Cant see the questionSolutionCant see the question.pdf
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1. Corporations, limited liability companies (LLCs), general and limited partnerships, and sole proprietorships are the more common legal entities used for operating a business. 2.Corporations and LLCs offer owners limited liability. General partners and sole proprietors may be held personally responsible for the debts of the general partnership and sole proprietorship. However,limited partners are not responsible for the partnership’s liabilities. 3.A business entity may be legally classified as a corporation, limited liability company (LLC), a general partnership (GP), a limited partnership (LP), or a sole proprietorship under state law. However, for tax purposes a business entity can be classified as either a separate taxpaying entity or as a flow-through entity. Separate taxpaying entities pay tax on their own income. In contrast, flow-through entities generally don’t pay taxes because income from these entities flows through to their business owners who are responsible for paying tax on the income. C corporations are separate taxpaying entities and if elected some flow-through entities may be treated as separate taxpaying entities. Flow-through entities are usually taxed as either partnerships, sole proprietorships, or disregarded entities. 4.A taxation principle referring to income taxes that are paid twice on the same source of earned income. Double taxation occurs because corporations are considered separate legal entities from their shareholders. As such, corporations pay taxes on their annual earnings, just as individuals do. When corporations pay out dividends to shareholders, those dividend payments incur income-tax liabilities for the shareholders who receive them, even though the earnings that provided the cash to pay the dividends were already taxed at the corporate level. 5. Double tax is actually more favorable if you, 1. Pay reasonable salaries to shareholders. 2. Lease property from shareholders. 3. Defer or eliminate dividend payments. 4. Defer capital gains taxes on shares by making lifetime gifts of appreciated stock Solution 1. Corporations, limited liability companies (LLCs), general and limited partnerships, and sole proprietorships are the more common legal entities used for operating a business. 2.Corporations and LLCs offer owners limited liability. General partners and sole proprietors may be held personally responsible for the debts of the general partnership and sole proprietorship. However,limited partners are not responsible for the partnership’s liabilities. 3.A business entity may be legally classified as a corporation, limited liability company (LLC), a general partnership (GP), a limited partnership (LP), or a sole proprietorship under state law. However, for tax purposes a business entity can be classified as either a separate taxpaying entity or as a flow-through entity. Separate taxpaying entities pay tax on their own income. In contrast, flow-through entities generally don’t pay taxes because income fr.
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System is a set of interacting or interdependent entities forming an.pdf
Hi,
I have updated the code. Highlighted the code changes below. we can end the loop by break
statement.
#include
using namespace std;
int main()
{
int x = 10;
int y =20;
int z = x + y;
while(z == z){
if(y ==z){
do{
for(int i=0; i<2; i++){
cout<<\"This is the answer \"<
Solution
Hi,
I have updated the code. Highlighted the code changes below. we can end the loop by break
statement.
#include
using namespace std;
int main()
{
int x = 10;
int y =20;
int z = x + y;
while(z == z){
if(y ==z){
do{
for(int i=0; i<2; i++){
cout<<\"This is the answer \"<.
Solution Given thatBy definition of logarithm, we getSolution.pdf
Given
F1 = -5i+3j-4k
F2 = 5i-5j-2k
The position vectors of the forces F1 and F2 are r1 and r2 respectively
From figure we can write
r1 = 4 i -0.15 j + 0.25 k
r2 = 4 i +0.15 j + 0.25 k
Now the equivalent force at O is
FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k
FR = -2 j -6 k
Therefore
FRx, FRy,FRz = (0, -2, -6)
Now the resultant moment about O is
MR,O = r1XF1 + r2X F2
MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k)
MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i
MR,O = 0.8 i + 24 j -9.5 k
so
(MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5)
Solution
Given
F1 = -5i+3j-4k
F2 = 5i-5j-2k
The position vectors of the forces F1 and F2 are r1 and r2 respectively
From figure we can write
r1 = 4 i -0.15 j + 0.25 k
r2 = 4 i +0.15 j + 0.25 k
Now the equivalent force at O is
FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k
FR = -2 j -6 k
Therefore
FRx, FRy,FRz = (0, -2, -6)
Now the resultant moment about O is
MR,O = r1XF1 + r2X F2
MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k)
MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i
MR,O = 0.8 i + 24 j -9.5 k
so
(MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5).
C is the abbrevation for Carbon .pdf
C is the abbrevation for Carbon
Solution
C is the abbrevation for Carbon.
Percent error actual - experimental actual x 100.AMU, atomic.pdf
EAR = (1 + .127/365)^365 - 1, or 13.2%
Solution
EAR = (1 + .127/365)^365 - 1, or 13.2%.
Please follow the code and comments for description and outputs C.pdf
Erythrocyte Protoporphyrin
Erythrocyte Protoporphyrin ( hemoglobin precursor) begin to accumulate as hemoglobin as
hematocrit declineduring the 3rd stage of iron deficiency occurs when their is lack of iron limits
hemoglobin production.
World wide,how many people are estimated to have iron deficiency anemia
Ans; 2.0 to 2.5billion ( over 30% 0f world population according to WHO)
Solution
Erythrocyte Protoporphyrin
Erythrocyte Protoporphyrin ( hemoglobin precursor) begin to accumulate as hemoglobin as
hematocrit declineduring the 3rd stage of iron deficiency occurs when their is lack of iron limits
hemoglobin production.
