35. Most of the spinal nerves are mixed (having both a sensory and a.pdf
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35. Most of the spinal nerves are mixed (having both a sensory and a motor aspect.). Answer is C. Mixed Solution 35. Most of the spinal nerves are mixed (having both a sensory and a motor aspect.). Answer is C. Mixed.
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10. The push button is pushed to the second stop when we pipette out.pdf
10. The push button is pushed to the second stop when we pipette out or transfer the sample, and
to completely void the sample without leaving any sample in the pipette tips
11. The button is to be released after it is withdrawn from the solution/liquid. If it is released
prior to withdrawal, the solution reenters the pipette tip.
12. If the push button is pressed to the second stop and is released even before the pipette tip is
withdrawn from the solution, the solution enters the pipette.
13. When concentrations or dilutions of the sample that are to be transferred changes, when
different solutions or chemicals are transferred, when containment is required, tips need to be
changed with each transfer.
14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes.
They are still useful because linkers or adapters can be attached to the blunt end to make them
useful. Moreover, new restriction sites can be created by attaching linkers or adapters.
15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it
recognizes the same restriction site in both the DNA. Therefore, the two different DNA can
basepair at the sticky ends.
Solution
10. The push button is pushed to the second stop when we pipette out or transfer the sample, and
to completely void the sample without leaving any sample in the pipette tips
11. The button is to be released after it is withdrawn from the solution/liquid. If it is released
prior to withdrawal, the solution reenters the pipette tip.
12. If the push button is pressed to the second stop and is released even before the pipette tip is
withdrawn from the solution, the solution enters the pipette.
13. When concentrations or dilutions of the sample that are to be transferred changes, when
different solutions or chemicals are transferred, when containment is required, tips need to be
changed with each transfer.
14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes.
They are still useful because linkers or adapters can be attached to the blunt end to make them
useful. Moreover, new restriction sites can be created by attaching linkers or adapters.
15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it
recognizes the same restriction site in both the DNA. Therefore, the two different DNA can
basepair at the sticky ends..
1. Working capital is the capital that a company uses for its day to.pdf
1. Working capital is the capital that a company uses for its day to day operations. Working
capital is computed using the formula: Working capital = current assets – current liabilities.
2. Determinants of working capital requirements are – nature of business, seasonality of
operations, production policy, market conditions and conditions of supply. A service company
will have a lower working capital requirement than a manufacturing company. Firms with
seasonality in operations will have higher fluctuations with regards to their working capital
requirements. Market conditions in the form of degree of competition will affect working capital
requirements. Higher competitive pressure will increase working capital requirements.
3. Three major decisions that managers have to take while performing the finance function are:
(i) Investment decision – Managers have to select those assets that the business will invest in. It
is also known as capital budgeting decisions.
(ii) Financing decision – These decisions pertain to determining how the total funds that are
required for the business will be obtained – will it be through debt or equity or a mix of debt or
equity.
(iii) Dividend decision – This pertains to determining what quantum of earnings should be
distributed to shareholders as dividends and what quantum should be retained for meeting future
growth requirements of the company.
4. Financial management involves planning, directing, monitoring and controlling the monetary
resources of a company in such a manner that the goals and objectives of the company are
achieved in an efficient manner. It involves management of capital budgeting, capital structure,
working capital management, etc. Financial management is important as it helps an organization
to set clarity towards its financial goals and helps in efficient utilization of resources.
5. Cash flow is not a suitable judge of profitability as it merely shows the changes in cash
position of a firm in a financial year from its operating activities, from its financing activities and
from its investing activities. Cash flow statement does not help us to determine gross profit
margins, operating profit margins and net profit margins. We can just gauge the reason for
changes in cash position in a year.
6. Financial risk refers to all kinds of risk associated with a financial transaction and an
investment transaction. It can be in the form of credit risk, asset backed risk, investment risk etc.
These risks arise due to the fact that there is a probability of loss that is inherent in any financing
and investment method and this probability of loss cannot be avoided.
Solution
1. Working capital is the capital that a company uses for its day to day operations. Working
capital is computed using the formula: Working capital = current assets – current liabilities.
2. Determinants of working capital requirements are – nature of business, seasonality of
operations, production policy, market conditions and condi.
1. no2.The Waterfall model has some disadvantages.Agile software.pdf
1. no
2.The Waterfall model has some disadvantages.
Agile software development evolved to eliminate the issues the Waterfall model has. It has a
completely new framework. While the Waterfall model has a sequential design, the Agile model
followed an incremental approach.When clients who used to follow the Waterfall model
switched to Agile, the transition brought many issues with it. The reason being inadaptability to a
different approach to software development. The end product turned out to be a disaster.
3. yes
4. The initialization step creates a base version of the system. The goal for this initial
implementation is to create a product to which the user can react. It should offer a sampling of
the key aspects of the problem and provide a solution that is simple enough to understand and
implement easily. To guide the iteration process, a project control list is created that contains a
record of all tasks that need to be performed. It includes such items as new features to be
implemented and areas of redesign of the existing solution.
5.yes
6.the mainly reason is to testing should be conducted faulty elements of the software can be
quickly identified because few changes are made within any single iteration.Customer can
respond to features and review the product for any needed or useful changes., a plan is developed
for the next increments, and modifications are made accordingly. This process continues, with
increments being delivered until the complete product is delivered.
