Mean can be computed for variables with ordinal-level measurements..pdf
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Mean can be computed for variables with ordinal-level measurements. True or False? Solution Sol) False Mean can not be computed for variables with ordinal-level measurements. we can find median for ordinal measurement. Ex: Gender, Intelligence and taste. These are qualitative variables so for these type of variables we can not find mean. we use median.
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Write an equation for the height of the point P above the ground as .pdf
Write an equation for the height of the point P above the ground as a function of t, time in
seconds.
h(t) =
Hint: Start by thinking: What is the height of the point P above the center? How is the center
changing?
Solution
height of center at ant time = R + 7(sinA)t = 2 + 7(100/700)t = (2 + t) cm
now height of P wrt center = Rcos(x/2) where x = 7t
height of P = 2+ t + Rcos(7t/2)
= 2 + t + 2cos(7t/2)
7t/2 is value in radian. oon putting t in second
for t= 0,(starting point)
height = 2 + 2cos0 = 4 cm
at the end, at = 100 cm,
h = 2 + 100 + 2cos(350) = 101.433 cm.
Wild-type (WT) mice are irradiated, destroying all of their bone marr.pdf
Wild-type (WT) mice are irradiated, destroying all of their bone marrow cells including CD34+
hematopoietic stem cells (HSC). CD34+ HSC is isolated from (1) an untreated wild-type
syngeneic control mouse, and (2) various immune gene mutant (-/- null) donor mice. The CD34+
HSC from the different mouse genotypes is injected into the wild-type irradiated recipient mice.
Predict the reconstitution of the hematopoietic system in each of the recipient mice (+ or -).
Solution
Answer -
Hematopoietic cells reconstituted in recipient
CD34+HSC
Doner GenotypeErythrocytes macrophages T cells B cells
Wild Type mouse + + + +
CD19(-/-)mouse + + + -
Rag1/2(-/-)mouse + + - -
erythropoietin EPO(-/-) - + + +
AIRE(-/-)mouse + + + +
explaination -Wild type mouse has normal CD34+HSC which is responsible to diffentiate into
all type of immune cells in bone marrow. Whwn this CD34+HSC from wild type mouse injected
into recipient mouse its bone marrow produce all type of immune cells.
CD19 is a B cell lymphocyte antigen It is a protein which is coded by CD19 gene. it is found on
the surface of B cells and helps in its proliferation and differenciation.CD34+HSC from doner
who lack CD19 gene when injected into recipient mouse,It will be unable to produce B cell.
Rag 1/2 also known as recombination activating gene codes a protein which paly an important
role in recombination of the gene of immunoglobuline and T cell receptor marker.two gene
products of recombination activating gene are Rag 1 and Rag 2 protein.These two rag protein are
essential to the generation of mature Bcells and Tcells. .CD34+Hsc from doner who lack the
gene codes gor Rag1/2 protein when injected to the recipient mouse, recipient mousewill be
unable to produce mature B and T cells.
Erythropoietin EPO is a protein harmone secreated by kidney in responce to less
oxygen/Heamoglobine in blood.erythropoietin stimulates bone marrow to produce more Red
blood cells (erythrocytes) which contain Heamoglobine ( an Oxygen carrier) and supply required
amount of oxygen to the body by mean of blood flow through the body. when CD34+HSC from
doner lack the gene codes for Erythropoietin injected into recipient mouse,it bone marrow will
be unable to produce erythrocytes.
AIRE is an auto immune regulator protein encoded by AIRE gene.It controles a mechanism that
prevent T cells (an important cell for immune response) from attacking the self body antigen.T
cells recognizes antigen which are expresed on the surface of antigen presenting cell by T cell
receptors present on T cells and produce Immune responce against that antigen. In autoimmune
diseases T cell Recognize self antigen and produce immune response against self antigen which
leads to sever problem in self body.AIRE protein play an important role to controle these kind of
autoimmune disorders by the removel of T cell through apoptosis while T cells are still in
thymus.
When CD34+HSC from doner WHICh lake the gene for AIRE protein injected into recipient
mouse.Reci.
Which of the following is considered to be a distributed denial of s.pdf
Which of the following is considered to be a distributed denial of service attack because it comes
from numerous sources?a.Smurf
Solution
The Distributed denial of service (DDOS) is an attack where the attacker repeatedly pings the
server from different computers/servers to restrict the server from handling the requests of the
legitimate users. This way all the requests of the authorized users will get be declined.
The SYN flood attack and the death ping send multiple requests to the server from a single
source and the sequence prediction attack is used to analyze the traffic to identify the data
packets.
The smurf attack uses a broadcast network to send multiple requests to the server from different
sources. Hence, the correct options is a. Smurf..
Which of the folllowing statements about the solubility of 1- propan.pdf
Which of the folllowing statements about the solubility of 1- propanol, ch3ch2ch2oh, and
ethanethiol, ch3ch2sh, are true?
A) Ethanethiol is more soluble in water because sulfer has agreater electronegativity value than
oxygen.
B) Ethanethiol is more soluble in water because it has a lower molecular mass than 1-propanol.
C) 1-propanol is more soluble in water because it has a greater molecular mass than ethanethiol.
D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value than
sulfer.
Solution
D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value
than sulfer.
\\\ eason
propanol has a hydrogen bonding by itself while ethanethiol does not have any hydrogen
bonding by itself that is why propanol is more soluble in water. because oxygen has a greater
electronegativity value than sulfer.
the more hydrogen bonding by itself the more soluble in water!.
What type of mutation that impacts splicing is a significant contrib.pdf
What type of mutation that impacts splicing is a significant contributor to human disease?
Solution
Disruptions in splicing are major contibutor of human diseases.Most genes in higher eukaryotes
are composed of introns (non coding segments) and exons (coding segments). The majority of
human intron removals ae catalyzed by a large and dynamic ribonucleoprotein (RNP) complex
called the spliceosome.This process involves two sequential transesterification reactions which
ligate the exons and release the intron as a lariat.
Alternate splicing which allows for the productioin of many proteins from a single genomic
locous.This greatly augments the repertorie of the protein.This greatly augments the repertoire of
proteins that can be produced by a given gene and plays an important role in
evolution,development and disease..
What data indicate that all three germ layers are specified in the b.pdf
What data indicate that all three germ layers are specified in the blastula? What are the
differences between dorsal and ventral mesodermal derivatives and what cellular interactions are
required for their specification?
Solution
Three germ layers of amphibians are specified in the blastula is determined by isolating these
tissues in vitro, and they are able to form into specific germ layers. The animal pole cap cells
make ectoderm, marginal region cells make mesoderm, and vegetal cells make endoderm.
Theyare specified but not determined. It is also interesting to note that if animal cap cells are
place co-cultured with vegetal cells, the animal cap cells with become mesoderm. This indicates
that vegetal cells induce other cells to form mesoderm. Experiment paired animal cap cells in 4
different sections of vegetal blastomeres to see if they induce different dorsal-ventral
mesodermal fates. Result showed that different sections of vegetal blastomere have specific
inductive capacities, which is crucial for dorsal-ventral mesoderm determination. Difference
between dorsal and ventral mesoderm derivatives: dorsal mesoderm is the notochord and somite.
In all bilaterian animals, the mesoderm is one of the three primary germ layers in the very early
embryo. The other two layers are the ectoderm (outside layer) and endoderm (inside layer), with
the mesoderm as the middle layer between them.
The mesoderm forms mesenchyme, mesothelium, non-epithelial blood cells and coelomocytes.
Some of the mesoderm derivatives include the muscle (smooth, cardiac and skeletal), the
muscles of the tongue (occipital somites), the pharyngeal arches muscle (muscles of mastication,
muscles of facial expressions), connective tissue, dermis and subcutaneous layer of the skin
At mid-blastula two signaling centers are present on the dorsal side: The prospective
neuroectoderm expresses bone morphogenetic protein (BMP) antagonists, and the future dorsal
endoderm secretes Nodal-related mesoderm-inducing factors. When dorsal mesoderm is formed
at gastrula, a cocktail of growth factor antagonists is secreted by the Spemann organizer and
further patterns the embryo. A ventral gastrula signaling center opposes the actions of the dorsal
organizer, and another set of secreted antagonists is produced ventrally under the control of
BMP4. The early dorsal -Catenin signal inhibits BMP expression at the transcriptional level and
promotes expression of secreted BMP antagonists in the prospective central nervous system
(CNS). In the absence of mesoderm, expression of Chordin and Noggin in ectoderm is required
for anterior CNS formation. FGF (fibroblast growth factor) and IGF (insulin-like growth factor)
signals are also potent neural inducers. Neural induction by anti-BMPs such as Chordin requires
mitogen-activated protein kinase (MAPK) activation mediated by FGF and IGF. These multiple
signals can be integrated at the level of Smad1. Phosphorylation by BMP receptor stimulates
Smad1 transcrip.
Unlike RNA, DNA containsA. adenine.B. uracil.C. phosphate grou.pdf
Unlike RNA, DNA contains
A. adenine.
B. uracil.
C. phosphate groups.
D. thymine.
Solution
Answer is: Thymine
Explanation:
Base Pairing in DNA: A-T (Adenine-Thymine), G-C (Guanine-Cytosine)
Base Pairing in RNA: A-U (Adenine-Uracil), G-C (Guanine-Cytosine)
The phosphate group is present in both RNA and DNA..
Use provided HTML file.Create JavaScript file to validate user i.pdf
Use provided HTML file.
