The significance of a notochord in the evolution of chordates is that.pdf
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The Ruritanian Chemical Company shows the following account balances for the month of May 2017 in Ruritanian dollars (R$). Calculate the cost of goods manufactured for May 2017. Also, prepare an income statement for May 2017. R$ Customer service cost 800 Direct manufacturing labour 5000 Depreciation plant and equipment 3000 Revenue 51000 Marketing and advertising 1800 Direct materials inventory 1.5.17 3000 Finished goods inventory 31.5.17 6000 Plant repairs and maintenance 3400 Plant utilities 5600 Direct material purchases 7000 Work in progress inventory 31.5.17 4000 Fire insurance plant 800 Indirect materials used 1000 General administrative costs 3000 Depreciation plant building 2000 Work in progress inventory 1.5.17 500 Finished goods inventory 1.5.17 2000 Direct materials inventory 31.5.17 1000 Indirect manufacturing labour 1200 Plant supervision 4000 Miscellaneous plant overhead 1000 Solution Above is the Cost sheet , Cost of Production is manufacturing cost for the month of May- 2017Calaculation of cost of goods manufactured in the month of May-2017ParticularsAmount ($)Amount ($)Direct Materials$ 9,000.00Direct material purchases$ 7,000.00ADD- Opening Direct materials inventory$ 3,000.00LESS- Closing Direct materials inventory$(1,000.00)Direct Labour$ 5,000.00Prime cost$ 14,000.00Factory Over heads$ 22,000.00Depreciation plant and equipment$ 3,000.00Depreciation plant building$ 2,000.00Plant repairs and maintenance$ 3,400.00Fire insurance plant$ 800.00Indirect materials used$ 1,000.00Indirect manufacturing labour$ 1,200.00Plant utilities$ 5,600.00Miscellaneous plant overhead$ 1,000.00Plant supervision$ 4,000.00Factory Cost$ 36,000.00Work in progress inventory$ (3,500.00)ADD- Opening Work in progress inventory$ 500.00LESS- Closing Work in progress inventory$(4,000.00)Finished goods$ (4,000.00)ADD- Opening Finished goods inventory$ 2,000.00LESS- Closing Finished goods inventory$(6,000.00)Cost of Production$ 28,500.00Cost of goods sold$ 28,500.00.
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on dry soil and they are very adapted to it. In case of alga, the surrounding water supports it and
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survive.
2)
Animal Cell
Plant cell
Bacteria
These are eukaryotes
These ar.
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In nonvascular plants the _ is usually no photosynthetic and is_ dependent on the garnet
zoophyte nutritionally, sporophyte sporophyte reproducing. gametophyte gametophyte
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gametophyte
Solution
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Reset the sum variable to 0
Using another for loop (which needs its own loop variable), run through all valid elements of the
array, starting from index 0. Inside the loop:
Use the sum variable to accumulate the numerator of the variance formula shown on page 1.
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this is the entire function
bool GenerateStatistics(FILE *fp2, double myArray[], int dataElementsToProcess)
{
if (fp2 == NULL && myArray == NULL && dataElementsToProcess < 1)
{
return false;
}
else
{
int i = 0,o = 0;
int sum = 0, sum1 = 0;
double maxValue = 0;
double minValue = DBL_MAX;
double mean, variance, stdDeviation;
for (int i = 0; i < 600; i++)
{
sum += myArray[i];
if (myArray[i] > maxValue)
{
maxValue = myArray[i];
}
if (myArray[i] < minValue)
{
minValue = myArray[i]; }
}
mean = sum / dataElementsToProcess;
sum = 0;
for (int o = 0; o < 600; o++)
{
sum += myArray[o];
}
variance = sqrt(pow((sum - mean), 2) / (dataElementsToProcess));
stdDeviation = sqrt(variance);
fprintf(fp2, \"Reaction Rate Statistics:\ \ \");
fprintf(fp2, \"Number of experiments = %d\ \", dataElementsToProcess);
fprintf(fp2, \"Minimum rate\\t\\t= %f\ \", minValue);
fprintf(fp2, \"Maximum rate\\t\\t= %f\ \", maxValue);
fprintf(fp2, \"Arithmetic mean\\t\\t= %f\ \", mean);
fprintf(fp2, \"Variance\\t\\t= %f\ \", variance);
fprintf(fp2, \"Standard deviation\\t= %f\ \ \ \", stdDeviation);
}
return true;
}
Solution
Here in the function you have
average - mean
Noofelements - dataElementsTo Process
Variance formula is
V = summation over i=1 to n ((xi - mean )*(xi - mean))
So perform like this using a for loop over my_array
double V=0.0
for(int i=0;i<600;i++)
{
V=V+pow((my_Array[i]-mean),2)
}
Now calculate standard deviation using sqrt of variance
sd=0.0
sd=sqrt(V).
