Multiple ChoiceQuestion 1What features are associated with the l.pdf
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Multiple Choice Question 1 What features are associated with the leaf epidermis? A= guard cells B= trichomes C= stomata D= all of the above are correct Question 2 The main function of root hairs is protection against herbivores? True or False Solution 1.A all of the above are correct. The epidermis of the leaf contains stomata which are the openings through which exchange of gases takes place and the guard cells control the opening and closing of the stomata. Trichomes are hair like appendages which prevent transpiration, and discourage herbivores to feed on the plant. 2.False. The main function of the root hairs is to collect water and minerals from the surroundings.
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Do markets erode moral valuesDo markets erode moral values.pdf
Do markets erode moral values?
Do markets erode moral values?
Solution
Marekets affects moral values as their are many examples we can consider, for instance many
people take stand against child labor, exploitations done to the worforce, cruelty against animals
for meat etc. Therefore its a long lasting, debatable topic. It also depends on what kind of market
it is individual, bilateral or multilateral.
Many researches and studies shows that when there is only one party then that particular party
get more guilt while in bilateral market the guilt is shared so it diminishes and in multilateral
market the percentage of eroding moral values are much higher than that in bilateral and
individual. In markets, many people ignore their own ethics, morals, rules etc.
Concept Simulation 25.2 illustrates the concepts pertinent to this pr.pdf
Concept Simulation 25.2 illustrates the concepts pertinent to this problem. A 2.50-cm-high
object is situated 17.1 cm in front of a concave mirror that has a radius of curvature of 8.39 cm.
Calculate the location and the height of the image.
Solution
h = 2.50 cm, u = - 17.1 cm
f = R/2 = -8.39 / 2 = - 4.195 cm
1/v + 1/u = 1/f
v = [f * u] / [u - f] = [- 17.1 *- 4.195] / [- 17.1 + 4.195]
= - 5.56 cm [the left of itself ]
--------------------------------------------------
magnification = m = h\'/ h = - v/u
h\' / 2.50 = - [- 5.56 /- 17.1]
h\' = - 0.813 cm [image is inverted].
Based on what you have learned about the aquatic environment and the .pdf
Based on what you have learned about the aquatic environment and the life needs and
adaptations of planktonic and benthic organisms found in it, What changes do you expect in
abiotic conditions in or just above the sediments to those in the water column at the same site?
How might the species of organisms and their abundance differ between those habitats, and
might any differences be related to differences in abiotic conditions? What differences might
you expect in the water column at a deep site versus a shallow site?
Solution
The variations in the abiotic components in a n aquatic ecosytem between the sedimentary zone
and the water column basically lies in the difference in the functionality of the components of the
ecosystem. These variations result due to the activity of the biotic components also.
The sedimentation zones are analogous to the to soil of the terrestrial ecosystem and tthey are the
sources of substrate nutrients.aquatic sediments are derived from and comprise of natural ,
physical,chemical and biologicaal components related to their water sheds.
fast moving waters have coarse grained sediments while the quiscent or the still waters contain
fine grained sediments. Organic matter are derived from the decomposing of the plant and the
animal tissue and hence is rich in nutrients like the phosphates and nitrates . It is the zone where
maximum decompositional activity takes place.. The dissolved chemicals reflect the nature of the
organic and inorganic matter of the sediments.
While the column of the water at the same site may not be rich in the sediments but may have
floating sediments and the nutrients like the nitrates and phosphates may be at a lower
concentration when compared to the sedimental zone due to lack of the decomposing matter.
2. DIversity of organisms and their abundance :
The organisms that occur in the water column and the sedimental zone varies.This is mainly
because of the differences in their habitat and nutritional requirements.
The sedimental zone is rich in decomposing bacteria and other detritus organisms while the
water column may support planktons both phyto and zoo plankton. free floating algae, the
plankton diversity also varies according to their light requirement , oxygen requirement etc..
A) If A, then B. If B, then C. Therefore, if A, then C There are 350.pdf
A) If A, then B. If B, then C. Therefore, if A, then C There are 350 million Americans. Only
50,000 Americans have ever been to Siberia. Sarah is a randomly selected American. Therefore,
Sarah has never been to Siberia. Can you expalin your answer. 1. Deductive, valid 2. Inductive,
reasonable 3. Inductive, unreasonable
Solution.
Describe the therapeutic goal in treating acid-peptic disease.So.pdf
Describe the therapeutic goal in treating acid-peptic disease.
Solution
The therapeutic goal in treating acid-peptic disease include the following:
Relief from the symptoms such as pain that occurs in between meals in the middle or upper
portion of stomach , bloating, heart burn anf nausea.
Healing of the ulcer that has affected the mucosal part of esophagus, stomach and duodenum.
Prevention of reoccurrence of the ulcer.
Synthesis of drugs such as histamine H2-receptor antagonists.
Synthesis of drugs such as proton-pump inhibitors (PPIs) that bind to H+,K+-ATPase
irreversibly thereby blocking the final step of gastric acid secretion.
Synthesis of antibiotics against H. pylori which can cause peptic ulcers..
Compare structure and culture of two or more firms. Which would you .pdf
Compare structure and culture of two or more firms. Which would you prefer to work for?
no spam or plagiarism complete answer.... ..
Solution
S.No.
Particulars
Firm A (MNC in FMCG industry)
Firm B ( Domestic Manufacturer in FMCG industry)
1
Organization structure
Line Structure with functional Breakup and Project Structure. Both structure are interwoven with
each other. Here, organizational objective is at prime focus. More attention is given on team
work.
Line structure with functional breakup. People follow their regular job and functional interest is
given due consideration.
2
Flow of Information
Top-Down, Bottom-up and parallel
Top-Down
3
Culture
Strong organizational Culture. It binds people through shared assumption and knowledge. Core
is developed through Strong values.
Weak Organizational Culture. It is less binding in nature and people rely more on rule books.
4
Reliance on Written Rules
Lesser Reliance
Strong Reliance
5
Technology
Excessive use of Technology
Lesser use of Technology
6
Change management approach
Emergent change approach
Planned change approach
I would like to work for MNC in FMCG industry because it gives me delegation of authority and
encourages me to work as a team that is the necessary ground of any successful initiative.
Note:
Actual Name of the company cannot be disclosed due to violation of the right to use the name /
brand and related issues.
S.No.
Particulars
Firm A (MNC in FMCG industry)
Firm B ( Domestic Manufacturer in FMCG industry)
1
Organization structure
Line Structure with functional Breakup and Project Structure. Both structure are interwoven
with each other. Here, organizational objective is at prime focus. More attention is given on
team work.
Line structure with functional breakup. People follow their regular job and functional interest is
given due consideration.
2
Flow of Information
Top-Down, Bottom-up and parallel
Top-Down
3
Culture
Strong organizational Culture. It binds people through shared assumption and knowledge. Core
is developed through Strong values.
Weak Organizational Culture. It is less binding in nature and people rely more on rule books.