Create *JavaScript file* to validate user input and process user input.
Show the conformation window for Valid Submission.
Title is required field.
Description is required field.
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Rate this photo => positive numeric value 1 – 10
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User must accept the software license.
User must select “Creative Commons Type
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script.js
/**************format.css**************************************/.
Recommended
Write an equation for the height of the point P above the ground as .pdf
Write an equation for the height of the point P above the ground as a function of t, time in
seconds.
h(t) =
Hint: Start by thinking: What is the height of the point P above the center? How is the center
changing?
Solution
height of center at ant time = R + 7(sinA)t = 2 + 7(100/700)t = (2 + t) cm
now height of P wrt center = Rcos(x/2) where x = 7t
height of P = 2+ t + Rcos(7t/2)
= 2 + t + 2cos(7t/2)
7t/2 is value in radian. oon putting t in second
for t= 0,(starting point)
height = 2 + 2cos0 = 4 cm
at the end, at = 100 cm,
h = 2 + 100 + 2cos(350) = 101.433 cm.
Wild-type (WT) mice are irradiated, destroying all of their bone marr.pdf
Wild-type (WT) mice are irradiated, destroying all of their bone marrow cells including CD34+
hematopoietic stem cells (HSC). CD34+ HSC is isolated from (1) an untreated wild-type
syngeneic control mouse, and (2) various immune gene mutant (-/- null) donor mice. The CD34+
HSC from the different mouse genotypes is injected into the wild-type irradiated recipient mice.
Predict the reconstitution of the hematopoietic system in each of the recipient mice (+ or -).
Solution
Answer -
Hematopoietic cells reconstituted in recipient
CD34+HSC
Doner GenotypeErythrocytes macrophages T cells B cells
Wild Type mouse + + + +
CD19(-/-)mouse + + + -
Rag1/2(-/-)mouse + + - -
erythropoietin EPO(-/-) - + + +
AIRE(-/-)mouse + + + +
explaination -Wild type mouse has normal CD34+HSC which is responsible to diffentiate into
all type of immune cells in bone marrow. Whwn this CD34+HSC from wild type mouse injected
into recipient mouse its bone marrow produce all type of immune cells.
CD19 is a B cell lymphocyte antigen It is a protein which is coded by CD19 gene. it is found on
the surface of B cells and helps in its proliferation and differenciation.CD34+HSC from doner
who lack CD19 gene when injected into recipient mouse,It will be unable to produce B cell.
Rag 1/2 also known as recombination activating gene codes a protein which paly an important
role in recombination of the gene of immunoglobuline and T cell receptor marker.two gene
products of recombination activating gene are Rag 1 and Rag 2 protein.These two rag protein are
essential to the generation of mature Bcells and Tcells. .CD34+Hsc from doner who lack the
gene codes gor Rag1/2 protein when injected to the recipient mouse, recipient mousewill be
unable to produce mature B and T cells.
Erythropoietin EPO is a protein harmone secreated by kidney in responce to less
oxygen/Heamoglobine in blood.erythropoietin stimulates bone marrow to produce more Red
blood cells (erythrocytes) which contain Heamoglobine ( an Oxygen carrier) and supply required
amount of oxygen to the body by mean of blood flow through the body. when CD34+HSC from
doner lack the gene codes for Erythropoietin injected into recipient mouse,it bone marrow will
be unable to produce erythrocytes.
AIRE is an auto immune regulator protein encoded by AIRE gene.It controles a mechanism that
prevent T cells (an important cell for immune response) from attacking the self body antigen.T
cells recognizes antigen which are expresed on the surface of antigen presenting cell by T cell
receptors present on T cells and produce Immune responce against that antigen. In autoimmune
diseases T cell Recognize self antigen and produce immune response against self antigen which
leads to sever problem in self body.AIRE protein play an important role to controle these kind of
autoimmune disorders by the removel of T cell through apoptosis while T cells are still in
thymus.
When CD34+HSC from doner WHICh lake the gene for AIRE protein injected into recipient
mouse.Reci.
Which of the following is considered to be a distributed denial of s.pdf
Which of the following is considered to be a distributed denial of service attack because it comes
from numerous sources?a.Smurf
Solution
The Distributed denial of service (DDOS) is an attack where the attacker repeatedly pings the
server from different computers/servers to restrict the server from handling the requests of the
legitimate users. This way all the requests of the authorized users will get be declined.
The SYN flood attack and the death ping send multiple requests to the server from a single
source and the sequence prediction attack is used to analyze the traffic to identify the data
packets.
The smurf attack uses a broadcast network to send multiple requests to the server from different
sources. Hence, the correct options is a. Smurf..
Which of the folllowing statements about the solubility of 1- propan.pdf
Which of the folllowing statements about the solubility of 1- propanol, ch3ch2ch2oh, and
ethanethiol, ch3ch2sh, are true?
A) Ethanethiol is more soluble in water because sulfer has agreater electronegativity value than
oxygen.
B) Ethanethiol is more soluble in water because it has a lower molecular mass than 1-propanol.
C) 1-propanol is more soluble in water because it has a greater molecular mass than ethanethiol.
D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value than
sulfer.
Solution
D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value
than sulfer.
\\\ eason
propanol has a hydrogen bonding by itself while ethanethiol does not have any hydrogen
bonding by itself that is why propanol is more soluble in water. because oxygen has a greater
electronegativity value than sulfer.
the more hydrogen bonding by itself the more soluble in water!.
What type of mutation that impacts splicing is a significant contrib.pdf
What type of mutation that impacts splicing is a significant contributor to human disease?
Solution
Disruptions in splicing are major contibutor of human diseases.Most genes in higher eukaryotes
are composed of introns (non coding segments) and exons (coding segments). The majority of
human intron removals ae catalyzed by a large and dynamic ribonucleoprotein (RNP) complex
called the spliceosome.This process involves two sequential transesterification reactions which
ligate the exons and release the intron as a lariat.
Alternate splicing which allows for the productioin of many proteins from a single genomic
locous.This greatly augments the repertorie of the protein.This greatly augments the repertoire of
proteins that can be produced by a given gene and plays an important role in
evolution,development and disease..
What data indicate that all three germ layers are specified in the b.pdf
What data indicate that all three germ layers are specified in the blastula? What are the
differences between dorsal and ventral mesodermal derivatives and what cellular interactions are
required for their specification?
Solution
Three germ layers of amphibians are specified in the blastula is determined by isolating these
tissues in vitro, and they are able to form into specific germ layers. The animal pole cap cells
make ectoderm, marginal region cells make mesoderm, and vegetal cells make endoderm.
Theyare specified but not determined. It is also interesting to note that if animal cap cells are
place co-cultured with vegetal cells, the animal cap cells with become mesoderm. This indicates
that vegetal cells induce other cells to form mesoderm. Experiment paired animal cap cells in 4
different sections of vegetal blastomeres to see if they induce different dorsal-ventral
mesodermal fates. Result showed that different sections of vegetal blastomere have specific
inductive capacities, which is crucial for dorsal-ventral mesoderm determination. Difference
between dorsal and ventral mesoderm derivatives: dorsal mesoderm is the notochord and somite.
In all bilaterian animals, the mesoderm is one of the three primary germ layers in the very early
embryo. The other two layers are the ectoderm (outside layer) and endoderm (inside layer), with
the mesoderm as the middle layer between them.
The mesoderm forms mesenchyme, mesothelium, non-epithelial blood cells and coelomocytes.
Some of the mesoderm derivatives include the muscle (smooth, cardiac and skeletal), the
muscles of the tongue (occipital somites), the pharyngeal arches muscle (muscles of mastication,
muscles of facial expressions), connective tissue, dermis and subcutaneous layer of the skin
At mid-blastula two signaling centers are present on the dorsal side: The prospective
neuroectoderm expresses bone morphogenetic protein (BMP) antagonists, and the future dorsal
endoderm secretes Nodal-related mesoderm-inducing factors. When dorsal mesoderm is formed
at gastrula, a cocktail of growth factor antagonists is secreted by the Spemann organizer and
further patterns the embryo. A ventral gastrula signaling center opposes the actions of the dorsal
organizer, and another set of secreted antagonists is produced ventrally under the control of
BMP4. The early dorsal -Catenin signal inhibits BMP expression at the transcriptional level and
promotes expression of secreted BMP antagonists in the prospective central nervous system
(CNS). In the absence of mesoderm, expression of Chordin and Noggin in ectoderm is required
for anterior CNS formation. FGF (fibroblast growth factor) and IGF (insulin-like growth factor)
signals are also potent neural inducers. Neural induction by anti-BMPs such as Chordin requires
mitogen-activated protein kinase (MAPK) activation mediated by FGF and IGF. These multiple
signals can be integrated at the level of Smad1. Phosphorylation by BMP receptor stimulates
Smad1 transcrip.
Unlike RNA, DNA containsA. adenine.B. uracil.C. phosphate grou.pdf
Unlike RNA, DNA contains
A. adenine.
B. uracil.
C. phosphate groups.
D. thymine.
Solution
Answer is: Thymine
Explanation:
Base Pairing in DNA: A-T (Adenine-Thymine), G-C (Guanine-Cytosine)
Base Pairing in RNA: A-U (Adenine-Uracil), G-C (Guanine-Cytosine)
The phosphate group is present in both RNA and DNA..
Use provided HTML file.Create JavaScript file to validate user i.pdf
Use provided HTML file.
Create *JavaScript file* to validate user input and process user input.