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This document compares and contrasts prokaryotic and eukaryotic cells. It begins by introducing that eukaryotic cells contain organelles and a nucleus, while prokaryotic cells do not. The main differences are that prokaryotes lack a nucleus and organelles, while eukaryotes have linear DNA, mitochondria, and membrane-bound organelles. However, both share some similarities like DNA, vesicles, and being single or multicellular. The document then discusses characteristics of prokaryotes, including bacteria and archaea, followed by characteristics of eukaryotes like organelles, size, and being the domain of life that includes animals and plants.
Bae 495r Stream Ecology
Streams are ecosystems that provide the basic requirements for life including water, nutrients, energy, and an electron acceptor like oxygen. Organisms living in streams must find ways to obtain these necessities in the flowing water environment. The riffle habitat within streams provides advantages for both autotrophs and heterotrophs due to higher oxygen, light levels, and habitat structure compared to pools. Woody debris and other obstacles are important for creating microhabitats and retaining organic matter that fuels the food web.
Part ABriefly, explain the current hypotheses about how brown alg.pdf
Part A:
Briefly, explain the current hypotheses about how brown algae, red algae, chlorophytes and
charophytes (couple of groups of green algae), and land plants are related. What are the
characteristics of the land plants that distinguish it from the most closely related algal group, and
(if applicable) how do each of these characteristics benefit plants in a terrestrial environment?
Part B:
What are the differences between plant cells, animal cells, and bacterial cells. And what are the
characteristics of different plant cell types.
Solution
(1)Evolution of plants has facilitated the survival of life on earth and these land plants has their
monophyletic lineage embedded with that of green algae and these green algae are believed to be
the oldest eukaryotic lineages.
Hypothesis states that land plants are related to brown algae, red algae, chlorophytes and
charophytes. There are so many evidences to prove that these land plants share similar
characteristics with that of brown algae, red algae, chlorophytes and charophytes. Plants and
algae are eukaryotic and both posses the chlorophyll, which allows them to make their own food,
both plants and algae have similar cell walls, stores energy in the form of starch, follows two
stage life cycle.
Sporopollenin, chlorophyll a, cellulose, chlorophyll b are common to both charophytes and land
plants, but plants has lignin and is absent in case of charophytes. Some green algae exhibit
alternation of generations like plants. All land plants exhibit alternation of generations. But no
charophytes exhibit alternation of generations.
Charophytes are very similar to plants as they both have homologous chloroplasts (chlorophyll b,
beta carotene is present, thylakoid are arranged as grana, DNA as genetic material), with respect
to biochemical similarity, both has cellulose cell wall, similar enzymes in peroxisomes, when it
comes to mitosis and cytokinesis, follow similar mechanisms like disappearance of nuclear
envelope and till cytokinesis the spindle will remain. Based on the above features the
Charophytes are considered as the closest living algal relatives of land plants
Most of the varieties of algal species live in extremely damp terrestrial environment, whereas
most of the plants live on land and some plants can live in water also, but algae cannot survive
on dry soil and they are very adapted to it. In case of alga, the surrounding water supports it and
the whole alga is involved in helping it in the process of photosynthesis, absorption of water and
minerals, holdfast will provides the anchoring. But in case of plants, each and every organ is
specialized for different functions, like leaves for photosynthesis, cuticle to prevent water loss,
stomata for gaseous exchange, stem to support, root to provide anchorage. But alternation of
generation is absent in charophytes, but it is found in all plants providing them extra benefit to
survive.
