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This short document contains a mathematical inequality comparing the values of 4, 1, 2, and 3. It states that 4 is greater than 1, which is greater than 2, which is greater than 3.
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1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
1. C.
HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS.
2. D.
3.C.
4.B.
5.A.
6.B.
Solution
1. C.
HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS.
2. D.
3.C.
4.B.
5.A.
6.B..
#include iostream #include cstddefusing namespace std;temp.pdf
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct node *ptr = start;
cout<data;
ptr=ptr->next;
}
}
}
//Prints total number of elements
int getSize()
{
struct node *ptr, *s;
int count=0;
if (start == NULL)
{
cout<<\"The List is empty\"<next;
count=1;
while (ptr != NULL)
{
count++;
ptr=ptr->next;
}
return count;
}
//Returns true if there is no element in the list.
bool empty()
{
if(start==NULL)
{
return empty;
}
}
};
Solution
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct .
omparative historical research is a method of social science that e.pdf
omparative historical research is a method of social science that examines historical events in
order to create explanations that are valid beyond a particular time and place, either by direct
comparison to other historical events, theory building, or reference to the present
day.[1]Generally, it involves comparisons of social processes across times and places. It overlaps
with historical sociology. While the disciplines ofhistory and sociology have always been
connected, they have connected in different ways at different times (see \'Major researchers\'
below). This form of research may use any of several theoretical orientations. It is distinguished
by the types of questions it asks, not the theoretical framework it employs (see \'Illustrations\'
below)
Some commentators have identified three waves of historical comparative research.[2] The first
wave of historical comparative research concerned how societies came to be modern, i.e. based
on individual and rational action, with exact definitions varying widely. Some of the major
researchers in this mode were Alexis de Tocqueville,[3]Karl Marx,[4]Emile Durkheim,[5]Max
Weber,[6] and W.E.B. Du Bois.[7] The second wave reacted to a perceived ahistorical body of
theory and sought to show how social systems were not static, but developed over time.[8]
Notable authors of this wave include Barrington Moore, Jr.,[9]Theda Skocpol,[10]Charles
Tilly,[11]Michael Mann,[12] and Mark Gould.[13] Some have placed the Annales school and
Pierre Bourdieu in this general group, despite their stylistic differences.[14] The current wave of
historical comparative research sociology is often but not exclusively post-structural in its
theoretical orientation. Influential current authors include Julia Adams,[15] Anne Laura
Stoler,[16]Philip Gorski,[17] and James Mahoney.[18]
There are four major methods that researchers use to collect historical data. These are archival
data, secondary sources, running records, and recollections. The archival data, or primary
sources, are typically the resources that researchers rely most heavily on. Archival data includes
official documents and other items that would be found in archives, museums, etc. Secondary
sources are the works of other historians who have written history. Running records are ongoing
series of statistical or other sorts of data, such as census data, ship\'s registries, property deeds,
etc. Finally recollections include sources such as autobiographies, memoirs or diaries.[19]
There are four stages, as discussed by Schutt, to systematic qualitative comparative historical
studies: (1) develop the premise of the investigation, identifying events, concepts, etc., that may
explain the phenomena; (2) choose the case(s) (location- nation, region) to examine; (3) use what
Theda Skocpol has termed as \"interpretive historical sociology\" and examine the similarities
and the differences; and (4) based on the information gathered, propose a causal explanation for
the pheno.
where is it .pdf
This very short document appears to be asking "where is it" but does not provide any other context or information. It repeats the phrase "where is it" twice but gives no indication of what "it" refers to or where one might expect to find it. The document on its own does not give enough information to form a meaningful summary.
The name of the reaction is aldol condesation (Cl.pdf
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a
enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form
the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The
functional group present in the product is a ketone andthe hydroxi groups that were eliminated.
Solution
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a
enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form
the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The
functional group present in the product is a ketone andthe hydroxi groups that were eliminated..
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will
react with Cr = 2x3.17/3 =6.14/3 =2.057
Solution
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will
react with Cr = 2x3.17/3 =6.14/3 =2.057.
Recommended
1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
1. C.
HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS.
2. D.
3.C.
4.B.
5.A.
6.B.
Solution
1. C.
HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS.
2. D.
3.C.
4.B.
5.A.
6.B..
