When the butanol is protonated, it is ionic and hence more soluble .pdf
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When the butanol is protonated, it is ionic and hence more soluble in water. C4H9-OH + H2SO4 ===> C4H9-OH2(+) + HSO4- The conjugate acid is a polar cation more compatible with the watery than with the organic phase of the mix. Solution When the butanol is protonated, it is ionic and hence more soluble in water. C4H9-OH + H2SO4 ===> C4H9-OH2(+) + HSO4- The conjugate acid is a polar cation more compatible with the watery than with the organic phase of the mix..
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Where is the reaction See what is the no. of m.pdf
Where is the reaction?? See what is the no. of moles.Which has more moles is and
Excess and other is limiting.
Solution
Where is the reaction?? See what is the no. of moles.Which has more moles is and
Excess and other is limiting..
viscous increases due to bulkiness and interlocki.pdf
viscous increases due to bulkiness and interlockings og liquids 1 3 diol maKes H
bond with neighbouring molecules hence all over the movement gets restrictied . and everything
move in bulkness hence viscous
Solution
viscous increases due to bulkiness and interlockings og liquids 1 3 diol maKes H
bond with neighbouring molecules hence all over the movement gets restrictied . and everything
move in bulkness hence viscous.
The factors which affect are H bond Electronegat.pdf
The factors which affect are: H bond Electronegativity Solvent These factors are
dependent on structure.
Solution
The factors which affect are: H bond Electronegativity Solvent These factors are
dependent on structure..
The compund with the lowest value of Ksp will pre.pdf
The compund with the lowest value of Ksp will precipitate first as it will need the
least amount of Cl- t reach its Ksp. Thus, AuCl will precipitate first.
Solution
The compund with the lowest value of Ksp will precipitate first as it will need the
least amount of Cl- t reach its Ksp. Thus, AuCl will precipitate first..
SO4 2- ---sulphate ion H2SO4---2H+ +SO4 2- conce.pdf
SO4 2- ---sulphate ion H2SO4--->2H+ +SO4 2- concentration of H+ is half of SO$
2- PH=-log[H+] =-log1/2[SO42-]
Solution
SO4 2- ---sulphate ion H2SO4--->2H+ +SO4 2- concentration of H+ is half of SO$
2- PH=-log[H+] =-log1/2[SO42-].
pKb = 4.7 pOH = 4.7 + log 0.20 0.25 =4.6 pH = 1.pdf
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4
Solution
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4.
Pouring the product on ice has two effects Firs.pdf
Pouring the product on ice has two effects: Firstly, this reduces the temperature of
the mixture. Reducing the temperature of any solution will reduce the solubility of dissolved
solutes (such as the product). Secondly, the ice will slightly dissolve at room temperature,
increasing the amount of water in your product solution. Methyl m-nitrobenzoate is not soluble
in water (which is why you did the reaction in glacial acid instead of dilute water solutions), so
this will aid in its precipitation out of solution.
Solution
Pouring the product on ice has two effects: Firstly, this reduces the temperature of
the mixture. Reducing the temperature of any solution will reduce the solubility of dissolved
solutes (such as the product). Secondly, the ice will slightly dissolve at room temperature,
increasing the amount of water in your product solution. Methyl m-nitrobenzoate is not soluble
in water (which is why you did the reaction in glacial acid instead of dilute water solutions), so
this will aid in its precipitation out of solution..
pH = pka + log ( [salt[[acid]) = 4.76 + log ( 0..pdf
pH = pka + log ( [salt[/[acid]) = 4.76 + log ( 0.452 /0.4) = 4.813
Solution
pH = pka + log ( [salt[/[acid]) = 4.76 + log ( 0.452 /0.4) = 4.813.
Recommended
Where is the reaction See what is the no. of m.pdf
Where is the reaction?? See what is the no. of moles.Which has more moles is and
Excess and other is limiting.
Solution
Where is the reaction?? See what is the no. of moles.Which has more moles is and
Excess and other is limiting..
viscous increases due to bulkiness and interlocki.pdf
viscous increases due to bulkiness and interlockings og liquids 1 3 diol maKes H
bond with neighbouring molecules hence all over the movement gets restrictied . and everything
move in bulkness hence viscous
Solution
viscous increases due to bulkiness and interlockings og liquids 1 3 diol maKes H
bond with neighbouring molecules hence all over the movement gets restrictied . and everything
move in bulkness hence viscous.
The factors which affect are H bond Electronegat.pdf
The factors which affect are: H bond Electronegativity Solvent These factors are
dependent on structure.
Solution
The factors which affect are: H bond Electronegativity Solvent These factors are
dependent on structure..
The compund with the lowest value of Ksp will pre.pdf
The compund with the lowest value of Ksp will precipitate first as it will need the
least amount of Cl- t reach its Ksp. Thus, AuCl will precipitate first.
Solution
The compund with the lowest value of Ksp will precipitate first as it will need the
least amount of Cl- t reach its Ksp. Thus, AuCl will precipitate first..
SO4 2- ---sulphate ion H2SO4---2H+ +SO4 2- conce.pdf
SO4 2- ---sulphate ion H2SO4--->2H+ +SO4 2- concentration of H+ is half of SO$
2- PH=-log[H+] =-log1/2[SO42-]
Solution
SO4 2- ---sulphate ion H2SO4--->2H+ +SO4 2- concentration of H+ is half of SO$
2- PH=-log[H+] =-log1/2[SO42-].
pKb = 4.7 pOH = 4.7 + log 0.20 0.25 =4.6 pH = 1.pdf
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4
Solution
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4.
Pouring the product on ice has two effects Firs.pdf
Pouring the product on ice has two effects: Firstly, this reduces the temperature of
the mixture. Reducing the temperature of any solution will reduce the solubility of dissolved
solutes (such as the product). Secondly, the ice will slightly dissolve at room temperature,
increasing the amount of water in your product solution. Methyl m-nitrobenzoate is not soluble
in water (which is why you did the reaction in glacial acid instead of dilute water solutions), so
this will aid in its precipitation out of solution.
Solution
Pouring the product on ice has two effects: Firstly, this reduces the temperature of
the mixture. Reducing the temperature of any solution will reduce the solubility of dissolved
solutes (such as the product). Secondly, the ice will slightly dissolve at room temperature,
increasing the amount of water in your product solution. Methyl m-nitrobenzoate is not soluble
in water (which is why you did the reaction in glacial acid instead of dilute water solutions), so
this will aid in its precipitation out of solution..
pH = pka + log ( [salt[[acid]) = 4.76 + log ( 0..pdf
pH = pka + log ( [salt[/[acid]) = 4.76 + log ( 0.452 /0.4) = 4.813
Solution
pH = pka + log ( [salt[/[acid]) = 4.76 + log ( 0.452 /0.4) = 4.813.