7.Agile Software Development
Because a framework that is used to structure, plan, and control the process of developing an
information system.The methods attempt to minimize risk by developing software in short
timeboxes, called iterations, Each iteration is like a miniature software project of its own, and
includes all the tasks necessary to release the mini increment of new functionality: planning,
requirements analysis, design, coding, testing, and documentation. While iteration may not add
enough functionality to warrant releasing the product, an agile software project intends to be
capable of releasing new software at the end of every iteration emphasize working software as
the primary measure of progress. Combined with the preference for face-to-face communication,
agile methods produce very little written documentation relative to other methods.
Solution
1. no
2.The Waterfall model has some disadvantages.
Agile software development evolved to eliminate the issues the Waterfall model has. It has a
completely new framework. While the Waterfall model has a sequential design, the Agile model
followed an incremental approach.When clients who used to follow the Waterfall model
switched to Agile, the transition brought many issues with it. The reason being inadaptability to a
different approach to software development. The end product turned out to be a disaster.
3. yes
4. The initialization step creates a base version of the system. The goal for this initial
implementat.
True ; this is known as Tyndal effect .pdf
True ; this is known as Tyndal effect
Solution
True ; this is known as Tyndal effect.
TLC is used to separate the compounds.the benzoic.pdf
TLC is used to separate the compounds.the benzoic acid and methanol are more
polar so these have small rf value.so we can easyly separate the ester
Solution
TLC is used to separate the compounds.the benzoic acid and methanol are more
polar so these have small rf value.so we can easyly separate the ester.
The answer to this question is (E) If you look at.pdf
The answer to this question is (E) If you look at N2, you can see that as 2 moles of
N2 is formed 4 moles of NH3 is reacting. Thus, since the rate of formation of N2 is 2.0
mol/(L*s), the rate of formation of NH3 is double that (2*2.0mol/(L*s))=4.0 mol/(L*s). I hope
this helped!
Solution
The answer to this question is (E) If you look at N2, you can see that as 2 moles of
N2 is formed 4 moles of NH3 is reacting. Thus, since the rate of formation of N2 is 2.0
mol/(L*s), the rate of formation of NH3 is double that (2*2.0mol/(L*s))=4.0 mol/(L*s). I hope
this helped!.
Recommended
10. The push button is pushed to the second stop when we pipette out.pdf
10. The push button is pushed to the second stop when we pipette out or transfer the sample, and
to completely void the sample without leaving any sample in the pipette tips
11. The button is to be released after it is withdrawn from the solution/liquid. If it is released
prior to withdrawal, the solution reenters the pipette tip.
12. If the push button is pressed to the second stop and is released even before the pipette tip is
withdrawn from the solution, the solution enters the pipette.
13. When concentrations or dilutions of the sample that are to be transferred changes, when
different solutions or chemicals are transferred, when containment is required, tips need to be
changed with each transfer.
14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes.
They are still useful because linkers or adapters can be attached to the blunt end to make them
useful. Moreover, new restriction sites can be created by attaching linkers or adapters.
15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it
recognizes the same restriction site in both the DNA. Therefore, the two different DNA can
basepair at the sticky ends.
Solution
10. The push button is pushed to the second stop when we pipette out or transfer the sample, and
to completely void the sample without leaving any sample in the pipette tips
11. The button is to be released after it is withdrawn from the solution/liquid. If it is released
prior to withdrawal, the solution reenters the pipette tip.
12. If the push button is pressed to the second stop and is released even before the pipette tip is
withdrawn from the solution, the solution enters the pipette.
13. When concentrations or dilutions of the sample that are to be transferred changes, when
different solutions or chemicals are transferred, when containment is required, tips need to be
changed with each transfer.
14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes.
They are still useful because linkers or adapters can be attached to the blunt end to make them
useful. Moreover, new restriction sites can be created by attaching linkers or adapters.
15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it
recognizes the same restriction site in both the DNA. Therefore, the two different DNA can
basepair at the sticky ends..
1. Working capital is the capital that a company uses for its day to.pdf
1. Working capital is the capital that a company uses for its day to day operations. Working
capital is computed using the formula: Working capital = current assets – current liabilities.
2. Determinants of working capital requirements are – nature of business, seasonality of
operations, production policy, market conditions and conditions of supply. A service company
will have a lower working capital requirement than a manufacturing company. Firms with
seasonality in operations will have higher fluctuations with regards to their working capital
requirements. Market conditions in the form of degree of competition will affect working capital
requirements. Higher competitive pressure will increase working capital requirements.
3. Three major decisions that managers have to take while performing the finance function are:
(i) Investment decision – Managers have to select those assets that the business will invest in. It
is also known as capital budgeting decisions.
(ii) Financing decision – These decisions pertain to determining how the total funds that are
required for the business will be obtained – will it be through debt or equity or a mix of debt or
equity.
(iii) Dividend decision – This pertains to determining what quantum of earnings should be
distributed to shareholders as dividends and what quantum should be retained for meeting future
growth requirements of the company.
4. Financial management involves planning, directing, monitoring and controlling the monetary
resources of a company in such a manner that the goals and objectives of the company are
achieved in an efficient manner. It involves management of capital budgeting, capital structure,
working capital management, etc. Financial management is important as it helps an organization
to set clarity towards its financial goals and helps in efficient utilization of resources.
5. Cash flow is not a suitable judge of profitability as it merely shows the changes in cash
position of a firm in a financial year from its operating activities, from its financing activities and
from its investing activities. Cash flow statement does not help us to determine gross profit
margins, operating profit margins and net profit margins. We can just gauge the reason for
changes in cash position in a year.
6. Financial risk refers to all kinds of risk associated with a financial transaction and an
investment transaction. It can be in the form of credit risk, asset backed risk, investment risk etc.
These risks arise due to the fact that there is a probability of loss that is inherent in any financing
and investment method and this probability of loss cannot be avoided.