2)
Animal Cell
Plant cell
Bacteria
These are eukaryotes
These ar.
Period 2 (phyla) 2016
The document discusses several animal phyla including Cnidaria, Platyhelminthes, Nematoda, Mollusca, Annelida, Arthropoda, Echinodermata, and Chordata. It provides information on the characteristics, habitats, reproduction, nutrition, cell structure, examples of organisms, and importance to humans for each phylum.
Animal Kingdom.pdf
The document discusses the classification and characteristics of animals in the kingdom Animalia. It describes 11 animal phyla and the key characteristics used to classify them, including their level of organization, symmetry, germ layers, coelom, and digestive/circulatory systems. Examples are provided of representative species from major phyla like porifera, cnidaria, mollusca, arthropoda, and chordata.
Similar to The significance of a notochord in the evolution of chordates is that.pdf (20)
4 chap 10 (kingdom animalia) f.sc 1st year biology helping notes
4 chap 10 (kingdom animalia) f.sc 1st year biology helping notes
Part ABriefly, explain the current hypotheses about how brown alg.pdf
Part ABriefly, explain the current hypotheses about how brown alg.pdf
More from irshadoptical
In practice, a RAM is used to simulate the operation of a stack by m.pdf
In nonvascular plants the _ is usually no photosynthetic and is_ dependent on the garnet
zoophyte nutritionally, sporophyte sporophyte reproducing. gametophyte gametophyte
structurally, sporophyte gametophyte reproductively. sporophyte sporophyte. nutritional.
gametophyte
Solution
sporophyte,nutrionally,gametophyte
In non vascular plants the sporophyte is usually non-photosynthetic and is nutrionally dependant
on the gametophyte
Explination
In bryophytes like mosses gametophyte n is the dominant generation and sporophyte 2n lives on
the gametophyte .....
i need help to edit the two methods below so that i can have them wo.pdf
I have to make a program for my C class and im stuck on the part where i have to calculate the
variance with elements from my array. This is exactly what im suppose to do:
Reset the sum variable to 0
Using another for loop (which needs its own loop variable), run through all valid elements of the
array, starting from index 0. Inside the loop:
Use the sum variable to accumulate the numerator of the variance formula shown on page 1.
After the end of this loop, compute the variance by dividing the new sum by the
dataElementsToProcess parameter.
this is the entire function
bool GenerateStatistics(FILE *fp2, double myArray[], int dataElementsToProcess)
{
if (fp2 == NULL && myArray == NULL && dataElementsToProcess < 1)
{
return false;
}
else
{
int i = 0,o = 0;
int sum = 0, sum1 = 0;
double maxValue = 0;
double minValue = DBL_MAX;
double mean, variance, stdDeviation;
for (int i = 0; i < 600; i++)
{
sum += myArray[i];
if (myArray[i] > maxValue)
{
maxValue = myArray[i];
}
if (myArray[i] < minValue)
{
minValue = myArray[i]; }
}
mean = sum / dataElementsToProcess;
sum = 0;
for (int o = 0; o < 600; o++)
{
sum += myArray[o];
}
variance = sqrt(pow((sum - mean), 2) / (dataElementsToProcess));
stdDeviation = sqrt(variance);
fprintf(fp2, \"Reaction Rate Statistics:\ \ \");
fprintf(fp2, \"Number of experiments = %d\ \", dataElementsToProcess);
fprintf(fp2, \"Minimum rate\\t\\t= %f\ \", minValue);
fprintf(fp2, \"Maximum rate\\t\\t= %f\ \", maxValue);
fprintf(fp2, \"Arithmetic mean\\t\\t= %f\ \", mean);
fprintf(fp2, \"Variance\\t\\t= %f\ \", variance);
fprintf(fp2, \"Standard deviation\\t= %f\ \ \ \", stdDeviation);
}
return true;
}
Solution
Here in the function you have
average - mean
Noofelements - dataElementsTo Process
Variance formula is
V = summation over i=1 to n ((xi - mean )*(xi - mean))
So perform like this using a for loop over my_array
double V=0.0
for(int i=0;i<600;i++)
{
V=V+pow((my_Array[i]-mean),2)
}
Now calculate standard deviation using sqrt of variance
sd=0.0
sd=sqrt(V).