NH3 is lewis base according to lewis electron pai.pdf
NH3 is lewis base according to lewis electron pair donar is base, nitrogen contains a
lone pair of electrons. AlBr3 is lewis acid according to lewis electron pair acceptor is acid, Al
has an empty orbital so it gains electron pair from nitrogen of ammonia leads to the formation of
co ordinate covalent bond. Br3Al<............:NH3
Solution
NH3 is lewis base according to lewis electron pair donar is base, nitrogen contains a
lone pair of electrons. AlBr3 is lewis acid according to lewis electron pair acceptor is acid, Al
has an empty orbital so it gains electron pair from nitrogen of ammonia leads to the formation of
co ordinate covalent bond. Br3Al<............:NH3.
Li has high reduction potential than Cobalt ther.pdf
Li has high reduction potential than Cobalt therfore reaction is spontaneous
Solution
Li has high reduction potential than Cobalt therfore reaction is spontaneous.
Has two binding sites for oxygen .pdf
Has two binding sites for oxygen
Solution
Has two binding sites for oxygen.
Types of physical situations or occurrencesIt is important to und.pdf
Types of physical situations or occurrences:
It is important to understand the spectrum of threats to information systems so that responsible
administrators can make sure that anticipation measures are widespread.The three types of
physical situations or occurrences that can constitute a threat to information systems within an
enterprise are:
1. Environmental threats:
Natural disasters are the source of a wide range of environmental threats to the information
processing facilities. It is possible to evaluate the risk of different types of natural disasters and
take suitable precautions so that catastrophic loss from natural disaster is prevented.
2. Technical threats:
These threats include electrical power, electromagnetic interference. Electrical power is
important for the operation of information system. All the electrical and electronic devices
require power without interuptions. Power utility problems like low voltage, high voltage and
noise may occur.
Example: Most of the computers are designed to withstand long-drawn-out voltage reductions of
about 20% without shutting down and without operational error. Therefore no damage occurs but
service is interrupted.
3. Human caused threats:
The human-caused threats are unauthorized physical access, theft, destruction of equipment or
data, mis-use of the resources. These threats are difficult to handle compared to environmental
and technical threats. The human caused threats are less predictable than other types of physical
threats.
Solution
Types of physical situations or occurrences:
It is important to understand the spectrum of threats to information systems so that responsible
administrators can make sure that anticipation measures are widespread.The three types of
physical situations or occurrences that can constitute a threat to information systems within an
enterprise are:
1. Environmental threats:
Natural disasters are the source of a wide range of environmental threats to the information
processing facilities. It is possible to evaluate the risk of different types of natural disasters and
take suitable precautions so that catastrophic loss from natural disaster is prevented.
2. Technical threats:
These threats include electrical power, electromagnetic interference. Electrical power is
important for the operation of information system. All the electrical and electronic devices
require power without interuptions. Power utility problems like low voltage, high voltage and
noise may occur.
Example: Most of the computers are designed to withstand long-drawn-out voltage reductions of
about 20% without shutting down and without operational error. Therefore no damage occurs but
service is interrupted.
3. Human caused threats:
The human-caused threats are unauthorized physical access, theft, destruction of equipment or
data, mis-use of the resources. These threats are difficult to handle compared to environmental
and technical threats. The human caused threats are less predictable than other type.
The term illegal acts, refers to violations of laws or governmental .pdf
The term illegal acts, refers to violations of laws or governmental regulations. Illegal acts by
clients are acts attributable to the entity whose financial statements are under audit or acts by
management or employees acting on behalf of the entity. Illegal acts by clients do not include
personal misconduct by the entity\'s personnel unrelated to their business activities.
When the auditor concludes, based on information obtained and, if necessary, consultation with
legal counsel, that an illegal act has or is likely to have occurred, the auditor should consider the
effect on the financial statements as well as the implications for other aspects of the audit.
The auditor should assure himself that the audit committee is adequately informed as soon as
practicable and prior to the issuance of the auditor\'s report with respect to illegal acts that come
to the auditor’s attention. The auditor need not communicate matters that are clearly
inconsequential and may reach agreement in advance with the audit committee on the nature of
such matters to be communicated. The communication should describe the act, the circumstances
of its occurrence, and the effect on the financial statements. Senior management may wish to
have its remedial actions communicated to the audit committee simultaneously. Possible
remedial actions include disciplinary action against involved personnel, seeking restitution,
adoption of preventive or corrective company policies, and modifications of specific control
activities. If senior management is involved in an illegal act, the auditor should communicate
directly with the audit committee. The communication may be oral or written. If the
communication is oral, the auditor should document it.
Solution
The term illegal acts, refers to violations of laws or governmental regulations. Illegal acts by
clients are acts attributable to the entity whose financial statements are under audit or acts by
management or employees acting on behalf of the entity. Illegal acts by clients do not include
personal misconduct by the entity\'s personnel unrelated to their business activities.
When the auditor concludes, based on information obtained and, if necessary, consultation with
legal counsel, that an illegal act has or is likely to have occurred, the auditor should consider the
effect on the financial statements as well as the implications for other aspects of the audit.
The auditor should assure himself that the audit committee is adequately informed as soon as
practicable and prior to the issuance of the auditor\'s report with respect to illegal acts that come
to the auditor’s attention. The auditor need not communicate matters that are clearly
inconsequential and may reach agreement in advance with the audit committee on the nature of
such matters to be communicated. The communication should describe the act, the circumstances
of its occurrence, and the effect on the financial statements. Senior management may wish to
have its remedial a.
Strategic planning has long been used as a tool for transforming and.pdf
Strategic planning has long been used as a tool for transforming and revitalizing corporations,
government agencies and nonprofit organizations. Recently, however, skepticism about planning
has been on the rise. Political and economic uncertainty is the norm and the pace of
technological and social change has accelerated.
No organization exists in a static environment. Social, political and economic trends continually
impact the demand for its offerings and services. Even as advances in technology present new
opportunities, they also generate new expectations. Needs and community demographics are all
subject to change. So too are methods for delivering programs and services. It is thus essential
that a strategic plan reflect the external environment. Programs, services and operations should
be reexamined and reshaped in light of current realities and future projections.