Show the conformation window for Valid Submission.
Title is required field.
Description is required field.
City is required field.
Rate this photo => positive numeric value 1 – 10
User must provide date and time.
User must accept the software license.
User must select “Creative Commons Type
Use JavaScript to display User information.
//HTML
Chapter 6Photo Details
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Description
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Choose continentAfricaAsiaEuropeNorth AmericaSouth America
Country
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City
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Copyright?
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h1, h2, h3, nav, footer {
font-family: Georgia, Cambria, \"Times New Roman\", serif;
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table {
border: collapse;
border-spacing: 0;
width; 90%;
margin: 0 auto;
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table tbody td{
/* border: 1pt solid #95BEF0; */
line-height: 1.5em;
vertical-align: top;
padding: 0.5em 0.75em;
}
legend {
background-color: #EBF4FB ;
margin: 0 auto;
width: 90%;
padding: 0.25em;
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font-weight: bold;
font-size: 100%;
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fieldset {
margin: 1em auto;
background-color: #FAFCFF;
width: 60%;
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margin-top: 0.5em;
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background-color: #EBF4FB;
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}
.centered {
text-align: center;
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.rounded, .rounded:hover {
border: 1px solid #172d6e;
border-bottom: 1px solid #0e1d45;
border-radius: 5px;
text-align: center;
color: white;
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Solution
script.js
/**************format.css**************************************/.
Two four-sided dice are rolled. One die has the letters A through D..pdf
Two four-sided dice are rolled. One die has the letters A through D. The other has the numbers 1
through 4.
Express the event “a roll of the dice will contain the letter A or B” as a set (show each of the
possible combinations).
What is the probability that a roll will contain an A or a B?
Solution
The set
S = { A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4,D1D,D3D4}
Let event M =a roll of the dice will contain the letter A or B
M = {A1,A2,A3,A4,B1,B2,B3,B4}
Probability = 8/16 = 1/2.
The job of a CODEC (CoderDecoder) is to Convert an analog voice si.pdf
The job of a CODEC (Coder/Decoder) is to: Convert an analog voice signal into a digital signal
Convert a digital signal into an analog voice signal Convert a computer signal into a voice signal
Both a and b
Solution
Option c is true, Since the job of a codec is to encode a stream of data a signal for storage and
transmission, possibly in encrypted form, and the process is revered in decoder function for
playback or editing. Codec are used in videoconferencing, streaming media, and video editing
applications..
Question 4 Constructive retirement is when A. P buys Ss common stoc.pdf
Question 4 Constructive retirement is when A. P buys S\'s common stock O B. P buys S\'s bonds
payable O C. P retires bonds payable. O D. P retires shares of its own stock.
Solution
If a company buys back its own share, which is treasury stock,and if the management of the
company decides to retire them permanently, its called constructive retirement.
So when P retires shares of its own stock, it results in constructive retirement .
Option D is correct..
Private and Public Health are working together through many program .pdf
Private and Public Health are working together through many program initiatives to promote
cost-effective, quality, and compassionate patient care and population health. Under the
Affordable Care Act (2010), and the patient-centered medical home model (PCMH) and
programs such as the All-Inclusive Care for the Elderly (PACE), how are health professionals in
both private and public health working together to meet the critical needs of the elderly
populations?
Solution
Let us start with PACE. It is a program helping people meet their healthcare needs without going
to a hospital or nursing home. A team of healthcare professionals work together to get this done.
PACE organizations work towards providing healthcare and services through their specialists so
that the needs of the elderly are met. the team of specialists come from different locations and
services. Some of them come from Public health organizations and some from private. However,
they all work together towards a common cause of serving the elderly. Thus PACE is an ideal
platform where both private and public health professionals work together to serve the elderly.
PCMH today is transforming the way primary care is managed and delivered. This model meets
a majority of each individual\'s health care needs in a comprehensive manner. Such a
comprehensive coverage would mean that a large team of specialists is involved including
doctors, nurses, pharmacists, assistants and so on. PCMH model is more relationship oriented. It
involves creating a relationship with the patient, understanding them and their families,
respecting their individual needs. This requires a lot of coordination across the healthcare
system, from hospitals to nursing homes to home health care. PCMH model delivers its services
around the clock. this requires the availability of services and specialist at every time. This is
possible through their pool of both private and public health professionals. Services are provided
in the shortest time by the available specialists, both private and public. Thus the PCMH model
also shows a way through which public and private health work together to serve the elderly.
Under the Affordable Care Act and programs like PACE and PCMH, a large pool of specialists
is created who jointly serve the elderly. Such a pool would not be possible if it is made by only
private or public healthcare professionals. Thus, the creation of these pools necessitates the
involvement of both private and public healthcare professionals. The creation of large pools acts
as a strong catalyst in bringing all types of health professionals towards the common goal of
serving the elderly and also the nation at large..
Name the two standards that are supported by major DBMSs for distrib.pdf
Name the two standards that are supported by major DBMSs for distributing data? Describe the
standard set of protocols developed by ISO and ANSI for remote access to database.
Solution
The two standards that are supported by major DBMS\'s Are ODMG 3.0 and JDO .
ODMG 3.0 stands for Object Data Management Group (ODMG) and JDO stands for java data
objects .
Remote database access (RDA) is the standard protocol for database access .
this standard has been developed by ISO and ANSI.
RDA provides protocol for eastablishing remote connection between database client and server.
The client is acting on behalf of an application program while the server is interfacing to a
process that controls data transfers to and from a database.
The RDA Communication Element converts RDA service requests into underlying ACSE and
TP service requests as part of an open systems interconnection..
Material Manufacturing Name 3 ways that the properties of thermosett.pdf
Material Manufacturing Name 3 ways that the properties of thermosetting polymers differ from
those of thermoplastics.
Solution
Thermoplastics can be melted and reshaped over and over again, while thermosetting plastics can
only be heated and shaped once
Thermoset are More resistant to high temperatures than thermoplastics
Thermoset have highly flexible design.
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Sho.pdf
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Show that the sum u_1 + u_2
is direct if dim(u_1 +u_2) = dim(u_1) + dim (u_2)
Solution
we know that
dim(U1+U2)= dim(U1)+dim(U2)-dim(U1 intersection U2)
given that dim(U1+U2)= dim(U1)+dim(U2)
i.e dim(U1 intersection U2) =0
U1 intersection U2 = {0}
hence sum U1+U2 is direct.
IV. Which compound would have the higher water solubility, tetrahydro.pdf
IV. Which compound would have the higher water solubility, tetrahydropyran (THP) or ethyl
propyl ether? Explain using chemical drawings and no more than 2 sentences. (10 points) THP
ethyl propyl ether
Solution
Cyclic ether is miscible than acyclic ether because the oxygen atom of cyclic ether are more
exposed towards hydrogen bonding. Hence THP (cyclic) is more soluble than ethyl propyl ether
(acyclic)..
Interseting discussion topic,want to chime inDescribe a symbol .pdf
Interseting discussion topic,
want to chime in?
Describe a symbol in your life. It can be a religious one or not. But I want to know something in
your life that is communal (that is to say, it takes place with other people) or personal (you go
through it alone) that points \"beyond itself\". Especially good topics are symbols that you share
with friends, family, teams, scout troops, or whatever - just a meaningful group (important
places, annual parties, hand signs, code words, handshakes, etc).
I want you to deeply diagnose the thing - even if you never have before: What are the motions
involved? Do they have a significance? What is the connection between sign and symbolized?
What feeling or event is it meant to evoke? Just describe each and every element that is
consistently involved and explain its significance... Is there a \"mythology\" behind it? (that is to
say, a key founding story that\'s interpretive, rather than exact, history)
Solution
hand signs:
it involves acceptance, sometime ignorance, sometime victory, some time hapiness. Yes a lot. It
is beyond the language one speaks. It has grat power to communicate. I don\'t know of
mythology but yes its there..
IN JAVA Write a program to create a binary data file named RandomInt.pdf
IN JAVA Write a program to create a binary data file named RandomInt.dat that consists of 10
random integer numbers using DataOutputStream. Use DataInputStream to read the integers
from RandomInt.dat just created and print them on the console in the same program.
Solution
DataInputStreamTest.java
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Random;
public class DataInputStreamTest {
public static void main(String[] args) {
String strFilePath = \"D:\\\\RandomInt.dat\";
Random r = new Random();
try
{
FileOutputStream outputStr = new FileOutputStream(strFilePath);
DataOutputStream outputStrData = new DataOutputStream(outputStr);
for(int i=0; i<10; i++){
outputStrData.writeInt(r.nextInt(100)+1);
outputStrData.writeChar(\'\\t\');
}
outputStrData.flush();
outputStrData.close();
outputStr.close();
FileInputStream inputStr = new FileInputStream(strFilePath);
DataInputStream inputStrData = new DataInputStream(inputStr);
int count = 0;
while(count<10){
int i = inputStrData.readInt();
inputStrData.readChar();
count++;
System.out.println( i);
}
outputStrData.flush();
inputStrData.close();
outputStrData.close();
}
catch(FileNotFoundException fe)
{
System.out.println(\"FileNotFoundException : \" + fe);
}
catch(IOException ioe)
{
System.out.println(\"IOException : \" + ioe);
}
}
}
Output:
65
8
95
69
1
77
16
97
85
1.
I wanna add the shape creator like rectangular , square , circle etc.pdf
I wanna add the shape creator like rectangular , square , circle etc shape creator in this program. I
tried many ways to do it but couldnt figure out the right way. Please help me do this. this is a
painter program and I want to add shape creator for an extra point.. Thanks..