4
Reliance on Written Rules
Lesser Reliance
Strong Reliance
5
Technology
Excessive use of Technology
Lesser use of Technology
6
Change management approach
Emergent change approach
Planned change approach.
WRITE A program that displays the values in the list numbers in desc.pdf
WRITE A program that displays the values in the list numbers in descending order sorted by the
sum of their digits that odd numbers . sorted by sum of odd digits: [1169, 290, 865, 1243, 1208]
Solution
python program:
list1=[];
list2=[];
print \"Enter number of variables:\";
size=input();
for i in range(size):
print \"Enter variable\",i+1;
v=input();
list1.append(v);
for number in list1:
n=number;
m=0;
while n>0:
r=n%10;
if r%2==1:
m=m+r;
n=n/10;
list2.append(m);
print \"list before ordering\";
print list1;
for num1 in range(len(list2)):
for num2 in range(num1+1,len(list2)):
if list2[num1].
What is an oligopotent stem cell and what is an exampleSolution.pdf
What is an oligopotent stem cell and what is an example?
Solution
Oligopotent cell lines have the ability to transform in to quite a limited number of several other
types of cells,an example of which is a myeloid cells ,B cells,T cells,plasma cells or the cells
comprising the lymphoid system.An example of Oligopotent cells are myeloid cell can
differentiate in to any of the blood stem cells found in the lymphatic system.
Recommended
Do markets erode moral valuesDo markets erode moral values.pdf
Do markets erode moral values?
Do markets erode moral values?
Solution
Marekets affects moral values as their are many examples we can consider, for instance many
people take stand against child labor, exploitations done to the worforce, cruelty against animals
for meat etc. Therefore its a long lasting, debatable topic. It also depends on what kind of market
it is individual, bilateral or multilateral.
Many researches and studies shows that when there is only one party then that particular party
get more guilt while in bilateral market the guilt is shared so it diminishes and in multilateral
market the percentage of eroding moral values are much higher than that in bilateral and
individual. In markets, many people ignore their own ethics, morals, rules etc.
Concept Simulation 25.2 illustrates the concepts pertinent to this pr.pdf
Concept Simulation 25.2 illustrates the concepts pertinent to this problem. A 2.50-cm-high
object is situated 17.1 cm in front of a concave mirror that has a radius of curvature of 8.39 cm.
Calculate the location and the height of the image.
Solution
h = 2.50 cm, u = - 17.1 cm
f = R/2 = -8.39 / 2 = - 4.195 cm
1/v + 1/u = 1/f
v = [f * u] / [u - f] = [- 17.1 *- 4.195] / [- 17.1 + 4.195]
= - 5.56 cm [the left of itself ]
--------------------------------------------------
magnification = m = h\'/ h = - v/u
h\' / 2.50 = - [- 5.56 /- 17.1]
h\' = - 0.813 cm [image is inverted].
Based on what you have learned about the aquatic environment and the .pdf
Based on what you have learned about the aquatic environment and the life needs and
adaptations of planktonic and benthic organisms found in it, What changes do you expect in
abiotic conditions in or just above the sediments to those in the water column at the same site?
How might the species of organisms and their abundance differ between those habitats, and
might any differences be related to differences in abiotic conditions? What differences might
you expect in the water column at a deep site versus a shallow site?
Solution
The variations in the abiotic components in a n aquatic ecosytem between the sedimentary zone
and the water column basically lies in the difference in the functionality of the components of the
ecosystem. These variations result due to the activity of the biotic components also.
The sedimentation zones are analogous to the to soil of the terrestrial ecosystem and tthey are the
sources of substrate nutrients.aquatic sediments are derived from and comprise of natural ,
physical,chemical and biologicaal components related to their water sheds.
fast moving waters have coarse grained sediments while the quiscent or the still waters contain
fine grained sediments. Organic matter are derived from the decomposing of the plant and the
animal tissue and hence is rich in nutrients like the phosphates and nitrates . It is the zone where
maximum decompositional activity takes place.. The dissolved chemicals reflect the nature of the
organic and inorganic matter of the sediments.
While the column of the water at the same site may not be rich in the sediments but may have
floating sediments and the nutrients like the nitrates and phosphates may be at a lower
concentration when compared to the sedimental zone due to lack of the decomposing matter.
2. DIversity of organisms and their abundance :
The organisms that occur in the water column and the sedimental zone varies.This is mainly
because of the differences in their habitat and nutritional requirements.
The sedimental zone is rich in decomposing bacteria and other detritus organisms while the
water column may support planktons both phyto and zoo plankton. free floating algae, the
plankton diversity also varies according to their light requirement , oxygen requirement etc..
A) If A, then B. If B, then C. Therefore, if A, then C There are 350.pdf
A) If A, then B. If B, then C. Therefore, if A, then C There are 350 million Americans. Only
50,000 Americans have ever been to Siberia. Sarah is a randomly selected American. Therefore,
Sarah has never been to Siberia. Can you expalin your answer. 1. Deductive, valid 2. Inductive,
reasonable 3. Inductive, unreasonable
Solution.
Describe the therapeutic goal in treating acid-peptic disease.So.pdf
Describe the therapeutic goal in treating acid-peptic disease.
Solution
The therapeutic goal in treating acid-peptic disease include the following:
Relief from the symptoms such as pain that occurs in between meals in the middle or upper
portion of stomach , bloating, heart burn anf nausea.
Healing of the ulcer that has affected the mucosal part of esophagus, stomach and duodenum.
Prevention of reoccurrence of the ulcer.
Synthesis of drugs such as histamine H2-receptor antagonists.
Synthesis of drugs such as proton-pump inhibitors (PPIs) that bind to H+,K+-ATPase
irreversibly thereby blocking the final step of gastric acid secretion.
Synthesis of antibiotics against H. pylori which can cause peptic ulcers..
Compare structure and culture of two or more firms. Which would you .pdf
Compare structure and culture of two or more firms. Which would you prefer to work for?
no spam or plagiarism complete answer.... ..
Solution
S.No.
Particulars
Firm A (MNC in FMCG industry)
Firm B ( Domestic Manufacturer in FMCG industry)
1
Organization structure
Line Structure with functional Breakup and Project Structure. Both structure are interwoven with
each other. Here, organizational objective is at prime focus. More attention is given on team
work.
Line structure with functional breakup. People follow their regular job and functional interest is
given due consideration.
2
Flow of Information
Top-Down, Bottom-up and parallel
Top-Down
3
Culture
Strong organizational Culture. It binds people through shared assumption and knowledge. Core
is developed through Strong values.
Weak Organizational Culture. It is less binding in nature and people rely more on rule books.
4
Reliance on Written Rules
Lesser Reliance
Strong Reliance
5
Technology
Excessive use of Technology
Lesser use of Technology
6
Change management approach
Emergent change approach
Planned change approach
I would like to work for MNC in FMCG industry because it gives me delegation of authority and
encourages me to work as a team that is the necessary ground of any successful initiative.
Note:
Actual Name of the company cannot be disclosed due to violation of the right to use the name /
brand and related issues.
S.No.