World wide,how many people are estimated to have iron deficiency anemia
Ans; 2.0 to 2.5billion ( over 30% 0f world population according to WHO).
organization acted as the IdPThe fundamental concept introduced b.pdf
Dear,
The answer is 25.
Solution
Dear,
The answer is 25..
Let X = JohnSolutionLet X = John.pdf
Average concentration of atrazine in the inflow water = 8.0 mg/L
Volume of reservoir = 3.0 x 105 m3
Total annual flow = 1.5 x 105 m3
Annual outflow due to evaporation =1.2 x 105 m3
Average concentration of atrazine in the outflow = 6.2 mg/L
Total quantity of atrazine added into the reservoir = 8.0 x 1.5 x 105 x 1000 mg = 1200 kg
As,the volume of the reservoir is steady, i.e. it is constant,
Outflow = Inflow – Evaporation loss
= (1.5 x 105 m3) - (1.2 x 105 m3)
= 0.3 x 105 m3
Total quantity of atrazine leaving the system = 6.2 mg/L x 0.3 x 105 m3 x 1000L/m3 = 186 kg
Total loss of atrazine by mass in the reservoir in the whole year = 1200 = 1014 kg
Internal Sink rate = 1014 kg/(3 x105 m3 x 365) = 9.26 mg/m3-day
Please provide feedback.
Solution
Average concentration of atrazine in the inflow water = 8.0 mg/L
Volume of reservoir = 3.0 x 105 m3
Total annual flow = 1.5 x 105 m3
Annual outflow due to evaporation =1.2 x 105 m3
Average concentration of atrazine in the outflow = 6.2 mg/L
Total quantity of atrazine added into the reservoir = 8.0 x 1.5 x 105 x 1000 mg = 1200 kg
As,the volume of the reservoir is steady, i.e. it is constant,
Outflow = Inflow – Evaporation loss
= (1.5 x 105 m3) - (1.2 x 105 m3)
= 0.3 x 105 m3
Total quantity of atrazine leaving the system = 6.2 mg/L x 0.3 x 105 m3 x 1000L/m3 = 186 kg
Total loss of atrazine by mass in the reservoir in the whole year = 1200 = 1014 kg
Internal Sink rate = 1014 kg/(3 x105 m3 x 365) = 9.26 mg/m3-day
Please provide feedback..
In the United States, policy makers, health professionals, and patie.pdf
Answer: Muscle temparature.
Proprioreceptor involves in the detection of changes in tension in muscles and tendons and
fibrous capsules in the joints. But temperature is sensed by specific receptors.
Solution
Answer: Muscle temparature.
Proprioreceptor involves in the detection of changes in tension in muscles and tendons and
fibrous capsules in the joints. But temperature is sensed by specific receptors..
Recommended
System is a set of interacting or interdependent entities forming an.pdf
Hi,
I have updated the code. Highlighted the code changes below. we can end the loop by break
statement.
#include
using namespace std;
int main()
{
int x = 10;
int y =20;
int z = x + y;
while(z == z){
if(y ==z){
do{
for(int i=0; i<2; i++){
cout<<\"This is the answer \"<
Solution
Hi,
I have updated the code. Highlighted the code changes below. we can end the loop by break
statement.
#include
using namespace std;
int main()
{
int x = 10;
int y =20;
int z = x + y;
while(z == z){
if(y ==z){
do{
for(int i=0; i<2; i++){
cout<<\"This is the answer \"<.
Solution Given thatBy definition of logarithm, we getSolution.pdf
Given
F1 = -5i+3j-4k
F2 = 5i-5j-2k
The position vectors of the forces F1 and F2 are r1 and r2 respectively
From figure we can write
r1 = 4 i -0.15 j + 0.25 k
r2 = 4 i +0.15 j + 0.25 k
Now the equivalent force at O is
FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k
FR = -2 j -6 k
Therefore
FRx, FRy,FRz = (0, -2, -6)
Now the resultant moment about O is
MR,O = r1XF1 + r2X F2
MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k)
MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i
MR,O = 0.8 i + 24 j -9.5 k
so
(MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5)
Solution
Given
F1 = -5i+3j-4k
F2 = 5i-5j-2k
The position vectors of the forces F1 and F2 are r1 and r2 respectively
From figure we can write
r1 = 4 i -0.15 j + 0.25 k
r2 = 4 i +0.15 j + 0.25 k
Now the equivalent force at O is
FR = F1 + F2 = -5i+3j-4k+ 5i-5j-2k
FR = -2 j -6 k
Therefore
FRx, FRy,FRz = (0, -2, -6)
Now the resultant moment about O is
MR,O = r1XF1 + r2X F2
MR,O =(4 i -0.15 j + 0.25 k)X(-5i+3j-4k)+(4 i +0.15 j + 0.25 k)X( 5i-5j-2k)
MR,O =12 k +16 j -0.75 k+0.6 i -1.25 j -0.75 i -20 k+8 j -0.75 k -0.3 i +1.25 j + 1.25 i
MR,O = 0.8 i + 24 j -9.5 k
so
(MR,Ox,MR,Oy, MR,Oz) = (0.8, 24, -9.5).
C is the abbrevation for Carbon .pdf
C is the abbrevation for Carbon
Solution
C is the abbrevation for Carbon.
Percent error actual - experimental actual x 100.AMU, atomic.pdf
EAR = (1 + .127/365)^365 - 1, or 13.2%
Solution
EAR = (1 + .127/365)^365 - 1, or 13.2%.
