template strand.
2. Elongation. RNA polymerase moves along the template strand, synthesising an mRNA
molecule. In prokaryotes RNA polymerase is a holoenzyme consisting of a number of subunits,
including a sigma factor (transcription factor) that recognises the promoter. In eukaryotes there
are three RNA polymerases: I, II and III. The process includes a proofreading mechanism.
3. Termination. In prokaryotes there are two ways in which transcription is ter.
Dysfunctional intercellular communication may result in various cond.pdf
Dysfunctional intercellular communication may result in various conditions such as disturbances
in homeostasis, disturbed physiological conditions leading to various diseases, metastasis of
diseases, cancer promotion and reproductive dysfunction. Therefore, intercellular
communication plays an important role in various physiological and metabolic functions in cells.
Communication failure disturb entire cellular processes. But, in some types of cancers,
intercellular communication through gap junctions must be dysfunctional to stop te development
of cancer.
Solution
Dysfunctional intercellular communication may result in various conditions such as disturbances
in homeostasis, disturbed physiological conditions leading to various diseases, metastasis of
diseases, cancer promotion and reproductive dysfunction. Therefore, intercellular
communication plays an important role in various physiological and metabolic functions in cells.
Communication failure disturb entire cellular processes. But, in some types of cancers,
intercellular communication through gap junctions must be dysfunctional to stop te development
of cancer..
Dear,The answer is 2.SolutionDear,The answer is 2..pdf
The document is a short note stating that the answer to an unknown question is the number 2. It provides no context or explanation for how the answer was determined. In just two sentences, it directly states that the answer is 2 without any supporting details or calculations.
Correcty option. C. PhenotypeThe visible expressions (traits or ch.pdf
Correcty option. C. Phenotype
The visible expressions (traits or characteristics like shape, size, color, etc.) of a gene is called its
phenotype. So, the fur color (a visible expression) in dog is a phenotype.
The genetic constitution (make up) of an organism with respect to one or more specified traits is
called its genotype. For example, if gene color in dog is mediated by gene F, then gene F at both
loci (maternal and fraternal) constitute the genotype of the dog with respect to the trait “fur
color”. A diploid organism (here dog) interties two sets of haploid genome- one from its mother
and another from father. By “both loci”, it is meant that gene F on the maternally inherited
chromosome and gene F from fraternally inherited chromosome.
Two or more forms of a gene are called allele. For example, gene F responsible for fur color in
dogs may exist in two forms- F and f – one being on maternal chromosome (say, allele F) and the
other being on fraternal chromosome (say, allele f).
Genes and alleles are visible by eyes. They reside on DNA. It is the expression (product) of the
gene that becomes visible and is called a phenotype.
Solution
Correcty option. C. Phenotype
The visible expressions (traits or characteristics like shape, size, color, etc.) of a gene is called its
phenotype. So, the fur color (a visible expression) in dog is a phenotype.
The genetic constitution (make up) of an organism with respect to one or more specified traits is
called its genotype. For example, if gene color in dog is mediated by gene F, then gene F at both
loci (maternal and fraternal) constitute the genotype of the dog with respect to the trait “fur
color”. A diploid organism (here dog) interties two sets of haploid genome- one from its mother
and another from father. By “both loci”, it is meant that gene F on the maternally inherited
chromosome and gene F from fraternally inherited chromosome.
Two or more forms of a gene are called allele. For example, gene F responsible for fur color in
dogs may exist in two forms- F and f – one being on maternal chromosome (say, allele F) and the
other being on fraternal chromosome (say, allele f).
Genes and alleles are visible by eyes. They reside on DNA. It is the expression (product) of the
gene that becomes visible and is called a phenotype..
Code public class program2 { public static void main(String.pdf
Code:
public class program2 {
public static void main(String[] args) {
int arr[][]={{0,2},{2,1,2,7,8},{2,1,0,6}};
System.out.println(method(arr));
}
public static int method(int arr[][]){
int counter=0;
for(int i=0;i
Solution
Code:
public class program2 {
public static void main(String[] args) {
int arr[][]={{0,2},{2,1,2,7,8},{2,1,0,6}};
System.out.println(method(arr));
}
public static int method(int arr[][]){
int counter=0;
for(int i=0;i.
can you write the questionsSolutioncan you write the questio.pdf
This short document consists of a request to write questions, but provides no context or details about what types of questions should be written. The document simply states "can you write the questions??" without any other information.
C. If you haven’t partitioned and formatted the hard disk before ins.pdf
C. If you haven’t partitioned and formatted the hard disk before inserting the Windows 7
installation disc, you can perform both tasks from Windows 7 Setup.
Solution
C. If you haven’t partitioned and formatted the hard disk before inserting the Windows 7
installation disc, you can perform both tasks from Windows 7 Setup..
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The data is Quantitative as there are numbers , and its an observati.pdf
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Thank you everyone! The answer was O(n!)SolutionThank you ever.pdf
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Given that T is unitaryT = 1I-ieH =1det (-T) = (-1)ndet T.pdf
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Gene expressionGene expression is the process by which the genetic.pdf
Dysfunctional intercellular communication may result in various cond.pdf
Dysfunctional intercellular communication may result in various cond.pdf
Dear,The answer is 2.SolutionDear,The answer is 2..pdf
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Correcty option. C. PhenotypeThe visible expressions (traits or ch.pdf
Correcty option. C. PhenotypeThe visible expressions (traits or ch.pdf
Code public class program2 { public static void main(String.pdf
Code public class program2 { public static void main(String.pdf
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Agenda
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For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
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Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
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