Help!! Which of the following correctly lists the simple molecules t.pdf
Hello, I need assistance with question 7. I\'m not very good at double entry, so any explanations
would be helpful. However, liliu (ii) Trial balance after balancing off the alMUUTTUU ompany
u to Q7 R. Poon was having a whale of a time trying to get a trial balance for his como Redwar.
He had the following transactions during the month of May. He asks voor prepare the book
entries and a trial balance. 1 May Invests £8,000 in his business bank account. 2 May Purchases
£2,000 goods for cash from E. Skimo. £4,000 of goods on credit from S. Eal. £1,000 of goods on
credit from P. Olar. 4 May Sells £1,500 goods for cash to A.R.C. Tic. £5,000 goods on credit to
H. Unter. £3,000 goods on credit to M. Dick. 7 May M. Dick returns £2,000 goods unpaid saying
that they were rotting. 8 May R. Poon sends M. Dick\'s £2,000 goods (that he had originally
purchased from S. Eal for £1,800) back unpaid because he can\'t stand the smelt 10 May
Purchases another £1,200 of goods on credit from S. Eal. 11 May H. Unter buys on credit £1,600
of goods from R. Poon.
Solution
PURCHASE ACCOUNT S.EAL A/C (CREDITOR) P.OLAR A/C
(*CREDITOR) DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 2-May TO CASH 2000 2-May BY PURCH A/C 4000 2-May BY PURCH
A/C 1000 2-May TO S .EAL 4000 10-May BY PURCH A/C 1200 20-May TO BANK 750
2-May TO P.OLAR 1000 8-May TO PUR. RETURN 1800 31-May BY CL BALANCE
250 10-May TO SEAL 1200 20-May TO BANK 3400 31-May P&L A/C 8200 31-
May CL BAL - - SALES ACCOUNT H.UNTERA/C (DEBTOR) M.DICKI
A/C DEBTOR DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 4-May BY CASH 1500 4-May SALES A/C 5000 4-May SALES A/C
3000 4-May BY H.UNTER 5000 11-May SALES A/C 1600 7-May SALES RETURN
2000 4-May BY M.DICKI 3000 21-May BY BANK 2000 21-May BY BANK 800 11-
May TO H.UNTER 1600 31-May CL. BAL 4600 (5000+1600-2000) 31-May TO CL. BAL
200 31-May P&L A/C 11100 SALES RETURN ACCOUNT
DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 7-May TO DICKI 2000
P&L A/C 2000 PURCHASE RETURN ACCOUNT DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) 8-May BY S.EAL 1800 31-