Solution
1. Working capital is the capital that a company uses for its day to day operations. Working
capital is computed using the formula: Working capital = current assets – current liabilities.
2. Determinants of working capital requirements are – nature of business, seasonality of
operations, production policy, market conditions and condi.
1. no2.The Waterfall model has some disadvantages.Agile software.pdf
1. no
2.The Waterfall model has some disadvantages.
Agile software development evolved to eliminate the issues the Waterfall model has. It has a
completely new framework. While the Waterfall model has a sequential design, the Agile model
followed an incremental approach.When clients who used to follow the Waterfall model
switched to Agile, the transition brought many issues with it. The reason being inadaptability to a
different approach to software development. The end product turned out to be a disaster.
3. yes
4. The initialization step creates a base version of the system. The goal for this initial
implementation is to create a product to which the user can react. It should offer a sampling of
the key aspects of the problem and provide a solution that is simple enough to understand and
implement easily. To guide the iteration process, a project control list is created that contains a
record of all tasks that need to be performed. It includes such items as new features to be
implemented and areas of redesign of the existing solution.
5.yes
6.the mainly reason is to testing should be conducted faulty elements of the software can be
quickly identified because few changes are made within any single iteration.Customer can
respond to features and review the product for any needed or useful changes., a plan is developed
for the next increments, and modifications are made accordingly. This process continues, with
increments being delivered until the complete product is delivered.
7.Agile Software Development
Because a framework that is used to structure, plan, and control the process of developing an
information system.The methods attempt to minimize risk by developing software in short
timeboxes, called iterations, Each iteration is like a miniature software project of its own, and
includes all the tasks necessary to release the mini increment of new functionality: planning,
requirements analysis, design, coding, testing, and documentation. While iteration may not add
enough functionality to warrant releasing the product, an agile software project intends to be
capable of releasing new software at the end of every iteration emphasize working software as
the primary measure of progress. Combined with the preference for face-to-face communication,
agile methods produce very little written documentation relative to other methods.
Solution
1. no
2.The Waterfall model has some disadvantages.
Agile software development evolved to eliminate the issues the Waterfall model has. It has a
completely new framework. While the Waterfall model has a sequential design, the Agile model
followed an incremental approach.When clients who used to follow the Waterfall model
switched to Agile, the transition brought many issues with it. The reason being inadaptability to a
different approach to software development. The end product turned out to be a disaster.
3. yes
4. The initialization step creates a base version of the system. The goal for this initial
implementat.
True ; this is known as Tyndal effect .pdf
True ; this is known as Tyndal effect
Solution
True ; this is known as Tyndal effect.
TLC is used to separate the compounds.the benzoic.pdf
TLC is used to separate the compounds.the benzoic acid and methanol are more
polar so these have small rf value.so we can easyly separate the ester
Solution
TLC is used to separate the compounds.the benzoic acid and methanol are more
polar so these have small rf value.so we can easyly separate the ester.
The answer to this question is (E) If you look at.pdf
The answer to this question is (E) If you look at N2, you can see that as 2 moles of
N2 is formed 4 moles of NH3 is reacting. Thus, since the rate of formation of N2 is 2.0
mol/(L*s), the rate of formation of NH3 is double that (2*2.0mol/(L*s))=4.0 mol/(L*s). I hope
this helped!
Solution
The answer to this question is (E) If you look at N2, you can see that as 2 moles of
N2 is formed 4 moles of NH3 is reacting. Thus, since the rate of formation of N2 is 2.0
mol/(L*s), the rate of formation of NH3 is double that (2*2.0mol/(L*s))=4.0 mol/(L*s). I hope
this helped!.
Redox reactions are analogous in that they are el.pdf
Redox reactions are analogous in that they are electron transfer reactions. Certain
chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are
oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse
oxidizing agent it is.
Solution
Redox reactions are analogous in that they are electron transfer reactions. Certain
chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are
oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse
oxidizing agent it is..
moles NH4NO3 = 2.57 g 80.04 gmol=0.0321 moles H.pdf
moles NH4NO3 = 2.57 g /80.04 g/mol=0.0321 moles H2O = 4 x 0.0321 /2= 0.0642
moles O2 = 0.0321/2 =0.0161 moles N2 = 0.0321 total moles = 0.112 so p = 0.112 x 0.08206 x
520 K/ 4.50=1.06 atm
Solution
moles NH4NO3 = 2.57 g /80.04 g/mol=0.0321 moles H2O = 4 x 0.0321 /2= 0.0642
moles O2 = 0.0321/2 =0.0161 moles N2 = 0.0321 total moles = 0.112 so p = 0.112 x 0.08206 x
520 K/ 4.50=1.06 atm.
mole fraction of He = mole of He total mole now.pdf
mole fraction of He = mole of He/ total mole now mole of He = weight/M.wt = 2/4
= 0.5 mole of O2 = 5.99/32 = 0.187 total mole = 0.687 mole fraction of He = 0.5/0.687 = 0.727
Solution
mole fraction of He = mole of He/ total mole now mole of He = weight/M.wt = 2/4
= 0.5 mole of O2 = 5.99/32 = 0.187 total mole = 0.687 mole fraction of He = 0.5/0.687 = 0.727.