Please follow the code and comments for description and outputs C.pdf
Erythrocyte Protoporphyrin
Erythrocyte Protoporphyrin ( hemoglobin precursor) begin to accumulate as hemoglobin as
hematocrit declineduring the 3rd stage of iron deficiency occurs when their is lack of iron limits
hemoglobin production.
World wide,how many people are estimated to have iron deficiency anemia
Ans; 2.0 to 2.5billion ( over 30% 0f world population according to WHO)
Solution
Erythrocyte Protoporphyrin
Erythrocyte Protoporphyrin ( hemoglobin precursor) begin to accumulate as hemoglobin as
hematocrit declineduring the 3rd stage of iron deficiency occurs when their is lack of iron limits
hemoglobin production.
World wide,how many people are estimated to have iron deficiency anemia
Ans; 2.0 to 2.5billion ( over 30% 0f world population according to WHO).
organization acted as the IdPThe fundamental concept introduced b.pdf
Dear,
The answer is 25.
Solution
Dear,
The answer is 25..
Let X = JohnSolutionLet X = John.pdf
Average concentration of atrazine in the inflow water = 8.0 mg/L
Volume of reservoir = 3.0 x 105 m3
Total annual flow = 1.5 x 105 m3
Annual outflow due to evaporation =1.2 x 105 m3
Average concentration of atrazine in the outflow = 6.2 mg/L
Total quantity of atrazine added into the reservoir = 8.0 x 1.5 x 105 x 1000 mg = 1200 kg
As,the volume of the reservoir is steady, i.e. it is constant,
Outflow = Inflow – Evaporation loss
= (1.5 x 105 m3) - (1.2 x 105 m3)
= 0.3 x 105 m3
Total quantity of atrazine leaving the system = 6.2 mg/L x 0.3 x 105 m3 x 1000L/m3 = 186 kg
Total loss of atrazine by mass in the reservoir in the whole year = 1200 = 1014 kg
Internal Sink rate = 1014 kg/(3 x105 m3 x 365) = 9.26 mg/m3-day
Please provide feedback.
Solution
Average concentration of atrazine in the inflow water = 8.0 mg/L
Volume of reservoir = 3.0 x 105 m3
Total annual flow = 1.5 x 105 m3
Annual outflow due to evaporation =1.2 x 105 m3
Average concentration of atrazine in the outflow = 6.2 mg/L
Total quantity of atrazine added into the reservoir = 8.0 x 1.5 x 105 x 1000 mg = 1200 kg
As,the volume of the reservoir is steady, i.e. it is constant,
Outflow = Inflow – Evaporation loss
= (1.5 x 105 m3) - (1.2 x 105 m3)
= 0.3 x 105 m3
Total quantity of atrazine leaving the system = 6.2 mg/L x 0.3 x 105 m3 x 1000L/m3 = 186 kg
Total loss of atrazine by mass in the reservoir in the whole year = 1200 = 1014 kg
Internal Sink rate = 1014 kg/(3 x105 m3 x 365) = 9.26 mg/m3-day
Please provide feedback..
In the United States, policy makers, health professionals, and patie.pdf
Answer: Muscle temparature.
Proprioreceptor involves in the detection of changes in tension in muscles and tendons and
fibrous capsules in the joints. But temperature is sensed by specific receptors.
Solution
Answer: Muscle temparature.
Proprioreceptor involves in the detection of changes in tension in muscles and tendons and
fibrous capsules in the joints. But temperature is sensed by specific receptors..
holding period return = (2D1P1Po)Solutionholding period retur.pdf
Ac- + H2O ~~> OH + AcH
Kb = [OH-][AcH]/[Ac-]
Kb=10^-14/Ka = 5.556x10^-10
[OH-]______[AcH]________[Ac-]
initial_____0_______0_________1M
change___+x______+x_________-x
final______x_______x__________0.052-x
Kb = x2/(0.052-x) = 5.556x10-10
Solve for x: x=5-log(0.535) = [OH-]
pOH = -log[OH] = 5.27
pH=14-pOH=8.73
Solution
Ac- + H2O ~~> OH + AcH
Kb = [OH-][AcH]/[Ac-]
Kb=10^-14/Ka = 5.556x10^-10
[OH-]______[AcH]________[Ac-]
initial_____0_______0_________1M
change___+x______+x_________-x
final______x_______x__________0.052-x
Kb = x2/(0.052-x) = 5.556x10-10
Solve for x: x=5-log(0.535) = [OH-]
pOH = -log[OH] = 5.27
pH=14-pOH=8.73.
create file name as board.javawrite below codeimport java.aw.pdf
100 DIM C(50) (50 is max number of servers) 110 INPUT S,NSTOP (S,NSTOP = number of
servers, customers to be
simulated) 120 FOR D=1 TO NSTOP 130 IA= (IA = interarrival time) 140 A=A+IA (A =
arrival time)
150 J=0 160 J=J+1 (J = index of server being probed) 170 IF J=S+1 THEN K=K+1 (K =
number of customers that are blocked) 180 IF J=S+1 THEN 270 190 IF AA THEN AB=AB+M-
A (AB = cumulative time during which all servers
busy) 270 NEXT D 280 PRINT K/NSTOP,AB/A (fraction of customers blocked, fraction of
time all servers are simultaneously busy)
Solution
100 DIM C(50) (50 is max number of servers) 110 INPUT S,NSTOP (S,NSTOP = number of
servers, customers to be
simulated) 120 FOR D=1 TO NSTOP 130 IA= (IA = interarrival time) 140 A=A+IA (A =
arrival time)
150 J=0 160 J=J+1 (J = index of server being probed) 170 IF J=S+1 THEN K=K+1 (K =
number of customers that are blocked) 180 IF J=S+1 THEN 270 190 IF AA THEN AB=AB+M-
A (AB = cumulative time during which all servers
busy) 270 NEXT D 280 PRINT K/NSTOP,AB/A (fraction of customers blocked, fraction of
time all servers are simultaneously busy).