Reference: Ten Keys to Successful Strategic Planning for Nonprofit and Foundation Leaders by
Richard A. Mittenthal
Solution
Strategic planning has long been used as a tool for transforming and revitalizing corporations,
government agencies and nonprofit organizations. Recently, however, skepticism about planning
has been on the rise. Political and economic uncertainty is the norm and the pace of
technological and social change has accelerated.
No organization exists in a static environment. Social, political and economic trends continually
impact the demand for its offerings and services. Even as advances in technology present new
opportunities, they also generate new expectations. Needs and community demographics are all
subject to change. So too are methods for delivering programs and services. It is thus essential
that a strategic plan reflect the external environment. Programs, services and operations should
be reexamined and reshaped in light of current realities and future projections.
Reference: Ten Keys to Successful Strategic Planning for Nonprofit and Foundation Leaders by
Richard A. Mittenthal.
Sorry I didnt get the question what you postSolutionSorry I .pdf
Sorry I didn\'t get the question what you post
Solution
Sorry I didn\'t get the question what you post.
Sella turcica is the depression in the sphenoid bone of the skull. S.pdf
Sella turcica is the depression in the sphenoid bone of the skull. Sella turcica is also known as
hypophysial fossa. It is located behind the chiasmatic groove and tuberculum sellae of the
sphenoid bone which hold the pituitary gland.
Solution
Sella turcica is the depression in the sphenoid bone of the skull. Sella turcica is also known as
hypophysial fossa. It is located behind the chiasmatic groove and tuberculum sellae of the
sphenoid bone which hold the pituitary gland..
P(child does not have a cavity) = (420)0.41 + (620)0.32 + (1020.pdf
P(child does not have a cavity) = (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27
so,
P(this child is 14 years old) = [(10/20)*0.27]/[ (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27]
= 0.4313
Solution
P(child does not have a cavity) = (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27
so,
P(this child is 14 years old) = [(10/20)*0.27]/[ (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27]
= 0.4313.
Ans 1)Difference- DNA is double stranded where as RNA is single .pdf
Ans 1)
Difference
- DNA is double stranded where as RNA is single stranded
- Four Nitrogenous bases are adenine, thymine, cytosine, and guanine in DNA where as adenine,
uracil , guanine, and cytosine in RNA.
- DNA is found in nucleus where as RNA is found in cytiplasm
Similar
Both have phosphate group and 5 carbon sugar.
Both are formed from nucleotides
Solution
Ans 1)
Difference
- DNA is double stranded where as RNA is single stranded
- Four Nitrogenous bases are adenine, thymine, cytosine, and guanine in DNA where as adenine,
uracil , guanine, and cytosine in RNA.
- DNA is found in nucleus where as RNA is found in cytiplasm
Similar
Both have phosphate group and 5 carbon sugar.
Both are formed from nucleotides.
Mandibles refer to the feeding appendages (or jaws) using which food.pdf
Mandibles refer to the feeding appendages (or jaws) using which food is ingested. Depending on
the type of organism and the prey that has to be captured, the structure of the mandible has
evolved over the ages. There are evolutionary and phylogenetic relationships to be considered
with regard to common features in the development of mandibles in insects and vertebrates too.
This is because several insect and crustacean adult mandibles are built in such a way that food is
mainly handled with the lower portion of the jaw.
Insect mandibles have evolved in accordance with the type of food or prey different species need
to catch. Common food sources for insects include algae,other insects, fish, other plants etc.
Insects may pierce the surface of their food and suck out fluids,or graze their food to separate it
from mud and ingest it. They can even bite off some portions of their food before ingestion or
engulf their food as a whole.Some insects chew leaves and other parts of plants. Their mandibles
have evolved accordingly.
Vertebrates have a variety of food sources and do not depend on any single trophic level for their
survival. The muscles that enable mastication of food move the lower jaw of vertebrate
mandibles. In humans, the voice box also moves the mandible. Most vertebrates either engulf
their food or shred it. Many have teeth for feeding, and can be either herbivores or carnivores.
Octopuses are invertebrates having a clever feeding mechanism. They use their long tentacles to
explore crevices for their prey. On capturing their prey, they immediately inject their poisonous
saliva to immobilize it, after which they use their beaks to crush and ingest their prey. They
mostly eat crabs, mollusks, lobsters etc. If the shell of their prey is too tough to break, they use
salivary papilla to drill a pinhole and eat the meat of their prey. Thus, the octopus has evolved a
feeding mechanism suitable to it.
Solution
Mandibles refer to the feeding appendages (or jaws) using which food is ingested. Depending on
the type of organism and the prey that has to be captured, the structure of the mandible has
evolved over the ages. There are evolutionary and phylogenetic relationships to be considered
with regard to common features in the development of mandibles in insects and vertebrates too.
This is because several insect and crustacean adult mandibles are built in such a way that food is
mainly handled with the lower portion of the jaw.
Insect mandibles have evolved in accordance with the type of food or prey different species need
to catch. Common food sources for insects include algae,other insects, fish, other plants etc.
Insects may pierce the surface of their food and suck out fluids,or graze their food to separate it
from mud and ingest it. They can even bite off some portions of their food before ingestion or
engulf their food as a whole.Some insects chew leaves and other parts of plants. Their mandibles
have evolved accordingly.
Vertebrates have.
A and B. Half-duplex network communications describes two devices ne.pdf
A and B. Half-duplex network communications describes two devices networked together where
only one device may communicate at any one time. The other device must wait until the first
device is finished before initiating communications. Full-duplex describes a situation where two
networked devices may transmit information to each other simultaneously, much in the same
way two people can speak at the same time during a phone conversation.
Solution
A and B. Half-duplex network communications describes two devices networked together where
only one device may communicate at any one time. The other device must wait until the first
device is finished before initiating communications. Full-duplex describes a situation where two
networked devices may transmit information to each other simultaneously, much in the same
way two people can speak at the same time during a phone conversation..
ionic equation 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) - 2Li+(aq.pdf
ionic equation: 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) -> 2Li+(aq)+ 2Cl-(aq) + SrSO4(s)
net ionic equation: SO42-(aq)+ Sr2+(aq) -> SrSO4(s)
Solution
ionic equation: 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) -> 2Li+(aq)+ 2Cl-(aq) + SrSO4(s)
net ionic equation: SO42-(aq)+ Sr2+(aq) -> SrSO4(s).