It is a molecular process involving the exchange .pdf
It is a molecular process involving the exchange of bonds between the two reacting
chemical species, which results in the creation of products with similar or identical bonding
affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These
chemical species can either be ionic or covalent. When referring to precipitation reactions
between solutions of ions in inorganic chemistry, these were formerly referred to as double
displacement or double replacement reactions, and these terms are still encouraged. It seems that
the term double decomposition is more specifically used when at least one of the substances does
not dissolve in the liquor, as the ligand or ion exchange takes place in the solid state of the
reactant, i.e. AX(aq) + BY(s) ? AY(aq) + BX(s). Types: Neutralization A neutralization reaction
is a specific type of double displacement reaction. Neutralization occurs when an acid reacts with
an equal amount of a base. A neutralization reaction creates a solution of a salt and water. For
example, hydrochloric acid reacts with sodium hydroxide to produce salt and water: HCl (aq) +
NaOH (aq) ? NaCl (aq) + H2O (l) Aqueous metathesis (precipitation) Metathesis reactions can
occur between two inorganic salts when one product is insoluble in water, driving the reaction
forward. For example, the precipitation of silver chloride from a mixture of silver nitrate and
sodium chloride causes sodium nitrate to be left in solution: AgNO3 (aq) + NaCl (aq) ? AgCl (s)
+ NaNO3 (aq) The formation of an insoluble gas that bubbles out of the solution, or a molecular
compound such as water, also drives the reaction to completion. Therefore, a solubility chart (or
general knowledge of solubility rules) can be used to predict whether two aqueous solutions will
react. HSAB theory can also be used to predict the products of a metathesis reaction. Aqueous
metathesis (double decomposition) The reactants need not to be dissolved for metathesis
reactions to take place. An example of this is the formation of barium thiocyanate when boiling a
mixture of copper(I)thiocyanate and barium hydroxide in water: Ba(OH)2 (s) + 2 CuCNS (s) ?
Ba(CNS)2 (aq) + 2CuOH (s) Acid and carbonates A subcategory of aqueous metathesis
reactions is the reaction of an acid with a carbonate or bicarbonate. Such a reaction always yields
carbonic acid as a product, which spontaneously decomposes into carbon dioxide and water. The
release of carbon dioxide gas from the reaction mixture drives the reaction to completion. For
example, a common, science-fair \"volcano\" reaction involves the reaction of acetic acid with
sodium bicarbonate: HCH3COO (aq) + NaHCO3 (s) ? NaCH3COO (aq) + CO2 (g) + H2O (l)
Solution
It is a molecular process involving the exchange of bonds between the two reacting
chemical species, which results in the creation of products with similar or identical bonding
affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These
che.
Helium contains a 1s^2 orbital. .pdf
Helium contains a 1s^2 orbital.
Solution
Helium contains a 1s^2 orbital..
From the LCAO method The overlap of the 1s AO on.pdf
From the LCAO method: The overlap of the 1s AO on one H atom with the is AO
on a second gives one bonding MO (s) at lower energy to the constituent AOs and one
antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO
\"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5
so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is
promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the
excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H.
(H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on
the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4
in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding
electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two
bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the
antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 -
gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the
charge will be +1.
Solution
From the LCAO method: The overlap of the 1s AO on one H atom with the is AO
on a second gives one bonding MO (s) at lower energy to the constituent AOs and one
antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO
\"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5
so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is
promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the
excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H.
(H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on
the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4
in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding
electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two
bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the
antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 -
gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the
charge will be +1..
Ecell=E(nought) -RT(2F)ln[Mg2+][Zn2+] .pdf
Ecell=E(nought) -RT/(2F)*ln[Mg2+]/[Zn2+]
Solution
Ecell=E(nought) -RT/(2F)*ln[Mg2+]/[Zn2+].
Cis-trans isomers occur because it is impossible .pdf
Cis-trans isomers occur because it is impossible to rotate the double bond without
breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually
separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3
group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the
other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of
the compound you drew (you have the cis isomer). Now, imagine that instead of the
CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s
would result in the same compound, and cis-trans isomers would not be possible. In general,
cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two
DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double
bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two -
CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III.
Solution
Cis-trans isomers occur because it is impossible to rotate the double bond without
breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually
separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3
group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the
other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of
the compound you drew (you have the cis isomer). Now, imagine that instead of the
CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s
would result in the same compound, and cis-trans isomers would not be possible. In general,
cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two
DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double
bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two -
CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III..
When a heterozygote (Aa) is not an exact intermediate between the tw.pdf
When a heterozygote (Aa) is not an exact intermediate between the two homozygotes (AA and
aa), then not all of the genetic variance in the population can be accounted for by differences in
the average effect of the A and a alleles.
In this case, the total genetic variance of the population can be subdivided into two components:
Additive Genetic Variance (sa2): Genetic variation attributed to average differences between
different allelic characters.
Dominance Genetic Variance (sd2): Variance due to heterozygotes not being exact midpoints of
homozygote parents.
So that,
Sg2 = Sa2 + Sd2
And now, total phenotypic variance can be rewritten as:
Sp2 = Se2 + Sg2 = Se2 + Sa2 + Sd2
From this, we can define a new kind of heritability:
Heritability in the narrow sense (h2):Proportion of phenotypic variance due to additive genetic
variance alone
h2 = Sa2 / Sp2 = Sa2 / Se2 + Sg2 = Sa2 / Se2 + Sd2 + Sa2
The greater h2 is for a population, the greater the fraction of the difference between selected
parents and the population as a whole that will be preserved in the offspring of those parents.
______________________________________________________________
To see why it is h2 that matters for response to selection, consider an extreme case:
2 homozygotes are equal in phenotype, say with a value of 20. (AA=aa=20)
But the heterozygote is overdominant, maybe with a value of 24. (Aa=24)
So for this population, there is only dominance variance, and no additive genetic variance
because there is no average difference between the carriers of a alleles and the carriers of A
alleles.
