The Jannuschs operated Festival Foods, a business that served conces.pdf
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The Jannuschs operated Festival Foods, a business that served concessions at festivals and events throughout Illinois and Indiana from late April to late October each year. The assets of the business included a truck and servicing trailer and equipment such as refrigerators and freezers, roasters, chairs and tables, fountain service, and signs and lighting equipment. Naffziger and her mother were interested in purchasing the business. They met several times with the Jannuschs and observed the business in operation. According to Mr. Jannusch, the parties had entered into an oral agreement on August 13, 2005, to sell Festival Foods to the Naffzigers for $150,000. For the $150,000, the Naffzigers would receive the truck and trailer, all necessary equipment, and the opportunity to work at event locations secured by the Jannuschs. The Naffzigers paid $10,000 immediately, with the balance to be paid when they received their loan money from the bank. The Naffzigers took possession of Festival Foods the next day and operated Festival Foods for the remainder of the 2005 season. The insurance and titles to the truck and trailer remained in Mr. Jannusch\'s name because he had not yet received the purchase price from the Naffzigers. Naffziger acknowledged that an oral agreement to purchase Festival Foods for $150,000 existed but stated that she could not recall specifically making an oral agreement on any particular date. The parties agreed that the Naffzigers would run Festival Foods as they pursued buying the business. The Naffzigers took possession of Festival Foods, receiving the income from the business, purchasing inventory, replacing equipment, paying taxes on the business, and paying employees. The income from Festival Foods was lower than the Naffzigers expected, and two days after the business season ended, they returned Festival Foods to the storage facility where Mr. Jannusch had stored it in the past. The Jannuschs brought an action for breach of an oral contract against the Naffzigers. The Naffzigers claimed that the deal was not sufficiently definite to be enforceable. Are they correct? Solution The Naffzigers are not correc, t in this case UCC applies and under the UCC an enforceable agreement existed. Truck and trailer were the centerpiece of the contract there fore the Naffzigers need to pay the money to the Jannuschs.
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In 1998, Congress passed legislation concerning shifting the burden .pdf
In 1998, Congress passed legislation concerning shifting the burden of proof to the IRS. The
taxpayer must introduce \"credible evidence\" to shift the burden of proof to the IRS. What
constitutes \"credible evidence?\"
Solution
Credible evidence is not evidence which is necessarily true, but is evidence worthy of belief, that
is, worthy to be considered by the jury. It is often natural, reasonable and probable as to make it
easy to believe.
Pursuant to 89 Ill. (Adm. Code 300.20), \"Credible evidence of child abuse or neglect\" means
that the available facts, when viewed in light of surrounding circumstances, would cause a
reasonable person to believe that a child was abused or neglected.”
The 1998 tax law has provided that the burden of proof in tax court cases will be on the
government, not the taxpayer. However, it also provides that this can be shifted back to the
taxpayer in a number of circumstances, mostly involving lack of taxpayer cooperation with
examinations. The issue of who bears the burden of proof is important, even if your case, like
most, never goes as far as the tax court. This is because the burden will operate as a background
issue that will tend to help to affect the settlement posture of both the taxpayer and the IRS.
The burden of production, also known as the burden of going forward, is the burden of providing
sufficient evidence on an issue in order to prevail in the dispute. In Tax Court cases initiated by
the IRS, the government bears the burden of production. IRS is internal reserve service.
In order to avoid a shift-back in the burden of proof, the taxpayer must meet his burden of
production. That is, the
taxpayer must meet his record-keeping, substantiation, and cooperation requirements.
Conclusion:
The Restructuring and Reform Act of 1998 allows for the shift of the burden of proof from
taxpayers to the government, and generally serves to protect the interests of taxpayers who are
being audited by the IRS. This policy change benefits taxpayers, who previously had been
saddled with the burden of proof in almost all civil tax matters. Taxpayers must provide
substantial and credible evidence, meet record-keeping and substantiation requirements, and
cooperate with the IRS. The burden of proof oftentimes does not shift to the IRS because the
taxpayer feels to meet one of these important provisions. For this reason, the IRS Restructuring
and Reform Act of 1998 does not always provide taxpayers much relief from the previous
system. From a practical standpoint, meticulous and accurate record-keeping is of the most
important so that the taxpayer may be able to shift the burden of proof to IRS..
If A denotes some event, what does A denote If P(A)equals=0.997, wh.pdf
If A denotes some event, what does A denote? If P(A)equals=0.997, what is the value of P(A)? If
P(A)equals=0.997, is A unusual?
Solution
Sol)
P(not A)=0.003 (1-0.997=0.003) P(not A) , meaning that compliment of A consists of all
outcomes(results) in which event A does not occur. It is not unusual(only outcomes <0.05 are
unusual)..
How do neutrophils find and get to an infectionSolutionNeutro.pdf
How do neutrophils find and get to an infection?
Solution
Neutrophils nucleus has a complex, lobulated shape. In a cross-section, neutrophils look like they
have more than one nucleus. Neutrophils have this unique nuclear shape, it makes these cells
very easy to identify under the microscope. They are very abundant, making up about 60% of the
immune cells in our blood. They are formed from stem cells in the bone marrow. Neutrophils
derives from staining characteristics on hematoxylin and eosin (H&E) histological or cytological
preparations. Neutrophils are the first cells to migrate to a site of infection, and with the help of
other innate immune cells, they work hard to keep things under control until the powerful
adaptive immune system kicks..
How many times will the following code print Welcome to Java in.pdf
How many times will the following code print \"Welcome to Java\"? int count = 0; while
(count
Solution
Ans)
15) D.10
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int count = 0;
while(count < 10){
System.out.println(\"welcome to java\");
count++;
}
}
}
16) E.45
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int y = 0;
for(int i=0;i<10;++i){
y+=i;
}
System.out.println(y);
}
}
17) B.10
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int y = 0;
int i=0;
for(i=0;i<10;++i){
y+=i;
}
System.out.println(i);
}
}
18) B.n
There are n iterations for the given for loop because it is given that i=1 to i <= n so there
aren iterations.
19) C.n-1
Here for loop starts from i=1 to i.
From your own words not copy past from internet please Talked about.pdf
From your own words not copy past from internet please Talked about\" allergies\".
Thanks.
Solution
Dear User, you have not mentioned the words limit in your question. so, a brief description of
Allergies is provided below in my own words.