Particulars
Firm A (MNC in FMCG industry)
Firm B ( Domestic Manufacturer in FMCG industry)
1
Organization structure
Line Structure with functional Breakup and Project Structure. Both structure are interwoven
with each other. Here, organizational objective is at prime focus. More attention is given on
team work.
Line structure with functional breakup. People follow their regular job and functional interest is
given due consideration.
2
Flow of Information
Top-Down, Bottom-up and parallel
Top-Down
3
Culture
Strong organizational Culture. It binds people through shared assumption and knowledge. Core
is developed through Strong values.
Weak Organizational Culture. It is less binding in nature and people rely more on rule books.
4
Reliance on Written Rules
Lesser Reliance
Strong Reliance
5
Technology
Excessive use of Technology
Lesser use of Technology
6
Change management approach
Emergent change approach
Planned change approach.
WRITE A program that displays the values in the list numbers in desc.pdf
WRITE A program that displays the values in the list numbers in descending order sorted by the
sum of their digits that odd numbers . sorted by sum of odd digits: [1169, 290, 865, 1243, 1208]
Solution
python program:
list1=[];
list2=[];
print \"Enter number of variables:\";
size=input();
for i in range(size):
print \"Enter variable\",i+1;
v=input();
list1.append(v);
for number in list1:
n=number;
m=0;
while n>0:
r=n%10;
if r%2==1:
m=m+r;
n=n/10;
list2.append(m);
print \"list before ordering\";
print list1;
for num1 in range(len(list2)):
for num2 in range(num1+1,len(list2)):
if list2[num1].
What is an oligopotent stem cell and what is an exampleSolution.pdf
What is an oligopotent stem cell and what is an example?
Solution
Oligopotent cell lines have the ability to transform in to quite a limited number of several other
types of cells,an example of which is a myeloid cells ,B cells,T cells,plasma cells or the cells
comprising the lymphoid system.An example of Oligopotent cells are myeloid cell can
differentiate in to any of the blood stem cells found in the lymphatic system.
What are 4 different types of CVI systemsSolutionThe four typ.pdf
What are 4 different types of CVI systems?
Solution
The four type of CVI i.e, Chemical VapourInfiltration system are;
1. Carbon-Carbon with Kerosene/ Methane as common precursor.
2. Carbon-Silicon Carbide with CH3SiCl3H2 as common precursor.
3. SiliconCarbide-Silicon Carbide with CH3SiCl3H2 as common precursor.
4. Alumina-Alumina with AlCl3CO2H2 as common precursor.
Why is RNA more prone than DNA to forming secondary structures.pdf
Why is RNA more prone than DNA to forming secondary structures?
Solution
DNA has limited catalytic ability because it lacks functional groups that can participate in
catalysis and has a regular structure that is not conducive to forming shapes required for
catalysis.RNA molecules can catalyze some reactions because they have exposed hydroxyl
functional groups and can fold into shapes that then can function in catalysis.The increased
stability of RNA duplexes allows for highly diverse secondary and tertiary structures..
What are the advantages and disadvantage of using the metric system o.pdf
What are the advantages and disadvantage of using the metric system of measurements?
Solution
Answer:
Advantages:
(1) Its based on decimal system instead of fractions which makes it a simpler system.
(2) Has one unit for each quantity which eliminates the need of conversion.
(3) It is universally accepted system.
Disadvantages:
(1) Difficult to make one acquainted with the metric system who is already using imperial units.
(2) Difficult to use with fractions..
Which of the following model organisms would be the best choice if y.pdf
Which of the following model organisms would be the best choice if you wanted to to and in
vivo study of early animal cellular division?
Xenopus
Mouse
E. coli
Arabidopsis
Drosophila
Xenopus
Mouse
E. coli
Arabidopsis
Drosophila
Solution
Model organism is an organism selected for intensive scientific study based on features that
make it easy to work with in the hope that findings will apply to other species .
The correct answer is Arabidopsis. The aerial parts of the plants are generated by group of
dividing cells called shoot apical meristem. To analyse, cell behaviour in these structures, we
developed a technique to visualise living shoot apical meristems using confocal microscope. This
method combined with fluorescent marker lines and vital stains, allows us to follow dynamics of
cell proliferation and division..
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
What happens to jointly held stock between two siblings when one .pdf
What happens to jointly held stock between two siblings when one dies. How should this
property be reported
What happens to jointly held stock between two siblings when one dies. How should this
property be reported
Solution
All of the stocked are transferred in the name of survived sibling , after verification of death
certificate.
To write a program that implements the following C++ concepts 1. Dat.pdf
To write a program that implements the following C++ concepts 1. Data Encapsulation 2.
Instantiate classes 3. Composition Class 4. Aggregation Class 5. Dynamic Memory 6. File
Stream Make a program that reads a file and can generate reports. Each file will have phone calls
records (see section 4). The program will process the data gathering all minutes and total amount
for each report. The program should have a menu (see section 1): 1. New Report – this option
will ask a file name (TE??????.tel). This option will instantiate an object of Report class, loading
the whole file into memory. 2. Delete a report – this option will display all reports available in
memory and ask which one you would like to delete. 3. Display a report – this option will
display all reports available in memory and ask you which one you would like to generate a
report. 4. Exit – Program ends. For option 1, you do not know how many reports are, meaning
that you will create dynamically the reports. For long distance calls the formula will be (minutes
call x $1.50). For local phone calls the formula will be (minutes call x $0.25)
--------------------------------------------------------------------------------------------------------------------
----------------------------
//Main.cpp
*#define between(x,y,z) ((x>=y && x<=z)? true:false)
#include
#include
#include
#include \"Data.h\"
#include \"Report.h\"
#include \"Menu.h\"
using namespace std;
//uses class Report to create reports
//uses class Menu to display a menu to choose
int main() {
int idx = 0;
int count = idx;
Menu menu(\"Reports\");
Data temp;
string fileName;
Report* phoneCompany;
Report* newReport(Report*, int&);
Report* deleteReport(Report*, int&, int);
int displayReport(Report*, int);
bool fileNameExists(string);
menu.addMenuItem(\"New Report\");
menu.addMenuItem(\"Delete a Report\");
menu.addMenuItem(\"Display a Report\");
ofstream openFile;
phoneCompany = NULL;
phoneCompany = new Report[count];
do {
cout << menu;
cin >> menu;
switch (menu.getChoice()) {
case \'1\':
cout << \"\ \\tEnter report name (TE??????): \";
cin >> fileName;
openFile.open(fileName + \".tel\");
if (openFile.good()) {
cout << \"\ \\tReport succesfully created.\ \";
phoneCompany[count].setFileName(fileName);
count++;
}
else {
cout << \"\ \\terror: Could not create report.\ \";
}
break;
case \'2\':
if (count != 0) {
do {
system(\"cls\");
cout << \"\ \ \\tDelete a Report:\ \ \";
idx = displayReport(phoneCompany, count);
if (between(idx, 0, count - 1)) {
phoneCompany = deleteReport(phoneCompany, count, idx);
idx = 0;
}
else
cout << \"\ \\terror: Report could not be deleted.\ \";
} while (idx != count);
}
else {
cout << \"\ \ \\tNo reports to delete.\ \ \";
system(\"pause\");
}
break;
case \'3\':
if (count != 0) {
do {
system(\"cls\");
cout << \"Display a Report:\ \ \";
idx = displayReport(phoneCompany, count);
system(\"cls\");
cout << \"\ \\t\" << phoneCompany[idx].getFileName();
cout << \"\ \ \\tWould you like to edit the report?\";
cout << \.