Instruction FormatMachine instruction has an opcode and zero or m.pdf
Instruction Format:
Machine instruction has an opcode and zero or more operands. Architectures are differentiated
from one another by the number of bits allowed per instruction (16, 32, and 64 are the most
common), by the number of operands allowed per instruction, and by the types of instructions
and data each can process. More specifically, instruction sets are differentiated by the following
features:
• Operand storage in the CPU (data can be stored in a stack structure or in registers) • Number of
explicit operands per instruction (zero, one, two, and three being the most common)
• Operand location (instructions can be classified as register-to-register, register-to-memory or
memory-to-memory, which simply refer to the combinations of operands allowed per
instruction)
• Operations (including not only types of operations but also which instructions can access
memory and which cannot) • Type and size of operands (operands can be addresses, numbers, or
even characters)
The following are some common instruction formats:
• OPCODE only (zero addresses)
• OPCODE + 1 Address (usually a memory address)
• OPCODE + 2 Addresses (usually registers, or one register and one memory address)
• OPCODE + 3 Addresses (usually registers, or combinations of registers and memory)
Instruction Length:
The traditional method for describing a computer architecture is to specify the maximum number
of operands, or addresses, contained in each instruction. This has a direct impact on the length of
the instruction itself. Instructions on current architectures can be formatted in two ways: • Fixed
length-Wastes space but is fast and results in better performance when instruction-level
pipelining is used, as we see in Section 5.5. • Variable length-More complex to decode but saves
storage space.
Memory organization affects instruction format. If memory has, for example, 16 or 32-bit words
and is not byte-addressable, it is difficult to access a single character. For this reason, even
machines that have 16-, 32-, or 64-bit words are often byteaddressable, meaning every byte has a
unique address even though words are longer than 1 byte.
COMMON BUS Structure:
The basic computer has eight registers, a memory unit, and a control unit. Paths must be
provided to transfer information from one register to another and between memory and registers.
The number of wires will be excessive if connections are made between the outputs of each
register and the inputs of the other registers. A more efficient scheme for transferring
information in a system with many registers is to use a common bus. It is known that how to
construct a bus PC AR 11 0 11 0 15 0 15 0 7 0 7 0 IR TR INPR AC DR 15 0 15 0 Memory 4096
words 16 bits per word 6 system using multiplexers or three-state buffer gates. The outputs of
seven registers and memory are connected to the common bus. The specific output that is
selected for the bus lines at any given time is determined from the binary value of the selection
vari.
Human infant.Human infant takes time to learn and gradually captur.pdf
Human infant.
Human infant takes time to learn and gradually captures learning of things.
Solution
Human infant.
Human infant takes time to learn and gradually captures learning of things..
Grit and other organic material interfere with the Coliform count. T.pdf
Grit and other organic material interfere with the Coliform count. The organisms may remain
viable but without division within the grit. The abrasive grit particles may cause damage to the
cells and lead to less coliform count than the actual count.
In general, grit is removed by placing grids that specifically remove grit. Or else, differential
sedimentation is applied to remove grit.
At the upstream source, generally the organisms enter the stream at a point source and hence are
more concentrated. When they travel down, the cells gradually get dilution since the volume of
water wherein the cells are spread increases with time and space. Hence, lesser density of
coliforms is observed at a downstream site.
Solution
Grit and other organic material interfere with the Coliform count. The organisms may remain
viable but without division within the grit. The abrasive grit particles may cause damage to the
cells and lead to less coliform count than the actual count.
In general, grit is removed by placing grids that specifically remove grit. Or else, differential
sedimentation is applied to remove grit.
At the upstream source, generally the organisms enter the stream at a point source and hence are
more concentrated. When they travel down, the cells gradually get dilution since the volume of
water wherein the cells are spread increases with time and space. Hence, lesser density of
coliforms is observed at a downstream site..
Hi, Please find my code.I have correted all of your classes.Plea.pdf
Hi, Please find my code.
I have correted all of your classes.
Please let me know in case if do not get concpt.
######## Node.java ###########
public class Node {
private E element;
private Node next;
public Node() {
this.element = null;
this.next = null;
}
public Node(E e) {
this.element = e;
this.next = null;
}
public E getElement() {
return this.element;
}
public void setElement(E element) {
this.element= element;
}
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
}
########## Stack.java #############
public class Stack {
private int N;
private Node top;
public Stack() {
N = 0;
this.top =null;
}
/*
* places element on the top of the stack
*/
public void push(E element){
Node temp = new Node(element);
temp.setNext(top);
top = temp;
N++;
}
/*
* remove the top node and return its contents
*/
public E pop(){
if(top == null)
return null;
E e = top.getElement();
top = top.getNext();
N--;
return e;
}
/*
* Look at the top element of the Stack and return it, without removing
*/
public E peek(){
if(top == null)
return null;
return top.getElement();
}
//returns the size of the stack
public int size(){
return N; //replace
}
public boolean isEmpty(){
return top==null;
}
}
############ Queue.java ############
public class Queue {
private Node front;
private Node back;
private int N;
public Queue() {
this.front = null;
this.back = null;
N = 0;
}
/*
* places element in the back of the Queue
*/
public void enqueue(E element){
Node temp = new Node(element);
if(front == null) {
front = temp;
back = temp;
}
else{
back.setNext(temp);
back = temp;
}
N++;
}
/*
* remove the front node of the queue and return it
*/
public E dequeue(){
if(front == null) {
return null;
}
E item = front.getElement();
front = front.getNext();
// if we had only one element
if(front == null){
back = null;
}
N--;
return item;
}
/*
* Look at the front of the queue and return it, without removing
*/
public E peek(){
if(back == null)
return null;
return back.getElement();
}
//returns the size of the queue
public int size(){
return N;
}
public boolean isEmpty(){
return front==null;
}
}
########### Palindrome.java ############
import java.util.Scanner;
public class Palindrome
{
public static void main(String[ ] args)
{
Scanner input = new Scanner(System.in);
String inputString;
System.out.print(\"Please enter your string: \");
inputString = input.next( );
if (isPalindrome( inputString )){
System.out.println(\"Yes it is a palindrome.\");
}
else{
System.out.println(\"No this is not a palindrome.\");
}
}
public static boolean isPalindrome(String input)
{
Queue q = new Queue ();
Stack s = new Stack ();
char letter;
int i;
for (i = 0; i < input.length( ); i++)
{
letter = input.charAt(i);
q.enqueue(letter);
s.push(letter);
}
while (!q.isEmpty( ))
{
if (q.dequeue() != s.pop( ))
return false;
}
return true;
}
}
/*
Sample Output:
Please enter your string: MadaM
Yes it is a palindrome.
*/
Solution
Hi, Please find my code.
I have correted all of your classes.
Pl.