#include iostream #include cstddefusing namespace std;temp.pdf
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct node *ptr = start;
cout<data;
ptr=ptr->next;
}
}
}
//Prints total number of elements
int getSize()
{
struct node *ptr, *s;
int count=0;
if (start == NULL)
{
cout<<\"The List is empty\"<next;
count=1;
while (ptr != NULL)
{
count++;
ptr=ptr->next;
}
return count;
}
//Returns true if there is no element in the list.
bool empty()
{
if(start==NULL)
{
return empty;
}
}
};
Solution
#include
#include
using namespace std;
template class Node{
private:
T value;
Node *next;
public:
Node();
Node (T data);
T remove();
T getValue();
Node* getNext();
void setValue(T data);
void setNext(Node *nextNode);
};
template
Node::Node(){
T value =NULL;
}
template
Node::Node(T data){
T value=data;
}
template
T Node::getValue(){
return value;
}
template
Node* Node::getNext(){
return next;
}
template
void Node::setValue(T data){
T value=data;
}
template
void Node::setNext(Node *nextNode){
next=nextNode;
}
template class MyLinkedList{
Node start;
//Define a constructor
MyLinkedList()
{
start->setValue(0);
start->setNext(NULL);
}
//Member Functions:
// Adds a new element to the list
void add (T val)
{
/* creating a new node to hold value*/
Nodetemp(val,NULL);
/* If no start node till now, this node is made as start node*/
if (start->next == NULL)
{
start.setNext(temp);
}
/* if start node exists, insert new node at end of list*/
else
{
s = start->next;
/* locating end of list*/
while (s->next != NULL)
{
s= s->next;
}
temp->next = NULL;
/* attaching new node */
s->next = temp;
cout<<\"Element Inserted\"<next->next != NULL)
{
s = s->next;
}
//After locating the end node
ptr=s->next;
s->next=NULL;
//deletting end node
return ptr;
}
}
//Prints elements of the list
void printContent()
{
struct .
omparative historical research is a method of social science that e.pdf
omparative historical research is a method of social science that examines historical events in
order to create explanations that are valid beyond a particular time and place, either by direct
comparison to other historical events, theory building, or reference to the present
day.[1]Generally, it involves comparisons of social processes across times and places. It overlaps
with historical sociology. While the disciplines ofhistory and sociology have always been
connected, they have connected in different ways at different times (see \'Major researchers\'
below). This form of research may use any of several theoretical orientations. It is distinguished
by the types of questions it asks, not the theoretical framework it employs (see \'Illustrations\'
below)
Some commentators have identified three waves of historical comparative research.[2] The first
wave of historical comparative research concerned how societies came to be modern, i.e. based
on individual and rational action, with exact definitions varying widely. Some of the major
researchers in this mode were Alexis de Tocqueville,[3]Karl Marx,[4]Emile Durkheim,[5]Max
Weber,[6] and W.E.B. Du Bois.[7] The second wave reacted to a perceived ahistorical body of
theory and sought to show how social systems were not static, but developed over time.[8]
Notable authors of this wave include Barrington Moore, Jr.,[9]Theda Skocpol,[10]Charles
Tilly,[11]Michael Mann,[12] and Mark Gould.[13] Some have placed the Annales school and
Pierre Bourdieu in this general group, despite their stylistic differences.[14] The current wave of
historical comparative research sociology is often but not exclusively post-structural in its
theoretical orientation. Influential current authors include Julia Adams,[15] Anne Laura
Stoler,[16]Philip Gorski,[17] and James Mahoney.[18]
There are four major methods that researchers use to collect historical data. These are archival
data, secondary sources, running records, and recollections. The archival data, or primary
sources, are typically the resources that researchers rely most heavily on. Archival data includes
official documents and other items that would be found in archives, museums, etc. Secondary
sources are the works of other historians who have written history. Running records are ongoing
series of statistical or other sorts of data, such as census data, ship\'s registries, property deeds,
etc. Finally recollections include sources such as autobiographies, memoirs or diaries.[19]
There are four stages, as discussed by Schutt, to systematic qualitative comparative historical
studies: (1) develop the premise of the investigation, identifying events, concepts, etc., that may
explain the phenomena; (2) choose the case(s) (location- nation, region) to examine; (3) use what
Theda Skocpol has termed as \"interpretive historical sociology\" and examine the similarities
and the differences; and (4) based on the information gathered, propose a causal explanation for
the pheno.
where is it .pdf
This very short document appears to be asking "where is it" but does not provide any other context or information. It repeats the phrase "where is it" twice but gives no indication of what "it" refers to or where one might expect to find it. The document on its own does not give enough information to form a meaningful summary.
