4-Propane because a saturated hydrocarbon has no .pdf
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4-Propane because a saturated hydrocarbon has no double bonds and no triple bonds. The suffix -ane means alkane, which refers to saturated hydrocarbons. Solution 4-Propane because a saturated hydrocarbon has no double bonds and no triple bonds. The suffix -ane means alkane, which refers to saturated hydrocarbons..
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3. Data Link Layer is the answer.This is because, SAPs act as an i.pdf
3. Data Link Layer is the answer.
This is because, SAPs act as an interface between data link layer and network layer.
Solution
3. Data Link Layer is the answer.
This is because, SAPs act as an interface between data link layer and network layer..
1. According to most surveys and reports the most infectious disease.pdf
1. According to most surveys and reports the most infectious disease in New York city which is a
greatest public health concern is HIV/AIDS. HIV is a silent disease when first acquired, though
this latency varies in length. The progression from HIV infection to AIDS varies from 5–12
years. AIDS is a condition in humans in which progressive failure of the immune system allows
life-threatening opportunistic infections and cancers to thrive. The overall death rate among
persons diagnosed with HIV/AIDS in New York City decreased by 62% from 2001 to 2012.
According to CDC estimates, 1,218,400 persons aged 13 years and older are living with HIV
infection, including 156,300 (12.8%) who are unaware of their infection.
3. The work \'salient\' means most noticeable or prominent features.
\"Diseases may be classified or recognized by salient features.\"
4. Steps in investigating an infectious disease outbreak;
(a) Identify the source and try to eliminate it as soon as possible
(b) Isolate people those infected with the disease to avaoid its spread.
(c) Early treatment of infected people.
(d) Development of strategies to prevent future outbreaks, calculations of risks and potential
hazard in future
(e) Create public awareness about the spread of disease/ its symptoms, causees and prevention
etc.
(f) Evaluate existing prevention strategies
(g) Address public concern
5. Worldwide deaths occur even by diseases which are vaccine-reventable. This mainly occurs
due to the lack of public awareness. More and more people should be made aware through
awareness programmes and informed about the development of new vaccines which can prevent
major diseases. This is of prime importance to reduce the death toll.
Solution
1. According to most surveys and reports the most infectious disease in New York city which is a
greatest public health concern is HIV/AIDS. HIV is a silent disease when first acquired, though
this latency varies in length. The progression from HIV infection to AIDS varies from 5–12
years. AIDS is a condition in humans in which progressive failure of the immune system allows
life-threatening opportunistic infections and cancers to thrive. The overall death rate among
persons diagnosed with HIV/AIDS in New York City decreased by 62% from 2001 to 2012.
According to CDC estimates, 1,218,400 persons aged 13 years and older are living with HIV
infection, including 156,300 (12.8%) who are unaware of their infection.
3. The work \'salient\' means most noticeable or prominent features.
\"Diseases may be classified or recognized by salient features.\"
4. Steps in investigating an infectious disease outbreak;
(a) Identify the source and try to eliminate it as soon as possible
(b) Isolate people those infected with the disease to avaoid its spread.
(c) Early treatment of infected people.
(d) Development of strategies to prevent future outbreaks, calculations of risks and potential
hazard in future
(e) Create public awareness about the spread of disease/ its sympt.
1) D. Platelets are the blood cells which aid in blood clotting.2).pdf
1) D. Platelets are the blood cells which aid in blood clotting.
2) C. Intestinal epithelial cells lines the intestine and helps in absorption of nutrients. ( Simple
columnar epithelium)
Solution
1) D. Platelets are the blood cells which aid in blood clotting.
2) C. Intestinal epithelial cells lines the intestine and helps in absorption of nutrients. ( Simple
columnar epithelium).
Vertical take-off and landing (VTOL) aircraft uses powered rotors fo.pdf
Vertical take-off and landing (VTOL) aircraft uses powered rotors for take off and landing
operations. For fixed wing aircraft, it should accelerate horizontally before its take-off. In VTOL
aircraft all moving stabilizer are positioned at a tail end and wing are positioned halfway
between the distal ends of the wing. Wing configuration – Substantially straight wing and canted
wing configuration. VTOL aircraft also includes pogo support is configured to deploy from each
of the wing’s distal ends. Harrier – unique design feature. It is capable of vertical/short takeoff
and landing (V/STOL) operations. In Harrier’s exhaust nozzle from the engine are able to swivel.
When the aircraft is take off/landing nozzle in downward position to create thrust to take off.
However, when the aircraft reached a desired altitude, nozzle turns to point backward to move
forward. For better of Harrier aircraft design, we need to study all forces – engine thrust, air
resistance and the plane own weight.
