r=(13)d(I-)dt .pdf
•
0 likes•2 views
This 3 sentence document provides a mathematical formula for calculating the rate of change of angular momentum (r) as equal to one-third times the derivative of the moment of inertia (I) with respect to time. No other context or explanation is provided about the physical scenario or variables represented in the formula.
Report
Share
Report
Share
Download to read offline
Recommended
Use of the plates within a specific time is necessary as time of inc.pdf
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition.
Solution
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition..
The five terrestrial adaptations of the seed plants include, reducti.pdf
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today.
Solution
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today..
SELF THEORY OF LATE ADULT HOOD it describes the changes of a adult.pdf
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing.
Solution
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing..
PV = PMTiThe goal is to solve the formula for i, doing so by isol.pdf
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV
Solution
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV.
Normal gene regulation in MAT locus YeastThere are 3 regulatory p.pdf
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus
Solution
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus.
Ionic compounds have electropositive and electronegative elements pr.pdf
Ionic compounds have electropositive and electronegative elements present and high melting
points due to strong ionic bonding.
Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and
low melting points.
Nonpolar molecules have weak intermolecular forces, and low melting points.
LiF - ionic compound
ionic bonding
BeF2 - ionic compound
ionic bonding
BF3 - nonpolar molecule
dispersion force
CF4 - nonpolar molecule
dispersion force
NF3 - polar molecule
dispersion force
dipole-dipole interaction
OF2 - polar molecule
dispersion force
dipole-dipole interaction
F2 - nonpolar molecule
dispersion force
Solution
Ionic compounds have electropositive and electronegative elements present and high melting
points due to strong ionic bonding.
Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and
low melting points.
Nonpolar molecules have weak intermolecular forces, and low melting points.
LiF - ionic compound
ionic bonding
BeF2 - ionic compound
ionic bonding
BF3 - nonpolar molecule
dispersion force
CF4 - nonpolar molecule
dispersion force
NF3 - polar molecule
dispersion force
dipole-dipole interaction
OF2 - polar molecule
dispersion force
dipole-dipole interaction
F2 - nonpolar molecule
dispersion force.
Grey matter in the spinal cord is known as the grey column which tra.pdf
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system.
Solution
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system..
Recommended
Use of the plates within a specific time is necessary as time of inc.pdf
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition.
Solution
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition..
The five terrestrial adaptations of the seed plants include, reducti.pdf
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today.
Solution
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today..
SELF THEORY OF LATE ADULT HOOD it describes the changes of a adult.pdf
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing.
Solution
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing..
PV = PMTiThe goal is to solve the formula for i, doing so by isol.pdf
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV
Solution
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV.
Normal gene regulation in MAT locus YeastThere are 3 regulatory p.pdf
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus
Solution
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus.
Ionic compounds have electropositive and electronegative elements pr.pdf
Ionic compounds have electropositive and electronegative elements present and high melting
points due to strong ionic bonding.
Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and
low melting points.
Nonpolar molecules have weak intermolecular forces, and low melting points.
LiF - ionic compound
ionic bonding
BeF2 - ionic compound
ionic bonding
BF3 - nonpolar molecule
dispersion force
CF4 - nonpolar molecule
dispersion force
NF3 - polar molecule
dispersion force
dipole-dipole interaction
OF2 - polar molecule
dispersion force
dipole-dipole interaction
F2 - nonpolar molecule
dispersion force
Solution
Ionic compounds have electropositive and electronegative elements present and high melting
points due to strong ionic bonding.
Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and
low melting points.
Nonpolar molecules have weak intermolecular forces, and low melting points.
LiF - ionic compound
ionic bonding
BeF2 - ionic compound
ionic bonding
BF3 - nonpolar molecule
dispersion force
CF4 - nonpolar molecule
dispersion force
NF3 - polar molecule
dispersion force
dipole-dipole interaction
OF2 - polar molecule
dispersion force
dipole-dipole interaction
F2 - nonpolar molecule
dispersion force.
Grey matter in the spinal cord is known as the grey column which tra.pdf
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system.