The name of the reaction is aldol condesation (Cl.pdf
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a
enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form
the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The
functional group present in the product is a ketone andthe hydroxi groups that were eliminated.
Solution
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a
enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form
the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The
functional group present in the product is a ketone andthe hydroxi groups that were eliminated..
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will
react with Cr = 2x3.17/3 =6.14/3 =2.057
Solution
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will
react with Cr = 2x3.17/3 =6.14/3 =2.057.
Part A neutralizes acids - base Part B produces.pdf
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a
soapy feel - base Part D: turns litmus red - acid
Solution
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a
soapy feel - base Part D: turns litmus red - acid.
Lewis Acid Base Reaction .pdf
Lewis acid-base reactions involve the sharing of electron pairs between reactants. A Lewis acid is an electron pair acceptor that has an empty orbital, while a Lewis base is an electron pair donor that has a lone pair of electrons. When a Lewis acid and base interact, the base donates an electron pair to form a new bond with the empty orbital on the acid.
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 =
10 pH+pOH=14 pOh = 14-pH = 14-10=4
Solution
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 =
10 pH+pOH=14 pOh = 14-pH = 14-10=4.
It is a reaction for secondary amines .Option 2 i.pdf
It is a reaction for secondary amines .Option 2 is correct.
Solution
It is a reaction for secondary amines .Option 2 is correct..
fluorene is mobile phase and fluorenone is statio.pdf
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it
depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene
and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move
slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent
(pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind
of applies to chromatography wherein rather than dissolve just replace it with moves with, so like
moves with like. That being said the pentane will \"carry\" the fluorene through the alumina
slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that
being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity
(remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the
alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through
the alumina than would the fluorene, because there is no attraction between all these polar
compounds which will allow it to move faster, rather than a nonpolar and polar chemical having
an attration towards each other and thus moving more slowly.
Solution
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it
depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene
and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move
slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent
(pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind
of applies to chromatography wherein rather than dissolve just replace it with moves with, so like
moves with like. That being said the pentane will \"carry\" the fluorene through the alumina
slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that
being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity
(remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the
alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through
the alumina than would the fluorene, because there is no attraction between all these polar
compounds which will allow it to move faster, rather than a nonpolar and polar chemical having
an attration towards each other and thus moving more slowly..
equivalent mass is 27.9 . so Molecular massmole.pdf
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9
looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2
that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe )
Solution
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9
looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2
that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ).
There is no solutionSolutionThere is no solution.pdf
The document repeats the phrase "There is no solution" with no other context or information provided. It consists of three short sentences stating "There is no solution" with no explanation or details given about what problem is being referred to or discussed.
This is a case of allopatric speciation. In this case lice from gori.pdf
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two
different species. So this is case of geographical speciation where species become apart from
each other to an extent that they cannot interbreed.
Solution
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two
different species. So this is case of geographical speciation where species become apart from
each other to an extent that they cannot interbreed..
d. 8 The bonding electrons are the electrons taki.pdf
The document discusses the bonding electrons in a carbon dioxide (CO2) molecule. Carbon is centered between two hydrogen atoms and one oxygen atom. Carbon shares two electrons with each hydrogen atom, for a total of four electrons shared with hydrogen. Carbon also shares four electrons with the oxygen atom. Therefore, the total number of bonding electrons is the four shared with hydrogen plus the four shared with oxygen, which is eight.
The genetical changes of single locus are responsible for divergent .pdf
The genetical changes of single locus are responsible for divergent branches of Y chromosome
phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial
DNA sequences of Neanderthals and modern humans were compared and analyzed. The
Hominins species were found to be 30000 year old and were compared with Neanderthal
specimens.
The relationships between these populations have drawn with the help of the age of specimens
plotted against the average genetic distance. The Hominins type sequences are very rare when
compared with the modern European samples. The prehistoric Europeans had shown constant
number of differences when compared to today’s Europeans and Neanderthals.
The Neanderthals have shown the genetic continuity in modern human’s genealogy. The
Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to
the similar ages. The Neanderthals and early anatomical modern humans were thought to be
regional populations of the same evolving species connected by gene flow and both the archaic
and modern humans contribute in different proportions to the present human gene pool.
The transition and demographic replacement separated the population as Neanderthals and
modern humans are considered just as one population observed at different times. The multi
regional models predict that at least some level of genetic continuity can be observed from
archaic Neanderthal forms to the contemporary Hominins forms and up to today’s Europeans.