Solution
Vertical take-off and landing (VTOL) aircraft uses powered rotors for take off and landing
operations. For fixed wing aircraft, it should accelerate horizontally before its take-off. In VTOL
aircraft all moving stabilizer are positioned at a tail end and wing are positioned halfway
between the distal ends of the wing. Wing configuration – Substantially straight wing and canted
wing configuration. VTOL aircraft also includes pogo support is configured to deploy from each
of the wing’s distal ends. Harrier – unique design feature. It is capable of vertical/short takeoff
and landing (V/STOL) operations. In Harrier’s exhaust nozzle from the engine are able to swivel.
When the aircraft is take off/landing nozzle in downward position to create thrust to take off.
However, when the aircraft reached a desired altitude, nozzle turns to point backward to move
forward. For better of Harrier aircraft design, we need to study all forces – engine thrust, air
resistance and the plane own weight..
The degree of ionization is influenced by the con.pdf
The degree of ionization is influenced by the concentration of the solution that
means the dilution. more is the dilution more is the degree of ionization.
Solution
The degree of ionization is influenced by the concentration of the solution that
means the dilution. more is the dilution more is the degree of ionization..
Problem 1 H2SO4 is diprotic, so it gives [H+] o.pdf
Problem 1: H2SO4 is diprotic, so it gives [H+] of 2(0.255)M = 0.51M KOH is
monobasic, so it gives [OH-] of 0.255M M1V1 = M2V2 0.51(45) = 0.255(V2) V2 = 90mL
Problem 2: H2SO4 is diprotic, so it gives [H+] of 2(0.140)M = 0.28M KOH is monobasic, so it
gives [OH-] of 0.255M M1V1 = M2V2 0.28(195) = 0.255(V2) V2 = 214.12mL Problem 3:
H2SO4 is diprotic, so it gives [H+] of 2(0.110)M = 0.22M KOH is monobasic, so it gives [OH-]
of 0.255M M1V1 = M2V2 0.22(70) = 0.255(V2) V2 = 60.39mL
Solution
Problem 1: H2SO4 is diprotic, so it gives [H+] of 2(0.255)M = 0.51M KOH is
monobasic, so it gives [OH-] of 0.255M M1V1 = M2V2 0.51(45) = 0.255(V2) V2 = 90mL
Problem 2: H2SO4 is diprotic, so it gives [H+] of 2(0.140)M = 0.28M KOH is monobasic, so it
gives [OH-] of 0.255M M1V1 = M2V2 0.28(195) = 0.255(V2) V2 = 214.12mL Problem 3:
H2SO4 is diprotic, so it gives [H+] of 2(0.110)M = 0.22M KOH is monobasic, so it gives [OH-]
of 0.255M M1V1 = M2V2 0.22(70) = 0.255(V2) V2 = 60.39mL.
only one because it exist only one isomer .pdf
only one because it exist only one isomer
Solution
only one because it exist only one isomer.
when 71.1 moles of carbon dioxide gas react with .pdf
when 71.1 moles of carbon dioxide gas react with excess hydrogen gas 71.1 moles
of methane is produced
Solution
when 71.1 moles of carbon dioxide gas react with excess hydrogen gas 71.1 moles
of methane is produced.
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3. Data Link Layer is the answer.This is because, SAPs act as an i.pdf
3. Data Link Layer is the answer.
This is because, SAPs act as an interface between data link layer and network layer.
Solution
3. Data Link Layer is the answer.
This is because, SAPs act as an interface between data link layer and network layer..
1. According to most surveys and reports the most infectious disease.pdf
1. According to most surveys and reports the most infectious disease in New York city which is a
greatest public health concern is HIV/AIDS. HIV is a silent disease when first acquired, though
this latency varies in length. The progression from HIV infection to AIDS varies from 5–12
years. AIDS is a condition in humans in which progressive failure of the immune system allows
life-threatening opportunistic infections and cancers to thrive. The overall death rate among
persons diagnosed with HIV/AIDS in New York City decreased by 62% from 2001 to 2012.
According to CDC estimates, 1,218,400 persons aged 13 years and older are living with HIV
infection, including 156,300 (12.8%) who are unaware of their infection.
3. The work \'salient\' means most noticeable or prominent features.
\"Diseases may be classified or recognized by salient features.\"
4. Steps in investigating an infectious disease outbreak;
(a) Identify the source and try to eliminate it as soon as possible
(b) Isolate people those infected with the disease to avaoid its spread.
(c) Early treatment of infected people.
(d) Development of strategies to prevent future outbreaks, calculations of risks and potential
hazard in future
(e) Create public awareness about the spread of disease/ its symptoms, causees and prevention
etc.
(f) Evaluate existing prevention strategies
(g) Address public concern
5. Worldwide deaths occur even by diseases which are vaccine-reventable. This mainly occurs
due to the lack of public awareness. More and more people should be made aware through
awareness programmes and informed about the development of new vaccines which can prevent
major diseases. This is of prime importance to reduce the death toll.