//Painter.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Painter extends JFrame
{
private JButton blueButton, redButton, greenButton,eraserButton,clearButton;
private Container myCP;
public Paint panel;
private JPanel mainContainer;
public Painter()
{
mainContainer = new JPanel(new BorderLayout());
mainContainer.setBackground(Color.WHITE);
JPanel panel2 = new JPanel();
panel2.setLayout(new FlowLayout());
blueButton = new JButton(\"Blue\");
blueButton.setPreferredSize(new Dimension(75,50));
panel2.add(blueButton);
blueButton.addActionListener(new
ButtonListener());
redButton = new JButton(\"Red\");
redButton.setPreferredSize(new Dimension(75,50));
panel2.add(redButton);
redButton.addActionListener(new ButtonListener());
greenButton = new JButton(\"Green\");
greenButton.setPreferredSize(new Dimension(75,50));
panel2.add(greenButton);
greenButton.addActionListener(new ButtonListener());
eraserButton = new JButton(\"Eraser\");
eraserButton.setPreferredSize(new Dimension(75,50));
panel2.add(eraserButton);
eraserButton.addActionListener(new ButtonListener());
clearButton = new JButton(\"Clear\");
clearButton.setPreferredSize(new Dimension(75,50));
panel2.add(clearButton);
clearButton.addActionListener(new ButtonListener());
panel = new Paint();
mainContainer.add(panel2, BorderLayout.NORTH);
mainContainer.add(panel, BorderLayout.CENTER);
add(mainContainer);
setVisible(true);
}
private class ButtonListener implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
if (e.getSource() == blueButton)
{
panel.addCoordinateList(new ColorPoints(Color.BLUE,
new Point()));
}
if (e.getSource() == redButton)
{
panel.addCoordinateList(new ColorPoints(Color.RED,
new Point()));
}
if (e.getSource() == greenButton)
{
panel.addCoordinateList(new
ColorPoints(Color.GREEN,
new Point()));
}
if (e.getSource() == eraserButton)
{
panel.addCoordinateList(new ColorPoints(panel.getBackground(),
new Point()));
}
if (e.getSource() == clearButton)
{
panel.setCoordinateList(new ArrayList());
repaint();
}
}
}
public static void main(String args[])
{
Painter myAppF = new Painter();
myAppF.setTitle(\"Painter\");
myAppF.setVisible(true);
myAppF.setSize(500, 500);
myAppF.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
System.out.println(\"Programmed By Aashutosh Bajgain and Sujan Gautam!\");
}
}
//Paint.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Paint extends JPanel
{
private boolean lineBeingDrawn;
private Point startingPoint;
private ArrayList coordinateList;
private Color selectedColor;
private ColorPoints pwc;
public Paint()
{
lineBeingDrawn = false;
startingPoint .
I keep on get a redefinition error and an undefined error.Customer.pdf
I keep on get a redefinition error and an undefined error.
Customer List
(1) CustomerList() and ~CustomerList() - default constructor and destructor.
(2) bool addStore(Store*s) - Add an instance of store to the linked list. Return true if successful.
(3) Store *removeStore(int ID) - Locate a Store in the list if it exists, remove and return it.
(4) Store *getStore(int ID) - Locate a Store in the list and return a pointer to it.
(5) bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip) - Locate a
store in the list then update all its information. Return true if successful.
(6) void printStoreInfo() - Iterate over the list of store and have each print itself.
Employee Record
(1) Add a pointer to a class of type CustomerList called m_pCustomerList.
(2) Add a function CustomerList *getCustomerList to return the pointer to the
EmployeeRecord\'s CustomerList object.
-------------------------------------------------------------
CustomerList.h
#pragma once;
#include
#include \"Store.h\"
class CustomerList
{
public:
Store *m_pHead;
CustomerList();
~CustomerList();
bool addStore(Store *s);
Store *removeStore(int ID);
Store *getStore(int ID);
bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip);
void printStoresInfo();
};
----------------------------------------------------------------------------
CustomerList.cpp
#include
#include \"CustomerList.h\"
#include \"Store.h\"
using namespace std;
CustomerList::CustomerList()
{
// Default constructor
}
CustomerList::~CustomerList()
{
// Destructor
}
bool CustomerList:: addStore(Store *s)
{
//creating a new instance
s = new Store();
if(s==NULL)
return true;
}
Store *CustomerList::removeStore(int ID)
{
Store *temp, *back;
temp = m_pHead;
back = NULL;
while((temp != NULL)&&(ID != temp ->getStoreID()))
{
back=temp;
temp=temp->m_pNext;
}
if(temp==NULL)
return NULL;
else if(back==NULL)
{
m_pHead = m_pHead ->m_pNext;
return temp;
}
else
{
back -> m_pNext = temp-> m_pNext;
return temp;
}
return NULL;
}
Store *CustomerList::getStore(int ID)
{
Store *temp;
temp = m_pHead;
while((temp != NULL) && (ID != temp ->getStoreID()))
{
temp = temp->m_pNext;
}
if(temp == NULL)
return NULL;
else
{
Store *retStore = new Store;
*retStore = *temp;
retStore->m_pNext = NULL;
return retStore;
}
}
bool CustomerList::updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip)
{
return false;
}
void CustomerList::printStoresInfo()
{
Store *temp;
cout << \"
=====================================================================
================ \" << endl;
if(m_pHead== NULL)
cout << \" The List is currently empty.\ \" ;
else
{
temp = m_pHead;
while(temp != NULL)
{
cout << temp->m_pNext << endl;
}
}
}
--------------------------------------------------------------------------------
Store.h
#pragma once;
#include
#include
using namespace std;
class Store
{
private:
int m_iStoreID;
char m_sStoreName[64];
char m_sAddress[64];
char m_sCity[32];
char m_sState[32];
char m_sZip[.
For this homework, you will write a program to create and manipulate.pdf
For this homework, you will write a program to create and manipulate a simple linked list. For
each node in the linked list you will generate a random number, create a node that holds that
number, and insert that node into its sorted position in the linked list. Once the nodes are
inserted, you will traverse the list, printing the value held in each node. Then you will clean up
the list (deleting all the nodes) and exit the program. You will also learn about a tool that you can
use to help you check for memory errors in your code.
Specifications:
Define a struct appropriate for holding a random integer. This struct will also contain a \"next\"
pointer to reference a separate instance of the struct.
You may use the typedef keyword to give your struct a separate typename if you wish, but this is
not a requirement.
The program will read command line arguments. The first argument is the program name (as it
will always be) followed by a random number seed, the number of random numbers to generate
and ending with the Maximum Possible Value of the Random numbers generated (i.e. argc
should be 4).
Your program will use a \"dummy node\" for the linked list. In other words, the head of the list
will be an actual node, but the data field in that node will not be used for sorting purposes.
Your program must contain the following functions:
main() (for obvious reasons)
insertNodeSorted() - You may implement this either of two ways.
One way is to use the head of the list as one parameter and the integer value as the second. The
function will allocate memory for the node, initialize it and then insert it in the proper location in
the list.
The other way is to pass the head of the list as one parameter and a previously allocated and
initialized node as the other. The existing node is inserted in the proper location in the list.
printList() - Takes the head of the list as its only parameter, traverses the list, printing out the
data in sorted order.
deleteList() - Takes the head of the list as its only parameter, traverses the list, deleting all nodes.
The basic algorithm of your program is
Use a loop (based upon command line input values) that:
Generates a random number
Prints the number to stdout
Creates a node that contains that random number
Inserts the new node in sorted order into the existing list
Print the sorted linked list
Delete the linked list
You will use a Makefile to compile your program
Solution
#include
#include
#include
struct test_struct
{
int val;
struct test_struct *next;
};
struct test_struct *head = NULL;
struct test_struct *curr = NULL;
struct test_struct* create_list(int val)
{
printf(\"\ creating list with headnode as [%d]\ \",val);
struct test_struct *ptr = (struct test_struct*)malloc(sizeof(struct test_struct));
if(NULL == ptr)
{
printf(\"\ Node creation failed \ \");
return NULL;
}
ptr->val = val;
ptr->next = NULL;
head = curr = ptr;
return ptr;
}
struct test_struct* add_to_list(int val, bool add_to_end)
{
if(NULL == head)
{
retur.
How are the three basic skill sets identified and described by Rober.pdf
How are the three basic skill sets identified and described by Robert Katz (i.e., technical skills,
human skills, and conceptual skills) related to the execution of the five management functions
and Mintzberg\'s ten management roles?
Solution
Robert Katz 3 basic skills:
Human Skill: This is the ability to work with, understand and motivate other people
Technical Skill: The ability is to apply specialized knowledge or expertise
Conceptual skill: This is an ability to critically analyze, diagnose a situation and forward a
feasible solution
Management functions:
Planning: Planning is referred to the first step on management, planning is critical because these
are the directions that will be given in order to execute the strategic plan of an organization
Organizing: Once the planning is set, we need to organize who is going to be doing what tasks,
this is where organizing comes
Staffing: When we know what to do and how are going to do we need to allocate the correct
people to each function, make sure everything is cover according to schedules
Coordinating: This is the day to day work, we need to have supervisors or managers making sure
all the operation is running as planned
Controlling: controlling basically means making sure we are meeting our targets for an specific
time period, we cannot control what we don’t measure this is why reporting is so important
Mintzberg\'s ten management roles
Informational:
Monitor: On this role the person seeks for work related information
Disseminator: Communication of information to the organization
Spokesperson: This role communicates the information to outsiders
Interpersonal
Figurehead: Performs social and legal duties
Leader: Motivates employees and directs them on the correct path
Liaison: Maintains contact within and outside the organization
Decisional
Entrepreneur: Looks for new ideas and starts new projects
Disturbance handler: Deals with problems
Resources Allocator: Decides where to apply resources
Negotiator: Defends business best interests
All of these previous described skills related to each other because they all are important on
managers, these are knowledge that are critical for strategic planning and daily operation.