Answer) Allergies are referred to several reactions or response as an outcome of hypersensitive
reaction of our body’s immune system, when exposed to several allergic substances. these
allergic substances are called as allergen. there are number of common allergens such as Dust
mites, Pollens, Insect stings, Pet dander, Mold spores, Food, household chemicals and
Medicines. these substances are not always allergic as the other people are not allergic to them,
but some people are allergic to them. our immune system makes antibodies against the foreign
particles entering our body. in case, when our immune system is weak (particularly in children) it
is possible that encounter of such allergic substances during weak immunity may result in
hypersensitive reaction. this first-time hypersensitive response is called as acute response when
our immune system starts making antibodies against those substances. further multiple exposures
to same allergens result in a higher response and the person becomes allergic to those substances.
symptoms of allergies are sneezing, wheezing and coughing, a running nose or blocked nose, a
red itchy rash, red itchy watery eyes. these symptoms can be observed within minutes of
exposure to allergens. generally, these responses are mild but a prolonged hypersensitive reaction
may result in anaphylaxis or anaphylactic shock which is a medical emergency.
As per studies, one out of five American is affected with some sort of allergies. and the allergies
are more commonly observed in the developed countries. there are different diagnostic tests to
identify whether a person has an allergic response to possible allergens or not. these tests include
patch testing, skin prick testing, blood testing etc. in these tests, a much smaller dose of allergen
is provided to the person to observe the immune response. once a person is identified allergic to
certain substance, it is consulted to avoid the exposure to those substances. while, in case of
sudden exposure they are provided with the medications such as antihistamines, epinephrine and
glucocorticoids etc. under certain limitations, immunotherapy is also provided as treatment
against and asthma and allergies to insect bites..
Forms an important part of the subunits of the ribosome. The _ are sp.pdf
Forms an important part of the subunits of the ribosome. The _ are spliced back together to form
the mRNA molecule. Is a nucleic acid that directs the production of proteins?
Solution
Protein forms an important part in the subunits of ribosomes.
The precursor messenger RNA is spliced back together to form the mRNA molecule.
Yes,nucleic acid (DNA or RNA) directs the production of proteins. They have informations for
the process of gene expression..
During DNA replication, a series of steps are required to ensure tha.pdf
During DNA replication, a series of steps are required to ensure that the genetic information in a
cell is correctly and accurately duplicated. Place the following phrases in the correct order.
Binding proteins stabilize the unwound double helix.
DNA helicase unwinds the DNA double helix.
Topoisomerase IV separates linked chromosomes.
DNA pol III adds complementary nucleotides.
DNA ligase seals any nicks in the DNA.
DNA pol I replaces the RNA primer.
Primase generates a RNA primer that binds to the origin.
Solution
During DNA replication, a correct order of steps takes place to ensure that the genetic
information in a cell is correctly and accurately duplicated..
Disease X occurs when someone has the xx genotype. However, only 20 .pdf
Disease X occurs when someone has the xx genotype. However, only 20 of 50 people with the
xx genotype actually get the disease. Of the 20 who get disease X, 10 severely display the
disease while 10 have a milder form. Explain using the appropriate terms.
Solution
Answer: XX genotype indicate that it is a autosomal recessively inherited trait and it is female.
sickle cell anaemia is one best example which match these results.
XhXh = Homozygous dominnant (10 people are severely effected).
XXh = heterozygous dominant (10 people may act as carriers) which are milder forms..
Recommended
In 1998, Congress passed legislation concerning shifting the burden .pdf
In 1998, Congress passed legislation concerning shifting the burden of proof to the IRS. The
taxpayer must introduce \"credible evidence\" to shift the burden of proof to the IRS. What
constitutes \"credible evidence?\"
Solution
Credible evidence is not evidence which is necessarily true, but is evidence worthy of belief, that
is, worthy to be considered by the jury. It is often natural, reasonable and probable as to make it
easy to believe.
Pursuant to 89 Ill. (Adm. Code 300.20), \"Credible evidence of child abuse or neglect\" means
that the available facts, when viewed in light of surrounding circumstances, would cause a
reasonable person to believe that a child was abused or neglected.”
The 1998 tax law has provided that the burden of proof in tax court cases will be on the
government, not the taxpayer. However, it also provides that this can be shifted back to the
taxpayer in a number of circumstances, mostly involving lack of taxpayer cooperation with
examinations. The issue of who bears the burden of proof is important, even if your case, like
most, never goes as far as the tax court. This is because the burden will operate as a background
issue that will tend to help to affect the settlement posture of both the taxpayer and the IRS.
The burden of production, also known as the burden of going forward, is the burden of providing
sufficient evidence on an issue in order to prevail in the dispute. In Tax Court cases initiated by
the IRS, the government bears the burden of production. IRS is internal reserve service.
In order to avoid a shift-back in the burden of proof, the taxpayer must meet his burden of
production. That is, the
taxpayer must meet his record-keeping, substantiation, and cooperation requirements.
Conclusion:
The Restructuring and Reform Act of 1998 allows for the shift of the burden of proof from
taxpayers to the government, and generally serves to protect the interests of taxpayers who are
being audited by the IRS. This policy change benefits taxpayers, who previously had been
saddled with the burden of proof in almost all civil tax matters. Taxpayers must provide
substantial and credible evidence, meet record-keeping and substantiation requirements, and
cooperate with the IRS. The burden of proof oftentimes does not shift to the IRS because the
taxpayer feels to meet one of these important provisions. For this reason, the IRS Restructuring
and Reform Act of 1998 does not always provide taxpayers much relief from the previous
system. From a practical standpoint, meticulous and accurate record-keeping is of the most
important so that the taxpayer may be able to shift the burden of proof to IRS..
If A denotes some event, what does A denote If P(A)equals=0.997, wh.pdf
If A denotes some event, what does A denote? If P(A)equals=0.997, what is the value of P(A)? If
P(A)equals=0.997, is A unusual?
Solution
Sol)
P(not A)=0.003 (1-0.997=0.003) P(not A) , meaning that compliment of A consists of all
outcomes(results) in which event A does not occur. It is not unusual(only outcomes <0.05 are
unusual)..
How do neutrophils find and get to an infectionSolutionNeutro.pdf
How do neutrophils find and get to an infection?
Solution
Neutrophils nucleus has a complex, lobulated shape. In a cross-section, neutrophils look like they
have more than one nucleus. Neutrophils have this unique nuclear shape, it makes these cells
very easy to identify under the microscope. They are very abundant, making up about 60% of the
immune cells in our blood. They are formed from stem cells in the bone marrow. Neutrophils
derives from staining characteristics on hematoxylin and eosin (H&E) histological or cytological
preparations. Neutrophils are the first cells to migrate to a site of infection, and with the help of
other innate immune cells, they work hard to keep things under control until the powerful
adaptive immune system kicks..
How many times will the following code print Welcome to Java in.pdf
How many times will the following code print \"Welcome to Java\"? int count = 0; while
(count
Solution
Ans)
15) D.10
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int count = 0;
while(count < 10){
System.out.println(\"welcome to java\");
count++;
}
}
}
16) E.45
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int y = 0;
for(int i=0;i<10;++i){
y+=i;
}
System.out.println(y);
}
}
17) B.10
public class MultiQues {
public static void main(String[] args) {
// TODO Auto-generated method stub
int y = 0;
int i=0;
for(i=0;i<10;++i){
y+=i;
}
System.out.println(i);
}
}
18) B.n
There are n iterations for the given for loop because it is given that i=1 to i <= n so there
aren iterations.