Solution
1. According to most surveys and reports the most infectious disease in New York city which is a
greatest public health concern is HIV/AIDS. HIV is a silent disease when first acquired, though
this latency varies in length. The progression from HIV infection to AIDS varies from 5–12
years. AIDS is a condition in humans in which progressive failure of the immune system allows
life-threatening opportunistic infections and cancers to thrive. The overall death rate among
persons diagnosed with HIV/AIDS in New York City decreased by 62% from 2001 to 2012.
According to CDC estimates, 1,218,400 persons aged 13 years and older are living with HIV
infection, including 156,300 (12.8%) who are unaware of their infection.
3. The work \'salient\' means most noticeable or prominent features.
\"Diseases may be classified or recognized by salient features.\"
4. Steps in investigating an infectious disease outbreak;
(a) Identify the source and try to eliminate it as soon as possible
(b) Isolate people those infected with the disease to avaoid its spread.
(c) Early treatment of infected people.
(d) Development of strategies to prevent future outbreaks, calculations of risks and potential
hazard in future
(e) Create public awareness about the spread of disease/ its sympt.
1) D. Platelets are the blood cells which aid in blood clotting.2).pdf
1) D. Platelets are the blood cells which aid in blood clotting.
2) C. Intestinal epithelial cells lines the intestine and helps in absorption of nutrients. ( Simple
columnar epithelium)
Solution
1) D. Platelets are the blood cells which aid in blood clotting.
2) C. Intestinal epithelial cells lines the intestine and helps in absorption of nutrients. ( Simple
columnar epithelium).
Vertical take-off and landing (VTOL) aircraft uses powered rotors fo.pdf
Vertical take-off and landing (VTOL) aircraft uses powered rotors for take off and landing
operations. For fixed wing aircraft, it should accelerate horizontally before its take-off. In VTOL
aircraft all moving stabilizer are positioned at a tail end and wing are positioned halfway
between the distal ends of the wing. Wing configuration – Substantially straight wing and canted
wing configuration. VTOL aircraft also includes pogo support is configured to deploy from each
of the wing’s distal ends. Harrier – unique design feature. It is capable of vertical/short takeoff
and landing (V/STOL) operations. In Harrier’s exhaust nozzle from the engine are able to swivel.
When the aircraft is take off/landing nozzle in downward position to create thrust to take off.
However, when the aircraft reached a desired altitude, nozzle turns to point backward to move
forward. For better of Harrier aircraft design, we need to study all forces – engine thrust, air
resistance and the plane own weight.
Solution
Vertical take-off and landing (VTOL) aircraft uses powered rotors for take off and landing
operations. For fixed wing aircraft, it should accelerate horizontally before its take-off. In VTOL
aircraft all moving stabilizer are positioned at a tail end and wing are positioned halfway
between the distal ends of the wing. Wing configuration – Substantially straight wing and canted
wing configuration. VTOL aircraft also includes pogo support is configured to deploy from each
of the wing’s distal ends. Harrier – unique design feature. It is capable of vertical/short takeoff
and landing (V/STOL) operations. In Harrier’s exhaust nozzle from the engine are able to swivel.
When the aircraft is take off/landing nozzle in downward position to create thrust to take off.
However, when the aircraft reached a desired altitude, nozzle turns to point backward to move
forward. For better of Harrier aircraft design, we need to study all forces – engine thrust, air
resistance and the plane own weight..
The degree of ionization is influenced by the con.pdf
The degree of ionization is influenced by the concentration of the solution that
means the dilution. more is the dilution more is the degree of ionization.
Solution
The degree of ionization is influenced by the concentration of the solution that
means the dilution. more is the dilution more is the degree of ionization..
Problem 1 H2SO4 is diprotic, so it gives [H+] o.pdf
Problem 1: H2SO4 is diprotic, so it gives [H+] of 2(0.255)M = 0.51M KOH is
monobasic, so it gives [OH-] of 0.255M M1V1 = M2V2 0.51(45) = 0.255(V2) V2 = 90mL
Problem 2: H2SO4 is diprotic, so it gives [H+] of 2(0.140)M = 0.28M KOH is monobasic, so it
gives [OH-] of 0.255M M1V1 = M2V2 0.28(195) = 0.255(V2) V2 = 214.12mL Problem 3:
H2SO4 is diprotic, so it gives [H+] of 2(0.110)M = 0.22M KOH is monobasic, so it gives [OH-]
of 0.255M M1V1 = M2V2 0.22(70) = 0.255(V2) V2 = 60.39mL
Solution
Problem 1: H2SO4 is diprotic, so it gives [H+] of 2(0.255)M = 0.51M KOH is
monobasic, so it gives [OH-] of 0.255M M1V1 = M2V2 0.51(45) = 0.255(V2) V2 = 90mL
Problem 2: H2SO4 is diprotic, so it gives [H+] of 2(0.140)M = 0.28M KOH is monobasic, so it
gives [OH-] of 0.255M M1V1 = M2V2 0.28(195) = 0.255(V2) V2 = 214.12mL Problem 3:
H2SO4 is diprotic, so it gives [H+] of 2(0.110)M = 0.22M KOH is monobasic, so it gives [OH-]
of 0.255M M1V1 = M2V2 0.22(70) = 0.255(V2) V2 = 60.39mL.