The results have not matched with the view that the Neanderthal were genetically related with
the anatomically modern ancestors of current Europeans or contributed to the present day human
gene pool.
Solution
The genetical changes of single locus are responsible for divergent branches of Y chromosome
phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial
DNA sequences of Neanderthals and modern humans were compared and analyzed. The
Hominins species were found to be 30000 year old and were compared with Neanderthal
specimens.
The relationships between these populations have drawn with the help of the age of specimens
plotted against the average genetic distance. The Hominins type sequences are very rare when
compared with the modern European samples. The prehistoric Europeans had shown constant
number of differences when compared to today’s Europeans and Neanderthals.
The Neanderthals have shown the genetic continuity in modern human’s genealogy. The
Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to
the similar ages. The Neanderthals and early anatomical modern humans were thought to be
regional populations of the same evolving species connected by gene flow and both the archaic
and modern humans contribute in different proportions to the present human gene pool.
The transition and demographic replacement separated the population as Neanderthals and
modern humans .
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
The corrects answers are:
Thr and Asn are polar amino acids.
Isoleucine has more than one stereocenter (chiral center)
The Leu side chain does not from hydrogen bonds with other amino acids
Solution
The corrects answers are:
Thr and Asn are polar amino acids.
Isoleucine has more than one stereocenter (chiral center)
The Leu side chain does not from hydrogen bonds with other amino acids.
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4
i.e Z4 = ( 0,1,2,3)
U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2
So, isomorphic Z4 and U(8) are not isomorphic.
Solution
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4
i.e Z4 = ( 0,1,2,3)
U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2
So, isomorphic Z4 and U(8) are not isomorphic..
selected palntsxerophyte verus mesophyte========================.pdf
selected palnts
xerophyte verus mesophyte
=========================================================
A xerophyte is a types of plant that has adjusted to get by in a domain with minimal fluid water,
for example, a forsake or an ice-or snow-canvassed locale in the Alps or the Arctic.
The morphology and physiology of xerophytes are differently adjusted to moderate water, and
usually likewise to store substantial amounts of water, amid dry periods. Different species might
be adjusted to survive long stretches of parching of their tissues, amid which their metabolic
action may viably close down. Plants with such morphological and physiological adjustments are
xeromorphic.
Xerophytic plants may have comparable shapes, structures, and structures and look
fundamentally the same as, regardless of the possibility that the plants are not firmly related,
through a procedure called concurrent development. For instance, a few types of desert flora
(individuals from the family Cactaceae), which advanced just in the Americas, may seem like
Euphorbias, which are dispersed around the world. A random types of caudiciforms, plants with
swollen bases that are utilized to store water, may likewise show such likenesses.
Xerophytic plants can have less general surface territory than different plants, so diminishing the
range that is presented to the air and lessening water misfortune by vanishing. Xerophytes can
have littler leaves or less branches than different plants. A case of leaf surface decrease are the
spines of a desert flora. A case of compaction and diminishment of spreading are the barrel
desert flora. Different xerophytes may have their leaves compacted at the base, as in a basal
rosette, which might be littler than the plant\'s blossom. This adjustment is displayed by some
Agave and Eriogonum species, which can be discovered developing close Death Valley.
A few xerophytes have minor hairs on their surface to give a wind break and decrease wind
current, along these lines diminishing the rate of dissipation. At the point when a plant surface is
secured with minor hairs, it is called tomentose.
In a still domain, the regions under the leaves/spines where transpiration is occurring structure a
little limited environment that is more soaked than typical with water vapor. In the event that this
is not overwhelmed by wind, the water vapor potential angle is diminished as is transpiration.
Subsequently, in a windier circumstance, this confinement is not held thus the angle stays high,
which helps the loss of water vapor. Spines trap a layer of dampness furthermore moderate air
development over tissues.
===================
Mesophytes are earthbound plants which are adjusted to neither an especially dry nor especially
wet environment. A case of a mesophytic living space would be a country calm glade, which
may contain goldenrod, clover, oxeye daisy, and Rosa multiflora.
Mesophytic plants have unbending, tough, openly expanded stems and stringy, all around create.