19) C.n-1
Here for loop starts from i=1 to i.
From your own words not copy past from internet please Talked about.pdf
From your own words not copy past from internet please Talked about\" allergies\".
Thanks.
Solution
Dear User, you have not mentioned the words limit in your question. so, a brief description of
Allergies is provided below in my own words.
Answer) Allergies are referred to several reactions or response as an outcome of hypersensitive
reaction of our body’s immune system, when exposed to several allergic substances. these
allergic substances are called as allergen. there are number of common allergens such as Dust
mites, Pollens, Insect stings, Pet dander, Mold spores, Food, household chemicals and
Medicines. these substances are not always allergic as the other people are not allergic to them,
but some people are allergic to them. our immune system makes antibodies against the foreign
particles entering our body. in case, when our immune system is weak (particularly in children) it
is possible that encounter of such allergic substances during weak immunity may result in
hypersensitive reaction. this first-time hypersensitive response is called as acute response when
our immune system starts making antibodies against those substances. further multiple exposures
to same allergens result in a higher response and the person becomes allergic to those substances.
symptoms of allergies are sneezing, wheezing and coughing, a running nose or blocked nose, a
red itchy rash, red itchy watery eyes. these symptoms can be observed within minutes of
exposure to allergens. generally, these responses are mild but a prolonged hypersensitive reaction
may result in anaphylaxis or anaphylactic shock which is a medical emergency.
As per studies, one out of five American is affected with some sort of allergies. and the allergies
are more commonly observed in the developed countries. there are different diagnostic tests to
identify whether a person has an allergic response to possible allergens or not. these tests include
patch testing, skin prick testing, blood testing etc. in these tests, a much smaller dose of allergen
is provided to the person to observe the immune response. once a person is identified allergic to
certain substance, it is consulted to avoid the exposure to those substances. while, in case of
sudden exposure they are provided with the medications such as antihistamines, epinephrine and
glucocorticoids etc. under certain limitations, immunotherapy is also provided as treatment
against and asthma and allergies to insect bites..
Forms an important part of the subunits of the ribosome. The _ are sp.pdf
Forms an important part of the subunits of the ribosome. The _ are spliced back together to form
the mRNA molecule. Is a nucleic acid that directs the production of proteins?
Solution
Protein forms an important part in the subunits of ribosomes.
The precursor messenger RNA is spliced back together to form the mRNA molecule.
Yes,nucleic acid (DNA or RNA) directs the production of proteins. They have informations for
the process of gene expression..
During DNA replication, a series of steps are required to ensure tha.pdf
During DNA replication, a series of steps are required to ensure that the genetic information in a
cell is correctly and accurately duplicated. Place the following phrases in the correct order.
Binding proteins stabilize the unwound double helix.
DNA helicase unwinds the DNA double helix.
Topoisomerase IV separates linked chromosomes.
DNA pol III adds complementary nucleotides.
DNA ligase seals any nicks in the DNA.
DNA pol I replaces the RNA primer.
Primase generates a RNA primer that binds to the origin.
Solution
During DNA replication, a correct order of steps takes place to ensure that the genetic
information in a cell is correctly and accurately duplicated..
Disease X occurs when someone has the xx genotype. However, only 20 .pdf
Disease X occurs when someone has the xx genotype. However, only 20 of 50 people with the
xx genotype actually get the disease. Of the 20 who get disease X, 10 severely display the
disease while 10 have a milder form. Explain using the appropriate terms.
Solution
Answer: XX genotype indicate that it is a autosomal recessively inherited trait and it is female.
sickle cell anaemia is one best example which match these results.
XhXh = Homozygous dominnant (10 people are severely effected).
XXh = heterozygous dominant (10 people may act as carriers) which are milder forms..
Darwins Theory had five main tenets (or components), what are they.pdf
Darwin\'s Theory had five main tenets (or components), what are they? ). These are not to be
confused with the logic of Natural Selection, which is sometimes referred to as the tenets of
Darwin’s theory
Solution
5 major tenets of Darwins theory are :
1. The evolution stating that the world is directionally getting transformed continuously with tha
passage of time,
2. The common descent which states that every group of the branching organism has decended
from a common ancestor
3. The gradualism which states that during evolution, the transformation happens very gradually
not at once,
4. The multiplication of species which states that the sympatric speciation leads to the
diversification in species, and
5. The natural selection which states that the evolutionary change happens due to the survival of
the fittest because the most fit would have a reproductive advantage, generation after generation..
Define toxic shock syndrome (TSS) impetigo MRSASolutionAnswer.pdf
Define: toxic shock syndrome (TSS) impetigo MRSA
Solution
Answer:
1. Toxic Shock Syndrome (TSS): It is a severe human disease caused by bacteria Staphylococcus
aureus. The bacteria produced superantigens which act as a toxins (mainly Toxic Shock
Syndrome Toxin-1) to cause this syndrome. Superantigens are bacterial proteins that stimulate
the immune system much more extensively than do normal antigens. As they provoke such a
drastic immune response thus, they are termed superantigens. Their action is to stimulate T cells
to proliferate nonspecifically through simultaneous interaction with class II MHC proteins on
antigen-presenting cells and variable regions on the chain of the T-cell receptor complex.
It is most commonly seen in females who use superabsorbent tampons during menstruation.
However, the toxin associated with this syndrome is also produced in men and in
nonmenstruating women by S. aureus present at sites other than the genital area such as surgical
wound infections.
Symptoms: Low blood pressure, fever, diarrhea, an extensive skin rash (like sun burn), and
ultimately shedding of the skin.
2. Impetigo: It is the most frequently diagnosed skin (superficial cutaneous) infection caused by
bacteria Staphylococcuspyogenes and S. aureus. It is most commonly seen in children.
Symptoms: The encrusted pustules, crusty lesions and vesicles surrounded by a red border,
usually located on the face.
3. MRSA (Methicillin Resistant Staphylococcusaureus): Bacteria Staphylococcus aureus is a
cause of considerable morbidity and mortality as a nosocomial or hospital-acquired pathogen
since late 1950s and early 1960s. However, penicillinase-resistant, semisynthetic penicillins have
proved to be successful antimicrobial agents in the treatment of staphylococcal infections. But it
is unfortunate that the staphylococci become resistant through procuring of a chromosomal gene
(mecA) that encodes an alternate target protein which is not inactivated by methicillin. The
majority of the strains are resistant to several of the most commonly used antimicrobial agents,
including macrolides, aminoglycosides, and the beta-lactam antibiotics, including the latest
generation of cephalosporins. Thus, MSRA strains have recently emerged as a major, difficult to
treat, hospital-acquired infection in humans, therefore considered as super-bug..