Technical skills are related with planning organizing and controlling because these are
specialized areas.
Human skills are related to staffing because it involves dealing directly with people when are
hiring them or letting them go
Conceptual skills is related to controlling because you have to analyze date to make decisions..
how does control impact culture and vice versaSolutionThere.pdf
how does control impact culture and vice versa
Solution
There is a cause and effect relationship between culture and control, for example: an attitude of
“it’s not my problem” is a cultural aspect that is likely to enhance the omission of necessary date
entries which are port of control.
“Poor managerial supervision” can lead to control deficiencies simply because people will not
give adequate attention and effort when executing control.
Companies that have major control failures often replace their senior management people, and
when the new senior management have the duty of changing the culture and assuring effective
control
Reference: Managing Organizational Culture for Effective Internal Control: From Practice to
Theory.
Getting the following errorsError 1 error C2436 {ctor} mem.pdf
Getting the following errors:
Error 1 error C2436: \'{ctor}\' : member function or nested class in constructor initializer list on
line 565
Error 2 error C2436: \'{ctor}\' : member function or nested class in constructor initializer list on
line 761
I need this code to COMPILE and RUN, but I cannot get rid of this error. Please Help!!
#include
#include
#include
#include
using namespace std;
enum contactGroupType
{// used in extPersonType
FAMILY,
FRIEND,
BUSINESS,
UNFILLED
};
class addressType
{
private:
string st_address;
string city;
string state;
int zip;
public:
void print(string, string, string, int)const;
void setStreet(string);
string getStreet()const;
void setCity(string);
string getCity()const;
void setState(string);
string getState()const;
void setZip(int);
int getZip()const;
void set(string, string, string, int);// set all address fields
string get()const;// get address as one concatenated string
addressType();
// ~addressType();
};
class personType
{
private:
string firstName;
string lastName;
public:
void print()const;
void setName(string first, string last);
string getFirstName()const;
string getLastName()const;
string get()const;// return First Last names concatenated
personType & operator=(const personType &);
personType(string, string);
personType();
};
class dateType
{
private:
int dMonth;
int dDay;
int dYear;
public:
void setDate(int month, int day, int year);
int getDay()const;
int getMonth()const;
int getYear()const;
void print()const;
string get()const;// return string representation as DD/MM/YYYY
dateType & operator=(const dateType & d);
dateType(int, int, int);
dateType();
};
class extPersonType :public personType {
private:
addressType address;// added members
dateType birthday;
contactGroupType group;
string phone;
public:
// methods
void setPhone(string);
string getPhone()const;
void setGroup(contactGroupType);
contactGroupType getGroup()const;
void setBirthday(int, int, int);
dateType getBirthday()const;
void print();
string get()const;// return string representation of ext person type
extPersonType & operator=(const extPersonType & p);
string groupToString(contactGroupType)const;
contactGroupType stringToGroup(string)const;
// constructors
extPersonType();
extPersonType(string first, string last);
};
// because we have no arrayListType, we are using our own
// implementation with a small subset of functions
class arrayListType
{
extPersonType array[500];
int size;
public:
arrayListType();
extPersonType & operator[](int i);
void removeLast();// remove last element
void add(const extPersonType &);// add new element
int getSize()const;// get array size
};
class addressBookType :public arrayListType
{
private:
static const char FS = \'\\t\';// field separator in file (TAB char)
int current;// current position
string fileName;// filename
fstream fileStream;// file as fstream
/* filiters */
contactGroupType fltGroup;
string fltFromLast, fltToLast;
dateType fltFromDate, fltTiDate;
/* flags for effective filters */
bool fl.
ezt Connect ACCOUNTING nline .1.4 Journal EntriesT-Accounts Questio.pdf
ezt Connect ACCOUNTING nline .1.4: Journal Entries/T-Accounts Questions 4-6 (of 8) The
following information applies to the questions displayed below Laser Delivery Services, Inc.
(LDS) was Incorporated the year January 1. The following transactions occurred during a.
Recelved $40,000 cash from the company\'s founders in exchange for common stock. b.
Purchased land for $15,500, signing a two-year note Ignore Interest). Bought two used delivery
trucks at the start of the year at a cost of $12,000 each; pald $5,000 cash and signed a note due in
three years for $19,000 (ignore Interest). C. d. Paid $1,500 cash to a truck repair shop for a new
motor, which increased the cost of one of the trucks e. Stockholder Jonah Lee paid $350,000
cash for a house for his personal use. value 3.00 points Required: 1. Analyze each item for its
effects on the accounting equation of LDS. (Enter any decreases balances with a minus sign.)
TIP: The new motor in transaction (d) is treated as an increase to the cost of the truck
Stockholders Assets Liabilities Equ Accounts Payable Payable Common Stock Notes Cash
EquipmentLand Beg. a. b. 40,000 40,000 15 500
Solution
Journal entries
Laser Delivery services Inc
Transaction
General Journal
Debit
Credit
a
Cash
$ 40,000.00
Common Stock
$ 40,000.00
(common stock issued to Founders )
b
Land
$ 15,500.00
Notes Payable
$ 15,500.00
(land Purchased against notes payable)
c
Delivery Truck
$ 24,000.00
Cash
$ 5,000.00
Notes Payable
$ 19,000.00
(Delivery truck purchased and partially paid in cash)
d
Delivery truck
$ 1,500.00
Cash
$ 1,500.00
(Capital expenditure done on truck to make it fit for working)
e
No journal Entry in Companies accounts
$ 81,000.00
$ 81,000.00
1)
Effect of each item on the Accounting Equation
Assets =
Liabilities +
Share Holders Equity
Cash
Equipment
Land
Accounts Payable
Notes Payable
Common stock
Beginning
$ -
a
$ 40,000.00
$ 40,000.00
b
$ 15,500.00
$ 15,500.00
c
$ (5,000.00)
$ 24,000.00
$ 19,000.00
d
$ (1,500.00)
$ 1,500.00
e
No effect
Total
$ 33,500.00
$ 25,500.00
$ 15,500.00
$ -
$ 34,500.00
$ 40,000.00
3)
T accounts
Cash
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
a
$ 40,000.00
$ 40,000.00
c
$ 5,000.00
$ 35,000.00
d
$ 1,500.00
$ 33,500.00
Closing
$ 33,500.00
Land
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
b
$ 15,500.00
$ 15,500.00
Closing
$ 15,500.00
Common Stock
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
a
$ 40,000.00
$ 40,000.00
Closing
$ 40,000.00
Equipment
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
c
$ 24,000.00
$ 24,000.00
d
$ 1,500.00
$ 25,500.00
Closing
$ 25,500.00
Notes Payable
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
b
$ 15,500.00
$ 15,500.00
c
$ 19,000.00
$ 34,500.00
Closing
$ 34,500.00
Balance sheet
Laser Delivery services Inc
Liabilities
Assets
Current Liabilities
Fixed Assets
Notes Payable
$ 34,500.00
Equipment
$ 25,500.00
Land
$ 15,500.00
Shareholders’ equity
Common stock
$ 40,000.00
Current assets
Cash
$ 33,500.00
Total
$ 74,500.00
$ 74,500.00
Journal .
est.phpaction-skip&to-16; hm Fady Ibrahim User Settings My Classes .pdf
est.php?action-skip&to-16; hm Fady Ibrahim| User Settings My Classes LogOut Forums
Calendar Gradebook m # 4-Chapter 4 on OHM-Due 12/06/2017 at 1 1:59 PM Due in 4 hours, 11
minutes. Due Wed 12/06/2017 11:59 pr Show Intro/lnstruction A population numbers 17,000
organisms initially and grows by 83% each year. Suppose P represents population, and t the
number of years of growth. An exponential model for the population can be written in the form P
= a , where Preview This is attempt 1 of 1. Submit
Solution
P= a bt
After1 year, population becomes 17000+0.083(17000)=18411
18411= 17000 b1
b= 18411/17000=1.083
P= 17000(1.083)t.
Discuss the challenges for intelligence analysts when considering th.pdf
Discuss the challenges for intelligence analysts when considering the usage of dual-use material.