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
52000
Ruth & Associates
Statement of Changes in Stockholders’ Equity
For the Year Ended December 31, 2010
Beginning Common Stock
25000
Plus: Common Stock Issued
15000
Ending Common Stock
40000
Beginning Retained Earnings
18000
Plus: Net Income
52000
Less: Dividends
10000
Ending Retained Earnings
60000
Total Stockholders’ Equity
100000
Ruth & Associates
Balance Sheet
As of the December 31, 2010
Assets
Cash
52000
Accounts Receivable
26000
Supplies
3000
Prepaid Rent
5000
Land
63000
Total Assets
149000
Liabilities
Interest Payable
2000
Salaries Payable
7000
Unearned Revenue
8000
Notes Payable
32000
Total Liabilities
49000
Stockholders’ Equity
Common Stock
40000
Retained Earnings
60000
Total Stockholders’ Equity
100000
Total Liab. and Stockholders’ Equity
149000
Ruth & Associates
Statement of Cash Flows
For the Year Ended December 31, 2010
Cash Flow From Operating Activities
52000
Cash Flow From Investing Activities
-30000
Cash Flow From Financing Activities
20000
Net Change in Cash Plus:
42000
Beginning Cash Balance
10000
Ending Cash Balance
52000
Ending Cash Balance
52000
Net Change in Cash
42000
= Beginning Cash Balance
10000
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
52000
Ruth & Associates
Statement of Changes in Stockholders’ Equity
For the Year Ended December 31, 2010
Beginning Common Stock
25000
Plus: Common Stock Issued
15000
Ending Common Stock
40000
Beginning Retained Earnings
18000
Plus: Net Income
52000
Less: Dividends
10000
Ending Retained Earnings
60000
Total Stockholders’ Equity
100000
Ruth & Associates
Balance Sheet
As of the December 31, 2010
Assets
Cash
52000
Accounts Receivable
26000
Supplies
3000
Prepaid Rent
5000
Land
63000
Total Assets
149000
Liabilities
Interest Payable
2000
Salaries Payable
7000
Unearned Revenue
8000
Notes Payable
32000
Total Liabilities
49000
Stockholders’ Equity
Common Stock
40000
Retained Earnings
60000
Total Stockholders’ Equity
100000
Total Liab. and Stockholders’ Equity
149000
Ruth & Associates
Statement of Cash Flows
For the Year Ended December 31, 2010
Cash Flow From Operating Activities
52000
Cash Flow From Investing Activities
-30000
Cash Flow From Financing Activities
20000
Net Change in Cash Plus:
42000
Beginning Cash Balance
10000
Ending Cash Balance
52000
Ending Cash Balance
52000
Net Change in Cash
42000
= Beginning Cash Balance
10000
Solution
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
5200.
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296
P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720
Solution
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296
P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720.
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
Part A: blood solutes, swell
Part B: blood osmotic pressure, blood osmotic pressure
Part C: concentrations.
Part D: passive
Part E: number, increases
Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH)
Part G: Aldosterone
Part H: over hydration,water intoxication.
Part I: low, high
Part J: swell, increased.
Solution
Part A: blood solutes, swell
Part B: blood osmotic pressure, blood osmotic pressure
Part C: concentrations.
Part D: passive
Part E: number, increases
Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH)
Part G: Aldosterone
Part H: over hydration,water intoxication.
Part I: low, high
Part J: swell, increased..
no of units producedno of units of component A requiredprice per.pdf
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16800
june
1250
14
17500
total direct labor cost in process department -April to june
48300
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16.
benzene is less reactive than toluene towardselec.pdf
benzene is less reactive than toluene towardselectrophilic substitution reactions
because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates
ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating
capacity is decreased because nitrogroup is strongly electron withdrawing and will with
drawelectrondensity.
Solution
benzene is less reactive than toluene towardselectrophilic substitution reactions
because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates
ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating
capacity is decreased because nitrogroup is strongly electron withdrawing and will with
drawelectrondensity..
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets
Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued
Income Taxes + Other Current Liabilites)
Non Operating Assets = Marketable securities current + Marketable securities non current +
Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets
2008:
Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202
Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682
Net non operating obligations = 11202 - 9682 = $1520
2007:
Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486
Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853
Net non operating obligations = 8486 - 8853 = -$367
Solution
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets
Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued
Income Taxes + Other Current Liabilites)
Non Operating Assets = Marketable securities current + Marketable securities non current +
Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets
2008:
Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202
Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682
Net non operating obligations = 11202 - 9682 = $1520
2007:
Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486
Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853
Net non operating obligations = 8486 - 8853 = -$367.