A system was set up in parallel with two components, A and B. The pr.pdf
A system was set up in parallel with two components, A and B. The probability of failure for A
is .20 and for B is .10. In order for the system to fail, both must fail. What is the probability of a
system failure?
Solution
P( system to fail, both must fail ) = 1 - P( A Fail) * P( B Fail) = 1 - [ (1-0.20) ( 1-0.10) ] = 1 -
0.72 = 0.28.
All content is made permanent for future use. All content will be lo.pdf
All content is made permanent for future use. All content will be lost. How would a port most
likely be used? A. To connect additional input, output, communication devices to a PC
Solution
Question 7
A bay would most likely hold a HARD DRIVE.
A hard drive is a component that is installed in a bay inside the system case..
a. A waitress is serving 5 people at a table. She has the 5 dishes t.pdf
a. A waitress is serving 5 people at a table. She has the 5 dishes they ordered (all 5 different), but
she can\'t remember who gets what. How many different possible way are there for her to give
the 5 dishes to the 5 people? Explain you answer
b. Based on part (a), if the waitress just hands the dishes out randomly, what is the probability
that she will hand the dishes out correctly
Solution
a. A waitress is serving 5 people at a table. She has the 5 dishes they ordered (all 5 different),
but she can\'t remember who gets what. How many different possible way are there for her to
give the 5 dishes to the 5 people? Explain you answer different possible ways = 5! =120
WAYS different possible ways first person get dish =5 different possible ways second
person get dish =4 different possible ways third person get dish =3 different possible ways
fourth person get dish =2 different possible way fifth person get dish =1 different possible
ways = 5*4*3*2*1 =120 b. Based on part (a), if the waitress just hands the dishes out
randomly, what is the probability that she will hand the dishes out correctly probability =
1/120 =0.0083.
Biochemistry homework Focus concept ripening fruit and the activity.pdf
Biochemistry homework: Focus concept ripening fruit and the activity tin methyl- The link
between the production of methanol in wild type and transgenic tomato esterase, the enzyme
responsible for methanol production, is examined in fruit. Prerequisite Enzyme kinetics and
inhibition. Background major source ofthis methanol is likely Methanol is produced by fruit
during the ripening process. The to be from the enzyme p methylesterase which acts upon pectin
methyl esters to de-esterify them and produce methanol and other pectic substances One might
assume that the production of methanol is correlated with PME activity, but this has not been
firmly established. In this case, the investigators at the Plant Science Department at Rutgers
University developed a transgenic tomato which was deficient in PME activity. By comparing
the behavior of the transgenic tomato with the wild type tomato, the production of methanol in
the ripening tomato fruit may be better understood. Questions breaker (Br), turning (Tru) and red
ripe (RR) tomatoes. They also determined the amount of PME protein present by using a
Western blot assay, which uses a specific antibody to detect the PME protein. The results are
shown in Figure 12.1. What is your interpretation ofthese results? Transgenio ING MG Be
Figure 12.1: PME protein levels (A) and activ- ity (B) in wild-type and transgenic tomatoes.
(Modified from Frenkel, et al. 1998.)
Solution
As given in A the protein level of PME is completely negligible in transgenic variety and present
in wild type. In wild type variety the protein level in less in the IMG stage as compared to the
subsequent stages this is because the expression of PME increases as the ripening of tomato
increases.
in fig. B : In the case of activity of the PME in the wild type variety increase is shown until it
reaches the Tu stage and it starts decreasing in the fully ripening stage since the PME activity is
not required to do the ripening.
little activity in the transgenic variety is also shown because of the endogenous expression of
PME because the PME is only altered not fully knockout.
A process that has its representative PCB in a device queue is in th.pdf
A process that has its representative PCB in a device queue is in the _____ state.
New
ready
running
terminated
blocked or waiting
Solution
When a process in currently running in the cpu and suddenly have to wait for i/o then scheduler
save its state ( program counter registers) into pcb and then add to the device queue which
process Wait for i/o from it. So here answer is waiting.
Which of the following statements is (are) trueI Cash flow is a.pdf
Which of the following statements is (are) true:
I : Cash flow is about predicting when money will come in and will go out.
II: Cash flow is not about making sure that there is enough money to pay your debts.
Both statements are true
Solution
Answer: Both statements are false
Explanation:
Both the Statements are false because cash flow can be termed as total net amount of money
being transfered into and out of business, specifically as effecting liquidity. So, Cash neither
predicting nor making sure..
What is the relationship of the epithelium of the epidermis to the co.pdf
What is the relationship of the epithelium of the epidermis to the connective tissues layers of the
dermis? Describe the boundary between the epidermis and dermis and the various types of
connective tissues that compose the dermis.
Solution
Epidermis is a type of epithelium tissue containing typically thin sheets of cells. Epidermis forms
the outermost layer of the skin containing 4 to 5 layers of epithelial cells.The outermost part of
the epidermis is comprises of flattened cells in stratified layers overlying on a basal layer
(stratum basale) made up of columnar cells arranged perpendicularly. Beneath the epidermis lies
the middle layer called dermis layer. It is composed of dense irregular connective tissue and
areolar connective tissue such as a collagen with elastin arranged in bundled and woven pattern.
This layer comprise of the papillary and reticular layers. Dermis contains connective tissues,
vessels, glands, follicles, hair roots and sensory nerve endings.
This superficial paillary layer of the dermis, made up of areolar connective tissue like collagen
and elastin fibers in a loose mesh, projects itself into the stratum basale of the epidermis to form
finger-like dermal papillae. The dermis layer also consists of blood and lymph vessels and
nerves, which end in epidermal layer and also hair follicles and sweat glands which project out
through epidermis.
Thus dermis is the layer of the skin between the epidermic and the subcutaneous tissues,
primarily composed of dense irregulat connective tissue. These connective tissue provides
cushions to the body from stress.Its area adjoining the epidermis is called as papillary region and
the deeper area is known as reticular dermis. Dermis is tightly connected to epidermis through a
basement membrane. Papillary layercomprising of mainly collagen, intewines with the rete
ridges of the epidermis. Reticular layer consists of irregular connective tissue fibers, densely
packed collagen, elantin and reticular fibers. Structurally, dermis comprises of collagen, elastic
fibers and fibrillar matrix. It also provides the sence of touch, heat etc..
What is a linked listWhat is a linked lists general syntaxCan .pdf
What is a linked list?
What is a linked lists general syntax?
Can you give me any tips on using linked lists or any extra general info I may need to know?
Solution
Linked Lists: It is a collection of data elements. A single linked list points to the next element of
the linked list. A double linked list points to both the previous and next elements.
For every node in the linked list, it has 2 parts - data and pointer
Unlike arrays, linked lists are not stored at continguous location. As compared to arrays, linked
lists offer dynamic size (we can add or delete elements from the linked list easily). A
disadvantage as compared to arrays being, random access of the element is not possible in a
linked list.