The HER incentive programs incentive payments to eligible hospitals, .pdf
The HER incentive programs incentive payments to eligible hospitals, and CAHs: to adopt,
implement, upgrade, or demonstrate meaningful use of CEHRT to donate technology to
understand and rural areas to participate in clinical trials Whish of the following entities is
charged with managing the federal HER incentive programs? CMS ONCHIT congress Which
of the following organizations is the principal federal entity charged with coordination of
nationwide efforts to implement and use the most advanced HIT and the electronic of health
information? CMD AHRQ ONCHIT The law that ended preexisting condition exclusions for
children (health plans can no longer limit or deny benefits to children younger than 19 because
of a preexisting condition)is: HIPAA HITECH PPACA The CLIA of 1988: eliminates the
exception under the HIPAA privacy rule to an individual\'s right to access his or her protected
health information when it id held by a CLIA-certified or CLIA-exempt laboratory. Created the
exception under the security rule to an individual\'s right to access his or her protected health
information when it id held by a CLIA-certified or CLIA-exempt laboratory. determined that
patients may not access their protected health information when it is held by a CLIA-certified or
CLIA-exempt laboratory. Project HealthDesign was a groundbreaking national program of
Robert wood Johnson foundation\'s pioneer portfolio, designed to spark innovation in: Personal
health technology EHR robotics
Solution
8) CMS
7) a
9)c. ONCHIT
10)c. PPAC
11)a.e eliminated the exception
12)a. Personal health technology.
The comet Shoemaker-Levy 9 broke up into a number of pieces called t.pdf
The comet Shoemaker-Levy 9 broke up into a number of pieces called the ‘string of pearls’ as it
approached close to Jupiter. The breakup happened because of the
__________________________________ of Jupiter. (Fill in the blank)
Solution
Differential pull of the planet\'s enormous gravitational force on the near and far sides of the
comet fragmented it into 21 or more large pieces and an enormous amount of smaller debris.
The existence of universal values is still a matter of debate. But e.pdf
The existence of universal values is still a matter of debate. But even in situations where it is no
clear common ground, Donaldson’s ethical algorithm can help by allowing one to consider if the
economic development levels are different between two countries, whether or not the act would
be considered moral if the two countries have equivalent economies. TRUE or FALSE
Solution
TRUE.
The c program will implement the Caesar Cipher. Your program shou.pdf
\" The c program will implement the Caesar Cipher. Your program should take a command line
argument, which will be the offset, while the input through standard in will be the part that will
be rotated(encrypted/decrypted). To simplify this assignment you will only need to rotate
characters that are alpha characters (letters). No other characters should be rotated. You can also
assume that the input will only be a single line long. Your script will take a single command line
argument, which will be the rotation argument to the Caesar c program you wrote above. Your
script should then call the \"Program2_2.\" you wrote each of the .enc files. Your script should
output a .dec file for each .enc file that exists. \"
Solution
#include
#include
#define MAXSIZE 1024
void encrypt(char*);
void decrypt(char*);
int menu();
int
main(void)
{
char c,
choice[2],
s[MAXSIZE];
while(1)
{
menu();
gets(choice);
if((choice[0]==\'e\')||(choice[0]==\'E\'))
{
puts(\"Input text to encrypt->\");
gets(s);
encrypt(s);
}
else if((choice[0]==\'d\')||(choice[0]==\'D\'))
{
puts(\"Input text to decrypt->\");
gets(s);
decrypt(s);
}
else
break;
}
return 0;
}
void encrypt(char*str)
{
int n=0;
char *p=str,
q[MAXSIZE];
while(*p)
{
if(islower(*p))
{
if((*p>=\'a\')&&(*p<\'x\'))
q[n]=toupper(*p + (char)3);
else if(*p==\'x\')
q[n]=\'A\';
else if(*p==\'y\')
q[n]=\'B\';
else
q[n]=\'C\';
}
else
{
q[n]=*p;
}
n++; p++;
}
q[n++]=\'\\0\';
puts(q);
}
void decrypt(char*str)
{
int n=0;
char *p=str,
q[MAXSIZE];
while(*p)
{
if(isupper(*p))
{
if((*p>=\'D\')&&(*p<=\'Z\'))
q[n]=tolower(*p - (char)3);
else if(*p==\'A\')
q[n]=\'x\';
else if(*p==\'B\')
q[n]=\'y\';
else
q[n]=\'z\';
}
else
{
q[n]=*p;
}
n++; p++;
}
q[n++]=\'\\0\';
puts(q);
}
int menu()
{
puts(\"To encrypt, input e or E\ \");
puts(\"To decrypt, input d or D\ \");
puts(\"To exit, input any other letter\ \");
puts(\"Your choice:->\ \");
return 0;
}.
Amino acids are acids because they contain which functional group S.pdf
Amino acids are acids because they contain which functional group? Select one: a. carboxyl b.
sulfhydryl c. carbonyl d. methyl e. amino
Solution
Amino acids possess acidic nature due to the presence of Carboxylic group I.e. CooH group.
The CooH group has the capability to donate H+ or proton, so it act as an acid.
According to Branstead lowry theory, compound that release H+ ion is called Acid. And due to
the presence of CooH group, Amino acid act like acid.
Suppose that theta is an acute angle of a right triangle and tan (the.pdf
Suppose that theta is an acute angle of a right triangle and tan (theta) = 4 Squareroot 2/7. Find
sin(theta) and cos(theta).
Solution
given tan =(42)/7
tan =oppositeside/adjacent side =(42)/7
oppositeside=42 ,adjacent side=7
hypotenuse2=opposite side2+adjacent side2
hypotenuse2=(42)2+72
hypotenuse2=81
hypotenuse=9
sin =oppositeside/hypotenuse=(42)/9
cos =adjacentside/hypotenuse=7/9.
Supplements of the water-soluble vitamins ______ and _______ are mos.pdf
Supplements of the water-soluble vitamins ______ and _______ are most likely to cause toxicity
symptoms. A.thiamin; riboflavin B.vitamin B-12; pantothenic acid C.vitamin B-6; vitamin C
D.biotin; vitamin B-12
Solution
Of the different water soluble vitamins, four i.e. folate, niacin, vitamin B6 and vitamin C have
shown the potential for toxicity. Hence the correct answer is option C. – vitamin B6 and vitamin
C..
Legionellae, the genus of bacteria that causes Legionnaires disease.pdf
Legionellae, the genus of bacteria that causes Legionnaire\'s disease, are common in __. Hot
water supplies water supplies Rivers and lakes above Drinking All of the What level of biosafety
is needed for laboratory activities involving TB cultures at American Thoracic Society levels II
and III?