Now suppose that the population mates randomly and the genotype proportions are:
If only individuals with the highest values were selected to breed, all parents would be
heterozygotes.
Offspring of heterozygotes would exactly reproduce the 1:2:1 distribution of genotypes seen in
the parental generation.
Thus, there would be no change in genetic composition of the F1 generation, despite the fact
that there is genetic variation in the population.
So, selection proves to be ineffective when all variance is due to dominance and there is no
additive variance.
It is important to consider that dominance variance arises from more than just the deviation of
heterozygotes from the exact midpoint of homozygotes.
Dominance variance (sd2) is considered to be any genetic variation that cannot be explained by
average allelic differences, or additive genetic variance (sa2)
For example, if a trait is affected by multiple loci then any epistatic interactions between the loci
will appear as variance not associated with additive variance.
In theory, this interaction variance (si2) could be separated from dominance variance (sd2).
In practice, however, it is not possible to achieve accurate measures of separating these two
parameters so all nonadditive variance is attributed to dominance variance (sd2).
_____________________________________________________________________________
________________
Important Mess.
Volume of water = masssolubility of calcium phosphate= 13.41 x 1.pdf
Volume of water = mass/solubility of calcium phosphate
= 1/3.41 x 10-5
= 2.93 x 104 L
Solution
Volume of water = mass/solubility of calcium phosphate
= 1/3.41 x 10-5
= 2.93 x 104 L.
Astatine is a halogen. Column VII(7) on the perio.pdf
Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements
that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br),
Chlorine(Cl), Fluorine(F)
Solution
Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements
that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br),
Chlorine(Cl), Fluorine(F).
The use of a generic data type is preferred over using Object as a g.pdf
The use of a generic data type is preferred over using Object as a general class in a collection
because it ensures that all the members of the collection are objects related by inheritance. Using
generic data type, the immediate benefit can be obtained without derivation from a base
collection and implementation of type specific members
Solution
The use of a generic data type is preferred over using Object as a general class in a collection
because it ensures that all the members of the collection are objects related by inheritance. Using
generic data type, the immediate benefit can be obtained without derivation from a base
collection and implementation of type specific members.
The Hofmann rearrangement is the organic reaction of a primary amide.pdf
The Hofmann rearrangement is the organic reaction of a primary amide to a primary amine with
one fewer carbon atom.
The reaction is named after its dicoverer.
Solution
The Hofmann rearrangement is the organic reaction of a primary amide to a primary amine with
one fewer carbon atom.
The reaction is named after its dicoverer..
RainFall.java public class RainFall { private double months.pdf
RainFall.java
public class RainFall {
private double months[];
String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public RainFall(double d[]){
months = d;
}
public void displayTotalRainFall(){
double total = 0;
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]);
}
}
RainFallTest.java
import java.util.Scanner;
public class RainFallTest {
static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
double month[] = new double[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i
Solution
RainFall.java
public class RainFall {
private double months[];
String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public RainFall(double d[]){
months = d;
}
public void displayTotalRainFall(){
double total = 0;
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]);
}
}
RainFallTest.java
import java.util.Scanner;
public class RainFallTest {
static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
double month[] = new double[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i.
Proximal convoluted tubule performs absorption of majority of water .pdf
Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney.
It is lined by cuboidal epithelium.
Urethral orifice is the region which is the external opening of the urethra.
Solution
Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney.
It is lined by cuboidal epithelium.
Urethral orifice is the region which is the external opening of the urethra..
Please find my implementationpublic KeyedItem remove() { DO.pdf
Please find my implementation:
public KeyedItem remove()
{
//DO THIS
//remove the smallest item
// since Linked List is in sorted order, so head is the samllest element
if(head == null)
return null;
// return head element and move head one step forward
KeyedItem item = head.getItem();
head = head.getNext();
return item;
}
private Node locateNodeAdd(KeyedItem item)
{
//DO THIS
//find the insertion location (remember FIFO for duplicates)
Node curr = head;
Node prev = null;
while(curr != null){
if((current.getItem().getKey()).compareTo(item.getKey) >= 0){
return prev;
}
prev = curr;
curr = curr.getNext();
}
return prev;
}
Solution
Please find my implementation:
public KeyedItem remove()
{
//DO THIS
//remove the smallest item
// since Linked List is in sorted order, so head is the samllest element
if(head == null)
return null;
// return head element and move head one step forward
KeyedItem item = head.getItem();
head = head.getNext();
return item;
}
private Node locateNodeAdd(KeyedItem item)
{
//DO THIS
//find the insertion location (remember FIFO for duplicates)
Node curr = head;
Node prev = null;
while(curr != null){
if((current.getItem().getKey()).compareTo(item.getKey) >= 0){
return prev;
}
prev = curr;
curr = curr.getNext();
}
return prev;
}.
Oxidation means loss of electrons.Metals are electropositive and l.pdf
Oxidation means loss of electrons.
Metals are electropositive and lose electrons easily to form positively charged ions with stable
octet structures, thus they tend to be oxidized.
Reduction means gain of electrons.
Non-metals are electronegative and gain electrons easily to form negatively charged ions with
stable octet structures, thus they tend to be reduced.
Solution
Oxidation means loss of electrons.
Metals are electropositive and lose electrons easily to form positively charged ions with stable
octet structures, thus they tend to be oxidized.
Reduction means gain of electrons.
Non-metals are electronegative and gain electrons easily to form negatively charged ions with
stable octet structures, thus they tend to be reduced..