Solution
Dual use commodities are “those that can be used in weapons of mass destruction (WMD)
programs or in legitimate civilian applications” (Fuhrmann, 2014, p.2). Those materials can be
used for practical or radical reasons hence the reason whythe intelligence community faces
challenges in determining the threat involved. We can see this taking place in China where its
leaders continue to pursue dual tracks of facilitating internet access to the country for economic
and commerce growth and policing certain unwanted online behaviors. China also continues to
expand its cyber-attacks around the world and continues its network exploitation as well as theft
of intellectual property (Clapper, 2014). This recently took place where several F-35 jet
prototypes ended up with in the hands of the Chinese government..
devryu.instructure.com ersity Topic Week 3 The Division of Power ul.pdf
devryu.instructure.com ersity Topic: Week 3: The Division of Power ules Week 3: The Division
of Power . due Jan 21 This is a graded discussion: 20 points possible WEEK 3: THE DIVISION
OF POWER As we have seen through our readings, governments differ on where the power of
the state should be held. Unitary systems concentrate the power within the central government
and little or no authority is granted to the component areas. In contrast, federal systems allow
first-order civil divisions to have some autonomy, while the central government maintains
authority over some areas. Take a moment to compare and contrast the strengths and weaknesses
of each system. Search entries or author Unread | subscribe Reply Joel Terwilliger (instructor)
Yesterday is the centralization of power in a nation\'s capital with little autonomy for & Jones,
2014, p. 55). This system has the control and makes the decisions over citizen\'s lives (taxes,
education, and laws) and the local leaders. With this type of system, The unitary system from our
text, subdivisions (Roskin, Cord, Medeiros,
Solution
Strengths
Unitary System - One of the main strengths in this system is the National Unity promotion and
this can get to help the system to witstand many troublesome incidents.
Federal System- The main strengths of this system is that it pops out to be successful in good,
large as well as culturally diverse countries where justice can be done to each and every area of
the place.
Weakness
Unitary system - The weakness is that the public policies may not get to fil the needs of all the
people all in all.
Federal system - Here, one of the main weakness is that there can be a conflict that can arise
between state and the federal governments which can create hectic problems on the whole..
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH ANSWERS.
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.
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Two four-sided dice are rolled. One die has the letters A through D..pdf
Two four-sided dice are rolled. One die has the letters A through D. The other has the numbers 1
through 4.
Express the event “a roll of the dice will contain the letter A or B” as a set (show each of the
possible combinations).
What is the probability that a roll will contain an A or a B?
Solution
The set
S = { A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4,D1D,D3D4}
Let event M =a roll of the dice will contain the letter A or B
M = {A1,A2,A3,A4,B1,B2,B3,B4}
Probability = 8/16 = 1/2.
The job of a CODEC (CoderDecoder) is to Convert an analog voice si.pdf
The job of a CODEC (Coder/Decoder) is to: Convert an analog voice signal into a digital signal
Convert a digital signal into an analog voice signal Convert a computer signal into a voice signal
Both a and b
Solution
Option c is true, Since the job of a codec is to encode a stream of data a signal for storage and
transmission, possibly in encrypted form, and the process is revered in decoder function for
playback or editing. Codec are used in videoconferencing, streaming media, and video editing
applications..
Question 4 Constructive retirement is when A. P buys Ss common stoc.pdf
Question 4 Constructive retirement is when A. P buys S\'s common stock O B. P buys S\'s bonds
payable O C. P retires bonds payable. O D. P retires shares of its own stock.
Solution
If a company buys back its own share, which is treasury stock,and if the management of the
company decides to retire them permanently, its called constructive retirement.
So when P retires shares of its own stock, it results in constructive retirement .
Option D is correct..
Private and Public Health are working together through many program .pdf
Private and Public Health are working together through many program initiatives to promote
cost-effective, quality, and compassionate patient care and population health. Under the
Affordable Care Act (2010), and the patient-centered medical home model (PCMH) and
programs such as the All-Inclusive Care for the Elderly (PACE), how are health professionals in
both private and public health working together to meet the critical needs of the elderly
populations?
Solution
Let us start with PACE. It is a program helping people meet their healthcare needs without going
to a hospital or nursing home. A team of healthcare professionals work together to get this done.
PACE organizations work towards providing healthcare and services through their specialists so
that the needs of the elderly are met. the team of specialists come from different locations and
services. Some of them come from Public health organizations and some from private. However,
they all work together towards a common cause of serving the elderly. Thus PACE is an ideal
platform where both private and public health professionals work together to serve the elderly.
PCMH today is transforming the way primary care is managed and delivered. This model meets
a majority of each individual\'s health care needs in a comprehensive manner. Such a
comprehensive coverage would mean that a large team of specialists is involved including
doctors, nurses, pharmacists, assistants and so on. PCMH model is more relationship oriented. It
involves creating a relationship with the patient, understanding them and their families,
respecting their individual needs. This requires a lot of coordination across the healthcare
system, from hospitals to nursing homes to home health care. PCMH model delivers its services
around the clock. this requires the availability of services and specialist at every time. This is
possible through their pool of both private and public health professionals. Services are provided
in the shortest time by the available specialists, both private and public. Thus the PCMH model
also shows a way through which public and private health work together to serve the elderly.
Under the Affordable Care Act and programs like PACE and PCMH, a large pool of specialists
is created who jointly serve the elderly. Such a pool would not be possible if it is made by only
private or public healthcare professionals. Thus, the creation of these pools necessitates the
involvement of both private and public healthcare professionals. The creation of large pools acts
as a strong catalyst in bringing all types of health professionals towards the common goal of
serving the elderly and also the nation at large..
Name the two standards that are supported by major DBMSs for distrib.pdf
Name the two standards that are supported by major DBMSs for distributing data? Describe the
standard set of protocols developed by ISO and ANSI for remote access to database.
Solution
The two standards that are supported by major DBMS\'s Are ODMG 3.0 and JDO .
ODMG 3.0 stands for Object Data Management Group (ODMG) and JDO stands for java data
objects .
Remote database access (RDA) is the standard protocol for database access .
this standard has been developed by ISO and ANSI.
RDA provides protocol for eastablishing remote connection between database client and server.
The client is acting on behalf of an application program while the server is interfacing to a
process that controls data transfers to and from a database.
The RDA Communication Element converts RDA service requests into underlying ACSE and
TP service requests as part of an open systems interconnection..
Material Manufacturing Name 3 ways that the properties of thermosett.pdf
Material Manufacturing Name 3 ways that the properties of thermosetting polymers differ from
those of thermoplastics.
Solution
Thermoplastics can be melted and reshaped over and over again, while thermosetting plastics can
only be heated and shaped once
Thermoset are More resistant to high temperatures than thermoplastics
Thermoset have highly flexible design.
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Sho.pdf
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Show that the sum u_1 + u_2
is direct if dim(u_1 +u_2) = dim(u_1) + dim (u_2)
Solution
we know that
dim(U1+U2)= dim(U1)+dim(U2)-dim(U1 intersection U2)
given that dim(U1+U2)= dim(U1)+dim(U2)
i.e dim(U1 intersection U2) =0
U1 intersection U2 = {0}
hence sum U1+U2 is direct.
IV. Which compound would have the higher water solubility, tetrahydro.pdf
IV. Which compound would have the higher water solubility, tetrahydropyran (THP) or ethyl
propyl ether? Explain using chemical drawings and no more than 2 sentences. (10 points) THP
ethyl propyl ether
Solution
Cyclic ether is miscible than acyclic ether because the oxygen atom of cyclic ether are more
exposed towards hydrogen bonding. Hence THP (cyclic) is more soluble than ethyl propyl ether
(acyclic)..
Interseting discussion topic,want to chime inDescribe a symbol .pdf
Interseting discussion topic,
want to chime in?
Describe a symbol in your life. It can be a religious one or not. But I want to know something in
your life that is communal (that is to say, it takes place with other people) or personal (you go
through it alone) that points \"beyond itself\". Especially good topics are symbols that you share
with friends, family, teams, scout troops, or whatever - just a meaningful group (important
places, annual parties, hand signs, code words, handshakes, etc).
I want you to deeply diagnose the thing - even if you never have before: What are the motions
involved? Do they have a significance? What is the connection between sign and symbolized?
What feeling or event is it meant to evoke? Just describe each and every element that is
consistently involved and explain its significance... Is there a \"mythology\" behind it? (that is to
say, a key founding story that\'s interpretive, rather than exact, history)
Solution
hand signs:
it involves acceptance, sometime ignorance, sometime victory, some time hapiness. Yes a lot. It
is beyond the language one speaks. It has grat power to communicate. I don\'t know of
mythology but yes its there..
IN JAVA Write a program to create a binary data file named RandomInt.pdf
IN JAVA Write a program to create a binary data file named RandomInt.dat that consists of 10
random integer numbers using DataOutputStream. Use DataInputStream to read the integers
from RandomInt.dat just created and print them on the console in the same program.
Solution
DataInputStreamTest.java
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Random;
public class DataInputStreamTest {
public static void main(String[] args) {
String strFilePath = \"D:\\\\RandomInt.dat\";
Random r = new Random();
try
{
FileOutputStream outputStr = new FileOutputStream(strFilePath);
DataOutputStream outputStrData = new DataOutputStream(outputStr);
for(int i=0; i<10; i++){
outputStrData.writeInt(r.nextInt(100)+1);
outputStrData.writeChar(\'\\t\');
}
outputStrData.flush();
outputStrData.close();
outputStr.close();
FileInputStream inputStr = new FileInputStream(strFilePath);
DataInputStream inputStrData = new DataInputStream(inputStr);
int count = 0;
while(count<10){
int i = inputStrData.readInt();
inputStrData.readChar();
count++;
System.out.println( i);
}
outputStrData.flush();
inputStrData.close();
outputStrData.close();
}
catch(FileNotFoundException fe)
{
System.out.println(\"FileNotFoundException : \" + fe);
}
catch(IOException ioe)
{
System.out.println(\"IOException : \" + ioe);
}
}
}
Output:
65
8
95
69
1
77
16
97
85
1.