General Syntax in C:
//node
struct node
{
int data; // data element
struct node *next; // pointer to next element
}
Best way to learn linked lists is to implement the data structure along with some basic operations
on it: Inserting an element (at the start, at the end, at the middle), Deleting an element (from the
start, from the end, from the middle), Counting the number of elements etc..
What is the output of running class GenericMethodDemo of the followi.pdf
What is the output of running class GenericMethodDemo of the following code?
public class GenericMethodDemo {
public static void main(String[] args ) {
Integer[] integers = {1, 2, 3, 4, 5};
String[] strings = {\"London\", \"Paris\", \"Dhaka\",
\"Austin\", \"Sydney\"};
GenericMethodDemo.print(integers);
GenericMethodDemo.print(strings);
}
public static void print(E[] list) {
for (int i = 0; i < list.length; i++){
System.out.print(list[i] + \" \");
}
System.out.println();
}
}
What is the output of running class GenericMethodDemo of the following code?
public class GenericMethodDemo {
public static void main(String[] args ) {
Integer[] integers = {1, 2, 3, 4, 5};
String[] strings = {\"London\", \"Paris\", \"Dhaka\",
\"Austin\", \"Sydney\"};
GenericMethodDemo.print(integers);
GenericMethodDemo.print(strings);
}
public static void print(E[] list) {
for (int i = 0; i < list.length; i++){
System.out.print(list[i] + \" \");
}
System.out.println();
}
}
Solution
output 1 :
London
Paris
Dhaka
Austin
Sydney
The above is the output of hte above program.
What do you think are the roles of professional societies in contemp.pdf
What do you think are the roles of professional societies in contemporary engineering? Do you
expect to join a professional engineering society in the future?
Solution
Roles of professional societies in contemporary engineering:
Engineers have the very unique role for solving social problems using the skills and talent of
theirs.They solve the problems using devices, systems and machines.They are responsible for
connecting, supplying and protecting the society.
One of the key area considered is known as CONNECT. The engineers links communities
through transport infrastructure. The roads,railways and aircraft transport systems allow people
to travel from one end of globe to other.Also there are technologies like 3G, 4G which enables
faster communication with people without any physical transportation.So the Engineers are
responsible for connecting people over the globe.
The other strand is SUPPLY. The society is not only contributed but also sustained by
Water,Gas,Electricity and Sewage Infrastructure.These system are developed by the Engineers
which serves the basic needs of everyone.
Also another considering aspect is the Protection. There are so many ways in which the engineer
is directly or indirectly responsible for protection of the society.The Engineers are responsible
for designing the flood defense and military systems.They are also responsible for maintenance
of these systems.By inventing such system they become able to protect their country.
In addition to these, Engineers are responsible for shaping the future.Their efforts of saving the
information and utilizing it on required times has revolutionized the era.
Due to the above mentioned roles we can join professional engineering to better shape our
society..
What is the compensation depth for phytoplankton How does it differ .pdf
What is the compensation depth for phytoplankton? How does it differ from the critical depth?
Solution
Planktons are two types phytoplanktons and these are weakly swimming and free floating
microalgae and photosynthetic whereas zooplanktons are fast swimming and feed on
phytoplanktons. The compensation depth for phytoplankton is mainly useful to denote the
growth of phytoplankton blooms and it is defined as the depth at which the net growth rate of
phytoplankton is zero. The compensation depth of phytoplankton bloom is 0.1 to 1% (above this
depth population grows) where solar radiation is going to reach adequately for their survival. The
critical depth is defined as the critical depths is differ from it in which the phytoplankton biomass
cannot grow adequately after a certain depth interval.
what are structured data types give an exampleSolutionDefini.pdf
what are structured data types ? give an example
Solution
Definition of structured data types:
A structure is a user defined data type that groups logically related data items of different data
types into a single unit.All the elements of a structure are stored at contiguous memory locations.
A variable of structure type can store multiple data items of different data types under the one
name.As the data of contact_information thst is firstname, lastname,homephone, mobilephone
are structured data types
Syntax:
struct name_of_the_structure
{
type_1 field_name1;
type_2 field_name2;
...
type_N field_nameN;
};
Example:
struct contact_information
{
char firstname[FIRST_SIZE];
char lastname[LAST_SIZE];
unsigned int homephone;
unsigned int mobilephone;
};.
VoIP uses which of the following because network congestion can be s.pdf
VoIP uses which of the following because network congestion can be such a critical problem?
A. Time-division multiplexing
B. TCP protocol
C. VLANs technology
D. Isochronous design
Solution
D. VoIP is very time sensitive and, as such, should be based on an isochronous design. This
means that the entire system must be engineered to deliver output with exactly the same timing
as the input. FireWire is another example of a device that contains an isochronous interface.
Answer A is incorrect because VoIP does not use time-division multiplexing. Answer B is
incorrect because VoIP uses UDP for the voice portion of the call, not TCP. Some
implementations of VoIP can use TCP for setup and call control. Answer C is incorrect because
VLANs are not used for timing and delay problems, but are used to separate the VoIP from
general traffic to make it more secure from sniffing..
Week #9 Nervous System IIIThe sequence of information flow in re.pdf
Week #9: Nervous System III:
The sequence of information flow in response to smelling popcorn at the movie theater would
be:
A. impulses are sent via Cranial Nerve I to the olfactory bulbs and analyzed
B. perception brain interprets as the Smell of hot, buttered popcorn
C. olfactory sensory receptor cells in the epithelium are activated
D. impulse sent to olfactory cortex of the temporary lobes
E. sensation (Recalled memory from limbic system)
F. impulse travels via olfactory tracts to limbic system
Solution
The sequence of information as follows:
C. olfactory sensory receptor cells in the epithelium are activated
D. impulse sent to olfactory cortex of the temporary lobes
A. impulses are sent via Cranial Nerve I to the olfactory bulbs and analyzed
F. impulse travels via olfactory tracts to limbic system
E. sensation (Recalled memory from limbic system)
B. perception brain interprets as the Smell of hot, buttered popcorn.
Under what assumptions would Multiple Regtression be usedSoluti.pdf
Under what assumptions would Multiple Regtression be used?
Solution
Firstly, multiple linear regression needs the relationship between the independent and dependent
variables to be linear.
Secondly, the multiple linear regression analysis requires all variables to be normal.
Thirdly, multiple linear regression assumes that there is little or no multicollinearity in the data
Fourthly, multiple linear regression analysis requires that there is little or no autocorrelation in
the data.
The last assumption the multiple linear regression analysis makes is homoscedasticity..