Mutualism 1. Two organisms live together and both gain from that.pdf
Mutualism : 1. Two organisms live together and both gain from that association
Chemotrophs : 2. Use chemicals to produce their own food
Commensalism : 3. One organism benefits from the association and the other one is unaffected
Autotrophs : 4. Produce their own food
Photoautotrophs : 5. Use light to produce their own food
Heterotrophs : 6. Feed on others
Parasitism : 7. Affects others negatively
Solution
Mutualism : 1. Two organisms live together and both gain from that association
Chemotrophs : 2. Use chemicals to produce their own food
Commensalism : 3. One organism benefits from the association and the other one is unaffected
Autotrophs : 4. Produce their own food
Photoautotrophs : 5. Use light to produce their own food
Heterotrophs : 6. Feed on others
Parasitism : 7. Affects others negatively.
a) The area of a disk of radius 1 is Pi.We need the integral over .pdf
a) The area of a disk of radius 1 is Pi.
We need the integral over the unit disk of f(x,y) to be 1 so
int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1
So c=1/Pi
b)
Fr(r)=P(R <= r)
= P(X^2+Y^2 <= r^2)
= int_(x^2+y^2<=r^2) 1/Pi dydx
= Pir^2/pi ( since the disk of radius r^2 has area pir^2 )
= r^2
c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2)
fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1
Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2)
Solution
a) The area of a disk of radius 1 is Pi.
We need the integral over the unit disk of f(x,y) to be 1 so
int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1
So c=1/Pi
b)
Fr(r)=P(R <= r)
= P(X^2+Y^2 <= r^2)
= int_(x^2+y^2<=r^2) 1/Pi dydx
= Pir^2/pi ( since the disk of radius r^2 has area pir^2 )
= r^2
c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2)
fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1
Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2).
Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Pavilions
NGV Architecture Commission Competition
MPavilion Commission Competition
Other Pavilion Designs
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
More Related Content
More from anjaliselectionahd
Redox reactions are analogous in that they are el.pdf
Redox reactions are analogous in that they are electron transfer reactions. Certain
chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are
oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse
oxidizing agent it is.
Solution
Redox reactions are analogous in that they are electron transfer reactions. Certain
chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are
oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse
oxidizing agent it is..
moles NH4NO3 = 2.57 g 80.04 gmol=0.0321 moles H.pdf
moles NH4NO3 = 2.57 g /80.04 g/mol=0.0321 moles H2O = 4 x 0.0321 /2= 0.0642
moles O2 = 0.0321/2 =0.0161 moles N2 = 0.0321 total moles = 0.112 so p = 0.112 x 0.08206 x
520 K/ 4.50=1.06 atm
Solution
moles NH4NO3 = 2.57 g /80.04 g/mol=0.0321 moles H2O = 4 x 0.0321 /2= 0.0642
moles O2 = 0.0321/2 =0.0161 moles N2 = 0.0321 total moles = 0.112 so p = 0.112 x 0.08206 x
520 K/ 4.50=1.06 atm.
mole fraction of He = mole of He total mole now.pdf
mole fraction of He = mole of He/ total mole now mole of He = weight/M.wt = 2/4
= 0.5 mole of O2 = 5.99/32 = 0.187 total mole = 0.687 mole fraction of He = 0.5/0.687 = 0.727
Solution
mole fraction of He = mole of He/ total mole now mole of He = weight/M.wt = 2/4
= 0.5 mole of O2 = 5.99/32 = 0.187 total mole = 0.687 mole fraction of He = 0.5/0.687 = 0.727.
It is a molecular process involving the exchange .pdf
It is a molecular process involving the exchange of bonds between the two reacting
chemical species, which results in the creation of products with similar or identical bonding
affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These
chemical species can either be ionic or covalent. When referring to precipitation reactions
between solutions of ions in inorganic chemistry, these were formerly referred to as double
displacement or double replacement reactions, and these terms are still encouraged. It seems that
the term double decomposition is more specifically used when at least one of the substances does
not dissolve in the liquor, as the ligand or ion exchange takes place in the solid state of the
reactant, i.e. AX(aq) + BY(s) ? AY(aq) + BX(s). Types: Neutralization A neutralization reaction
is a specific type of double displacement reaction. Neutralization occurs when an acid reacts with
an equal amount of a base. A neutralization reaction creates a solution of a salt and water. For
example, hydrochloric acid reacts with sodium hydroxide to produce salt and water: HCl (aq) +
NaOH (aq) ? NaCl (aq) + H2O (l) Aqueous metathesis (precipitation) Metathesis reactions can
occur between two inorganic salts when one product is insoluble in water, driving the reaction
forward. For example, the precipitation of silver chloride from a mixture of silver nitrate and
sodium chloride causes sodium nitrate to be left in solution: AgNO3 (aq) + NaCl (aq) ? AgCl (s)
+ NaNO3 (aq) The formation of an insoluble gas that bubbles out of the solution, or a molecular
compound such as water, also drives the reaction to completion. Therefore, a solubility chart (or
general knowledge of solubility rules) can be used to predict whether two aqueous solutions will
react. HSAB theory can also be used to predict the products of a metathesis reaction. Aqueous
metathesis (double decomposition) The reactants need not to be dissolved for metathesis
reactions to take place. An example of this is the formation of barium thiocyanate when boiling a
mixture of copper(I)thiocyanate and barium hydroxide in water: Ba(OH)2 (s) + 2 CuCNS (s) ?