Solution
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system..
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Solution
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delive.
c. A destructor to free any allocated objects.d. A copy constructo.pdf
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables.
Solution
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables..
Barrier to entryThe government intervention to the barriers to en.pdf
Barrier to entry:
The government intervention to the barriers to entry to exist as a result of, while others occur
naturally within the business world. Often, existing firms within an industry lobby for the
government to erect new barriers to entry. In this is done to done by the protect the integrity of
the industry and prevent fly-by-night operations from setting up shop and hawking inferior
products and services. In reality, firms favor barriers to entry when already comfortably
ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue.
Other barriers to entry occur naturally, often evolving over time as certain industry players
establish dominance.
A company can use this technology, for example, to build a barrier to entry, to build in switching
costs, and even, sometimes, to completely change the basis of competition. some companies
have seized the advantage, while others, more complacent, have ended up playing the difficult
and expensive game of catch-up ball. He also points out that it is important for executives to
make this competitive analysis in assessing where IS fits in their companies, since in some cases
it appropriately plays a support role and can add only modestly to the value of a company’s
products, while in other settings it is at the core of their competitive survival.
The computer’s main purpose is to cut order-entry costs and to provide more flexibility to
customers in the time and process of order submission. The system yields a larger competitive
advantage, adding value for customers and a substantial rise in their sales. The resulting sharp
increase in the company’s market share forces a primary competitor into a corporate
reorganization and a massive systems development effort to contain the damage.
Natural Barriers to Entry:
In industries where customers incur high costs switching from one brand to another, this
becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective
customers to pay the money required to make a chahange. Brand identity and customer loyalty
serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics
of an industry take shape of the natural barrier.
Barrier option:
A barrier option is a type of option whose payoff depends on whether or not the underlying asset
has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it
can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but
limiting losses for the writer. It can also be a knock-in, meaning it has no value until the
underlying reaches a certain price.
BREAKING DOWN \'Barrier Option\':
Barrier options are considered a type of exotic option because they are more complex than basic
American or European options. Barrier options are also considered a type of path-dependent
option because their value fluctuates as the underlying\'s value changes during the opt.
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules
Solution
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules.
An example of when a one-way ANOVA could be used is if you want to d.pdf
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA.
Solution
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA..
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn
Solution
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn.
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol
Solution
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol.
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis
Solution
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.
The intermediate is the compound A. There are a c.pdf
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration).
Solution
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration)..
When ionic solids dissolve in water theions that are adjacent to each.pdf
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in thesolution process. the solution process is governed bythe
solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have
only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent
interactions. When we consider the soluteinteractions we begin to see some of the problems that
can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to
overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a
solidis a measure of the strength of those electrostatic attractions.The lattice energy works to
keep the ions in the solid state andionic compounds with large lattice energies are insoluble in
waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves
in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g)
ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation;
H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two
equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3
is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms
as the energy required to separate the ions in thesolute and the heat of solution isendothermic.
Solution
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in.
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
You have the priority for 1 right (the OH), howev.pdf
You have the priority for 1 right (the OH), however the positions 2 and 3 as you
have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group,
however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority
2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H
is projected outwards so you get the opposite configuration of what you see (if it looks like an S
you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom
Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of
the molecule, therefore the R is really an S. Hope this helped!
Solution
You have the priority for 1 right (the OH), however the positions 2 and 3 as you
have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group,
however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority
2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H
is projected outwards so you get the opposite configuration of what you see (if it looks like an S
you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom
Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of
the molecule, therefore the R is really an S. Hope this helped!.
a electrolytic solution because it is an ionic co.pdf
a electrolytic solution because it is an ionic compound which dissociates in water
and conduct electricity
Solution
a electrolytic solution because it is an ionic compound which dissociates in water
and conduct electricity.
conjugated dienes are those in which two double b.pdf
conjugated dienes are those in which two double bonds are separated by a single
bond so answer is E
Solution
conjugated dienes are those in which two double bonds are separated by a single
bond so answer is E.