only one because it exist only one isomer .pdf
only one because it exist only one isomer
Solution
only one because it exist only one isomer.
when 71.1 moles of carbon dioxide gas react with .pdf
when 71.1 moles of carbon dioxide gas react with excess hydrogen gas 71.1 moles
of methane is produced
Solution
when 71.1 moles of carbon dioxide gas react with excess hydrogen gas 71.1 moles
of methane is produced.
rate at t=0 is R=0.042 Overall order of reaction=.pdf
rate at t=0 is R=0.042 Overall order of reaction=1+1=2
Solution
rate at t=0 is R=0.042 Overall order of reaction=1+1=2.
moles of H+ = 0.337 0.6303 = 0.2124 moles of OH.pdf
moles of H+ = 0.337 * 0.6303 = 0.2124 moles of OH- = 0.447 * 10^0.02 = 0.46807
Moles of OH- left = 0.46807 - 0.2124 = 0.25566 moles [OH-] = 0.25566 / (0.337 + 0.447) =
0.3261 pOH = - log [OH-] = 0.487 pH = 14 - pOH = 13.513
Solution
moles of H+ = 0.337 * 0.6303 = 0.2124 moles of OH- = 0.447 * 10^0.02 = 0.46807
Moles of OH- left = 0.46807 - 0.2124 = 0.25566 moles [OH-] = 0.25566 / (0.337 + 0.447) =
0.3261 pOH = - log [OH-] = 0.487 pH = 14 - pOH = 13.513.
lithium hydrogenphosphite LiH2PO3 lead(IV) hyd.pdf
lithium hydrogenphosphite : LiH2PO3 lead(IV) hydrogensulfide : Pb(HSO4)4
bismuth(V) sulfate : Bi2(SO4)5 sodium nitride: Na3N chloric acid: HClO3 perbromic acid:
HBrO4 hydrochloric acid: HCl hydroselenic acid: H2Se hypoiodous acid: HIO bromous acid :
HBrO2
Solution
lithium hydrogenphosphite : LiH2PO3 lead(IV) hydrogensulfide : Pb(HSO4)4
bismuth(V) sulfate : Bi2(SO4)5 sodium nitride: Na3N chloric acid: HClO3 perbromic acid:
HBrO4 hydrochloric acid: HCl hydroselenic acid: H2Se hypoiodous acid: HIO bromous acid :
HBrO2.
HClO3 HClO2 HClO HBrO Solution .pdf
HClO3 > HClO2 > HClO > HBrO
Solution
HClO3 > HClO2 > HClO > HBrO.
d. PO43- as can acceppt H+ but cant donate H+. .pdf
d. PO43- as can acceppt H+ but can\'t donate H+.
Solution
d. PO43- as can acceppt H+ but can\'t donate H+..
water + water = water 0.108 + 1x = 0.90(8 + x)0.8 + x = 7.2 + .pdf
water + water = water
0.10*8 + 1*x = 0.90(8 + x)
0.8 + x = 7.2 + 0.90x
0.10x = 6.4
x=64
Solution
water + water = water
0.10*8 + 1*x = 0.90(8 + x)
0.8 + x = 7.2 + 0.90x
0.10x = 6.4
x=64.
cooking of food, burning of paper, .pdf
cooking of food, burning of paper,
Solution
cooking of food, burning of paper,.
The rst time they get one of the two sexes. After that they are wait.pdf
The rst time they get one of the two sexes. After that they are waiting for the event of probability
1/2 and expected value of 2, that gives them the other one, so the mean is 1 + 2 = 3.
Solution
The rst time they get one of the two sexes. After that they are waiting for the event of probability
1/2 and expected value of 2, that gives them the other one, so the mean is 1 + 2 = 3..
System Thinking and Engineering-SolutionSystem Thinking and E.pdf
System Thinking and Engineering:-
Solution
System Thinking and Engineering:-.
Solution1.) False (access the modules in the VBA editor (ALT+F11).pdf
Solution
:
1.) False (access the modules in the VBA editor (ALT+F11)? If so, you can right-click then
choose \"Remove Module )
2.) False: (Other plots are, Pie, line, Area, Bar, Stock chart, surface chart etc)
3.) True
4.) False (it can be changed)
5.) False (it can be changed)
6.) d ( A function program in excel cannot have multiple outputs)
7.) Boolen
8.) Solve equations.