Solution
1. The genus Legionellae is a pathogenic group of gram-negative bacteria. It is common in many
environment including soil and aquatic systems. The correct answer is All the above.
2. Biosafety level 3..
In your own words, what is thw application of columbia CNA with 5 s.pdf
In your own words, what is thw application of columbia CNA with 5% sheep blood agar?
Solution
Columbia CNA with 5% Sheep blood agar is a kind of selective as well as differential
meadium. Mainly this is used for culture if Gram positive bacteria. In this media, differential
growth of Entericocci, Staphylococci , hemolytic streptococci occurs while some clinical isolates
like Klebsiella, Pseudomonas and Proteus are inhibited. Therefire this media is also used as
differential medium. The differential selection occurs on the basis of hemolytic reactions..
List two properties of water and discuss how two substances in our b.pdf
List two properties of water and discuss how two substances in our body dissolve based on this.
After listing properties, could you give 2-3 sentences per example please?
Solution
Water is a universal solvent and it has hydrogen bonding associated with highest heat capacity
and is referred as “specific heat capacity”. This specific heat is predominantly due to
intermolecular and intermolecular hydrogen bonding. In nature on land when heat is absorbed
result in breaking of hydrogen bonds specifically followed by production of heat. This specific
heat capacity of water is going to make the blood and lymph fluid to dissolve electrolytes within
the cellular environment to maintain homeostasis by keeping blood pH -buffer as 7.4
Water is a universal solvent and it has both polar because it has both positive charge (H+ or
H3O+) and negative charge (OH-). This polarity enables water to make substances to dissolve in
it as charged species. This polar nature of water is going to enable \"bicarbonate substance to
dissolve in blood\" in buffering system of the cellular respiration, finally involves an acid & base
homeostatic mechanism further involving the equilibrium balance of carbonic acid(H2CO3),
bicarbonate ion (HCO3-), and carbon dioxide (CO2) in order to maintain isotonic pH (7.4). On
the other hand, water molecules possess a higher attractive power to each other so that these
forces are referred as cohesive forces responsible for “surface tension” phenomenon” and these
attractions are predominantly due to hydrogen bonding.
1). Due of the presence of hydrogen bonds and strong intermolecular forces, water has high
specific heat index. Specific heat capacity of water is defined as; the amount of heat (energy in
joules) required for raise of temperature of water per unit mass by 10C.
Our body temperature is maintained higher than the surrounding temperature. The heat of
vaporization of the water is higher (40.65 kJ/mol), this is due to the hydrogen bonds that connect
water molecules. So, due to the temperature difference, water escapes out and carry the body
heat (the specific heat capacity of the water is 5 times more than the sand). So, during the process
of vaporization of water, water releases heat into the environment and cool the body.
2). Because of the hydrogen bonding, water exhibits high surface tension. So, water molecules
on the surface are more strongly attracted to each other than to the air above. This is necessary
for the functioning of alveoli. The layer of moisture over the alveoli is necessary for the gaseous
exchange.
3). Due to the polar nature of water, water can dissolve inorganic salts that dissociate into ions in
the presence of water (like dissolves like). So, the blood can carry many of the water soluble
substances throughout the body. Water cannot dissolve nonpolar substances; they are dissolved
in non-polar solvents.
4). Plasma contains approximately 91% of water; this maintains the viscosity of the blood..
Interview NotesDarcy is 45 years old, single, and a U.S. citizen w.pdf
Interview Notes
Darcy is 45 years old, single, and a U.S. citizen with a valid Social Security number. She had
$45,000 in wages.
During the interview with Darcy, you determine the following facts:
Darcy’s son Chris, age 21, is unmarried and was a full-time student working on a degree in
accounting during 2017.
Chris’ income was $8,500 in wages, which he used to pay his tuition. He did not provide more
than half his own support.
Chris lived on campus during the school year, but came home on breaks and for the summer.
Chris is in his third year of college.
Chris has never had a felony drug conviction.
Chris is a U.S. citizen with a valid Social Security number.
Questions:
1. Who can claim Chris\' personal or dependency exemption?
A. Chris can claim his personal exemption because he had earned income.
B. Chris can claim his personal exemption if Darcy does not to claim him.
C. Chris does not have a filing requirement so he cannot claim a personal
exemption.
D. Chris must file a tax return and claim zero personal exemptions because Darcy can
claim him as a dependent on her tax return.
2. Darcy can claim the expenses Chris paid as qualifying expenses for the American opportunity
credit if Darcy claims Chris as a dependent on her return.
True
False
Solution
Answer 1. To be dependent the following conditions should be met:
*Chris is independent. because Chris lived on campus during the school year.
A. Chris can claim his personal exemption because he had earned
Answer 2. True
Only qualifying educational expenses can be claimed on income taxes anyway.
if Darcy claims Chris as a dependent on her return..
Ligands bind reversibly to their receptors. Select the statements th.pdf
Ligands bind reversibly to their receptors. Select the statements that describe what would happen
if ligands bound irreversibly with their receptors.
A) Pathway will be permanently active.
B)The signal cannot be transduced in the cell.
C)The cell cannot receive new signals.
D)Cell metabolism cannot occur.
Solution
B) The signal cannot be transduced in the cell
They work as antagonists or otherwise called as blockers. They have affinity but they do not
have the efficacy for cognate receptors and hence the binding disrupts the interaction and inhibit
the function of the agonist. Therefore, the signal remains untranslated into the cell and signal
transduction does not occur..
In the rest of this lab, you will help build such a list class. The .pdf
In the rest of this lab, you will help build such a list class. The class definition is stored in
intlist.h - please study it to know its parts:
The public declarations show what a client program can do with an IntList object. Notice,
however, that the list node structure definition is in the private area. Clients can\'t directly use or
even refer to such nodes, but all that clients should care about are the values stored in these
nodes anyway! Also notice that every list object will have one node pointer, to point at the first
node in the list or at 0 if the list is empty.
Now look at intlist.cpp. All but sum, contains, max, average and insertFirst are already done.
You will implement those five methods in Step 4. The other parts are implemented at the bottom
of the file - do not change what is done.
An application file would create IntList objects and use them to solve problems. For this lab, our
application is just a testing program - testlist.cpp - and you should look at that file next. Two
IntList objects are created; then numbers read from the command line are appended to one of the
lists; and finally the methods are tested for each list.
You can use this Makefile to help you complile the program. If you store all four of these files in
the same directory, then you just type \"make\" (without the quotes) to compile them and create
the executable program.
Step 4: Implement five linked list functions
First: switch roles between pilot and navigator if you did not already do that. Always stay logged
into the original pilot\'s account.
You should be able to run the program now (assuming you compiled it in Step 3) - it requires
you to enter a starter list of integers on the command line:
Next time we\'ll run it properly. The contains method is checked for three values that ought to be
in the list, plus one value that should not (the sum). Here is a sample run with initial values of 5,
7, 9 and 11:
See that append, print and count all work. But the others need to be fixed.