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
Mobile malware is on the rise as more people use mobile devices. Attackers create malware to target mobile devices in order to steal personal information like passwords, bank details, and credit card numbers from infected phones and tablets. Security experts advise mobile users to only install apps from official app stores, keep devices updated with the latest software and security patches, and use caution when clicking links or attachments in messages from unknown sources.
How to Build a Module in Odoo 17 Using the Scaffold Method
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
PCOS corelations and management through Ayurveda.
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
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Part A neutralizes acids - base Part B produces.pdf
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a
soapy feel - base Part D: turns litmus red - acid
Solution
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a
soapy feel - base Part D: turns litmus red - acid.
Lewis Acid Base Reaction .pdf
Lewis acid-base reactions involve the sharing of electron pairs between reactants. A Lewis acid is an electron pair acceptor that has an empty orbital, while a Lewis base is an electron pair donor that has a lone pair of electrons. When a Lewis acid and base interact, the base donates an electron pair to form a new bond with the empty orbital on the acid.
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 =
10 pH+pOH=14 pOh = 14-pH = 14-10=4
Solution
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 =
10 pH+pOH=14 pOh = 14-pH = 14-10=4.
It is a reaction for secondary amines .Option 2 i.pdf
It is a reaction for secondary amines .Option 2 is correct.
Solution
It is a reaction for secondary amines .Option 2 is correct..
fluorene is mobile phase and fluorenone is statio.pdf
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it
depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene
and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move
slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent
(pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind
of applies to chromatography wherein rather than dissolve just replace it with moves with, so like
moves with like. That being said the pentane will \"carry\" the fluorene through the alumina
slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that
being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity
(remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the
alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through
the alumina than would the fluorene, because there is no attraction between all these polar
compounds which will allow it to move faster, rather than a nonpolar and polar chemical having
an attration towards each other and thus moving more slowly.
Solution
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it
depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene
and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move
slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent
(pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind
of applies to chromatography wherein rather than dissolve just replace it with moves with, so like
moves with like. That being said the pentane will \"carry\" the fluorene through the alumina
slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that
being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity
(remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the
alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through
the alumina than would the fluorene, because there is no attraction between all these polar
compounds which will allow it to move faster, rather than a nonpolar and polar chemical having
an attration towards each other and thus moving more slowly..
equivalent mass is 27.9 . so Molecular massmole.pdf
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9
looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2
that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe )
Solution
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9
looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2
that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ).
There is no solutionSolutionThere is no solution.pdf
The document repeats the phrase "There is no solution" with no other context or information provided. It consists of three short sentences stating "There is no solution" with no explanation or details given about what problem is being referred to or discussed.
This is a case of allopatric speciation. In this case lice from gori.pdf
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two
different species. So this is case of geographical speciation where species become apart from
each other to an extent that they cannot interbreed.
Solution
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two
different species. So this is case of geographical speciation where species become apart from
each other to an extent that they cannot interbreed..
d. 8 The bonding electrons are the electrons taki.pdf
The document discusses the bonding electrons in a carbon dioxide (CO2) molecule. Carbon is centered between two hydrogen atoms and one oxygen atom. Carbon shares two electrons with each hydrogen atom, for a total of four electrons shared with hydrogen. Carbon also shares four electrons with the oxygen atom. Therefore, the total number of bonding electrons is the four shared with hydrogen plus the four shared with oxygen, which is eight.
The genetical changes of single locus are responsible for divergent .pdf
The genetical changes of single locus are responsible for divergent branches of Y chromosome
phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial
DNA sequences of Neanderthals and modern humans were compared and analyzed. The
Hominins species were found to be 30000 year old and were compared with Neanderthal
specimens.
The relationships between these populations have drawn with the help of the age of specimens
plotted against the average genetic distance. The Hominins type sequences are very rare when
compared with the modern European samples. The prehistoric Europeans had shown constant
number of differences when compared to today’s Europeans and Neanderthals.
The Neanderthals have shown the genetic continuity in modern human’s genealogy. The
Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to
the similar ages. The Neanderthals and early anatomical modern humans were thought to be
regional populations of the same evolving species connected by gene flow and both the archaic
and modern humans contribute in different proportions to the present human gene pool.
The transition and demographic replacement separated the population as Neanderthals and
modern humans are considered just as one population observed at different times. The multi
regional models predict that at least some level of genetic continuity can be observed from
archaic Neanderthal forms to the contemporary Hominins forms and up to today’s Europeans.
The results have not matched with the view that the Neanderthal were genetically related with
the anatomically modern ancestors of current Europeans or contributed to the present day human
gene pool.