I wanna add the shape creator like rectangular , square , circle etc.pdf
I wanna add the shape creator like rectangular , square , circle etc shape creator in this program. I
tried many ways to do it but couldnt figure out the right way. Please help me do this. this is a
painter program and I want to add shape creator for an extra point.. Thanks..
//Painter.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Painter extends JFrame
{
private JButton blueButton, redButton, greenButton,eraserButton,clearButton;
private Container myCP;
public Paint panel;
private JPanel mainContainer;
public Painter()
{
mainContainer = new JPanel(new BorderLayout());
mainContainer.setBackground(Color.WHITE);
JPanel panel2 = new JPanel();
panel2.setLayout(new FlowLayout());
blueButton = new JButton(\"Blue\");
blueButton.setPreferredSize(new Dimension(75,50));
panel2.add(blueButton);
blueButton.addActionListener(new
ButtonListener());
redButton = new JButton(\"Red\");
redButton.setPreferredSize(new Dimension(75,50));
panel2.add(redButton);
redButton.addActionListener(new ButtonListener());
greenButton = new JButton(\"Green\");
greenButton.setPreferredSize(new Dimension(75,50));
panel2.add(greenButton);
greenButton.addActionListener(new ButtonListener());
eraserButton = new JButton(\"Eraser\");
eraserButton.setPreferredSize(new Dimension(75,50));
panel2.add(eraserButton);
eraserButton.addActionListener(new ButtonListener());
clearButton = new JButton(\"Clear\");
clearButton.setPreferredSize(new Dimension(75,50));
panel2.add(clearButton);
clearButton.addActionListener(new ButtonListener());
panel = new Paint();
mainContainer.add(panel2, BorderLayout.NORTH);
mainContainer.add(panel, BorderLayout.CENTER);
add(mainContainer);
setVisible(true);
}
private class ButtonListener implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
if (e.getSource() == blueButton)
{
panel.addCoordinateList(new ColorPoints(Color.BLUE,
new Point()));
}
if (e.getSource() == redButton)
{
panel.addCoordinateList(new ColorPoints(Color.RED,
new Point()));
}
if (e.getSource() == greenButton)
{
panel.addCoordinateList(new
ColorPoints(Color.GREEN,
new Point()));
}
if (e.getSource() == eraserButton)
{
panel.addCoordinateList(new ColorPoints(panel.getBackground(),
new Point()));
}
if (e.getSource() == clearButton)
{
panel.setCoordinateList(new ArrayList());
repaint();
}
}
}
public static void main(String args[])
{
Painter myAppF = new Painter();
myAppF.setTitle(\"Painter\");
myAppF.setVisible(true);
myAppF.setSize(500, 500);
myAppF.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
System.out.println(\"Programmed By Aashutosh Bajgain and Sujan Gautam!\");
}
}
//Paint.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Paint extends JPanel
{
private boolean lineBeingDrawn;
private Point startingPoint;
private ArrayList coordinateList;
private Color selectedColor;
private ColorPoints pwc;
public Paint()
{
lineBeingDrawn = false;
startingPoint .
I keep on get a redefinition error and an undefined error.Customer.pdf
I keep on get a redefinition error and an undefined error.
Customer List
(1) CustomerList() and ~CustomerList() - default constructor and destructor.
(2) bool addStore(Store*s) - Add an instance of store to the linked list. Return true if successful.
(3) Store *removeStore(int ID) - Locate a Store in the list if it exists, remove and return it.
(4) Store *getStore(int ID) - Locate a Store in the list and return a pointer to it.
(5) bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip) - Locate a
store in the list then update all its information. Return true if successful.
(6) void printStoreInfo() - Iterate over the list of store and have each print itself.
Employee Record
(1) Add a pointer to a class of type CustomerList called m_pCustomerList.
(2) Add a function CustomerList *getCustomerList to return the pointer to the
EmployeeRecord\'s CustomerList object.
-------------------------------------------------------------
CustomerList.h
#pragma once;
#include
#include \"Store.h\"
class CustomerList
{
public:
Store *m_pHead;
CustomerList();
~CustomerList();
bool addStore(Store *s);
Store *removeStore(int ID);
Store *getStore(int ID);
bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip);
void printStoresInfo();
};
----------------------------------------------------------------------------
CustomerList.cpp
#include
#include \"CustomerList.h\"
#include \"Store.h\"
using namespace std;
CustomerList::CustomerList()
{
// Default constructor
}
CustomerList::~CustomerList()
{
// Destructor
}
bool CustomerList:: addStore(Store *s)
{
//creating a new instance
s = new Store();
if(s==NULL)
return true;
}
Store *CustomerList::removeStore(int ID)
{
Store *temp, *back;
temp = m_pHead;
back = NULL;
while((temp != NULL)&&(ID != temp ->getStoreID()))
{
back=temp;
temp=temp->m_pNext;
}
if(temp==NULL)
return NULL;
else if(back==NULL)
{
m_pHead = m_pHead ->m_pNext;
return temp;
}
else
{
back -> m_pNext = temp-> m_pNext;
return temp;
}
return NULL;
}
Store *CustomerList::getStore(int ID)
{
Store *temp;
temp = m_pHead;
while((temp != NULL) && (ID != temp ->getStoreID()))
{
temp = temp->m_pNext;
}
if(temp == NULL)
return NULL;
else
{
Store *retStore = new Store;
*retStore = *temp;
retStore->m_pNext = NULL;
return retStore;
}
}
bool CustomerList::updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip)
{
return false;
}
void CustomerList::printStoresInfo()
{
Store *temp;
cout << \"
=====================================================================
================ \" << endl;
if(m_pHead== NULL)
cout << \" The List is currently empty.\ \" ;
else
{
temp = m_pHead;
while(temp != NULL)
{
cout << temp->m_pNext << endl;
}
}
}
--------------------------------------------------------------------------------
Store.h
#pragma once;
#include
#include
using namespace std;
class Store
{
private:
int m_iStoreID;
char m_sStoreName[64];
char m_sAddress[64];
char m_sCity[32];
char m_sState[32];
char m_sZip[.
For this homework, you will write a program to create and manipulate.pdf
For this homework, you will write a program to create and manipulate a simple linked list. For
each node in the linked list you will generate a random number, create a node that holds that
number, and insert that node into its sorted position in the linked list. Once the nodes are
inserted, you will traverse the list, printing the value held in each node. Then you will clean up
the list (deleting all the nodes) and exit the program. You will also learn about a tool that you can
use to help you check for memory errors in your code.
Specifications:
Define a struct appropriate for holding a random integer. This struct will also contain a \"next\"
pointer to reference a separate instance of the struct.
You may use the typedef keyword to give your struct a separate typename if you wish, but this is
not a requirement.
The program will read command line arguments. The first argument is the program name (as it
will always be) followed by a random number seed, the number of random numbers to generate
and ending with the Maximum Possible Value of the Random numbers generated (i.e. argc
should be 4).
Your program will use a \"dummy node\" for the linked list. In other words, the head of the list
will be an actual node, but the data field in that node will not be used for sorting purposes.
Your program must contain the following functions:
main() (for obvious reasons)
insertNodeSorted() - You may implement this either of two ways.
One way is to use the head of the list as one parameter and the integer value as the second. The
function will allocate memory for the node, initialize it and then insert it in the proper location in
the list.
The other way is to pass the head of the list as one parameter and a previously allocated and
initialized node as the other. The existing node is inserted in the proper location in the list.
printList() - Takes the head of the list as its only parameter, traverses the list, printing out the
data in sorted order.
deleteList() - Takes the head of the list as its only parameter, traverses the list, deleting all nodes.
The basic algorithm of your program is
Use a loop (based upon command line input values) that:
Generates a random number
Prints the number to stdout
Creates a node that contains that random number
Inserts the new node in sorted order into the existing list
Print the sorted linked list
Delete the linked list
You will use a Makefile to compile your program
Solution
#include
#include
#include
struct test_struct
{
int val;
struct test_struct *next;
};
struct test_struct *head = NULL;
struct test_struct *curr = NULL;
struct test_struct* create_list(int val)
{
printf(\"\ creating list with headnode as [%d]\ \",val);
struct test_struct *ptr = (struct test_struct*)malloc(sizeof(struct test_struct));
if(NULL == ptr)
{
printf(\"\ Node creation failed \ \");
return NULL;
}
ptr->val = val;
ptr->next = NULL;
head = curr = ptr;
return ptr;
}
struct test_struct* add_to_list(int val, bool add_to_end)
{
if(NULL == head)
{
retur.
How are the three basic skill sets identified and described by Rober.pdf
How are the three basic skill sets identified and described by Robert Katz (i.e., technical skills,
human skills, and conceptual skills) related to the execution of the five management functions
and Mintzberg\'s ten management roles?
Solution
Robert Katz 3 basic skills:
Human Skill: This is the ability to work with, understand and motivate other people
Technical Skill: The ability is to apply specialized knowledge or expertise
Conceptual skill: This is an ability to critically analyze, diagnose a situation and forward a
feasible solution
Management functions:
Planning: Planning is referred to the first step on management, planning is critical because these
are the directions that will be given in order to execute the strategic plan of an organization
Organizing: Once the planning is set, we need to organize who is going to be doing what tasks,
this is where organizing comes
Staffing: When we know what to do and how are going to do we need to allocate the correct
people to each function, make sure everything is cover according to schedules
Coordinating: This is the day to day work, we need to have supervisors or managers making sure
all the operation is running as planned
Controlling: controlling basically means making sure we are meeting our targets for an specific
time period, we cannot control what we don’t measure this is why reporting is so important
Mintzberg\'s ten management roles
Informational:
Monitor: On this role the person seeks for work related information
Disseminator: Communication of information to the organization
Spokesperson: This role communicates the information to outsiders
Interpersonal
Figurehead: Performs social and legal duties
Leader: Motivates employees and directs them on the correct path
Liaison: Maintains contact within and outside the organization
Decisional
Entrepreneur: Looks for new ideas and starts new projects
Disturbance handler: Deals with problems
Resources Allocator: Decides where to apply resources
Negotiator: Defends business best interests
All of these previous described skills related to each other because they all are important on
managers, these are knowledge that are critical for strategic planning and daily operation.