Solution If an infection led to significant deterioration of the .pdf
Solution
:
If an infection led to significant deterioration of the alveolar wall then exchange of oxygen and
carbon dioxide at the walls of alveoli reduced.It is because there is very less surface area for the
oxygen-carbon exchanging process.Due to result of this lungs expands.It effect breathing
process.And cause non repairable damage to the lung tissues.This is a life threatening damage..
Organizational structure refers to the way that an organization arra.pdf
Organizational structure refers to the way that an organization arranges people and jobs so that
its work can be performed and its goals can be met. When a work group is very small, and face-
to-face communication is frequent, formal structure may be unnecessary, but in a larger
organization decisions have to be made about the delegation of various tasks. Thus, procedures
are established that assign responsibilities for various functions. It is these decisions that
determine the organizational structure.
In an organization of any size or complexity, employees\' responsibilities typically are defined by
what they do, who they report to, and for managers, who reports to them. Over time these
definitions are assigned to positions in the organization rather than to specific individuals. The
relationships among these positions are illustrated graphically in an organizational chart. The
best organizational structure for any organization depends on many factors including the work it
does; its size in terms of employees, revenue, and the geographic dispersion of its facilities; and
the range of its businesses (the degree to which it is diversified across markets).
DEVELOPMENT OF THE TRADITIONAL ORGANIZATIONAL STRUCTURE
Understanding the historical context from which some of today\'s dominant organizational
structures have developed helps to explain why some structures are the way they are. For
instance, why are the old, but still operational steel mills such as U.S. Steel and Bethlehem Steel
structured using vertical hierarchies? Why are newer steel mini-mills such as Chaparral Steel
structured more horizontally, capitalizing on the innovativeness of their employees? Part of the
reason, as this section discusses, is that organizational structure has a certain inertia—the idea
borrowed from physics and chemistry that something in motion tends to continue on that same
path. Changing
an organization\'s structure is a daunting managerial task, and the immensity of such a project is
at least partly responsible for why organizational structures change infrequently.
At the beginning of the twentieth century the United States business sector was thriving. Industry
was shifting from job-shop manufacturing to mass production, and thinkers like Frederick Taylor
in the United States and Henri Fayol in France studied the new systems and developed principles
to determine how to structure organizations for the greatest efficiency and productivity, which in
their view was very much like a machine. Even before this, German sociologist and engineer
Max Weber had concluded that when societies embrace capitalism, bureaucracy is the inevitable
result. Yet, because his writings were not translated into English until 1949, Weber\'s work had
little influence on American management practice until the middle of the twentieth century.
Management thought during this period did match Weber\'s ideas of bureaucracy, where power
is ascribed to positions rather than to the individua.
II alone sufficient while I alone not sufficient to answerSoluti.pdf
II alone sufficient while I alone not sufficient to answer
Solution
II alone sufficient while I alone not sufficient to answer.
import java.util.Scanner;class BinaryNode{ BinaryNode left.pdf
import java.util.Scanner;
class BinaryNode
{
BinaryNode left, right;
int data;
public BinaryNode()
{
left = null;
right = null;
data = 0;
}
public BinaryNode(int n)
{
left = null;
right = null;
data = n;
}
public void setLeft(BinaryNode n)
{
left = n;
}
public void setRight(BinaryNode n)
{
right = n;
}
public BinaryNode getLeft()
{
return left;
}
public BinaryNode getRight()
{
return right;
}
public void setData(int d)
{
data = d;
}
public int getData()
{
return data;
}
}
class Binary
{
private BinaryNode root;
public Binary()
{
root = null;
}
public boolean isEmpty()
{
return root == null;
}
public void insert(int data)
{
root = insert(root, data);
}
private BinaryNode insert(BinaryNode node, int data)
{
if (node == null)
node = new BinaryNode(data);
else
{
if (data <= node.getData())
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
}
return node;
}
public void delete(int k)
{
if (isEmpty())
System.out.println(\"Tree Empty\");
else if (search(k) == false)
System.out.println(\"Sorry \"+ k +\" is not present\");
else
{
root = delete(root, k);
System.out.println(k+ \" deleted from the tree\");
}
}
private BinaryNode delete(BinaryNode root, int k)
{
BinaryNode p, p2, n;
if (root.getData() == k)
{
BinaryNode lt, rt;
lt = root.getLeft();
rt = root.getRight();
if (lt == null && rt == null)
return null;
else if (lt == null)
{
p = rt;
return p;
}
else if (rt == null)
{
p = lt;
return p;
}
else
{
p2 = rt;
p = rt;
while (p.getLeft() != null)
p = p.getLeft();
p.setLeft(lt);
return p2;
}
}
if (k < root.getData())
{
n = delete(root.getLeft(), k);
root.setLeft(n);
}
else
{
n = delete(root.getRight(), k);
root.setRight(n);
}
return root;
}
public boolean search(int val)
{
return search(root, val);
}
private boolean search(BinaryNode r, int val)
{
boolean found = false;
while ((r != null) && !found)
{
int rval = r.getData();
if (val < rval)
r = r.getLeft();
else if (val > rval)
r = r.getRight();
else
{
found = true;
break;
}
found = search(r, val);
}
return found;
}
public void inorder()
{
inorder(root);
}
private void inorder(BinaryNode r)
{
if (r != null)
{
inorder(r.getLeft());
System.out.print(r.getData() +\" \");
inorder(r.getRight());
}
}
public void preorder()
{
preorder(root);
}
private void preorder(BinaryNode r)
{
if (r != null)
{
System.out.print(r.getData() +\" \");
preorder(r.getLeft());
preorder(r.getRight());
}
}
public void postorder()
{
postorder(root);
}
private void postorder(BinaryNode r)
{
if (r != null)
{
postorder(r.getLeft());
postorder(r.getRight());
System.out.print(r.getData() +\" \");
}
}
}
public class BinarySearchTree
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
Binary Binary = new Binary();
System.out.println(\"Binary Search Tree Test\ \");
char ch;
do
{
System.out.println(\"\ Binary Search Tree Operations\ \");
System.out.println(\"1. insert \");
System.out.println(\"2. delete\");
System.out.println(\"3. search\");
System.out.println(\"4. check empty\");
int .