Solution
The genetical changes of single locus are responsible for divergent branches of Y chromosome
phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial
DNA sequences of Neanderthals and modern humans were compared and analyzed. The
Hominins species were found to be 30000 year old and were compared with Neanderthal
specimens.
The relationships between these populations have drawn with the help of the age of specimens
plotted against the average genetic distance. The Hominins type sequences are very rare when
compared with the modern European samples. The prehistoric Europeans had shown constant
number of differences when compared to today’s Europeans and Neanderthals.
The Neanderthals have shown the genetic continuity in modern human’s genealogy. The
Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to
the similar ages. The Neanderthals and early anatomical modern humans were thought to be
regional populations of the same evolving species connected by gene flow and both the archaic
and modern humans contribute in different proportions to the present human gene pool.
The transition and demographic replacement separated the population as Neanderthals and
modern humans .
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
The corrects answers are:
Thr and Asn are polar amino acids.
Isoleucine has more than one stereocenter (chiral center)
The Leu side chain does not from hydrogen bonds with other amino acids
Solution
The corrects answers are:
Thr and Asn are polar amino acids.
Isoleucine has more than one stereocenter (chiral center)
The Leu side chain does not from hydrogen bonds with other amino acids.
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4
i.e Z4 = ( 0,1,2,3)
U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2
So, isomorphic Z4 and U(8) are not isomorphic.
Solution
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4
i.e Z4 = ( 0,1,2,3)
U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2
So, isomorphic Z4 and U(8) are not isomorphic..
selected palntsxerophyte verus mesophyte========================.pdf
selected palnts
xerophyte verus mesophyte
=========================================================
A xerophyte is a types of plant that has adjusted to get by in a domain with minimal fluid water,
for example, a forsake or an ice-or snow-canvassed locale in the Alps or the Arctic.
The morphology and physiology of xerophytes are differently adjusted to moderate water, and
usually likewise to store substantial amounts of water, amid dry periods. Different species might
be adjusted to survive long stretches of parching of their tissues, amid which their metabolic
action may viably close down. Plants with such morphological and physiological adjustments are
xeromorphic.
Xerophytic plants may have comparable shapes, structures, and structures and look
fundamentally the same as, regardless of the possibility that the plants are not firmly related,
through a procedure called concurrent development. For instance, a few types of desert flora
(individuals from the family Cactaceae), which advanced just in the Americas, may seem like
Euphorbias, which are dispersed around the world. A random types of caudiciforms, plants with
swollen bases that are utilized to store water, may likewise show such likenesses.
Xerophytic plants can have less general surface territory than different plants, so diminishing the
range that is presented to the air and lessening water misfortune by vanishing. Xerophytes can
have littler leaves or less branches than different plants. A case of leaf surface decrease are the
spines of a desert flora. A case of compaction and diminishment of spreading are the barrel
desert flora. Different xerophytes may have their leaves compacted at the base, as in a basal
rosette, which might be littler than the plant\'s blossom. This adjustment is displayed by some
Agave and Eriogonum species, which can be discovered developing close Death Valley.
A few xerophytes have minor hairs on their surface to give a wind break and decrease wind
current, along these lines diminishing the rate of dissipation. At the point when a plant surface is
secured with minor hairs, it is called tomentose.
In a still domain, the regions under the leaves/spines where transpiration is occurring structure a
little limited environment that is more soaked than typical with water vapor. In the event that this
is not overwhelmed by wind, the water vapor potential angle is diminished as is transpiration.
Subsequently, in a windier circumstance, this confinement is not held thus the angle stays high,
which helps the loss of water vapor. Spines trap a layer of dampness furthermore moderate air
development over tissues.
===================
Mesophytes are earthbound plants which are adjusted to neither an especially dry nor especially
wet environment. A case of a mesophytic living space would be a country calm glade, which
may contain goldenrod, clover, oxeye daisy, and Rosa multiflora.
Mesophytic plants have unbending, tough, openly expanded stems and stringy, all around create.