Trying to figure out this sequencing quesiton. Please explain how yo.pdf
Trying to figure out this sequencing quesiton. Please explain how you would approach this and
what the sequence would be at the end. Thank you 2. Determine the sequence of the 14-mer
peptide given the following data: a. Amino acid composition: (4S, 2L, F, G, I, K, M, T, W, Y) b.
ment of peptide with 1-fluoro-2,4-dinitro benzene followed by hydrolysis DNP-S
Solution
(a) 4 serine, 2 leucine, phenyalanine, glycine, isoleucine, lysine, methoinine, thereonine,
tryptophan, tyrosine..
(g) 2 serine, leucine, tryptophan, phenyalanine, glycine, isoleucine, lysine, leucine, homoserine,
thereonine, tyrosine
(e) Glycine, lysine, leucine, phenyalanine, isoleucine, tryptophan, thereonine, methoinine.
To create a node for a binary search tree (BTNode, ) and to create a .pdf
To create a node for a binary search tree (BTNode, ) and to create a binary search tree of
comparable objects (Tree BTNode BTNode -data:E -left: BTNode -right: BTNode +BTNode
(newData:E, newLeft: BTNode, newRight: BTNode) +getData():E +getLeft():BTNode
+getRight():BTNode +getRightMostData():E +inOrderPrint():void +removeRightmost():
BTNode +setData(newData:E):void +setLeft(newleft: BTNode):void +setRight(newRight:
BTNode):void needed for remove needed for remove Tree for Lab 7 Tree -root: 8TNode -
numltems:int +Tree() +add(element:E):void +remove(element:E):boolean + remove
+size():int + printTree():void
Solution
//BTNode.java
public class BTNode
{
private E data;
private BTNode left;
private BTNode right;
public BTNode(E newData, BTNode newLeft, BTNode newRight)
{
data = newData;
left = newLeft;
right = newRight;
}//end constructor
/**
* Returns the data
* @return The data
*/
public E getData()
{
return data;
}//end getData
/**
* Returns the left link
* @return the left link
*/
public BTNode getLeft()
{
return left;
}//end getLeft
/**
* Returns the right link
* @return the right link
*/
public BTNode getRight()
{
return right;
}//end getRight
/**
* Prints the tree in order from left to right.
*/
public void inOrderPrint()
{
if (left != null){
left.inOrderPrint();
}//end left
System.out.println(data);
if(right != null){
right.inOrderPrint();
}//end right
}//end inOrderPrint
/**
* Sets the data in the node
* @param newData The new data to be passed
*/
public void setData(E newData)
{
data = newData;
}//end setData
/**
* Sets a new link to left
* @param newLeft the new link to go left
*/
public void setLeft(BTNode newLeft)
{
left = newLeft;
}//end setLeft
/**
* Sets a new link to the right
* @param newRight the new link to the right
*/
public void setRight(BTNode newRight)
{
right = newRight;
}//end setRight
public E getRightMostData()
{
if(right == null){
return data;
} else {
return right.getRightMostData();
}//end right == null if-else
}//end getRightMostData
public BTNode removeRightmost()
{
if (right == null){
return left;
} else {
right = right.removeRightmost();
return this;
}//end if right == null if=-else
}//end removeRightMost
}//end BTNode
=====================================================================
==
//Tree.java
public class Tree>
{
private BTNode root;
private int numItems;
/**
* No-arg constructor
*/
public Tree()
{
root = null;
numItems = 0;
}//end constructor
public void add(E element)
{
if (root == null) {
root = new BTNode(element, null, null);
}else{
BTNode cursor = root;
boolean done = false;
while(!done){
if (element.compareTo(cursor.getData()) <= 0){//cursor.getData().compareTo(element) > 0){
if (cursor.getLeft() == null) {
cursor.setLeft(new BTNode(element,null,null));
done = true;
} else {
cursor = cursor.getLeft();
} //end left is null if-else
}else{
if(cursor.getRight() == null){
cursor.setRight(new BTNode(element, null, null));
done = true;
}else{
cursor = cursor.getRight();
}//end cursor.getRight if-else
}/.
The Morse code (see Table 6.10 in book) is a common code that is use.pdf
The Morse code (see Table 6.10 in book) is a common code that is used to encode messages
consisted of letters and digits. Each letter consists of a series of dots and dashes; for example, the
code for the letter a is *- and the code for the letter b is -***. Store each letter of the alphabet in a
node of a binary tree of level 5. The root node is at level 1 and stores no letter. The left node at
level 2 stores the letter e (code is *), and the right node stores letter t (code is -). The 4 nodes at
level 3 store the letters with codes (**, *-, -*, --). To build the tree (see Figure 6.38 in book),
read a file in which each line consists of a letter followed by its code. The letters should be
ordered by tree level. To find the position for a letter in the tree, scan the code and branch left for
a dot and branch right for a dash. Encode a message by replacing each letter by its code symbol.
Then decode the message using the Morse code tree. Make sure you use a delimiter symbol
between coded letters.
Implementation Notes: You will need to complete the generic implementation of the BinaryTree
data structure first (see slides). Once it is complete, solve the problem above (note that the book
uses dots and dashes, we will use the asterisk (*) as the dot, hyphen (-) as a dash, and space ( ) as
the delimiter symbol between coded letters). Provide a menu program that asks the user to 1) test
output for all morse code letters with their respective translated alphabet letters (make sure you
are using the binary tree to do the actual translation, output as a nicely formatted table), 2) enter
an input file name to decode morse code and output the translated text to the screen, or 3) enter
in a line of morse code through the console to decode morse code and output the translated text
to the screen
relevenat picture of both the tree diagram and the table of morse code
Solution
JAVA CODE:
import java.io.*;
import java.util.*;
public class MorseCoder implements MorseCodeInterface {
private MorseNode root;
/**
* constructor to build the tree
*/
public MorseCoder() {
root = new MorseNode();
readTreeInfo();
}
/**
* reads in the tree info from the text file (helper method)
*/
private void readTreeInfo() {
Scanner input = null;
try {
input = new Scanner(new File(\"encodings.txt\"));
} catch (FileNotFoundException exception) {
System.out.println(\"File not found!\");
}
while (input.hasNextLine()) {
String data = input.nextLine().trim();
if (data.length() > 0) {
add(data.substring(1).trim(), data.charAt(0));
}
}
input.close();
}
/**
* adds the letter to the tree based on the mcode string (helper method)
* @param mcode the string being fed in
* @param ltr the letter being added at the node
*/
private void add(String mcode, char ltr) {
MorseNode current = root;
String signal = \" \";
for (int i = 0; i < mcode.length(); i++) {
signal = mcode.substring(i, i + 1);
if (signal.equals(\".\")) {
if (current.getLeft() != null) {
current = current.getLeft();
} else {
current.set.
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
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Solution
5 major tenets of Darwins theory are :
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2. The common descent which states that every group of the branching organism has decended
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Solution
Answer:
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A system was set up in parallel with two components, A and B. The pr.pdf
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Solution
P( system to fail, both must fail ) = 1 - P( A Fail) * P( B Fail) = 1 - [ (1-0.20) ( 1-0.10) ] = 1 -
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Solution
a. A waitress is serving 5 people at a table. She has the 5 dishes they ordered (all 5 different),
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Solution
As given in A the protein level of PME is completely negligible in transgenic variety and present
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little activity in the transgenic variety is also shown because of the endogenous expression of
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A process that has its representative PCB in a device queue is in the _____ state.