Functional dependenciesAn FD X-Y is satisfied in an instance r of .pdf
Functional dependencies:An FD X->Y is satisfied in an instance r of R if for every pair of tuples,
t and s if t and s agree on all attributes in X then they must agree on all attributes in Y
A key constraint is a special kind of functional dependency:all attributes of relation occur on the
right-hand side of the FD:
SSN ->SSN,Name,Address
FD:
A->ABC
1st NF:
A->B
A->C
BC->A
D->E
It has primary key is unique so it is sastify the 1st normal form.Using the First Normal Form,
data redundancy increases, as there will be many columns with same data in multiple rows but
each row as a whole will be unique.
2nd Normal Form:As per the Second Normal Form there must not be any partial dependency of
any column on primary key. It is 2nd Normal Form .
Third Normal form:Third Normal form applies that every non-prime attribute of table must be
dependent on primary key, or we can say that, there should not be the case that a non-prime
attribute is determined by another non-prime attribute.So this transitive functional dependency
should be removed from the table and also the table must be in Second Normal form.
Boyce and codd Normal Form: It is a higher version of the Third Normal form. This form deals
with certain type of anamoly that is not handled by 3NF.A 3NF table which does not have
multiple overlapping candidate keys is said to be in BCNF.
R(ABC)
A->B is super key.
Solution
Functional dependencies:An FD X->Y is satisfied in an instance r of R if for every pair of tuples,
t and s if t and s agree on all attributes in X then they must agree on all attributes in Y
A key constraint is a special kind of functional dependency:all attributes of relation occur on the
right-hand side of the FD:
SSN ->SSN,Name,Address
FD:
A->ABC
1st NF:
A->B
A->C
BC->A
D->E
It has primary key is unique so it is sastify the 1st normal form.Using the First Normal Form,
data redundancy increases, as there will be many columns with same data in multiple rows but
each row as a whole will be unique.
2nd Normal Form:As per the Second Normal Form there must not be any partial dependency of
any column on primary key. It is 2nd Normal Form .
Third Normal form:Third Normal form applies that every non-prime attribute of table must be
dependent on primary key, or we can say that, there should not be the case that a non-prime
attribute is determined by another non-prime attribute.So this transitive functional dependency
should be removed from the table and also the table must be in Second Normal form.
Boyce and codd Normal Form: It is a higher version of the Third Normal form. This form deals
with certain type of anamoly that is not handled by 3NF.A 3NF table which does not have
multiple overlapping candidate keys is said to be in BCNF.
R(ABC)
A->B is super key..
CNS fibers lack a neurilemma. The neurilemma plays an important role.pdf
CNS fibers lack a neurilemma. The neurilemma plays an important role in fiber regeneration.
Wallerian degeneration is the process of degeneration of the axon distal to a site of transection. It
happens in both the CNS and the PNS.
Neuroregeneration refers to the regrowth or repair of nervous tissues, cells or cell products. Such
mechanisms may include generation of new neurons, glia, axons, myelin, or synapses.
Solution
CNS fibers lack a neurilemma. The neurilemma plays an important role in fiber regeneration.
Wallerian degeneration is the process of degeneration of the axon distal to a site of transection. It
happens in both the CNS and the PNS.
Neuroregeneration refers to the regrowth or repair of nervous tissues, cells or cell products. Such
mechanisms may include generation of new neurons, glia, axons, myelin, or synapses..
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
More Related Content
More from pritikulkarni20
NH3 is lewis base according to lewis electron pai.pdf
NH3 is lewis base according to lewis electron pair donar is base, nitrogen contains a
lone pair of electrons. AlBr3 is lewis acid according to lewis electron pair acceptor is acid, Al
has an empty orbital so it gains electron pair from nitrogen of ammonia leads to the formation of
co ordinate covalent bond. Br3Al<............:NH3
Solution
NH3 is lewis base according to lewis electron pair donar is base, nitrogen contains a
lone pair of electrons. AlBr3 is lewis acid according to lewis electron pair acceptor is acid, Al
has an empty orbital so it gains electron pair from nitrogen of ammonia leads to the formation of
co ordinate covalent bond. Br3Al<............:NH3.
Li has high reduction potential than Cobalt ther.pdf
Li has high reduction potential than Cobalt therfore reaction is spontaneous
Solution
Li has high reduction potential than Cobalt therfore reaction is spontaneous.
Has two binding sites for oxygen .pdf
Has two binding sites for oxygen
Solution
Has two binding sites for oxygen.
Types of physical situations or occurrencesIt is important to und.pdf
Types of physical situations or occurrences:
It is important to understand the spectrum of threats to information systems so that responsible
administrators can make sure that anticipation measures are widespread.The three types of
physical situations or occurrences that can constitute a threat to information systems within an
enterprise are:
1. Environmental threats:
Natural disasters are the source of a wide range of environmental threats to the information
processing facilities. It is possible to evaluate the risk of different types of natural disasters and
take suitable precautions so that catastrophic loss from natural disaster is prevented.
2. Technical threats:
These threats include electrical power, electromagnetic interference. Electrical power is
important for the operation of information system. All the electrical and electronic devices
require power without interuptions. Power utility problems like low voltage, high voltage and
noise may occur.
Example: Most of the computers are designed to withstand long-drawn-out voltage reductions of
about 20% without shutting down and without operational error. Therefore no damage occurs but
service is interrupted.
3. Human caused threats:
The human-caused threats are unauthorized physical access, theft, destruction of equipment or
data, mis-use of the resources. These threats are difficult to handle compared to environmental
and technical threats. The human caused threats are less predictable than other types of physical
threats.
Solution
Types of physical situations or occurrences:
It is important to understand the spectrum of threats to information systems so that responsible
administrators can make sure that anticipation measures are widespread.The three types of
physical situations or occurrences that can constitute a threat to information systems within an
enterprise are:
1. Environmental threats:
Natural disasters are the source of a wide range of environmental threats to the information
processing facilities. It is possible to evaluate the risk of different types of natural disasters and
take suitable precautions so that catastrophic loss from natural disaster is prevented.
2. Technical threats:
These threats include electrical power, electromagnetic interference. Electrical power is
important for the operation of information system. All the electrical and electronic devices
require power without interuptions. Power utility problems like low voltage, high voltage and
noise may occur.
Example: Most of the computers are designed to withstand long-drawn-out voltage reductions of
about 20% without shutting down and without operational error. Therefore no damage occurs but
service is interrupted.
3. Human caused threats:
The human-caused threats are unauthorized physical access, theft, destruction of equipment or
data, mis-use of the resources. These threats are difficult to handle compared to environmental
and technical threats. The human caused threats are less predictable than other type.