May P&L A/C 1800 WAGES A/C DATE PARTICULAR
AMOUNT (DR) AMOUNT (CR) 31-May BANK A/C 800 31-May P&L A/C
800 RENT A/C DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 31-May BANK A/C 500 31-May P&L A/C 500
ELECTRICITY EXP A/C DATE PARTICULAR AMOUNT (DR) AMOUNT
(CR) 31-May BANK A/C 75 31-May P&L A/C 75
STATIONERY A/C BANK INTEREST A/C DATE PARTICULAR AMOUNT
(DR) AMOUNT (CR) 15-May BY BANK - 250 31-May BANK A/C 25 31-May P&L
A/C 250 31-May P&L A/C 25 BANK A/C CAPITAL A/C
DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 1-May BY BANK 8000 1-May
CAPITAL A/C 8000 3-May CLOSING BALANCE 8000 15-May BANK INT . A/C 250
20-May S.EAL 3400 20-May P.OLA 750 CASH A/C 21-May h.unter
2000 DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 21-May M.DICKI 800 2-
May PURCHASE 2000 31-May WAGES 800 4-May SALES 1500 31-May RENT
500 CL. BALNANCE 500 31-May ELEC .
How does feasibility differ in an agile environment in comparison to.pdf
Traditional development assumes systems are fully specifiable through extensive planning, while agile development develops high quality adaptive software through small teams that continuously improve design and testing using fast feedback. Management shifts from command and control to leadership and collaboration, while knowledge moves from explicit to tacit. Other differences include communication style, life cycle model, organizational structure, quality control approach, and client involvement. Both approaches have merits and demerits, and neither is inherently simpler - what matters is properly understanding the methodology for the given situation. In-house software provides knowledgeable support from dedicated developers but can lack scalability, while outsourcing decreases coding burden but may increase costs.
describe a real-life example of selection against a homozygous reces.pdf
describe a real-life example of selection against a homozygous recessive genotype
Solution
When both the alleles of a particular chromosome are recessive, it’s called homozygous
recessive for that chromosome pair. There are few real life examples of homozygous recessive
such as albinism, cystic fibrosis etc.
Albinism is congenital disorder in humans that is a characterized by absence of pigment in skin,
eyes, and hairs. Absence of pigment may be partial or complete. Lack of skin pigmentation in
albinism makes individual more sensitive to sunlight and susceptible skin sunburn and skin
cancers. Albinism is also associated with visions defects such as amblyopia, photophobia and
nystagmus. This disorder is due to inheritance of recessive alleles which has defect/ absence of
tyrosinase involved in the production of melanin..
Define a motor unit. What roles do motor units play in graded contrac.pdf
Define a motor unit. What roles do motor units play in graded contractions in skeletal muscles?
Solution
Motor unit is made of Motor Neuron, and its branches extended to the myofibrils of muscle
fibres. Signal is passed through the neuron to the muscle fibres. Motor neuron reaches to the
muscle fibre at neuromuscular junctions. Motor neurons work in group to assist the contraction
of single muscle in body. Force of Muscle contraction is controlled by the number of activated
motor neurons..
An antibody is a protein that has which level of protein structure .pdf
An antibody is a protein that has which level of protein structure? Primary Quaternary Tertiary
Secondary Which cell type does the HIV infect? Cytotoxic T lymphocytes B lymphocytes
Helper T lymphocytes Suppressor T lymphocytes
Solution
antibody is a protein which is in quaternary structure because it consist of more than one
polypeptide chains (two heavy chain and two light chain ) joint by disulphide bond.
HIV infect Helper T lymphocyte(CD4+T cells).It causes the lower level of CD4+ cells by
pyroptosis.When CD4 cell number is declined below a normal level than cell mediated immunity
is lost and body become suceptible to various infection..
C++ PROGRAMThis program builds on the code below#include iostr.pdf
C++ PROGRAM
This program builds on the code below:
#include
#include
#include \"shirtType.h\"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open(\"h:\\\\shirtlab.txt\");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<\"Enter the new quantity of shirts: \"<>quantity;
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
Solution
#include
#include
struct node
{
int data;
struct node *next;
}*head;
void append(int nkkm)
{
struct node *temp,*right;
temp= (struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
right=(struct node *)head;
while(right->next != NULL)
right=right->next;
right->next =temp;
right=temp;
right->next=NULL;
}
void add( int nkkm )
{
struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
if (head== NULL)
{
head=temp;
head->next=NULL;
}
else
{
temp->next=head;
head=temp;
}
}
void addafter(int nkkm, int loc)
{
int i;
struct node *temp,*left,*right;
right=head;
for(i=1;inext;
}
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
left->next=temp;
left=temp;
left->next=right;
return;
}
void addaShirt(int nkkm)
{
int c=0;
struct node *temp;
temp=head;
if(temp==NULL)
{
add(nkkm);
}
else
{
while(temp!=NULL)
{
if(temp->datanext;
}
if(c==0)
add(nkkm);
else if(cdata==nkkm)
{
if(temp==head)
{
head=temp->next;
free(temp);
return 1;
}
else
{
prev->next=temp->next;
free(temp);
return 1;
}
}
else
{
prev=temp;
temp= temp->next;
}
}
return 0;
}
void display(struct node *r)
{
r=head;
if(r==NULL)
{
return;
}
while(r!=NULL)
{
printf(\"%d \",r->data);
r=r->next;
}
printf(\"\ \");
}
int count()
{
struct node *n;
int c=0;
n=head;
while(n!=NULL)
{
n=n->next;
c++;
}
return c;
}
int main()
{
int i,nkkm;
struct node *n;
head=NULL;
while(1)
{
printf(\"1.Add a shirt\ \");
printf(\"2.Sell a shirt\ \");
printf(\"3.Res.