New
ready
running
terminated
blocked or waiting
Solution
When a process in currently running in the cpu and suddenly have to wait for i/o then scheduler
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Explanation:
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Solution
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This superficial paillary layer of the dermis, made up of areolar connective tissue like collagen
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Thus dermis is the layer of the skin between the epidermic and the subcutaneous tissues,
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What is a linked list?
What is a linked lists general syntax?
Can you give me any tips on using linked lists or any extra general info I may need to know?
Solution
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linked list.
General Syntax in C:
//node
struct node
{
int data; // data element
struct node *next; // pointer to next element
}
Best way to learn linked lists is to implement the data structure along with some basic operations
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What is the output of running class GenericMethodDemo of the followi.pdf
What is the output of running class GenericMethodDemo of the following code?
public class GenericMethodDemo {
public static void main(String[] args ) {
Integer[] integers = {1, 2, 3, 4, 5};
String[] strings = {\"London\", \"Paris\", \"Dhaka\",
\"Austin\", \"Sydney\"};
GenericMethodDemo.print(integers);
GenericMethodDemo.print(strings);
}
public static void print(E[] list) {
for (int i = 0; i < list.length; i++){
System.out.print(list[i] + \" \");
}
System.out.println();
}
}
What is the output of running class GenericMethodDemo of the following code?
public class GenericMethodDemo {
public static void main(String[] args ) {
Integer[] integers = {1, 2, 3, 4, 5};
String[] strings = {\"London\", \"Paris\", \"Dhaka\",
\"Austin\", \"Sydney\"};
GenericMethodDemo.print(integers);
GenericMethodDemo.print(strings);
}
public static void print(E[] list) {
for (int i = 0; i < list.length; i++){
System.out.print(list[i] + \" \");
}
System.out.println();
}
}
Solution
output 1 :
London
Paris
Dhaka
Austin
Sydney
The above is the output of hte above program.
What do you think are the roles of professional societies in contemp.pdf
What do you think are the roles of professional societies in contemporary engineering? Do you
expect to join a professional engineering society in the future?
Solution
Roles of professional societies in contemporary engineering:
Engineers have the very unique role for solving social problems using the skills and talent of
theirs.They solve the problems using devices, systems and machines.They are responsible for
connecting, supplying and protecting the society.
One of the key area considered is known as CONNECT. The engineers links communities
through transport infrastructure. The roads,railways and aircraft transport systems allow people
to travel from one end of globe to other.Also there are technologies like 3G, 4G which enables
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responsible for connecting people over the globe.
The other strand is SUPPLY. The society is not only contributed but also sustained by
Water,Gas,Electricity and Sewage Infrastructure.These system are developed by the Engineers
which serves the basic needs of everyone.
Also another considering aspect is the Protection. There are so many ways in which the engineer
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for designing the flood defense and military systems.They are also responsible for maintenance
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Due to the above mentioned roles we can join professional engineering to better shape our
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What is the compensation depth for phytoplankton How does it differ .pdf
What is the compensation depth for phytoplankton? How does it differ from the critical depth?
Solution
Planktons are two types phytoplanktons and these are weakly swimming and free floating
microalgae and photosynthetic whereas zooplanktons are fast swimming and feed on
phytoplanktons. The compensation depth for phytoplankton is mainly useful to denote the
growth of phytoplankton blooms and it is defined as the depth at which the net growth rate of
phytoplankton is zero. The compensation depth of phytoplankton bloom is 0.1 to 1% (above this
depth population grows) where solar radiation is going to reach adequately for their survival. The
critical depth is defined as the critical depths is differ from it in which the phytoplankton biomass
cannot grow adequately after a certain depth interval.
what are structured data types give an exampleSolutionDefini.pdf
what are structured data types ? give an example
Solution
Definition of structured data types:
A structure is a user defined data type that groups logically related data items of different data
types into a single unit.All the elements of a structure are stored at contiguous memory locations.
A variable of structure type can store multiple data items of different data types under the one
name.As the data of contact_information thst is firstname, lastname,homephone, mobilephone
are structured data types
Syntax:
struct name_of_the_structure
{
type_1 field_name1;
type_2 field_name2;
...
type_N field_nameN;
};
Example:
struct contact_information
{
char firstname[FIRST_SIZE];
char lastname[LAST_SIZE];
unsigned int homephone;
unsigned int mobilephone;
};.
VoIP uses which of the following because network congestion can be s.pdf
VoIP uses which of the following because network congestion can be such a critical problem?
A. Time-division multiplexing
B. TCP protocol
C. VLANs technology
D. Isochronous design
Solution
D. VoIP is very time sensitive and, as such, should be based on an isochronous design. This
means that the entire system must be engineered to deliver output with exactly the same timing
as the input. FireWire is another example of a device that contains an isochronous interface.
Answer A is incorrect because VoIP does not use time-division multiplexing. Answer B is
incorrect because VoIP uses UDP for the voice portion of the call, not TCP. Some
implementations of VoIP can use TCP for setup and call control. Answer C is incorrect because
VLANs are not used for timing and delay problems, but are used to separate the VoIP from
general traffic to make it more secure from sniffing..
Week #9 Nervous System IIIThe sequence of information flow in re.pdf
Week #9: Nervous System III:
The sequence of information flow in response to smelling popcorn at the movie theater would
be:
A. impulses are sent via Cranial Nerve I to the olfactory bulbs and analyzed
B. perception brain interprets as the Smell of hot, buttered popcorn
C. olfactory sensory receptor cells in the epithelium are activated
D. impulse sent to olfactory cortex of the temporary lobes
E. sensation (Recalled memory from limbic system)
F. impulse travels via olfactory tracts to limbic system
Solution
The sequence of information as follows:
C. olfactory sensory receptor cells in the epithelium are activated
D. impulse sent to olfactory cortex of the temporary lobes
A. impulses are sent via Cranial Nerve I to the olfactory bulbs and analyzed
F. impulse travels via olfactory tracts to limbic system
E. sensation (Recalled memory from limbic system)
B. perception brain interprets as the Smell of hot, buttered popcorn.
Under what assumptions would Multiple Regtression be usedSoluti.pdf
Under what assumptions would Multiple Regtression be used?
Solution
Firstly, multiple linear regression needs the relationship between the independent and dependent
variables to be linear.
Secondly, the multiple linear regression analysis requires all variables to be normal.
Thirdly, multiple linear regression assumes that there is little or no multicollinearity in the data
Fourthly, multiple linear regression analysis requires that there is little or no autocorrelation in
the data.