The term illegal acts, refers to violations of laws or governmental .pdf
The term illegal acts, refers to violations of laws or governmental regulations. Illegal acts by
clients are acts attributable to the entity whose financial statements are under audit or acts by
management or employees acting on behalf of the entity. Illegal acts by clients do not include
personal misconduct by the entity\'s personnel unrelated to their business activities.
When the auditor concludes, based on information obtained and, if necessary, consultation with
legal counsel, that an illegal act has or is likely to have occurred, the auditor should consider the
effect on the financial statements as well as the implications for other aspects of the audit.
The auditor should assure himself that the audit committee is adequately informed as soon as
practicable and prior to the issuance of the auditor\'s report with respect to illegal acts that come
to the auditor’s attention. The auditor need not communicate matters that are clearly
inconsequential and may reach agreement in advance with the audit committee on the nature of
such matters to be communicated. The communication should describe the act, the circumstances
of its occurrence, and the effect on the financial statements. Senior management may wish to
have its remedial actions communicated to the audit committee simultaneously. Possible
remedial actions include disciplinary action against involved personnel, seeking restitution,
adoption of preventive or corrective company policies, and modifications of specific control
activities. If senior management is involved in an illegal act, the auditor should communicate
directly with the audit committee. The communication may be oral or written. If the
communication is oral, the auditor should document it.
Solution
The term illegal acts, refers to violations of laws or governmental regulations. Illegal acts by
clients are acts attributable to the entity whose financial statements are under audit or acts by
management or employees acting on behalf of the entity. Illegal acts by clients do not include
personal misconduct by the entity\'s personnel unrelated to their business activities.
When the auditor concludes, based on information obtained and, if necessary, consultation with
legal counsel, that an illegal act has or is likely to have occurred, the auditor should consider the
effect on the financial statements as well as the implications for other aspects of the audit.
The auditor should assure himself that the audit committee is adequately informed as soon as
practicable and prior to the issuance of the auditor\'s report with respect to illegal acts that come
to the auditor’s attention. The auditor need not communicate matters that are clearly
inconsequential and may reach agreement in advance with the audit committee on the nature of
such matters to be communicated. The communication should describe the act, the circumstances
of its occurrence, and the effect on the financial statements. Senior management may wish to
have its remedial a.
Strategic planning has long been used as a tool for transforming and.pdf
Strategic planning has long been used as a tool for transforming and revitalizing corporations,
government agencies and nonprofit organizations. Recently, however, skepticism about planning
has been on the rise. Political and economic uncertainty is the norm and the pace of
technological and social change has accelerated.
No organization exists in a static environment. Social, political and economic trends continually
impact the demand for its offerings and services. Even as advances in technology present new
opportunities, they also generate new expectations. Needs and community demographics are all
subject to change. So too are methods for delivering programs and services. It is thus essential
that a strategic plan reflect the external environment. Programs, services and operations should
be reexamined and reshaped in light of current realities and future projections.
Reference: Ten Keys to Successful Strategic Planning for Nonprofit and Foundation Leaders by
Richard A. Mittenthal
Solution
Strategic planning has long been used as a tool for transforming and revitalizing corporations,
government agencies and nonprofit organizations. Recently, however, skepticism about planning
has been on the rise. Political and economic uncertainty is the norm and the pace of
technological and social change has accelerated.
No organization exists in a static environment. Social, political and economic trends continually
impact the demand for its offerings and services. Even as advances in technology present new
opportunities, they also generate new expectations. Needs and community demographics are all
subject to change. So too are methods for delivering programs and services. It is thus essential
that a strategic plan reflect the external environment. Programs, services and operations should
be reexamined and reshaped in light of current realities and future projections.
Reference: Ten Keys to Successful Strategic Planning for Nonprofit and Foundation Leaders by
Richard A. Mittenthal.
Sorry I didnt get the question what you postSolutionSorry I .pdf
Sorry I didn\'t get the question what you post
Solution
Sorry I didn\'t get the question what you post.
Sella turcica is the depression in the sphenoid bone of the skull. S.pdf
Sella turcica is the depression in the sphenoid bone of the skull. Sella turcica is also known as
hypophysial fossa. It is located behind the chiasmatic groove and tuberculum sellae of the
sphenoid bone which hold the pituitary gland.
Solution
Sella turcica is the depression in the sphenoid bone of the skull. Sella turcica is also known as
hypophysial fossa. It is located behind the chiasmatic groove and tuberculum sellae of the
sphenoid bone which hold the pituitary gland..
P(child does not have a cavity) = (420)0.41 + (620)0.32 + (1020.pdf
P(child does not have a cavity) = (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27
so,
P(this child is 14 years old) = [(10/20)*0.27]/[ (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27]
= 0.4313
Solution
P(child does not have a cavity) = (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27
so,
P(this child is 14 years old) = [(10/20)*0.27]/[ (4/20)*0.41 + (6/20)*0.32 + (10/20)*0.27]
= 0.4313.
Ans 1)Difference- DNA is double stranded where as RNA is single .pdf
Ans 1)
Difference
- DNA is double stranded where as RNA is single stranded
- Four Nitrogenous bases are adenine, thymine, cytosine, and guanine in DNA where as adenine,
uracil , guanine, and cytosine in RNA.
- DNA is found in nucleus where as RNA is found in cytiplasm
Similar
Both have phosphate group and 5 carbon sugar.
Both are formed from nucleotides
Solution
Ans 1)
Difference
- DNA is double stranded where as RNA is single stranded
- Four Nitrogenous bases are adenine, thymine, cytosine, and guanine in DNA where as adenine,
uracil , guanine, and cytosine in RNA.
- DNA is found in nucleus where as RNA is found in cytiplasm
Similar
Both have phosphate group and 5 carbon sugar.
Both are formed from nucleotides.
Mandibles refer to the feeding appendages (or jaws) using which food.pdf
Mandibles refer to the feeding appendages (or jaws) using which food is ingested. Depending on
the type of organism and the prey that has to be captured, the structure of the mandible has
evolved over the ages. There are evolutionary and phylogenetic relationships to be considered
with regard to common features in the development of mandibles in insects and vertebrates too.
This is because several insect and crustacean adult mandibles are built in such a way that food is
mainly handled with the lower portion of the jaw.
Insect mandibles have evolved in accordance with the type of food or prey different species need
to catch. Common food sources for insects include algae,other insects, fish, other plants etc.