Dear,The answer is 25.SolutionDear,The answer is 25..pdf
1
Suppose, annualized growth in inflation rate is I
4.5 = 1.2*(1+I) ^3
(1+I)^3 = 4.5/1.2 = 3.75= 1.553^ (1/3)
I = 55.3%
In an economy, when GDP is growing consistently and unemployment rate is also falling.
Though, growth rate in inflation rate is very high and it is approx. 55.3%. It means that demand
is increasing and it is more than the supply. It is also causing overheating of the economy. Here,
more money is chasing few goods. It leads to further increase in price. And it drives inflation on
the higher side. Data given the problem confirms the economic scenario mentioned. It propels
the inflation further and it is very high as a proof.
2
Inflation is growing with annualized rate of 55.3%. In this case, Federal Reserve should use
contraction policy. As a part of it, Federal Reserve should deploy right money policy. It will
discourage people to borrow and demand will come down. Thus, inflation will be controlled.
3.
Federal Reserve can increase reserve requirements. Further, Federal Reserve can also sell
government securities in open market. It will suck cash out of the system and contractionary
policy will be implemented. Reduced government expending will also help. It will help curbing
the inflation.
4
If recession happens and due to this reason, layoff takes place as well as demand decreases then
contractionary economic policy will be ineffective. In those circumstances, Federal Reserve has
to reverse the strategy.
Technological change can also make a reduction in demand of products. In that industry,
contractionary policy will be discouraging.
Also, for those product categories where demand is inelastic in nature. Contractionary policy will
not work.
Note: Already answered similar question on Chegg Q&A board. Thus, above answer is the
modified version as per the requirements mentioned in the question.
Solution
1
Suppose, annualized growth in inflation rate is I
4.5 = 1.2*(1+I) ^3
(1+I)^3 = 4.5/1.2 = 3.75= 1.553^ (1/3)
I = 55.3%
In an economy, when GDP is growing consistently and unemployment rate is also falling.
Though, growth rate in inflation rate is very high and it is approx. 55.3%. It means that demand
is increasing and it is more than the supply. It is also causing overheating of the economy. Here,
more money is chasing few goods. It leads to further increase in price. And it drives inflation on
the higher side. Data given the problem confirms the economic scenario mentioned. It propels
the inflation further and it is very high as a proof.
2
Inflation is growing with annualized rate of 55.3%. In this case, Federal Reserve should use
contraction policy. As a part of it, Federal Reserve should deploy right money policy. It will
discourage people to borrow and demand will come down. Thus, inflation will be controlled.
3.
Federal Reserve can increase reserve requirements. Further, Federal Reserve can also sell
government securities in open market. It will suck cash out of the system and contractionary
policy will be implemented. Red.
Cybrcoin or commonly known as Bitcoin is gaining ground from the las.pdf
11. SAMPLING FREQUENCY REQUIRED IS 3998 HZ
12. to provide equivalent digital binary output
13.half of maximum frequency content of analog input
14. MOSFET gate current is equivalent to ZERO
15. DC series motor IS STARTED WITHOUT NO LOAD
Solution
11. SAMPLING FREQUENCY REQUIRED IS 3998 HZ
12. to provide equivalent digital binary output
13.half of maximum frequency content of analog input
14. MOSFET gate current is equivalent to ZERO
15. DC series motor IS STARTED WITHOUT NO LOAD.
Baud rate states that the symbol of frequency in a communication cha.pdf
1) D because he earns 25 dollars plus 80*h (h being the number of overtime hours)
2) Each side is x+6 (Not like what you drew 2 sides with X and the other 2 side with 6)
b) G. because 4(x+6) = 2(x+4)+2(x+8)
3) F
b)yes he can buy the item
Solution
1) D because he earns 25 dollars plus 80*h (h being the number of overtime hours)
2) Each side is x+6 (Not like what you drew 2 sides with X and the other 2 side with 6)
b) G. because 4(x+6) = 2(x+4)+2(x+8)
3) F
b)yes he can buy the item.
b.) varies only in name from a one way analysis of varianceSolut.pdf
1) (-3xy^-3)^-4/(xy^-1)^-2 = (xy^-1)^2/(-3xy^-3)^4 ( revesring the num and denom.)
Now , (x^2*y^-2)/(81x^4*y^-12) = x^2-4*y^(-2+12)/81 = x^-2*y^10/81
2) x^2 +4x+3 by (x+3)
x^2 +4x +3 = x^2 +3x+x +3 ( Factorising)
x(x+3) +1(x+3) = (x+1)(x+3)
So, we have (x+1)(x+3)/(x+3) = (x+1)
3) 4b^2 -4b -3 by 2b-1
4b^ -6b+2b -3 = 2b(2b-3) +1(2b-3) ( factorising)
= (2b+1)(2b-3)
(2b+1)(2b-3)/(2b-1) = 2b-3 (cancelling out common term )
4) (7x^2 +x^3)/(x^2 +5x -14)
We have to look for points where denominator is zero and function is undefined
x^2 +5x -14 = x^2 +7x -2x -14 = x(x+7)-2(x+7)
= (x-2)(x+7)
Denominator is zero at x =2 , -7
Domain : All real except x = 2 and x =-7
Solution
1) (-3xy^-3)^-4/(xy^-1)^-2 = (xy^-1)^2/(-3xy^-3)^4 ( revesring the num and denom.)