homeinsertAnimationsDevelopmentSolutionhomeinsertAni.pdf
home
insert
Animations
Development
Solution
home
insert
Animations
Development.
Dear,The answer is possible.SolutionDear,The answer is pos.pdf
Dear,
The answer is possible.
Solution
Dear,
The answer is possible..
d. none of the above. The graph has a cycle. It is not a tree..pdf
d. none of the above. The graph has a cycle. It is not a tree.
Solution
d. none of the above. The graph has a cycle. It is not a tree..
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Class 11 CBSE Business Studies Project ( AIDS TO TRADE - INSURANCE)
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rate at t=0 is R=0.042 Overall order of reaction=.pdf
rate at t=0 is R=0.042 Overall order of reaction=1+1=2
Solution
rate at t=0 is R=0.042 Overall order of reaction=1+1=2.
moles of H+ = 0.337 0.6303 = 0.2124 moles of OH.pdf
moles of H+ = 0.337 * 0.6303 = 0.2124 moles of OH- = 0.447 * 10^0.02 = 0.46807
Moles of OH- left = 0.46807 - 0.2124 = 0.25566 moles [OH-] = 0.25566 / (0.337 + 0.447) =
0.3261 pOH = - log [OH-] = 0.487 pH = 14 - pOH = 13.513
Solution
moles of H+ = 0.337 * 0.6303 = 0.2124 moles of OH- = 0.447 * 10^0.02 = 0.46807
Moles of OH- left = 0.46807 - 0.2124 = 0.25566 moles [OH-] = 0.25566 / (0.337 + 0.447) =
0.3261 pOH = - log [OH-] = 0.487 pH = 14 - pOH = 13.513.
lithium hydrogenphosphite LiH2PO3 lead(IV) hyd.pdf
lithium hydrogenphosphite : LiH2PO3 lead(IV) hydrogensulfide : Pb(HSO4)4
bismuth(V) sulfate : Bi2(SO4)5 sodium nitride: Na3N chloric acid: HClO3 perbromic acid:
HBrO4 hydrochloric acid: HCl hydroselenic acid: H2Se hypoiodous acid: HIO bromous acid :
HBrO2
Solution
lithium hydrogenphosphite : LiH2PO3 lead(IV) hydrogensulfide : Pb(HSO4)4
bismuth(V) sulfate : Bi2(SO4)5 sodium nitride: Na3N chloric acid: HClO3 perbromic acid:
HBrO4 hydrochloric acid: HCl hydroselenic acid: H2Se hypoiodous acid: HIO bromous acid :
HBrO2.
HClO3 HClO2 HClO HBrO Solution .pdf
HClO3 > HClO2 > HClO > HBrO
Solution
HClO3 > HClO2 > HClO > HBrO.
d. PO43- as can acceppt H+ but cant donate H+. .pdf
d. PO43- as can acceppt H+ but can\'t donate H+.
Solution
d. PO43- as can acceppt H+ but can\'t donate H+..
water + water = water 0.108 + 1x = 0.90(8 + x)0.8 + x = 7.2 + .pdf
water + water = water
0.10*8 + 1*x = 0.90(8 + x)
0.8 + x = 7.2 + 0.90x
0.10x = 6.4
x=64
Solution
water + water = water
0.10*8 + 1*x = 0.90(8 + x)
0.8 + x = 7.2 + 0.90x
0.10x = 6.4
x=64.
cooking of food, burning of paper, .pdf
cooking of food, burning of paper,
Solution
cooking of food, burning of paper,.