You are in charge of the investigation regarding possible exposur.pdf
You are conducting a study to see if students do better when they study all at once or in intervals.
One group of 12 participants took a test after studying for one hour continuously. The other
group of 12 participants took a test after studying for three twenty minute sessions. The first
group had a mean score of75 and a variance of 120. The second group had a mean score of 86
and a variance of 100.
a. What is the calculated t value? Are the mean test scores of these two groups significantly
different at the .05 level?
b. What would the t value be if there were only 6 participants in each group? Would the scores
be significant at the .05 level?
Solution
A)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 75
X2 = 86
Calculating the standard deviations of each group,
s1 = 10.95445115
s2 = 10
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 12
n2 = sample size of group 2 = 12
Thus, df = n1 + n2 - 2 = 22
Also, sD = 4.281744193
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -2.569046516 [ANSWER, T VALUE]
********************************
b)
Calculating the means of each group,
X1 = 75
X2 = 86
Calculating the standard deviations of each group,
s1 = 10.95445115
s2 = 10
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 6
n2 = sample size of group 2 = 6
Also, sD = 6.055300708
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -1.816590212 [ANSWER]
For part A, the P value is
p = 0.01750305
For part B, the P value is
p = 0.099328388
Hence, Part A would be significant, but part B wouldn\'t. [ANSWER]
**********************************
*****************************************************
Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please
resubmit this question together with the formula/method you use in determining the degrees of
freedom. That way we can continue helping you! Thanks!
********************************************.
Which of the following is not a product of the light reactions of ph.pdf
Which of the following developments have not helped to increase the importance of packaging
cover time Which stage is the most costly one? In which stage are criteria such as adequacy of
the technical - financial -manpower resources of the company and suitability to company
objectives and mission are used to evaluate new product Which stage involves a feasibility study
Which stage is most appropriate for \"informative\" advertising? At which stage should
management make a choice between skimming and penetration In which stage is
\"rejuvenation\" most likely to occur (if h does)? At which stage is heaviest price competition
most likely to occur? Which stage is the right time for starting to plan a new product
introduction? At which stage should product differentiation be at its lowest level? At which
stage do profits reach their highest level? A small confectionery (tath, Sekerleme, recel)
manufacturer markets its jams and jel an agreed upon price every year provided that the label is
changed into \"\"Sultan Ha following would be true
Solution
(8) (a)
Need for better packaging has nothing to do with economic crises - the crises can impact the
purchase decision with or without better & more attractive packaging.
(9) (C)
Concept development & testing is costliest because if the prototype developed fails, entire cost
incurred goes waste.
(10) (B)
These are tested at Screening stage for suitability of ideas generated.
(11) (D)
It involves technical, economic & business feasibility analysis.
(12) (A)
(13) (B)
(14) (D)
(15) (C)
(16) (C)
(17) (A)
(18) (B)
(19) Question is partially visible & can\'t be answered properly..
Write a line of code to create a Print Writer to write to the file .pdf
Write a class named Deque20, placed in the file Deque20.java. It should behave exactly as
described in the Java interface Deque (Google: \"Java 8 Deque\") except that it should only
implement the following methods: a no-argument constructor, addFirst, addLast, removeFirst,
removeLast, peekFirst, peekLast, and size. Your class should not extend or implement any class
or interface. It must use an array as its storage, allow generic types (E), and be as efficient as
possible. The Deque javadoc mentions several exceptions, but you should only return
NullPointerException (if someone tries to add a null reference to the Deque) and
NoSuchElementException (if someone tries to remove an item from an empty Deque).