Now , (x^2*y^-2)/(81x^4*y^-12) = x^2-4*y^(-2+12)/81 = x^-2*y^10/81
2) x^2 +4x+3 by (x+3)
x^2 +4x +3 = x^2 +3x+x +3 ( Factorising)
x(x+3) +1(x+3) = (x+1)(x+3)
So, we have (x+1)(x+3)/(x+3) = (x+1)
3) 4b^2 -4b -3 by 2b-1
4b^ -6b+2b -3 = 2b(2b-3) +1(2b-3) ( factorising)
= (2b+1)(2b-3)
(2b+1)(2b-3)/(2b-1) = 2b-3 (cancelling out common term )
4) (7x^2 +x^3)/(x^2 +5x -14)
We have to look for points where denominator is zero and function is undefined
x^2 +5x -14 = x^2 +7x -2x -14 = x(x+7)-2(x+7)
= (x-2)(x+7)
Denominator is zero at x =2 , -7
Domain : All real except x = 2 and x =-7.
AnswerThe evolutionary history of humans since the most recent co.pdf
(a) Ka for C6H5NH3+ = Kw/Kb for C6H5NH2
= 10-14/4.27 x 10-10 = 2.342 x 10-5
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log([C6H5NH2]/[C6H5NH3+])
4.20 = -log(2.342 x 10-5) + log(0.50/[C6H5NH3+])
log(0.50/[C6H5NH3+]) = -0.4304
0.50/[C6H5NH3+] = 10-0.4304 = 0.3712
[C6H5NH3+] = 1.347 1.35 M
(b) Initial moles of C6H5NH2 = volume x concentration
= 1.00 x 0.50 = 0.50 mol
Initial moles of C6H5NH3+ = volume x concentration
= 1.00 x 1.35 = 1.35 mol
C6H5NH3+ + NaOH => C6H5NH2 + H2O + Na+
Moles of NaOH added = mass/molar mass
= 4.00/40.00 = 0.100 mol
Moles of C6H5NH2 = initial moles of C6H5NH2 + moles of NaOH added
= 0.50 + 0.100 = 0.600 mol
Moles of C6H5NH3+ = initial moles of C6H5NH3+ - moles of NaOH added
= 1.35 - 0.100 = 1.25 mol
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log(moles of C6H5NH2/moles of C6H5NH3+) since volume = 1.00 L for both
= -log(2.342 x 10-5) + log(0.600/1.25)
= 4.31
Solution
(a) Ka for C6H5NH3+ = Kw/Kb for C6H5NH2
= 10-14/4.27 x 10-10 = 2.342 x 10-5
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log([C6H5NH2]/[C6H5NH3+])
4.20 = -log(2.342 x 10-5) + log(0.50/[C6H5NH3+])
log(0.50/[C6H5NH3+]) = -0.4304
0.50/[C6H5NH3+] = 10-0.4304 = 0.3712
[C6H5NH3+] = 1.347 1.35 M
(b) Initial moles of C6H5NH2 = volume x concentration
= 1.00 x 0.50 = 0.50 mol
Initial moles of C6H5NH3+ = volume x concentration
= 1.00 x 1.35 = 1.35 mol
C6H5NH3+ + NaOH => C6H5NH2 + H2O + Na+
Moles of NaOH added = mass/molar mass
= 4.00/40.00 = 0.100 mol
Moles of C6H5NH2 = initial moles of C6H5NH2 + moles of NaOH added
= 0.50 + 0.100 = 0.600 mol
Moles of C6H5NH3+ = initial moles of C6H5NH3+ - moles of NaOH added
= 1.35 - 0.100 = 1.25 mol
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log(moles of C6H5NH2/moles of C6H5NH3+) since volume = 1.00 L for both
= -log(2.342 x 10-5) + log(0.600/1.25)
= 4.31.
Answer-Cultural relativism is a complex concept that has its inte.pdf
(1)
Q = 50 - 6.25P
When demand increases by 20%, the new demand Q2 = 1.2 x Q
So:
1.2 x Q = 1.2 x (50 - 6.25P)
or,
Q1 = 60 - 7.5P
(2)
Original Qd = 50 - 6.25P
Qs = - 10 + 12.5P
In equilibrium, Qd = Qs
50 - 6.25P = - 10 + 12.5P
18.75P = 60
P = 3.20
Q = - 10 + (12.5 x 3.2) = 30
(3)
New Qd = 60 - 7.5P
Equating with Qs:
60 - 7.5P = - 10 + 12.5P
20P = 70
P = 3.50, Q = - 10 + (12.5 x 3.5) = 33.75
(4)
(a) elasticity of demand = Change in quantity demanded / change in price
= 0.20 / [(3.5 - 3.2) / 3.2] **
= 2.13
** change in quantity demanded = 20% (0.20) given
(b) Elasticity of supply = change in quantity supplied / change in price
= [(33.75 - 30) / 30] / [(3.5 - 3.2) / 3.2]
= 0.125 / 0.0938 = 1.33
Solution
(1)
Q = 50 - 6.25P
When demand increases by 20%, the new demand Q2 = 1.2 x Q
So:
1.2 x Q = 1.2 x (50 - 6.25P)
or,
Q1 = 60 - 7.5P
(2)
Original Qd = 50 - 6.25P
Qs = - 10 + 12.5P
In equilibrium, Qd = Qs
50 - 6.25P = - 10 + 12.5P
18.75P = 60
P = 3.20
Q = - 10 + (12.5 x 3.2) = 30
(3)
New Qd = 60 - 7.5P
Equating with Qs:
60 - 7.5P = - 10 + 12.5P
20P = 70
P = 3.50, Q = - 10 + (12.5 x 3.5) = 33.75
(4)
(a) elasticity of demand = Change in quantity demanded / change in price
= 0.20 / [(3.5 - 3.2) / 3.2] **
= 2.13
** change in quantity demanded = 20% (0.20) given
(b) Elasticity of supply = change in quantity supplied / change in price
= [(33.75 - 30) / 30] / [(3.5 - 3.2) / 3.2]
= 0.125 / 0.0938 = 1.33.