Use an editor (e.g., emacs) to make the following changes to intlist.cpp - do not change any of
the other files.
Fix the comment at the top to show your name(s) and the date.
Implement the sum method. See the count method for guidance.
Save, and then test your sum implementation - compile and execute testlist again. Verify sum is
working before going on.
Then implement the contains method, save again and test again.
Continue with the other three functions in the same way: implement and test one at time.
Here are correct results for the same sample data as above:
FILES:
Solution
#include \"IntList.h\"
#include
using std::cout;
// return sum of values in list
int IntList::sum() const {
int result = 0;
Node *n = first;
while (n) {
result+= n->info;
n = n->next;
}
return result;
}
// returns true if value is in the list; false if not
bool IntList::contains(int value) const {
Node *n = first;
while (n) {
if(n->info == value) return true;
n = n->next;
}
return false; // REPLACE TH.
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Normal Labour/ Stages of Labour/ Mechanism of Labour
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
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What are 4 different types of CVI systemsSolutionThe four typ.pdf
What are 4 different types of CVI systems?
Solution
The four type of CVI i.e, Chemical VapourInfiltration system are;
1. Carbon-Carbon with Kerosene/ Methane as common precursor.
2. Carbon-Silicon Carbide with CH3SiCl3H2 as common precursor.
3. SiliconCarbide-Silicon Carbide with CH3SiCl3H2 as common precursor.
4. Alumina-Alumina with AlCl3CO2H2 as common precursor.
Why is RNA more prone than DNA to forming secondary structures.pdf
Why is RNA more prone than DNA to forming secondary structures?
Solution
DNA has limited catalytic ability because it lacks functional groups that can participate in
catalysis and has a regular structure that is not conducive to forming shapes required for
catalysis.RNA molecules can catalyze some reactions because they have exposed hydroxyl
functional groups and can fold into shapes that then can function in catalysis.The increased
stability of RNA duplexes allows for highly diverse secondary and tertiary structures..
What are the advantages and disadvantage of using the metric system o.pdf
What are the advantages and disadvantage of using the metric system of measurements?
Solution
Answer:
Advantages:
(1) Its based on decimal system instead of fractions which makes it a simpler system.
(2) Has one unit for each quantity which eliminates the need of conversion.
(3) It is universally accepted system.
Disadvantages:
(1) Difficult to make one acquainted with the metric system who is already using imperial units.
(2) Difficult to use with fractions..
Which of the following model organisms would be the best choice if y.pdf
Which of the following model organisms would be the best choice if you wanted to to and in
vivo study of early animal cellular division?
Xenopus
Mouse
E. coli
Arabidopsis
Drosophila
Xenopus
Mouse
E. coli
Arabidopsis
Drosophila
Solution
Model organism is an organism selected for intensive scientific study based on features that
make it easy to work with in the hope that findings will apply to other species .
The correct answer is Arabidopsis. The aerial parts of the plants are generated by group of
dividing cells called shoot apical meristem. To analyse, cell behaviour in these structures, we
developed a technique to visualise living shoot apical meristems using confocal microscope. This
method combined with fluorescent marker lines and vital stains, allows us to follow dynamics of
cell proliferation and division..
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
What happens to jointly held stock between two siblings when one .pdf
What happens to jointly held stock between two siblings when one dies. How should this
property be reported
What happens to jointly held stock between two siblings when one dies. How should this
property be reported
Solution
All of the stocked are transferred in the name of survived sibling , after verification of death
certificate.
To write a program that implements the following C++ concepts 1. Dat.pdf
To write a program that implements the following C++ concepts 1. Data Encapsulation 2.
Instantiate classes 3. Composition Class 4. Aggregation Class 5. Dynamic Memory 6. File
Stream Make a program that reads a file and can generate reports. Each file will have phone calls
records (see section 4). The program will process the data gathering all minutes and total amount
for each report. The program should have a menu (see section 1): 1. New Report – this option
will ask a file name (TE??????.tel). This option will instantiate an object of Report class, loading
the whole file into memory. 2. Delete a report – this option will display all reports available in
memory and ask which one you would like to delete. 3. Display a report – this option will
display all reports available in memory and ask you which one you would like to generate a
report. 4. Exit – Program ends. For option 1, you do not know how many reports are, meaning
that you will create dynamically the reports. For long distance calls the formula will be (minutes
call x $1.50). For local phone calls the formula will be (minutes call x $0.25)
--------------------------------------------------------------------------------------------------------------------
----------------------------
//Main.cpp
*#define between(x,y,z) ((x>=y && x<=z)? true:false)
#include
#include
#include
#include \"Data.h\"
#include \"Report.h\"
#include \"Menu.h\"
using namespace std;
//uses class Report to create reports
//uses class Menu to display a menu to choose
int main() {
int idx = 0;
int count = idx;
Menu menu(\"Reports\");
Data temp;
string fileName;
Report* phoneCompany;
Report* newReport(Report*, int&);
Report* deleteReport(Report*, int&, int);
int displayReport(Report*, int);
bool fileNameExists(string);
menu.addMenuItem(\"New Report\");
menu.addMenuItem(\"Delete a Report\");
menu.addMenuItem(\"Display a Report\");
ofstream openFile;
phoneCompany = NULL;
phoneCompany = new Report[count];
do {
cout << menu;
cin >> menu;
switch (menu.getChoice()) {
case \'1\':
cout << \"\ \\tEnter report name (TE??????): \";
cin >> fileName;
openFile.open(fileName + \".tel\");
if (openFile.good()) {
cout << \"\ \\tReport succesfully created.\ \";
phoneCompany[count].setFileName(fileName);
count++;
}
else {
cout << \"\ \\terror: Could not create report.\ \";
}
break;
case \'2\':
if (count != 0) {
do {
system(\"cls\");
cout << \"\ \ \\tDelete a Report:\ \ \";
idx = displayReport(phoneCompany, count);
if (between(idx, 0, count - 1)) {
phoneCompany = deleteReport(phoneCompany, count, idx);
idx = 0;
}
else
cout << \"\ \\terror: Report could not be deleted.\ \";
} while (idx != count);
}
else {
cout << \"\ \ \\tNo reports to delete.\ \ \";
system(\"pause\");
}
break;
case \'3\':
if (count != 0) {
do {
system(\"cls\");
cout << \"Display a Report:\ \ \";
idx = displayReport(phoneCompany, count);
system(\"cls\");
cout << \"\ \\t\" << phoneCompany[idx].getFileName();
cout << \"\ \ \\tWould you like to edit the report?\";
cout << \.