Your deque should store objects in a circular array.
Probably the simplest implementation has objects with fields
The meaning being that the elements currently in the deque are at positions list[first_index],
list[(first_index+1)%list.length], list[(first_index+2)%list.length],...,
list[(first_index+num_elems-1)%list.length]. The modulus is necessary because first_index
might be near the end of the array and adding to it might result in an index that\'s out of bounds.
When an add operation would cause the capacity of the array to be exceeded, you should double
the size of the array by creating a new array of twice the size and copying the old array contents
into the new. Do not worry about ever reducing the size of an array. Also, whenever a reference
is removed from your deque, the old array location should be set to null so that the garbage
collector knows that the old item is no longer referred to by the deque.
Solution
public class Deque20 { private int first_index; private int num_elems; private Object[]
list; Deque20(){ first_index = 0; num_elems = 0; list = new Object[20]; }
private void doubleSize(){ Object[] arr = new Object[2 * list.length]; for(int i = 0; i <
list.length; ++i){ arr[i] = list[(i + first_index) % list.length]; } list = arr; }
public void addFirst(E val){ if(list.length == num_elems){ doubleSize(); }
int index = (first_index + list.length - 1) % list.length; list[index] = val; ++num_elems;
} public void addLast(E val){ if(list.length == num_elems){ doubleSize();
} int index = (first_index + num_elems) % list.length; list[index] = val;
++num_elems; } public E removeFirst(){ int index = first_index; E ret = (E)
list[index]; first_index = (first_index + 1) % list.length; return ret; } public E
removeLast(){ int index = ((first_index + num_elems) % list.length) - 1; E ret = (E)
list[index]; --num_elems; return ret; } public E peekFirst(){ return (E)
list[first_index]; } public E peekLast(){ int index = ((first_index + num_elems) %
list.length) - 1; return (E) list[index]; } public int size(){ return num_elems; }
}.
what is the difference bettween cotranslational import and posttrans.pdf
What is meant by ecological fallacy?
Solution
Answer:
The term \"fallacy\" means the invalid. ecological fallacy tells about the individuals in the
ecosystem which are invalid, it tell statistically about the group of individuals that are not part of
the group which they should belong to.
Eg. The earnings of people in particular district is high, the assumption that every person in that
distict is rich would be fallacy. There are few people with low earnings..
What is the meaning of critical periods in neonatal development Wha.pdf
what is the difference bettween cotranslational import and posttranslational import of proteins
into organelles? of the organelles listed below, which make use of which type of import? 1. ER
2. Mitochondria 3. nucleus 4. peroxisome
Solution
Cotranslational transport involves translocation of a growing peptide chain within the cell
simultaneously like to the ER.
Where as, post translational transport involves initially protein synthesis in the cytosol followed
by transport to ER.
Cotranslational - ER,
Post translational- Mitochondria, nucleus and peroxisome.
What are the major differences between foreign bonds and Eurobonds .pdf
what are some necessary organizational steps to successfully offer acupressure as part of nursing
care in the labor and delivery setting?
Solution
Principles of acupuncture forms the basis of acupressure.
During labour it can be used to relieve pain and increase the efficiency of the labor. From 37th
week onwards all points are safe for acupressure. Lesser pressure as compared to the nirmal
times has to be applied during labor. It should be started during early stages of labor and shall
intermittently be given during the entire process.
There are few specific acupressure points during the delivery and labor. These are lower back
points, large intestine, urinary bladder, kidney and spleen..
What is meant by ecological fallacySolutionAnswerThe term .pdf
What is Incorporation by Reference? and How is it used in a contract?
Solution
Incorporation by reference is an act of including second document within the first document by
giving only reference in first document. Ie if in first document only a single line is kept giving
reference of second document. than the secound document will be part of first document.
Incorporation by reference is often done in creating law as well in contract law nad trust and
estate law
Incorporation by reference is used in contract by way of terms and condition. For example there
are general set of terms and condition to contarct now instead of including all the tems and
condtion to contract a company can just give reference to it, making it binding to the companies
entering the contract.
Thus it is very important to read document if it is included by reference.