The last assumption the multiple linear regression analysis makes is homoscedasticity..
Trying to figure out this sequencing quesiton. Please explain how yo.pdf
Trying to figure out this sequencing quesiton. Please explain how you would approach this and
what the sequence would be at the end. Thank you 2. Determine the sequence of the 14-mer
peptide given the following data: a. Amino acid composition: (4S, 2L, F, G, I, K, M, T, W, Y) b.
ment of peptide with 1-fluoro-2,4-dinitro benzene followed by hydrolysis DNP-S
Solution
(a) 4 serine, 2 leucine, phenyalanine, glycine, isoleucine, lysine, methoinine, thereonine,
tryptophan, tyrosine..
(g) 2 serine, leucine, tryptophan, phenyalanine, glycine, isoleucine, lysine, leucine, homoserine,
thereonine, tyrosine
(e) Glycine, lysine, leucine, phenyalanine, isoleucine, tryptophan, thereonine, methoinine.
To create a node for a binary search tree (BTNode, ) and to create a .pdf
To create a node for a binary search tree (BTNode, ) and to create a binary search tree of
comparable objects (Tree BTNode BTNode -data:E -left: BTNode -right: BTNode +BTNode
(newData:E, newLeft: BTNode, newRight: BTNode) +getData():E +getLeft():BTNode
+getRight():BTNode +getRightMostData():E +inOrderPrint():void +removeRightmost():
BTNode +setData(newData:E):void +setLeft(newleft: BTNode):void +setRight(newRight:
BTNode):void needed for remove needed for remove Tree for Lab 7 Tree -root: 8TNode -
numltems:int +Tree() +add(element:E):void +remove(element:E):boolean + remove
+size():int + printTree():void
Solution
//BTNode.java
public class BTNode
{
private E data;
private BTNode left;
private BTNode right;
public BTNode(E newData, BTNode newLeft, BTNode newRight)
{
data = newData;
left = newLeft;
right = newRight;
}//end constructor
/**
* Returns the data
* @return The data
*/
public E getData()
{
return data;
}//end getData
/**
* Returns the left link
* @return the left link
*/
public BTNode getLeft()
{
return left;
}//end getLeft
/**
* Returns the right link
* @return the right link
*/
public BTNode getRight()
{
return right;
}//end getRight
/**
* Prints the tree in order from left to right.
*/
public void inOrderPrint()
{
if (left != null){
left.inOrderPrint();
}//end left
System.out.println(data);
if(right != null){
right.inOrderPrint();
}//end right
}//end inOrderPrint
/**
* Sets the data in the node
* @param newData The new data to be passed
*/
public void setData(E newData)
{
data = newData;
}//end setData
/**
* Sets a new link to left
* @param newLeft the new link to go left
*/
public void setLeft(BTNode newLeft)
{
left = newLeft;
}//end setLeft
/**
* Sets a new link to the right
* @param newRight the new link to the right
*/
public void setRight(BTNode newRight)
{
right = newRight;
}//end setRight
public E getRightMostData()
{
if(right == null){
return data;
} else {
return right.getRightMostData();
}//end right == null if-else
}//end getRightMostData
public BTNode removeRightmost()
{
if (right == null){
return left;
} else {
right = right.removeRightmost();
return this;
}//end if right == null if=-else
}//end removeRightMost
}//end BTNode
=====================================================================
==
//Tree.java
public class Tree>
{
private BTNode root;
private int numItems;
/**
* No-arg constructor
*/
public Tree()
{
root = null;
numItems = 0;
}//end constructor
public void add(E element)
{
if (root == null) {
root = new BTNode(element, null, null);
}else{
BTNode cursor = root;
boolean done = false;
while(!done){
if (element.compareTo(cursor.getData()) <= 0){//cursor.getData().compareTo(element) > 0){
if (cursor.getLeft() == null) {
cursor.setLeft(new BTNode(element,null,null));
done = true;
} else {
cursor = cursor.getLeft();
} //end left is null if-else
}else{
if(cursor.getRight() == null){
cursor.setRight(new BTNode(element, null, null));
done = true;
}else{
cursor = cursor.getRight();
}//end cursor.getRight if-else
}/.
The Morse code (see Table 6.10 in book) is a common code that is use.pdf
The Morse code (see Table 6.10 in book) is a common code that is used to encode messages
consisted of letters and digits. Each letter consists of a series of dots and dashes; for example, the
code for the letter a is *- and the code for the letter b is -***. Store each letter of the alphabet in a
node of a binary tree of level 5. The root node is at level 1 and stores no letter. The left node at
level 2 stores the letter e (code is *), and the right node stores letter t (code is -). The 4 nodes at
level 3 store the letters with codes (**, *-, -*, --). To build the tree (see Figure 6.38 in book),
read a file in which each line consists of a letter followed by its code. The letters should be
ordered by tree level. To find the position for a letter in the tree, scan the code and branch left for
a dot and branch right for a dash. Encode a message by replacing each letter by its code symbol.
Then decode the message using the Morse code tree. Make sure you use a delimiter symbol
between coded letters.
Implementation Notes: You will need to complete the generic implementation of the BinaryTree
data structure first (see slides). Once it is complete, solve the problem above (note that the book
uses dots and dashes, we will use the asterisk (*) as the dot, hyphen (-) as a dash, and space ( ) as
the delimiter symbol between coded letters). Provide a menu program that asks the user to 1) test
output for all morse code letters with their respective translated alphabet letters (make sure you
are using the binary tree to do the actual translation, output as a nicely formatted table), 2) enter
an input file name to decode morse code and output the translated text to the screen, or 3) enter
in a line of morse code through the console to decode morse code and output the translated text
to the screen
relevenat picture of both the tree diagram and the table of morse code
Solution
JAVA CODE:
import java.io.*;
import java.util.*;
public class MorseCoder implements MorseCodeInterface {
private MorseNode root;
/**
* constructor to build the tree
*/
public MorseCoder() {
root = new MorseNode();
readTreeInfo();
}
/**
* reads in the tree info from the text file (helper method)
*/
private void readTreeInfo() {
Scanner input = null;
try {
input = new Scanner(new File(\"encodings.txt\"));
} catch (FileNotFoundException exception) {
System.out.println(\"File not found!\");
}
while (input.hasNextLine()) {
String data = input.nextLine().trim();
if (data.length() > 0) {
add(data.substring(1).trim(), data.charAt(0));
}
}
input.close();
}
/**
* adds the letter to the tree based on the mcode string (helper method)
* @param mcode the string being fed in
* @param ltr the letter being added at the node
*/
private void add(String mcode, char ltr) {
MorseNode current = root;
String signal = \" \";
for (int i = 0; i < mcode.length(); i++) {
signal = mcode.substring(i, i + 1);
if (signal.equals(\".\")) {
if (current.getLeft() != null) {
current = current.getLeft();
} else {
current.set.
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