Answer The answer is (C)DoubleSolutionAnswer The a.pdf
#include
using namespace std;
int main()
{
int n;
cout<<\"enter n between 1 and 10 : \";
cin>>n;
cout<<\"\ Roman numeral is : \";
switch(n)
{
case 1:
cout<<\"I\";
break;
case 2:
cout<<\"II\";
break;
case 3:
cout<<\"III\";
break;
case 4:
cout<<\"IV\";
break;
case 5:
cout<<\"V\";
break;
case 6:
cout<<\"VI\";
break;
case 7:
cout<<\"VII\";
break;
case 8:
cout<<\"VIII\";
break;
case 9:
cout<<\"IX\";
break;
case 10:
cout<<\"X\";
break;
}
return 0;
}
Solution
#include
using namespace std;
int main()
{
int n;
cout<<\"enter n between 1 and 10 : \";
cin>>n;
cout<<\"\ Roman numeral is : \";
switch(n)
{
case 1:
cout<<\"I\";
break;
case 2:
cout<<\"II\";
break;
case 3:
cout<<\"III\";
break;
case 4:
cout<<\"IV\";
break;
case 5:
cout<<\"V\";
break;
case 6:
cout<<\"VI\";
break;
case 7:
cout<<\"VII\";
break;
case 8:
cout<<\"VIII\";
break;
case 9:
cout<<\"IX\";
break;
case 10:
cout<<\"X\";
break;
}
return 0;
}.
Advantages1.Easy to share data over a network2.Better communica.pdf
a. The results are inconclusive because
the king size cigarettes are different sizes than the filtered cigarettes.
b. No, the filters do not appear to
make a difference because there is insufficient evidence to warrant the
rejection of the claim.
c. No, the filters do not appear to
make difference because there is sufficient evidence to warrant the rejection
of the claim.
d. Given the king size cigarettes have
the largest mean, it appears that the filters do make a difference(although
this conclusion is not justified by the results from analysis of variance)
Solution
a. The results are inconclusive because
the king size cigarettes are different sizes than the filtered cigarettes.
b. No, the filters do not appear to
make a difference because there is insufficient evidence to warrant the
rejection of the claim.
c. No, the filters do not appear to
make difference because there is sufficient evidence to warrant the rejection
of the claim.
d. Given the king size cigarettes have
the largest mean, it appears that the filters do make a difference(although
this conclusion is not justified by the results from analysis of variance).
An S corporation is formed by filing an Articles of Incorporation wi.pdf
let be A=x1i+y1j
B=x2i+y2j
A+B=(x1+x2)i+(y1+y2)j
lA+Bl^2=(x1+x2)^2+(y1+y2)^2
now lAl^2=x1^2+y1^2
lBl^2=x2^2+y2^2
and AB=x1x2+y1y2
now
lAl^2+2AB+lBl^2=x1^2+x2^2+2x1x2+2y1y2+y1^2+y2^2=(x1+x2)^2+(y1+y2)^2=lA+Bl^2
there fore lA+Bl^2=lAl^2+2AB+lBl^2
or
lA+Bl^2=lAl^2+lBl^2+2lAllBl=lAl^2+2AB+lBl^2
Solution
let be A=x1i+y1j
B=x2i+y2j
A+B=(x1+x2)i+(y1+y2)j
lA+Bl^2=(x1+x2)^2+(y1+y2)^2
now lAl^2=x1^2+y1^2
lBl^2=x2^2+y2^2
and AB=x1x2+y1y2
now
lAl^2+2AB+lBl^2=x1^2+x2^2+2x1x2+2y1y2+y1^2+y2^2=(x1+x2)^2+(y1+y2)^2=lA+Bl^2
there fore lA+Bl^2=lAl^2+2AB+lBl^2
or
lA+Bl^2=lAl^2+lBl^2+2lAllBl=lAl^2+2AB+lBl^2.