What are the two main branches of Non-Euclidean Geometry What is the.pdf
What are the major differences between foreign bonds and Eurobonds? Why do some US
companies prefer to issue Eurobonds to raise US dollars from foreign markets?
Solution
A Eurobond is long term bond issued and sold by the compnies outside the country in which it is
dominated. If US company issued a bond in dollar and selling it to Japanese investors in Japan
then it is know as Eurobond.
Foreign is also sold in the other courtry but it is issued in that country currency as well that\'s a
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US companies issue Eurobonds because in that way US companies can hedge its currency risk..
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By what transport method does oxygen enter the blood from the alveoli? By what method is
oxygen transported to the body tissues from the lungs? By what method does oxygen enter the
body tissues from the blood?
a. Diffusion, bulk flow, diffusion
b. Bulk flow, bulk flow, diffusion
c. Bulk flow, diffusion, carrier proteins
d. Diffusion, diffusion, carrier proteins
e. Bulk flow, diffusion, bulk flow
Solution
b. Bulk flow, bulk flow, diffusion
bulk flow carries air or water, relatively high in O2 and low in CO2, past a respiratory surface.
The flow is usually propelled by muscular movements. Bulk flow of blood transports gases
between the respiratory system and the tissues. Blood is pumped through the body by the heart.
Diffusion transfers O2 out of the capillaries and into nearby tissues, and transfers CO2 from the
tissues into capillaries.
All of these are examples of evidence used to support the hypothesis.pdf
All of these are examples of evidence used to support the hypothesis of endosymbiosis,
EXCEPT:a. Mitochondria and plastids contain circular DNA molecules similar to the circular
chromosomes of bacteriab. Mitochondria and plastids, like prokaryotes, only contain genetic
material in the form of RNAc. Mitochondria and plastids can replicate themselves in a process
similar to prokaryotesd. Mitochondria and plastids have their own ribosomes and enzymes that
can be used to transcribe and translate DNA into proteinse. Mitochondria and plastids have a
double membrane (inner and outer)
Solution
Out of the above options , the correct will be
B. Mitochondria and plastids a have their genetic material in the form of DNA not RNA.
They donot have any such machinary for their expression , so this option is incorrect regarding
endosymbiosis.
Otherwise rest all the sentenses are true.
They have circular genome like Prokaryotes, and can replicate and also have double membrane
layer..
A 28 year-old biologist makes a trip to study life forms in a distan.pdf
A 28 year-old biologist makes a trip to study life forms in a distant planet 17 light-years away.
The round trip including a stay of 1 year in the distant planet takes 38 years according to the
clock on earth. The biologist\'s son is 2.0 year-old when she left. Assume that the planet is not
moving with respect to earth. How fast is the spaceship traveling? Give your answer in terms of
the speed of light. You do not need to enter \"c\" as unit.
How old is she when she returns according to her clock?How old is her son according to the
clock on earth?
What is the total distance traveled according to the biologist?
Give your answer in light-years. You do not need to enter the units.
Solution
As it takes 38 - 1 = 37 years for the biologist to make the trip
Speed = 17 / 38 = 0.45 c
Time dialation that is occurred t\' = t * sqrt(1 - (v2/c2))
t\' = 37 * sqrt(1 - 0.452) = 33.04 years
Biologists age = 28 + 33.04 + 1 = 62.04 years
Her son\'s age = 2.0 + 38 = 40 years.
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4. Name the following compounds: CO2 CH CH3 CH20H CH3COOH 5. Water has a molecular
weight of 18. A milliliter of liquid water has a mass of one gram. Calculate the molar
concentration of liquid water: 6. What kind of bonds hold together the following compounds
(match): hydrogen bonds water ionic bonds table salt covalent bonds nucleotides in RNA james
bonds strands in dsDNA 7. Fill in the blanks: is transcribed by to make is translated by to make
Solution
Answer:
4)
CO2 : carbondioxide
O2: Oxygen
N2: NItrogen
CH4 : Methane
CH3CH2OH : Ethyl alcohol
CH3COOH: Acetic acid.
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