Fluorine is the most reactive and most electroneg.pdf
Fluorine is the most reactive and most electronegative of the elements. Fluorine also
has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner.
As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the
other hand is considerably less electronegative than fluorine and it\'s electron affinity is
considerably less exothermic than that of fluorine.
Solution
Fluorine is the most reactive and most electronegative of the elements. Fluorine also
has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner.
As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the
other hand is considerably less electronegative than fluorine and it\'s electron affinity is
considerably less exothermic than that of fluorine..
hey dear neutron and protons do not occupy orbita.pdf
hey dear neutron and protons do not occupy orbitals... they are in nucleus and
electron are present in orbitals
Solution
hey dear neutron and protons do not occupy orbitals... they are in nucleus and
electron are present in orbitals.
at 100 degree KOH needs less Ea which cannot be p.pdf
at 100 degree KOH needs less Ea which cannot be provided at that temperature. We
have Ea(KOH) < Ea(NaOH)
Solution
at 100 degree KOH needs less Ea which cannot be provided at that temperature. We
have Ea(KOH) < Ea(NaOH).
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.
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 inbox5
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Solution
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delive.
c. A destructor to free any allocated objects.d. A copy constructo.pdf
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables.
Solution
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables..
Barrier to entryThe government intervention to the barriers to en.pdf
Barrier to entry:
The government intervention to the barriers to entry to exist as a result of, while others occur
naturally within the business world. Often, existing firms within an industry lobby for the
government to erect new barriers to entry. In this is done to done by the protect the integrity of
the industry and prevent fly-by-night operations from setting up shop and hawking inferior
products and services. In reality, firms favor barriers to entry when already comfortably
ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue.
Other barriers to entry occur naturally, often evolving over time as certain industry players
establish dominance.
A company can use this technology, for example, to build a barrier to entry, to build in switching
costs, and even, sometimes, to completely change the basis of competition. some companies
have seized the advantage, while others, more complacent, have ended up playing the difficult
and expensive game of catch-up ball. He also points out that it is important for executives to
make this competitive analysis in assessing where IS fits in their companies, since in some cases
it appropriately plays a support role and can add only modestly to the value of a company’s
products, while in other settings it is at the core of their competitive survival.
The computer’s main purpose is to cut order-entry costs and to provide more flexibility to
customers in the time and process of order submission. The system yields a larger competitive
advantage, adding value for customers and a substantial rise in their sales. The resulting sharp
increase in the company’s market share forces a primary competitor into a corporate
reorganization and a massive systems development effort to contain the damage.
Natural Barriers to Entry:
In industries where customers incur high costs switching from one brand to another, this
becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective
customers to pay the money required to make a chahange. Brand identity and customer loyalty
serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics
of an industry take shape of the natural barrier.
Barrier option:
A barrier option is a type of option whose payoff depends on whether or not the underlying asset
has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it
can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but
limiting losses for the writer. It can also be a knock-in, meaning it has no value until the
underlying reaches a certain price.
BREAKING DOWN \'Barrier Option\':
Barrier options are considered a type of exotic option because they are more complex than basic
American or European options. Barrier options are also considered a type of path-dependent
option because their value fluctuates as the underlying\'s value changes during the opt.
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules
Solution
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules.
An example of when a one-way ANOVA could be used is if you want to d.pdf
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA.
Solution
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA..
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn
Solution
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn.
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol
Solution
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol.
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis
Solution
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.
The intermediate is the compound A. There are a c.pdf
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration).
Solution
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration)..
When ionic solids dissolve in water theions that are adjacent to each.pdf
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in thesolution process. the solution process is governed bythe
solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have
only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent
interactions. When we consider the soluteinteractions we begin to see some of the problems that
can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to
overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a
solidis a measure of the strength of those electrostatic attractions.The lattice energy works to
keep the ions in the solid state andionic compounds with large lattice energies are insoluble in
waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves
in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g)
ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation;
H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two
equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3
is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms
as the energy required to separate the ions in thesolute and the heat of solution isendothermic.