Insects may pierce the surface of their food and suck out fluids,or graze their food to separate it
from mud and ingest it. They can even bite off some portions of their food before ingestion or
engulf their food as a whole.Some insects chew leaves and other parts of plants. Their mandibles
have evolved accordingly.
Vertebrates have a variety of food sources and do not depend on any single trophic level for their
survival. The muscles that enable mastication of food move the lower jaw of vertebrate
mandibles. In humans, the voice box also moves the mandible. Most vertebrates either engulf
their food or shred it. Many have teeth for feeding, and can be either herbivores or carnivores.
Octopuses are invertebrates having a clever feeding mechanism. They use their long tentacles to
explore crevices for their prey. On capturing their prey, they immediately inject their poisonous
saliva to immobilize it, after which they use their beaks to crush and ingest their prey. They
mostly eat crabs, mollusks, lobsters etc. If the shell of their prey is too tough to break, they use
salivary papilla to drill a pinhole and eat the meat of their prey. Thus, the octopus has evolved a
feeding mechanism suitable to it.
Solution
Mandibles refer to the feeding appendages (or jaws) using which food is ingested. Depending on
the type of organism and the prey that has to be captured, the structure of the mandible has
evolved over the ages. There are evolutionary and phylogenetic relationships to be considered
with regard to common features in the development of mandibles in insects and vertebrates too.
This is because several insect and crustacean adult mandibles are built in such a way that food is
mainly handled with the lower portion of the jaw.
Insect mandibles have evolved in accordance with the type of food or prey different species need
to catch. Common food sources for insects include algae,other insects, fish, other plants etc.
Insects may pierce the surface of their food and suck out fluids,or graze their food to separate it
from mud and ingest it. They can even bite off some portions of their food before ingestion or
engulf their food as a whole.Some insects chew leaves and other parts of plants. Their mandibles
have evolved accordingly.
Vertebrates have.
A and B. Half-duplex network communications describes two devices ne.pdf
A and B. Half-duplex network communications describes two devices networked together where
only one device may communicate at any one time. The other device must wait until the first
device is finished before initiating communications. Full-duplex describes a situation where two
networked devices may transmit information to each other simultaneously, much in the same
way two people can speak at the same time during a phone conversation.
Solution
A and B. Half-duplex network communications describes two devices networked together where
only one device may communicate at any one time. The other device must wait until the first
device is finished before initiating communications. Full-duplex describes a situation where two
networked devices may transmit information to each other simultaneously, much in the same
way two people can speak at the same time during a phone conversation..
ionic equation 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) - 2Li+(aq.pdf
ionic equation: 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) -> 2Li+(aq)+ 2Cl-(aq) + SrSO4(s)
net ionic equation: SO42-(aq)+ Sr2+(aq) -> SrSO4(s)
Solution
ionic equation: 2Li+(aq)+ SO42-(aq) + Sr2+(aq) + 2Cl-(aq) -> 2Li+(aq)+ 2Cl-(aq) + SrSO4(s)
net ionic equation: SO42-(aq)+ Sr2+(aq) -> SrSO4(s).
Instruction FormatMachine instruction has an opcode and zero or m.pdf
Instruction Format:
Machine instruction has an opcode and zero or more operands. Architectures are differentiated
from one another by the number of bits allowed per instruction (16, 32, and 64 are the most
common), by the number of operands allowed per instruction, and by the types of instructions
and data each can process. More specifically, instruction sets are differentiated by the following
features:
• Operand storage in the CPU (data can be stored in a stack structure or in registers) • Number of
explicit operands per instruction (zero, one, two, and three being the most common)
• Operand location (instructions can be classified as register-to-register, register-to-memory or
memory-to-memory, which simply refer to the combinations of operands allowed per
instruction)
• Operations (including not only types of operations but also which instructions can access
memory and which cannot) • Type and size of operands (operands can be addresses, numbers, or
even characters)
The following are some common instruction formats:
• OPCODE only (zero addresses)
• OPCODE + 1 Address (usually a memory address)
• OPCODE + 2 Addresses (usually registers, or one register and one memory address)
• OPCODE + 3 Addresses (usually registers, or combinations of registers and memory)
Instruction Length:
The traditional method for describing a computer architecture is to specify the maximum number
of operands, or addresses, contained in each instruction. This has a direct impact on the length of
the instruction itself. Instructions on current architectures can be formatted in two ways: • Fixed
length-Wastes space but is fast and results in better performance when instruction-level
pipelining is used, as we see in Section 5.5. • Variable length-More complex to decode but saves
storage space.
Memory organization affects instruction format. If memory has, for example, 16 or 32-bit words
and is not byte-addressable, it is difficult to access a single character. For this reason, even
machines that have 16-, 32-, or 64-bit words are often byteaddressable, meaning every byte has a
unique address even though words are longer than 1 byte.
COMMON BUS Structure:
The basic computer has eight registers, a memory unit, and a control unit. Paths must be
provided to transfer information from one register to another and between memory and registers.
The number of wires will be excessive if connections are made between the outputs of each
register and the inputs of the other registers. A more efficient scheme for transferring
information in a system with many registers is to use a common bus. It is known that how to
construct a bus PC AR 11 0 11 0 15 0 15 0 7 0 7 0 IR TR INPR AC DR 15 0 15 0 Memory 4096
words 16 bits per word 6 system using multiplexers or three-state buffer gates. The outputs of
seven registers and memory are connected to the common bus. The specific output that is
selected for the bus lines at any given time is determined from the binary value of the selection
vari.
Human infant.Human infant takes time to learn and gradually captur.pdf
Human infant.
Human infant takes time to learn and gradually captures learning of things.
Solution
Human infant.
Human infant takes time to learn and gradually captures learning of things..
Grit and other organic material interfere with the Coliform count. T.pdf
Grit and other organic material interfere with the Coliform count. The organisms may remain
viable but without division within the grit. The abrasive grit particles may cause damage to the
cells and lead to less coliform count than the actual count.
In general, grit is removed by placing grids that specifically remove grit. Or else, differential
sedimentation is applied to remove grit.
At the upstream source, generally the organisms enter the stream at a point source and hence are
more concentrated. When they travel down, the cells gradually get dilution since the volume of
water wherein the cells are spread increases with time and space. Hence, lesser density of
coliforms is observed at a downstream site.
Solution
Grit and other organic material interfere with the Coliform count. The organisms may remain
viable but without division within the grit. The abrasive grit particles may cause damage to the
cells and lead to less coliform count than the actual count.
In general, grit is removed by placing grids that specifically remove grit. Or else, differential
sedimentation is applied to remove grit.