The rst time they get one of the two sexes. After that they are wait.pdf
The rst time they get one of the two sexes. After that they are waiting for the event of probability
1/2 and expected value of 2, that gives them the other one, so the mean is 1 + 2 = 3.
Solution
The rst time they get one of the two sexes. After that they are waiting for the event of probability
1/2 and expected value of 2, that gives them the other one, so the mean is 1 + 2 = 3..
System Thinking and Engineering-SolutionSystem Thinking and E.pdf
System Thinking and Engineering:-
Solution
System Thinking and Engineering:-.
Solution1.) False (access the modules in the VBA editor (ALT+F11).pdf
Solution
:
1.) False (access the modules in the VBA editor (ALT+F11)? If so, you can right-click then
choose \"Remove Module )
2.) False: (Other plots are, Pie, line, Area, Bar, Stock chart, surface chart etc)
3.) True
4.) False (it can be changed)
5.) False (it can be changed)
6.) d ( A function program in excel cannot have multiple outputs)
7.) Boolen
8.) Solve equations.
Solution If an infection led to significant deterioration of the .pdf
Solution
:
If an infection led to significant deterioration of the alveolar wall then exchange of oxygen and
carbon dioxide at the walls of alveoli reduced.It is because there is very less surface area for the
oxygen-carbon exchanging process.Due to result of this lungs expands.It effect breathing
process.And cause non repairable damage to the lung tissues.This is a life threatening damage..
Organizational structure refers to the way that an organization arra.pdf
Organizational structure refers to the way that an organization arranges people and jobs so that
its work can be performed and its goals can be met. When a work group is very small, and face-
to-face communication is frequent, formal structure may be unnecessary, but in a larger
organization decisions have to be made about the delegation of various tasks. Thus, procedures
are established that assign responsibilities for various functions. It is these decisions that
determine the organizational structure.
In an organization of any size or complexity, employees\' responsibilities typically are defined by
what they do, who they report to, and for managers, who reports to them. Over time these
definitions are assigned to positions in the organization rather than to specific individuals. The
relationships among these positions are illustrated graphically in an organizational chart. The
best organizational structure for any organization depends on many factors including the work it
does; its size in terms of employees, revenue, and the geographic dispersion of its facilities; and
the range of its businesses (the degree to which it is diversified across markets).
DEVELOPMENT OF THE TRADITIONAL ORGANIZATIONAL STRUCTURE
Understanding the historical context from which some of today\'s dominant organizational
structures have developed helps to explain why some structures are the way they are. For
instance, why are the old, but still operational steel mills such as U.S. Steel and Bethlehem Steel
structured using vertical hierarchies? Why are newer steel mini-mills such as Chaparral Steel
structured more horizontally, capitalizing on the innovativeness of their employees? Part of the
reason, as this section discusses, is that organizational structure has a certain inertia—the idea
borrowed from physics and chemistry that something in motion tends to continue on that same
path. Changing
an organization\'s structure is a daunting managerial task, and the immensity of such a project is
at least partly responsible for why organizational structures change infrequently.
At the beginning of the twentieth century the United States business sector was thriving. Industry
was shifting from job-shop manufacturing to mass production, and thinkers like Frederick Taylor
in the United States and Henri Fayol in France studied the new systems and developed principles
to determine how to structure organizations for the greatest efficiency and productivity, which in
their view was very much like a machine. Even before this, German sociologist and engineer
Max Weber had concluded that when societies embrace capitalism, bureaucracy is the inevitable
result. Yet, because his writings were not translated into English until 1949, Weber\'s work had
little influence on American management practice until the middle of the twentieth century.
Management thought during this period did match Weber\'s ideas of bureaucracy, where power
is ascribed to positions rather than to the individua.
II alone sufficient while I alone not sufficient to answerSoluti.pdf
II alone sufficient while I alone not sufficient to answer
Solution
II alone sufficient while I alone not sufficient to answer.