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
52000
Ruth & Associates
Statement of Changes in Stockholders’ Equity
For the Year Ended December 31, 2010
Beginning Common Stock
25000
Plus: Common Stock Issued
15000
Ending Common Stock
40000
Beginning Retained Earnings
18000
Plus: Net Income
52000
Less: Dividends
10000
Ending Retained Earnings
60000
Total Stockholders’ Equity
100000
Ruth & Associates
Balance Sheet
As of the December 31, 2010
Assets
Cash
52000
Accounts Receivable
26000
Supplies
3000
Prepaid Rent
5000
Land
63000
Total Assets
149000
Liabilities
Interest Payable
2000
Salaries Payable
7000
Unearned Revenue
8000
Notes Payable
32000
Total Liabilities
49000
Stockholders’ Equity
Common Stock
40000
Retained Earnings
60000
Total Stockholders’ Equity
100000
Total Liab. and Stockholders’ Equity
149000
Ruth & Associates
Statement of Cash Flows
For the Year Ended December 31, 2010
Cash Flow From Operating Activities
52000
Cash Flow From Investing Activities
-30000
Cash Flow From Financing Activities
20000
Net Change in Cash Plus:
42000
Beginning Cash Balance
10000
Ending Cash Balance
52000
Ending Cash Balance
52000
Net Change in Cash
42000
= Beginning Cash Balance
10000
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
52000
Ruth & Associates
Statement of Changes in Stockholders’ Equity
For the Year Ended December 31, 2010
Beginning Common Stock
25000
Plus: Common Stock Issued
15000
Ending Common Stock
40000
Beginning Retained Earnings
18000
Plus: Net Income
52000
Less: Dividends
10000
Ending Retained Earnings
60000
Total Stockholders’ Equity
100000
Ruth & Associates
Balance Sheet
As of the December 31, 2010
Assets
Cash
52000
Accounts Receivable
26000
Supplies
3000
Prepaid Rent
5000
Land
63000
Total Assets
149000
Liabilities
Interest Payable
2000
Salaries Payable
7000
Unearned Revenue
8000
Notes Payable
32000
Total Liabilities
49000
Stockholders’ Equity
Common Stock
40000
Retained Earnings
60000
Total Stockholders’ Equity
100000
Total Liab. and Stockholders’ Equity
149000
Ruth & Associates
Statement of Cash Flows
For the Year Ended December 31, 2010
Cash Flow From Operating Activities
52000
Cash Flow From Investing Activities
-30000
Cash Flow From Financing Activities
20000
Net Change in Cash Plus:
42000
Beginning Cash Balance
10000
Ending Cash Balance
52000
Ending Cash Balance
52000
Net Change in Cash
42000
= Beginning Cash Balance
10000
Solution
Ruth & Associates
Income Statement
For the Year Ended December 31, 2010
Revenue
Consulting Revenue
100000
Interest Revenue
4000
Total Revenue
104000
Expenses
Salary Expense
47000
Interest Expense
5000
Total Expenses
52000
Net Income
5200.
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296
P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720
Solution
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296
P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720.
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
Part A: blood solutes, swell
Part B: blood osmotic pressure, blood osmotic pressure
Part C: concentrations.
Part D: passive
Part E: number, increases
Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH)
Part G: Aldosterone
Part H: over hydration,water intoxication.
Part I: low, high
Part J: swell, increased.
Solution
Part A: blood solutes, swell
Part B: blood osmotic pressure, blood osmotic pressure
Part C: concentrations.
Part D: passive
Part E: number, increases
Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH)
Part G: Aldosterone
Part H: over hydration,water intoxication.
Part I: low, high
Part J: swell, increased..
no of units producedno of units of component A requiredprice per.pdf
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16800
june
1250
14
17500
total direct labor cost in process department -April to june
48300
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16.
benzene is less reactive than toluene towardselec.pdf
benzene is less reactive than toluene towardselectrophilic substitution reactions
because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates
ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating
capacity is decreased because nitrogroup is strongly electron withdrawing and will with
drawelectrondensity.
Solution
benzene is less reactive than toluene towardselectrophilic substitution reactions
because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates
ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating
capacity is decreased because nitrogroup is strongly electron withdrawing and will with
drawelectrondensity..
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets
Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued
Income Taxes + Other Current Liabilites)
Non Operating Assets = Marketable securities current + Marketable securities non current +
Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets
2008:
Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202
Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682
Net non operating obligations = 11202 - 9682 = $1520
2007:
Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486
Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853
Net non operating obligations = 8486 - 8853 = -$367
Solution
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets
Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued
Income Taxes + Other Current Liabilites)
Non Operating Assets = Marketable securities current + Marketable securities non current +
Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets
2008:
Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202
Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682
Net non operating obligations = 11202 - 9682 = $1520
2007:
Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486
Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853
Net non operating obligations = 8486 - 8853 = -$367.
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