At the upstream source, generally the organisms enter the stream at a point source and hence are
more concentrated. When they travel down, the cells gradually get dilution since the volume of
water wherein the cells are spread increases with time and space. Hence, lesser density of
coliforms is observed at a downstream site..
Hi, Please find my code.I have correted all of your classes.Plea.pdf
Hi, Please find my code.
I have correted all of your classes.
Please let me know in case if do not get concpt.
######## Node.java ###########
public class Node {
private E element;
private Node next;
public Node() {
this.element = null;
this.next = null;
}
public Node(E e) {
this.element = e;
this.next = null;
}
public E getElement() {
return this.element;
}
public void setElement(E element) {
this.element= element;
}
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
}
########## Stack.java #############
public class Stack {
private int N;
private Node top;
public Stack() {
N = 0;
this.top =null;
}
/*
* places element on the top of the stack
*/
public void push(E element){
Node temp = new Node(element);
temp.setNext(top);
top = temp;
N++;
}
/*
* remove the top node and return its contents
*/
public E pop(){
if(top == null)
return null;
E e = top.getElement();
top = top.getNext();
N--;
return e;
}
/*
* Look at the top element of the Stack and return it, without removing
*/
public E peek(){
if(top == null)
return null;
return top.getElement();
}
//returns the size of the stack
public int size(){
return N; //replace
}
public boolean isEmpty(){
return top==null;
}
}
############ Queue.java ############
public class Queue {
private Node front;
private Node back;
private int N;
public Queue() {
this.front = null;
this.back = null;
N = 0;
}
/*
* places element in the back of the Queue
*/
public void enqueue(E element){
Node temp = new Node(element);
if(front == null) {
front = temp;
back = temp;
}
else{
back.setNext(temp);
back = temp;
}
N++;
}
/*
* remove the front node of the queue and return it
*/
public E dequeue(){
if(front == null) {
return null;
}
E item = front.getElement();
front = front.getNext();
// if we had only one element
if(front == null){
back = null;
}
N--;
return item;
}
/*
* Look at the front of the queue and return it, without removing
*/
public E peek(){
if(back == null)
return null;
return back.getElement();
}
//returns the size of the queue
public int size(){
return N;
}
public boolean isEmpty(){
return front==null;
}
}
########### Palindrome.java ############
import java.util.Scanner;
public class Palindrome
{
public static void main(String[ ] args)
{
Scanner input = new Scanner(System.in);
String inputString;
System.out.print(\"Please enter your string: \");
inputString = input.next( );
if (isPalindrome( inputString )){
System.out.println(\"Yes it is a palindrome.\");
}
else{
System.out.println(\"No this is not a palindrome.\");
}
}
public static boolean isPalindrome(String input)
{
Queue q = new Queue ();
Stack s = new Stack ();
char letter;
int i;
for (i = 0; i < input.length( ); i++)
{
letter = input.charAt(i);
q.enqueue(letter);
s.push(letter);
}
while (!q.isEmpty( ))
{
if (q.dequeue() != s.pop( ))
return false;
}
return true;
}
}
/*
Sample Output:
Please enter your string: MadaM
Yes it is a palindrome.
*/
Solution
Hi, Please find my code.
I have correted all of your classes.
Pl.
Functional dependenciesAn FD X-Y is satisfied in an instance r of .pdf
Functional dependencies:An FD X->Y is satisfied in an instance r of R if for every pair of tuples,
t and s if t and s agree on all attributes in X then they must agree on all attributes in Y
A key constraint is a special kind of functional dependency:all attributes of relation occur on the
right-hand side of the FD:
SSN ->SSN,Name,Address
FD:
A->ABC
1st NF:
A->B
A->C
BC->A
D->E
It has primary key is unique so it is sastify the 1st normal form.Using the First Normal Form,
data redundancy increases, as there will be many columns with same data in multiple rows but
each row as a whole will be unique.
2nd Normal Form:As per the Second Normal Form there must not be any partial dependency of
any column on primary key. It is 2nd Normal Form .
Third Normal form:Third Normal form applies that every non-prime attribute of table must be
dependent on primary key, or we can say that, there should not be the case that a non-prime
attribute is determined by another non-prime attribute.So this transitive functional dependency
should be removed from the table and also the table must be in Second Normal form.
Boyce and codd Normal Form: It is a higher version of the Third Normal form. This form deals
with certain type of anamoly that is not handled by 3NF.A 3NF table which does not have
multiple overlapping candidate keys is said to be in BCNF.
R(ABC)
A->B is super key.
Solution
Functional dependencies:An FD X->Y is satisfied in an instance r of R if for every pair of tuples,
t and s if t and s agree on all attributes in X then they must agree on all attributes in Y
A key constraint is a special kind of functional dependency:all attributes of relation occur on the
right-hand side of the FD:
SSN ->SSN,Name,Address
FD:
A->ABC
1st NF:
A->B
A->C
BC->A
D->E
It has primary key is unique so it is sastify the 1st normal form.Using the First Normal Form,
data redundancy increases, as there will be many columns with same data in multiple rows but
each row as a whole will be unique.
2nd Normal Form:As per the Second Normal Form there must not be any partial dependency of
any column on primary key. It is 2nd Normal Form .
Third Normal form:Third Normal form applies that every non-prime attribute of table must be
dependent on primary key, or we can say that, there should not be the case that a non-prime
attribute is determined by another non-prime attribute.So this transitive functional dependency
should be removed from the table and also the table must be in Second Normal form.
Boyce and codd Normal Form: It is a higher version of the Third Normal form. This form deals
with certain type of anamoly that is not handled by 3NF.A 3NF table which does not have
multiple overlapping candidate keys is said to be in BCNF.
R(ABC)
A->B is super key..
CNS fibers lack a neurilemma. The neurilemma plays an important role.pdf
CNS fibers lack a neurilemma. The neurilemma plays an important role in fiber regeneration.
Wallerian degeneration is the process of degeneration of the axon distal to a site of transection. It
happens in both the CNS and the PNS.
Neuroregeneration refers to the regrowth or repair of nervous tissues, cells or cell products. Such
mechanisms may include generation of new neurons, glia, axons, myelin, or synapses.
Solution
CNS fibers lack a neurilemma. The neurilemma plays an important role in fiber regeneration.
Wallerian degeneration is the process of degeneration of the axon distal to a site of transection. It
happens in both the CNS and the PNS.
Neuroregeneration refers to the regrowth or repair of nervous tissues, cells or cell products. Such
mechanisms may include generation of new neurons, glia, axons, myelin, or synapses..
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