A)depository institutionsB) contractual savings institutionsC)in.pdf
Income statement Net Sales / Income from Operations Rs m 8,741 Other
Operating Income Rs m 58 Total Income Rs m 8,799 Expenditure
(Increase) / Decrease in Finished / Trading Goods Rs m (74) Consumption of
Raw Materials and Movement in Work-in-Progress Rs m 5,686 Purchase of Trading
Goods Rs m 223 Employees Cost Rs m 532 Depreciation Rs m
207 Other Expenditure Rs m 1,370 Total Rs m 7,944
Profit / (Loss) from Operations Before Other Income, Financial Expenses and Exceptional Item
Rs m 855 Other Income Rs m - Profit Before Financial
Expenses and Exceptional Item Rs m 855 Interest Rs m 147
Profit / (Loss) After Financial Expenses But Before Exceptional Item Rs m 708
Exceptional Items Rs m - Profit from Ordinary Activities Before Tax Rs m
708 Tax Expense - Income Tax Rs m 241 Net Profit for the
Period Rs m 468 Balance Sheet ( Rs m) Sources of Funds Shareholders\'
Fund Share Capital 383 Equity Share Warrants - Reserves and
Surplus 5,956 Total 6,340 Minority - Loan Fund
Secured Loans 3,810 Unsecured Loans 3,690 Total 7,500
Deferred Tax Liability Net 1,052 Total 14,892 Application of
Funds Fixed Assets Gross Block 13,106 Less: Depreciation 4,699
Net Block 8,407 Capital Work-in-Progress 779 Share in Joint
Venture - Goodwill on Consolidation - Investments
4 Current Assets, Loans and Advances Inventories 4,194
Sundry Debtors 1,752 Cash and Bank Balances 2,315 Other Current
Assets 2 Loans and Advances 1,844 Total Current Assets, Loans
and Advances 10,106 Less: Current Liabilities and Provisions Current
Liabilities 3,886 Provisions 520 Total Current Liabilities and
Provisions 4,407 Net Current Assets 5,699 Deferred Revenue
Expenditure 3 Total 14,892 Change in Retained earning
Opening Balance Rs m 1,844 Add Transfer from Profit and Loss Account Rs m
500 Less Transfer to Profit & Loss Account Rs m (436) Common
Stock Repurchased and Cancelled Rs m - Closing Balance Rs m
1,908
Solution
Income statement Net Sales / Income from Operations Rs m 8,741 Other
Operating Income Rs m 58 Total Income Rs m 8,799 Expenditure
(Increase) / Decrease in Finished / Trading Goods Rs m (74) Consumption of
Raw Materials and Movement in Work-in-Progress Rs m 5,686 Purchase of Trading
Goods Rs m 223 Employees Cost Rs m 532 Depreciation Rs m
207 Other Expenditure Rs m 1,370 Total Rs m 7,944
Profit / (Loss) from Operations Before Other Income, Financial Expenses and Exceptional Item
Rs m 855 Other Income Rs m - Profit Before Financial
Expenses and Exceptional Item Rs m 855 Interest Rs m 147
Profit / (Loss) After Financial Expenses But Before Exceptional Item Rs m 708
Exceptional Items Rs m - Profit from Ordinary Activities Before Tax Rs m
708 Tax Expense - Income Tax Rs m 241 Net Profit for the
Period Rs m 468 Balance Sheet ( Rs m) Sources of Funds Shareholders\'
Fund Share Capital 383 Equity Share Warrants - Reserves and
Surplus 5,956 Total 6,340 Minority - Loan Fund
Secured Loans 3,810 Unsecured Loans 3,690 Total 7,500
Deferred Tax Liability Net 1,052 Total 14,892 Application of
F.
A is correct. History holds ten commands in the buffer by default..pdf
whats the right answer ??
Solution
whats the right answer ??.
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4.pdf
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4:5. moles of O2 = (0.221 mol/4) x
5 = 0.276 mol
Solution
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4:5. moles of O2 = (0.221 mol/4) x
5 = 0.276 mol.
1. Corporations, limited liability companies (LLCs), general and lim.pdf
resulting solution is acidic as CH3NH3Cl is weak acid.
Solution
resulting solution is acidic as CH3NH3Cl is weak acid..
Using the dilution formula M1V1 = M2V2 12 20 = .pdf
Co has +3 ionic charge
Solution
Co has +3 ionic charge.
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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holding period return = (2D1P1Po)Solutionholding period retur.pdf
Ac- + H2O ~~> OH + AcH
Kb = [OH-][AcH]/[Ac-]
Kb=10^-14/Ka = 5.556x10^-10
[OH-]______[AcH]________[Ac-]
initial_____0_______0_________1M
change___+x______+x_________-x
final______x_______x__________0.052-x
Kb = x2/(0.052-x) = 5.556x10-10
Solve for x: x=5-log(0.535) = [OH-]
pOH = -log[OH] = 5.27
pH=14-pOH=8.73
Solution
Ac- + H2O ~~> OH + AcH
Kb = [OH-][AcH]/[Ac-]
Kb=10^-14/Ka = 5.556x10^-10
[OH-]______[AcH]________[Ac-]
initial_____0_______0_________1M
change___+x______+x_________-x
final______x_______x__________0.052-x
Kb = x2/(0.052-x) = 5.556x10-10
Solve for x: x=5-log(0.535) = [OH-]
pOH = -log[OH] = 5.27
pH=14-pOH=8.73.