Ba(CNS)2 (aq) + 2CuOH (s) Acid and carbonates A subcategory of aqueous metathesis
reactions is the reaction of an acid with a carbonate or bicarbonate. Such a reaction always yields
carbonic acid as a product, which spontaneously decomposes into carbon dioxide and water. The
release of carbon dioxide gas from the reaction mixture drives the reaction to completion. For
example, a common, science-fair \"volcano\" reaction involves the reaction of acetic acid with
sodium bicarbonate: HCH3COO (aq) + NaHCO3 (s) ? NaCH3COO (aq) + CO2 (g) + H2O (l)
Solution
It is a molecular process involving the exchange of bonds between the two reacting
chemical species, which results in the creation of products with similar or identical bonding
affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These
che.
Helium contains a 1s^2 orbital. .pdf
Helium contains a 1s^2 orbital.
Solution
Helium contains a 1s^2 orbital..
From the LCAO method The overlap of the 1s AO on.pdf
From the LCAO method: The overlap of the 1s AO on one H atom with the is AO
on a second gives one bonding MO (s) at lower energy to the constituent AOs and one
antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO
\"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5
so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is
promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the
excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H.
(H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on
the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4
in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding
electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two
bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the
antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 -
gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the
charge will be +1.
Solution
From the LCAO method: The overlap of the 1s AO on one H atom with the is AO
on a second gives one bonding MO (s) at lower energy to the constituent AOs and one
antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO
\"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5
so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is
promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the
excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H.
(H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on
the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4
in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding
electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two
bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the
antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 -
gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the
charge will be +1..
Ecell=E(nought) -RT(2F)ln[Mg2+][Zn2+] .pdf
Ecell=E(nought) -RT/(2F)*ln[Mg2+]/[Zn2+]
Solution
Ecell=E(nought) -RT/(2F)*ln[Mg2+]/[Zn2+].
Cis-trans isomers occur because it is impossible .pdf
Cis-trans isomers occur because it is impossible to rotate the double bond without
breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually
separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3
group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the
other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of
the compound you drew (you have the cis isomer). Now, imagine that instead of the
CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s
would result in the same compound, and cis-trans isomers would not be possible. In general,
cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two
DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double
bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two -
CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III.
Solution
Cis-trans isomers occur because it is impossible to rotate the double bond without
breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually
separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3
group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the
other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of
the compound you drew (you have the cis isomer). Now, imagine that instead of the
CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s
would result in the same compound, and cis-trans isomers would not be possible. In general,
cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two
DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double
bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two -
CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III..
When a heterozygote (Aa) is not an exact intermediate between the tw.pdf
When a heterozygote (Aa) is not an exact intermediate between the two homozygotes (AA and
aa), then not all of the genetic variance in the population can be accounted for by differences in
the average effect of the A and a alleles.
In this case, the total genetic variance of the population can be subdivided into two components:
Additive Genetic Variance (sa2): Genetic variation attributed to average differences between
different allelic characters.
Dominance Genetic Variance (sd2): Variance due to heterozygotes not being exact midpoints of
homozygote parents.
So that,
Sg2 = Sa2 + Sd2
And now, total phenotypic variance can be rewritten as:
Sp2 = Se2 + Sg2 = Se2 + Sa2 + Sd2
From this, we can define a new kind of heritability:
Heritability in the narrow sense (h2):Proportion of phenotypic variance due to additive genetic
variance alone
h2 = Sa2 / Sp2 = Sa2 / Se2 + Sg2 = Sa2 / Se2 + Sd2 + Sa2
The greater h2 is for a population, the greater the fraction of the difference between selected
parents and the population as a whole that will be preserved in the offspring of those parents.
______________________________________________________________
To see why it is h2 that matters for response to selection, consider an extreme case:
2 homozygotes are equal in phenotype, say with a value of 20. (AA=aa=20)
But the heterozygote is overdominant, maybe with a value of 24. (Aa=24)
So for this population, there is only dominance variance, and no additive genetic variance
because there is no average difference between the carriers of a alleles and the carriers of A
alleles.
Now suppose that the population mates randomly and the genotype proportions are:
If only individuals with the highest values were selected to breed, all parents would be
heterozygotes.
Offspring of heterozygotes would exactly reproduce the 1:2:1 distribution of genotypes seen in
the parental generation.
Thus, there would be no change in genetic composition of the F1 generation, despite the fact
that there is genetic variation in the population.
So, selection proves to be ineffective when all variance is due to dominance and there is no
additive variance.
It is important to consider that dominance variance arises from more than just the deviation of
heterozygotes from the exact midpoint of homozygotes.
Dominance variance (sd2) is considered to be any genetic variation that cannot be explained by
average allelic differences, or additive genetic variance (sa2)
For example, if a trait is affected by multiple loci then any epistatic interactions between the loci
will appear as variance not associated with additive variance.
In theory, this interaction variance (si2) could be separated from dominance variance (sd2).
In practice, however, it is not possible to achieve accurate measures of separating these two
parameters so all nonadditive variance is attributed to dominance variance (sd2).
_____________________________________________________________________________
________________
Important Mess.
Volume of water = masssolubility of calcium phosphate= 13.41 x 1.pdf
Volume of water = mass/solubility of calcium phosphate
= 1/3.41 x 10-5
= 2.93 x 104 L
Solution
Volume of water = mass/solubility of calcium phosphate
= 1/3.41 x 10-5
= 2.93 x 104 L.
Astatine is a halogen. Column VII(7) on the perio.pdf
Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements
that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br),
Chlorine(Cl), Fluorine(F)
Solution
Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements
that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br),
Chlorine(Cl), Fluorine(F).