The HER incentive programs incentive payments to eligible hospitals, .pdf
The HER incentive programs incentive payments to eligible hospitals, and CAHs: to adopt,
implement, upgrade, or demonstrate meaningful use of CEHRT to donate technology to
understand and rural areas to participate in clinical trials Whish of the following entities is
charged with managing the federal HER incentive programs? CMS ONCHIT congress Which
of the following organizations is the principal federal entity charged with coordination of
nationwide efforts to implement and use the most advanced HIT and the electronic of health
information? CMD AHRQ ONCHIT The law that ended preexisting condition exclusions for
children (health plans can no longer limit or deny benefits to children younger than 19 because
of a preexisting condition)is: HIPAA HITECH PPACA The CLIA of 1988: eliminates the
exception under the HIPAA privacy rule to an individual\'s right to access his or her protected
health information when it id held by a CLIA-certified or CLIA-exempt laboratory. Created the
exception under the security rule to an individual\'s right to access his or her protected health
information when it id held by a CLIA-certified or CLIA-exempt laboratory. determined that
patients may not access their protected health information when it is held by a CLIA-certified or
CLIA-exempt laboratory. Project HealthDesign was a groundbreaking national program of
Robert wood Johnson foundation\'s pioneer portfolio, designed to spark innovation in: Personal
health technology EHR robotics
Solution
8) CMS
7) a
9)c. ONCHIT
10)c. PPAC
11)a.e eliminated the exception
12)a. Personal health technology.
The comet Shoemaker-Levy 9 broke up into a number of pieces called t.pdf
The comet Shoemaker-Levy 9 broke up into a number of pieces called the ‘string of pearls’ as it
approached close to Jupiter. The breakup happened because of the
__________________________________ of Jupiter. (Fill in the blank)
Solution
Differential pull of the planet\'s enormous gravitational force on the near and far sides of the
comet fragmented it into 21 or more large pieces and an enormous amount of smaller debris.
The existence of universal values is still a matter of debate. But e.pdf
The existence of universal values is still a matter of debate. But even in situations where it is no
clear common ground, Donaldson’s ethical algorithm can help by allowing one to consider if the
economic development levels are different between two countries, whether or not the act would
be considered moral if the two countries have equivalent economies. TRUE or FALSE
Solution
TRUE.
The c program will implement the Caesar Cipher. Your program shou.pdf
\" The c program will implement the Caesar Cipher. Your program should take a command line
argument, which will be the offset, while the input through standard in will be the part that will
be rotated(encrypted/decrypted). To simplify this assignment you will only need to rotate
characters that are alpha characters (letters). No other characters should be rotated. You can also
assume that the input will only be a single line long. Your script will take a single command line
argument, which will be the rotation argument to the Caesar c program you wrote above. Your
script should then call the \"Program2_2.\" you wrote each of the .enc files. Your script should
output a .dec file for each .enc file that exists. \"
Solution
#include
#include
#define MAXSIZE 1024
void encrypt(char*);
void decrypt(char*);
int menu();
int
main(void)
{
char c,
choice[2],
s[MAXSIZE];
while(1)
{
menu();
gets(choice);
if((choice[0]==\'e\')||(choice[0]==\'E\'))
{
puts(\"Input text to encrypt->\");
gets(s);
encrypt(s);
}
else if((choice[0]==\'d\')||(choice[0]==\'D\'))
{
puts(\"Input text to decrypt->\");
gets(s);
decrypt(s);
}
else
break;
}
return 0;
}
void encrypt(char*str)
{
int n=0;
char *p=str,
q[MAXSIZE];
while(*p)
{
if(islower(*p))
{
if((*p>=\'a\')&&(*p<\'x\'))
q[n]=toupper(*p + (char)3);
else if(*p==\'x\')
q[n]=\'A\';
else if(*p==\'y\')
q[n]=\'B\';
else
q[n]=\'C\';
}
else
{
q[n]=*p;
}
n++; p++;
}
q[n++]=\'\\0\';
puts(q);
}
void decrypt(char*str)
{
int n=0;
char *p=str,
q[MAXSIZE];
while(*p)
{
if(isupper(*p))
{
if((*p>=\'D\')&&(*p<=\'Z\'))
q[n]=tolower(*p - (char)3);
else if(*p==\'A\')
q[n]=\'x\';
else if(*p==\'B\')
q[n]=\'y\';
else
q[n]=\'z\';
}
else
{
q[n]=*p;
}
n++; p++;
}
q[n++]=\'\\0\';
puts(q);
}
int menu()
{
puts(\"To encrypt, input e or E\ \");
puts(\"To decrypt, input d or D\ \");
puts(\"To exit, input any other letter\ \");
puts(\"Your choice:->\ \");
return 0;
}.
Amino acids are acids because they contain which functional group S.pdf
Amino acids are acids because they contain which functional group? Select one: a. carboxyl b.
sulfhydryl c. carbonyl d. methyl e. amino
Solution
Amino acids possess acidic nature due to the presence of Carboxylic group I.e. CooH group.
The CooH group has the capability to donate H+ or proton, so it act as an acid.
According to Branstead lowry theory, compound that release H+ ion is called Acid. And due to
the presence of CooH group, Amino acid act like acid.
Suppose that theta is an acute angle of a right triangle and tan (the.pdf
Suppose that theta is an acute angle of a right triangle and tan (theta) = 4 Squareroot 2/7. Find
sin(theta) and cos(theta).
Solution
given tan =(42)/7
tan =oppositeside/adjacent side =(42)/7
oppositeside=42 ,adjacent side=7
hypotenuse2=opposite side2+adjacent side2
hypotenuse2=(42)2+72
hypotenuse2=81
hypotenuse=9
sin =oppositeside/hypotenuse=(42)/9
cos =adjacentside/hypotenuse=7/9.
Supplements of the water-soluble vitamins ______ and _______ are mos.pdf
Supplements of the water-soluble vitamins ______ and _______ are most likely to cause toxicity
symptoms. A.thiamin; riboflavin B.vitamin B-12; pantothenic acid C.vitamin B-6; vitamin C
D.biotin; vitamin B-12
Solution
Of the different water soluble vitamins, four i.e. folate, niacin, vitamin B6 and vitamin C have
shown the potential for toxicity. Hence the correct answer is option C. – vitamin B6 and vitamin
C..
Legionellae, the genus of bacteria that causes Legionnaires disease.pdf
Legionellae, the genus of bacteria that causes Legionnaire\'s disease, are common in __. Hot
water supplies water supplies Rivers and lakes above Drinking All of the What level of biosafety
is needed for laboratory activities involving TB cultures at American Thoracic Society levels II
and III?
Solution
1. The genus Legionellae is a pathogenic group of gram-negative bacteria. It is common in many
environment including soil and aquatic systems. The correct answer is All the above.
2. Biosafety level 3..
In your own words, what is thw application of columbia CNA with 5 s.pdf
In your own words, what is thw application of columbia CNA with 5% sheep blood agar?
Solution
Columbia CNA with 5% Sheep blood agar is a kind of selective as well as differential
meadium. Mainly this is used for culture if Gram positive bacteria. In this media, differential
growth of Entericocci, Staphylococci , hemolytic streptococci occurs while some clinical isolates
like Klebsiella, Pseudomonas and Proteus are inhibited. Therefire this media is also used as
differential medium. The differential selection occurs on the basis of hemolytic reactions..