import java.util.Scanner;class BinaryNode{ BinaryNode left.pdf
import java.util.Scanner;
class BinaryNode
{
BinaryNode left, right;
int data;
public BinaryNode()
{
left = null;
right = null;
data = 0;
}
public BinaryNode(int n)
{
left = null;
right = null;
data = n;
}
public void setLeft(BinaryNode n)
{
left = n;
}
public void setRight(BinaryNode n)
{
right = n;
}
public BinaryNode getLeft()
{
return left;
}
public BinaryNode getRight()
{
return right;
}
public void setData(int d)
{
data = d;
}
public int getData()
{
return data;
}
}
class Binary
{
private BinaryNode root;
public Binary()
{
root = null;
}
public boolean isEmpty()
{
return root == null;
}
public void insert(int data)
{
root = insert(root, data);
}
private BinaryNode insert(BinaryNode node, int data)
{
if (node == null)
node = new BinaryNode(data);
else
{
if (data <= node.getData())
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
}
return node;
}
public void delete(int k)
{
if (isEmpty())
System.out.println(\"Tree Empty\");
else if (search(k) == false)
System.out.println(\"Sorry \"+ k +\" is not present\");
else
{
root = delete(root, k);
System.out.println(k+ \" deleted from the tree\");
}
}
private BinaryNode delete(BinaryNode root, int k)
{
BinaryNode p, p2, n;
if (root.getData() == k)
{
BinaryNode lt, rt;
lt = root.getLeft();
rt = root.getRight();
if (lt == null && rt == null)
return null;
else if (lt == null)
{
p = rt;
return p;
}
else if (rt == null)
{
p = lt;
return p;
}
else
{
p2 = rt;
p = rt;
while (p.getLeft() != null)
p = p.getLeft();
p.setLeft(lt);
return p2;
}
}
if (k < root.getData())
{
n = delete(root.getLeft(), k);
root.setLeft(n);
}
else
{
n = delete(root.getRight(), k);
root.setRight(n);
}
return root;
}
public boolean search(int val)
{
return search(root, val);
}
private boolean search(BinaryNode r, int val)
{
boolean found = false;
while ((r != null) && !found)
{
int rval = r.getData();
if (val < rval)
r = r.getLeft();
else if (val > rval)
r = r.getRight();
else
{
found = true;
break;
}
found = search(r, val);
}
return found;
}
public void inorder()
{
inorder(root);
}
private void inorder(BinaryNode r)
{
if (r != null)
{
inorder(r.getLeft());
System.out.print(r.getData() +\" \");
inorder(r.getRight());
}
}
public void preorder()
{
preorder(root);
}
private void preorder(BinaryNode r)
{
if (r != null)
{
System.out.print(r.getData() +\" \");
preorder(r.getLeft());
preorder(r.getRight());
}
}
public void postorder()
{
postorder(root);
}
private void postorder(BinaryNode r)
{
if (r != null)
{
postorder(r.getLeft());
postorder(r.getRight());
System.out.print(r.getData() +\" \");
}
}
}
public class BinarySearchTree
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
Binary Binary = new Binary();
System.out.println(\"Binary Search Tree Test\ \");
char ch;
do
{
System.out.println(\"\ Binary Search Tree Operations\ \");
System.out.println(\"1. insert \");
System.out.println(\"2. delete\");
System.out.println(\"3. search\");
System.out.println(\"4. check empty\");
int .
homeinsertAnimationsDevelopmentSolutionhomeinsertAni.pdf
home
insert
Animations
Development
Solution
home
insert
Animations
Development.
Dear,The answer is possible.SolutionDear,The answer is pos.pdf
Dear,
The answer is possible.
Solution
Dear,
The answer is possible..
d. none of the above. The graph has a cycle. It is not a tree..pdf
d. none of the above. The graph has a cycle. It is not a tree.
Solution
d. none of the above. The graph has a cycle. It is not a tree..
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rate at t=0 is R=0.042 Overall order of reaction=.pdf
rate at t=0 is R=0.042 Overall order of reaction=.pdf
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moles of H+ = 0.337 0.6303 = 0.2124 moles of OH.pdf
lithium hydrogenphosphite LiH2PO3 lead(IV) hyd.pdf
lithium hydrogenphosphite LiH2PO3 lead(IV) hyd.pdf
d. PO43- as can acceppt H+ but cant donate H+. .pdf
d. PO43- as can acceppt H+ but cant donate H+. .pdf
water + water = water 0.108 + 1x = 0.90(8 + x)0.8 + x = 7.2 + .pdf
water + water = water 0.108 + 1x = 0.90(8 + x)0.8 + x = 7.2 + .pdf
The rst time they get one of the two sexes. After that they are wait.pdf
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System Thinking and Engineering-SolutionSystem Thinking and E.pdf
System Thinking and Engineering-SolutionSystem Thinking and E.pdf
Solution1.) False (access the modules in the VBA editor (ALT+F11).pdf
Solution1.) False (access the modules in the VBA editor (ALT+F11).pdf
Solution If an infection led to significant deterioration of the .pdf
Solution If an infection led to significant deterioration of the .pdf
Organizational structure refers to the way that an organization arra.pdf
Organizational structure refers to the way that an organization arra.pdf
II alone sufficient while I alone not sufficient to answerSoluti.pdf
II alone sufficient while I alone not sufficient to answerSoluti.pdf
import java.util.Scanner;class BinaryNode{ BinaryNode left.pdf
import java.util.Scanner;class BinaryNode{ BinaryNode left.pdf
homeinsertAnimationsDevelopmentSolutionhomeinsertAni.pdf
homeinsertAnimationsDevelopmentSolutionhomeinsertAni.pdf
Dear,The answer is possible.SolutionDear,The answer is pos.pdf
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