Technical skills are related with planning organizing and controlling because these are
specialized areas.
Human skills are related to staffing because it involves dealing directly with people when are
hiring them or letting them go
Conceptual skills is related to controlling because you have to analyze date to make decisions..
how does control impact culture and vice versaSolutionThere.pdf
how does control impact culture and vice versa
Solution
There is a cause and effect relationship between culture and control, for example: an attitude of
“it’s not my problem” is a cultural aspect that is likely to enhance the omission of necessary date
entries which are port of control.
“Poor managerial supervision” can lead to control deficiencies simply because people will not
give adequate attention and effort when executing control.
Companies that have major control failures often replace their senior management people, and
when the new senior management have the duty of changing the culture and assuring effective
control
Reference: Managing Organizational Culture for Effective Internal Control: From Practice to
Theory.
Getting the following errorsError 1 error C2436 {ctor} mem.pdf
Getting the following errors:
Error 1 error C2436: \'{ctor}\' : member function or nested class in constructor initializer list on
line 565
Error 2 error C2436: \'{ctor}\' : member function or nested class in constructor initializer list on
line 761
I need this code to COMPILE and RUN, but I cannot get rid of this error. Please Help!!
#include
#include
#include
#include
using namespace std;
enum contactGroupType
{// used in extPersonType
FAMILY,
FRIEND,
BUSINESS,
UNFILLED
};
class addressType
{
private:
string st_address;
string city;
string state;
int zip;
public:
void print(string, string, string, int)const;
void setStreet(string);
string getStreet()const;
void setCity(string);
string getCity()const;
void setState(string);
string getState()const;
void setZip(int);
int getZip()const;
void set(string, string, string, int);// set all address fields
string get()const;// get address as one concatenated string
addressType();
// ~addressType();
};
class personType
{
private:
string firstName;
string lastName;
public:
void print()const;
void setName(string first, string last);
string getFirstName()const;
string getLastName()const;
string get()const;// return First Last names concatenated
personType & operator=(const personType &);
personType(string, string);
personType();
};
class dateType
{
private:
int dMonth;
int dDay;
int dYear;
public:
void setDate(int month, int day, int year);
int getDay()const;
int getMonth()const;
int getYear()const;
void print()const;
string get()const;// return string representation as DD/MM/YYYY
dateType & operator=(const dateType & d);
dateType(int, int, int);
dateType();
};
class extPersonType :public personType {
private:
addressType address;// added members
dateType birthday;
contactGroupType group;
string phone;
public:
// methods
void setPhone(string);
string getPhone()const;
void setGroup(contactGroupType);
contactGroupType getGroup()const;
void setBirthday(int, int, int);
dateType getBirthday()const;
void print();
string get()const;// return string representation of ext person type
extPersonType & operator=(const extPersonType & p);
string groupToString(contactGroupType)const;
contactGroupType stringToGroup(string)const;
// constructors
extPersonType();
extPersonType(string first, string last);
};
// because we have no arrayListType, we are using our own
// implementation with a small subset of functions
class arrayListType
{
extPersonType array[500];
int size;
public:
arrayListType();
extPersonType & operator[](int i);
void removeLast();// remove last element
void add(const extPersonType &);// add new element
int getSize()const;// get array size
};
class addressBookType :public arrayListType
{
private:
static const char FS = \'\\t\';// field separator in file (TAB char)
int current;// current position
string fileName;// filename
fstream fileStream;// file as fstream
/* filiters */
contactGroupType fltGroup;
string fltFromLast, fltToLast;
dateType fltFromDate, fltTiDate;
/* flags for effective filters */
bool fl.
ezt Connect ACCOUNTING nline .1.4 Journal EntriesT-Accounts Questio.pdf
ezt Connect ACCOUNTING nline .1.4: Journal Entries/T-Accounts Questions 4-6 (of 8) The
following information applies to the questions displayed below Laser Delivery Services, Inc.
(LDS) was Incorporated the year January 1. The following transactions occurred during a.
Recelved $40,000 cash from the company\'s founders in exchange for common stock. b.
Purchased land for $15,500, signing a two-year note Ignore Interest). Bought two used delivery
trucks at the start of the year at a cost of $12,000 each; pald $5,000 cash and signed a note due in
three years for $19,000 (ignore Interest). C. d. Paid $1,500 cash to a truck repair shop for a new
motor, which increased the cost of one of the trucks e. Stockholder Jonah Lee paid $350,000
cash for a house for his personal use. value 3.00 points Required: 1. Analyze each item for its
effects on the accounting equation of LDS. (Enter any decreases balances with a minus sign.)
TIP: The new motor in transaction (d) is treated as an increase to the cost of the truck
Stockholders Assets Liabilities Equ Accounts Payable Payable Common Stock Notes Cash
EquipmentLand Beg. a. b. 40,000 40,000 15 500
Solution
Journal entries
Laser Delivery services Inc
Transaction
General Journal
Debit
Credit
a
Cash
$ 40,000.00
Common Stock
$ 40,000.00
(common stock issued to Founders )
b
Land
$ 15,500.00
Notes Payable
$ 15,500.00
(land Purchased against notes payable)
c
Delivery Truck
$ 24,000.00
Cash
$ 5,000.00
Notes Payable
$ 19,000.00
(Delivery truck purchased and partially paid in cash)
d
Delivery truck
$ 1,500.00
Cash
$ 1,500.00
(Capital expenditure done on truck to make it fit for working)
e
No journal Entry in Companies accounts
$ 81,000.00
$ 81,000.00
1)
Effect of each item on the Accounting Equation
Assets =
Liabilities +
Share Holders Equity
Cash
Equipment
Land
Accounts Payable
Notes Payable
Common stock
Beginning
$ -
a
$ 40,000.00
$ 40,000.00
b
$ 15,500.00
$ 15,500.00
c
$ (5,000.00)
$ 24,000.00
$ 19,000.00
d
$ (1,500.00)
$ 1,500.00
e
No effect
Total
$ 33,500.00
$ 25,500.00
$ 15,500.00
$ -
$ 34,500.00
$ 40,000.00
3)
T accounts
Cash
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
a
$ 40,000.00
$ 40,000.00
c
$ 5,000.00
$ 35,000.00
d
$ 1,500.00
$ 33,500.00
Closing
$ 33,500.00
Land
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
b
$ 15,500.00
$ 15,500.00
Closing
$ 15,500.00
Common Stock
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
a
$ 40,000.00
$ 40,000.00
Closing
$ 40,000.00
Equipment
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
c
$ 24,000.00
$ 24,000.00
d
$ 1,500.00
$ 25,500.00
Closing
$ 25,500.00
Notes Payable
Transaction
Debit
Credit
Balance
Beginning
$ -
$ -
b
$ 15,500.00
$ 15,500.00
c
$ 19,000.00
$ 34,500.00
Closing
$ 34,500.00
Balance sheet
Laser Delivery services Inc
Liabilities
Assets
Current Liabilities
Fixed Assets
Notes Payable
$ 34,500.00
Equipment
$ 25,500.00
Land
$ 15,500.00
Shareholders’ equity
Common stock
$ 40,000.00
Current assets
Cash
$ 33,500.00
Total
$ 74,500.00
$ 74,500.00
Journal .
est.phpaction-skip&to-16; hm Fady Ibrahim User Settings My Classes .pdf
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Calendar Gradebook m # 4-Chapter 4 on OHM-Due 12/06/2017 at 1 1:59 PM Due in 4 hours, 11
minutes. Due Wed 12/06/2017 11:59 pr Show Intro/lnstruction A population numbers 17,000
organisms initially and grows by 83% each year. Suppose P represents population, and t the
number of years of growth. An exponential model for the population can be written in the form P
= a , where Preview This is attempt 1 of 1. Submit
Solution
P= a bt
After1 year, population becomes 17000+0.083(17000)=18411
18411= 17000 b1
b= 18411/17000=1.083
P= 17000(1.083)t.
Discuss the challenges for intelligence analysts when considering th.pdf
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Dual use commodities are “those that can be used in weapons of mass destruction (WMD)
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Yesterday is the centralization of power in a nation\'s capital with little autonomy for & Jones,
2014, p. 55). This system has the control and makes the decisions over citizen\'s lives (taxes,
education, and laws) and the local leaders. With this type of system, The unitary system from our
text, subdivisions (Roskin, Cord, Medeiros,
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Strengths
Unitary System - One of the main strengths in this system is the National Unity promotion and
this can get to help the system to witstand many troublesome incidents.
Federal System- The main strengths of this system is that it pops out to be successful in good,
large as well as culturally diverse countries where justice can be done to each and every area of
the place.
Weakness
Unitary system - The weakness is that the public policies may not get to fil the needs of all the
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Federal system - Here, one of the main weakness is that there can be a conflict that can arise
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