List two properties of water and discuss how two substances in our b.pdf
List two properties of water and discuss how two substances in our body dissolve based on this.
After listing properties, could you give 2-3 sentences per example please?
Solution
Water is a universal solvent and it has hydrogen bonding associated with highest heat capacity
and is referred as “specific heat capacity”. This specific heat is predominantly due to
intermolecular and intermolecular hydrogen bonding. In nature on land when heat is absorbed
result in breaking of hydrogen bonds specifically followed by production of heat. This specific
heat capacity of water is going to make the blood and lymph fluid to dissolve electrolytes within
the cellular environment to maintain homeostasis by keeping blood pH -buffer as 7.4
Water is a universal solvent and it has both polar because it has both positive charge (H+ or
H3O+) and negative charge (OH-). This polarity enables water to make substances to dissolve in
it as charged species. This polar nature of water is going to enable \"bicarbonate substance to
dissolve in blood\" in buffering system of the cellular respiration, finally involves an acid & base
homeostatic mechanism further involving the equilibrium balance of carbonic acid(H2CO3),
bicarbonate ion (HCO3-), and carbon dioxide (CO2) in order to maintain isotonic pH (7.4). On
the other hand, water molecules possess a higher attractive power to each other so that these
forces are referred as cohesive forces responsible for “surface tension” phenomenon” and these
attractions are predominantly due to hydrogen bonding.
1). Due of the presence of hydrogen bonds and strong intermolecular forces, water has high
specific heat index. Specific heat capacity of water is defined as; the amount of heat (energy in
joules) required for raise of temperature of water per unit mass by 10C.
Our body temperature is maintained higher than the surrounding temperature. The heat of
vaporization of the water is higher (40.65 kJ/mol), this is due to the hydrogen bonds that connect
water molecules. So, due to the temperature difference, water escapes out and carry the body
heat (the specific heat capacity of the water is 5 times more than the sand). So, during the process
of vaporization of water, water releases heat into the environment and cool the body.
2). Because of the hydrogen bonding, water exhibits high surface tension. So, water molecules
on the surface are more strongly attracted to each other than to the air above. This is necessary
for the functioning of alveoli. The layer of moisture over the alveoli is necessary for the gaseous
exchange.
3). Due to the polar nature of water, water can dissolve inorganic salts that dissociate into ions in
the presence of water (like dissolves like). So, the blood can carry many of the water soluble
substances throughout the body. Water cannot dissolve nonpolar substances; they are dissolved
in non-polar solvents.
4). Plasma contains approximately 91% of water; this maintains the viscosity of the blood..
Interview NotesDarcy is 45 years old, single, and a U.S. citizen w.pdf
Interview Notes
Darcy is 45 years old, single, and a U.S. citizen with a valid Social Security number. She had
$45,000 in wages.
During the interview with Darcy, you determine the following facts:
Darcy’s son Chris, age 21, is unmarried and was a full-time student working on a degree in
accounting during 2017.
Chris’ income was $8,500 in wages, which he used to pay his tuition. He did not provide more
than half his own support.
Chris lived on campus during the school year, but came home on breaks and for the summer.
Chris is in his third year of college.
Chris has never had a felony drug conviction.
Chris is a U.S. citizen with a valid Social Security number.
Questions:
1. Who can claim Chris\' personal or dependency exemption?
A. Chris can claim his personal exemption because he had earned income.
B. Chris can claim his personal exemption if Darcy does not to claim him.
C. Chris does not have a filing requirement so he cannot claim a personal
exemption.
D. Chris must file a tax return and claim zero personal exemptions because Darcy can
claim him as a dependent on her tax return.
2. Darcy can claim the expenses Chris paid as qualifying expenses for the American opportunity
credit if Darcy claims Chris as a dependent on her return.
True
False
Solution
Answer 1. To be dependent the following conditions should be met:
*Chris is independent. because Chris lived on campus during the school year.
A. Chris can claim his personal exemption because he had earned
Answer 2. True
Only qualifying educational expenses can be claimed on income taxes anyway.
if Darcy claims Chris as a dependent on her return..
Ligands bind reversibly to their receptors. Select the statements th.pdf
Ligands bind reversibly to their receptors. Select the statements that describe what would happen
if ligands bound irreversibly with their receptors.
A) Pathway will be permanently active.
B)The signal cannot be transduced in the cell.
C)The cell cannot receive new signals.
D)Cell metabolism cannot occur.
Solution
B) The signal cannot be transduced in the cell
They work as antagonists or otherwise called as blockers. They have affinity but they do not
have the efficacy for cognate receptors and hence the binding disrupts the interaction and inhibit
the function of the agonist. Therefore, the signal remains untranslated into the cell and signal
transduction does not occur..
In the rest of this lab, you will help build such a list class. The .pdf
In the rest of this lab, you will help build such a list class. The class definition is stored in
intlist.h - please study it to know its parts:
The public declarations show what a client program can do with an IntList object. Notice,
however, that the list node structure definition is in the private area. Clients can\'t directly use or
even refer to such nodes, but all that clients should care about are the values stored in these
nodes anyway! Also notice that every list object will have one node pointer, to point at the first
node in the list or at 0 if the list is empty.
Now look at intlist.cpp. All but sum, contains, max, average and insertFirst are already done.
You will implement those five methods in Step 4. The other parts are implemented at the bottom
of the file - do not change what is done.
An application file would create IntList objects and use them to solve problems. For this lab, our
application is just a testing program - testlist.cpp - and you should look at that file next. Two
IntList objects are created; then numbers read from the command line are appended to one of the
lists; and finally the methods are tested for each list.
You can use this Makefile to help you complile the program. If you store all four of these files in
the same directory, then you just type \"make\" (without the quotes) to compile them and create
the executable program.
Step 4: Implement five linked list functions
First: switch roles between pilot and navigator if you did not already do that. Always stay logged
into the original pilot\'s account.
You should be able to run the program now (assuming you compiled it in Step 3) - it requires
you to enter a starter list of integers on the command line:
Next time we\'ll run it properly. The contains method is checked for three values that ought to be
in the list, plus one value that should not (the sum). Here is a sample run with initial values of 5,
7, 9 and 11:
See that append, print and count all work. But the others need to be fixed.
Use an editor (e.g., emacs) to make the following changes to intlist.cpp - do not change any of
the other files.
Fix the comment at the top to show your name(s) and the date.
Implement the sum method. See the count method for guidance.
Save, and then test your sum implementation - compile and execute testlist again. Verify sum is
working before going on.
Then implement the contains method, save again and test again.
Continue with the other three functions in the same way: implement and test one at time.
Here are correct results for the same sample data as above:
FILES:
Solution
#include \"IntList.h\"
#include
using std::cout;
// return sum of values in list
int IntList::sum() const {
int result = 0;
Node *n = first;
while (n) {
result+= n->info;
n = n->next;
}
return result;
}
// returns true if value is in the list; false if not
bool IntList::contains(int value) const {
Node *n = first;
while (n) {
if(n->info == value) return true;
n = n->next;
}
return false; // REPLACE TH.
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