Solution
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in.
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
You have the priority for 1 right (the OH), howev.pdf
You have the priority for 1 right (the OH), however the positions 2 and 3 as you
have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group,
however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority
2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H
is projected outwards so you get the opposite configuration of what you see (if it looks like an S
you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom
Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of
the molecule, therefore the R is really an S. Hope this helped!
Solution
You have the priority for 1 right (the OH), however the positions 2 and 3 as you
have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group,
however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority
2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H
is projected outwards so you get the opposite configuration of what you see (if it looks like an S
you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom
Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of
the molecule, therefore the R is really an S. Hope this helped!.
a electrolytic solution because it is an ionic co.pdf
a electrolytic solution because it is an ionic compound which dissociates in water
and conduct electricity
Solution
a electrolytic solution because it is an ionic compound which dissociates in water
and conduct electricity.
conjugated dienes are those in which two double b.pdf
conjugated dienes are those in which two double bonds are separated by a single
bond so answer is E
Solution
conjugated dienes are those in which two double bonds are separated by a single
bond so answer is E.
Fluorine is the most reactive and most electroneg.pdf
Fluorine is the most reactive and most electronegative of the elements. Fluorine also
has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner.
As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the
other hand is considerably less electronegative than fluorine and it\'s electron affinity is
considerably less exothermic than that of fluorine.
Solution
Fluorine is the most reactive and most electronegative of the elements. Fluorine also
has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner.
As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the
other hand is considerably less electronegative than fluorine and it\'s electron affinity is
considerably less exothermic than that of fluorine..
hey dear neutron and protons do not occupy orbita.pdf
hey dear neutron and protons do not occupy orbitals... they are in nucleus and
electron are present in orbitals
Solution
hey dear neutron and protons do not occupy orbitals... they are in nucleus and
electron are present in orbitals.
at 100 degree KOH needs less Ea which cannot be p.pdf
at 100 degree KOH needs less Ea which cannot be provided at that temperature. We
have Ea(KOH) < Ea(NaOH)
Solution
at 100 degree KOH needs less Ea which cannot be provided at that temperature. We
have Ea(KOH) < Ea(NaOH).
More from inbox5 (18)
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
c. A destructor to free any allocated objects.d. A copy constructo.pdf
c. A destructor to free any allocated objects.d. A copy constructo.pdf
Barrier to entryThe government intervention to the barriers to en.pdf
Barrier to entryThe government intervention to the barriers to en.pdf
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
An example of when a one-way ANOVA could be used is if you want to d.pdf
An example of when a one-way ANOVA could be used is if you want to d.pdf
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
The intermediate is the compound A. There are a c.pdf
The intermediate is the compound A. There are a c.pdf
When ionic solids dissolve in water theions that are adjacent to each.pdf
When ionic solids dissolve in water theions that are adjacent to each.pdf
#include iostream #include fstream #include map #include.pdf
#include iostream #include fstream #include map #include.pdf
You have the priority for 1 right (the OH), howev.pdf
You have the priority for 1 right (the OH), howev.pdf
a electrolytic solution because it is an ionic co.pdf
a electrolytic solution because it is an ionic co.pdf
conjugated dienes are those in which two double b.pdf
conjugated dienes are those in which two double b.pdf
Fluorine is the most reactive and most electroneg.pdf
Fluorine is the most reactive and most electroneg.pdf
hey dear neutron and protons do not occupy orbita.pdf
hey dear neutron and protons do not occupy orbita.pdf
at 100 degree KOH needs less Ea which cannot be p.pdf
at 100 degree KOH needs less Ea which cannot be p.pdf
Recently uploaded
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.
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
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Class 11 CBSE Business Studies Project ( AIDS TO TRADE - INSURANCE)
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
The Roman Empire A Historical Colossus.pdf
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Recently uploaded (20)
Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Overview on Edible Vaccine: Pros & Cons with Mechanism
Overview on Edible Vaccine: Pros & Cons with Mechanism
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...