The use of a generic data type is preferred over using Object as a g.pdf
The use of a generic data type is preferred over using Object as a general class in a collection
because it ensures that all the members of the collection are objects related by inheritance. Using
generic data type, the immediate benefit can be obtained without derivation from a base
collection and implementation of type specific members
Solution
The use of a generic data type is preferred over using Object as a general class in a collection
because it ensures that all the members of the collection are objects related by inheritance. Using
generic data type, the immediate benefit can be obtained without derivation from a base
collection and implementation of type specific members.
The Hofmann rearrangement is the organic reaction of a primary amide.pdf
The Hofmann rearrangement is the organic reaction of a primary amide to a primary amine with
one fewer carbon atom.
The reaction is named after its dicoverer.
Solution
The Hofmann rearrangement is the organic reaction of a primary amide to a primary amine with
one fewer carbon atom.
The reaction is named after its dicoverer..
RainFall.java public class RainFall { private double months.pdf
RainFall.java
public class RainFall {
private double months[];
String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public RainFall(double d[]){
months = d;
}
public void displayTotalRainFall(){
double total = 0;
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]);
}
}
RainFallTest.java
import java.util.Scanner;
public class RainFallTest {
static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
double month[] = new double[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i
Solution
RainFall.java
public class RainFall {
private double months[];
String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public RainFall(double d[]){
months = d;
}
public void displayTotalRainFall(){
double total = 0;
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]);
}
}
RainFallTest.java
import java.util.Scanner;
public class RainFallTest {
static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
double month[] = new double[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i.
Proximal convoluted tubule performs absorption of majority of water .pdf
Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney.
It is lined by cuboidal epithelium.
Urethral orifice is the region which is the external opening of the urethra.
Solution
Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney.
It is lined by cuboidal epithelium.
Urethral orifice is the region which is the external opening of the urethra..
Please find my implementationpublic KeyedItem remove() { DO.pdf
Please find my implementation:
public KeyedItem remove()
{
//DO THIS
//remove the smallest item
// since Linked List is in sorted order, so head is the samllest element
if(head == null)
return null;
// return head element and move head one step forward
KeyedItem item = head.getItem();
head = head.getNext();
return item;
}
private Node locateNodeAdd(KeyedItem item)
{
//DO THIS
//find the insertion location (remember FIFO for duplicates)
Node curr = head;
Node prev = null;
while(curr != null){
if((current.getItem().getKey()).compareTo(item.getKey) >= 0){
return prev;
}
prev = curr;
curr = curr.getNext();
}
return prev;
}
Solution
Please find my implementation:
public KeyedItem remove()
{
//DO THIS
//remove the smallest item
// since Linked List is in sorted order, so head is the samllest element
if(head == null)
return null;
// return head element and move head one step forward
KeyedItem item = head.getItem();
head = head.getNext();
return item;
}
private Node locateNodeAdd(KeyedItem item)
{
//DO THIS
//find the insertion location (remember FIFO for duplicates)
Node curr = head;
Node prev = null;
while(curr != null){
if((current.getItem().getKey()).compareTo(item.getKey) >= 0){
return prev;
}
prev = curr;
curr = curr.getNext();
}
return prev;
}.
Oxidation means loss of electrons.Metals are electropositive and l.pdf
Oxidation means loss of electrons.
Metals are electropositive and lose electrons easily to form positively charged ions with stable
octet structures, thus they tend to be oxidized.
Reduction means gain of electrons.
Non-metals are electronegative and gain electrons easily to form negatively charged ions with
stable octet structures, thus they tend to be reduced.
Solution
Oxidation means loss of electrons.
Metals are electropositive and lose electrons easily to form positively charged ions with stable
octet structures, thus they tend to be oxidized.
Reduction means gain of electrons.
Non-metals are electronegative and gain electrons easily to form negatively charged ions with
stable octet structures, thus they tend to be reduced..
Mutualism 1. Two organisms live together and both gain from that.pdf
Mutualism : 1. Two organisms live together and both gain from that association
Chemotrophs : 2. Use chemicals to produce their own food
Commensalism : 3. One organism benefits from the association and the other one is unaffected
Autotrophs : 4. Produce their own food
Photoautotrophs : 5. Use light to produce their own food
Heterotrophs : 6. Feed on others
Parasitism : 7. Affects others negatively
Solution
Mutualism : 1. Two organisms live together and both gain from that association
Chemotrophs : 2. Use chemicals to produce their own food
Commensalism : 3. One organism benefits from the association and the other one is unaffected
Autotrophs : 4. Produce their own food
Photoautotrophs : 5. Use light to produce their own food
Heterotrophs : 6. Feed on others
Parasitism : 7. Affects others negatively.
a) The area of a disk of radius 1 is Pi.We need the integral over .pdf
a) The area of a disk of radius 1 is Pi.
We need the integral over the unit disk of f(x,y) to be 1 so
int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1
So c=1/Pi
b)
Fr(r)=P(R <= r)
= P(X^2+Y^2 <= r^2)
= int_(x^2+y^2<=r^2) 1/Pi dydx
= Pir^2/pi ( since the disk of radius r^2 has area pir^2 )
= r^2
c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2)
fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1
Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2)
Solution
a) The area of a disk of radius 1 is Pi.
We need the integral over the unit disk of f(x,y) to be 1 so
int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1
So c=1/Pi
b)
Fr(r)=P(R <= r)
= P(X^2+Y^2 <= r^2)
= int_(x^2+y^2<=r^2) 1/Pi dydx
= Pir^2/pi ( since the disk of radius r^2 has area pir^2 )
= r^2
c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2)
fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1
Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2).
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