create file name as board.javawrite below codeimport java.aw.pdf
100 DIM C(50) (50 is max number of servers) 110 INPUT S,NSTOP (S,NSTOP = number of
servers, customers to be
simulated) 120 FOR D=1 TO NSTOP 130 IA= (IA = interarrival time) 140 A=A+IA (A =
arrival time)
150 J=0 160 J=J+1 (J = index of server being probed) 170 IF J=S+1 THEN K=K+1 (K =
number of customers that are blocked) 180 IF J=S+1 THEN 270 190 IF AA THEN AB=AB+M-
A (AB = cumulative time during which all servers
busy) 270 NEXT D 280 PRINT K/NSTOP,AB/A (fraction of customers blocked, fraction of
time all servers are simultaneously busy)
Solution
100 DIM C(50) (50 is max number of servers) 110 INPUT S,NSTOP (S,NSTOP = number of
servers, customers to be
simulated) 120 FOR D=1 TO NSTOP 130 IA= (IA = interarrival time) 140 A=A+IA (A =
arrival time)
150 J=0 160 J=J+1 (J = index of server being probed) 170 IF J=S+1 THEN K=K+1 (K =
number of customers that are blocked) 180 IF J=S+1 THEN 270 190 IF AA THEN AB=AB+M-
A (AB = cumulative time during which all servers
busy) 270 NEXT D 280 PRINT K/NSTOP,AB/A (fraction of customers blocked, fraction of
time all servers are simultaneously busy).
Dear,The answer is 25.SolutionDear,The answer is 25..pdf
1
Suppose, annualized growth in inflation rate is I
4.5 = 1.2*(1+I) ^3
(1+I)^3 = 4.5/1.2 = 3.75= 1.553^ (1/3)
I = 55.3%
In an economy, when GDP is growing consistently and unemployment rate is also falling.
Though, growth rate in inflation rate is very high and it is approx. 55.3%. It means that demand
is increasing and it is more than the supply. It is also causing overheating of the economy. Here,
more money is chasing few goods. It leads to further increase in price. And it drives inflation on
the higher side. Data given the problem confirms the economic scenario mentioned. It propels
the inflation further and it is very high as a proof.
2
Inflation is growing with annualized rate of 55.3%. In this case, Federal Reserve should use
contraction policy. As a part of it, Federal Reserve should deploy right money policy. It will
discourage people to borrow and demand will come down. Thus, inflation will be controlled.
3.
Federal Reserve can increase reserve requirements. Further, Federal Reserve can also sell
government securities in open market. It will suck cash out of the system and contractionary
policy will be implemented. Reduced government expending will also help. It will help curbing
the inflation.
4
If recession happens and due to this reason, layoff takes place as well as demand decreases then
contractionary economic policy will be ineffective. In those circumstances, Federal Reserve has
to reverse the strategy.
Technological change can also make a reduction in demand of products. In that industry,
contractionary policy will be discouraging.
Also, for those product categories where demand is inelastic in nature. Contractionary policy will
not work.
Note: Already answered similar question on Chegg Q&A board. Thus, above answer is the
modified version as per the requirements mentioned in the question.
Solution
1
Suppose, annualized growth in inflation rate is I
4.5 = 1.2*(1+I) ^3
(1+I)^3 = 4.5/1.2 = 3.75= 1.553^ (1/3)
I = 55.3%
In an economy, when GDP is growing consistently and unemployment rate is also falling.
Though, growth rate in inflation rate is very high and it is approx. 55.3%. It means that demand
is increasing and it is more than the supply. It is also causing overheating of the economy. Here,
more money is chasing few goods. It leads to further increase in price. And it drives inflation on
the higher side. Data given the problem confirms the economic scenario mentioned. It propels
the inflation further and it is very high as a proof.
2
Inflation is growing with annualized rate of 55.3%. In this case, Federal Reserve should use
contraction policy. As a part of it, Federal Reserve should deploy right money policy. It will
discourage people to borrow and demand will come down. Thus, inflation will be controlled.
3.
Federal Reserve can increase reserve requirements. Further, Federal Reserve can also sell
government securities in open market. It will suck cash out of the system and contractionary
policy will be implemented. Red.
Cybrcoin or commonly known as Bitcoin is gaining ground from the las.pdf
11. SAMPLING FREQUENCY REQUIRED IS 3998 HZ
12. to provide equivalent digital binary output
13.half of maximum frequency content of analog input
14. MOSFET gate current is equivalent to ZERO
15. DC series motor IS STARTED WITHOUT NO LOAD
Solution
11. SAMPLING FREQUENCY REQUIRED IS 3998 HZ
12. to provide equivalent digital binary output
13.half of maximum frequency content of analog input
14. MOSFET gate current is equivalent to ZERO
15. DC series motor IS STARTED WITHOUT NO LOAD.
Baud rate states that the symbol of frequency in a communication cha.pdf
1) D because he earns 25 dollars plus 80*h (h being the number of overtime hours)
2) Each side is x+6 (Not like what you drew 2 sides with X and the other 2 side with 6)
b) G. because 4(x+6) = 2(x+4)+2(x+8)
3) F
b)yes he can buy the item
Solution
1) D because he earns 25 dollars plus 80*h (h being the number of overtime hours)
2) Each side is x+6 (Not like what you drew 2 sides with X and the other 2 side with 6)
b) G. because 4(x+6) = 2(x+4)+2(x+8)
3) F
b)yes he can buy the item.
b.) varies only in name from a one way analysis of varianceSolut.pdf
1) (-3xy^-3)^-4/(xy^-1)^-2 = (xy^-1)^2/(-3xy^-3)^4 ( revesring the num and denom.)
Now , (x^2*y^-2)/(81x^4*y^-12) = x^2-4*y^(-2+12)/81 = x^-2*y^10/81
2) x^2 +4x+3 by (x+3)
x^2 +4x +3 = x^2 +3x+x +3 ( Factorising)
x(x+3) +1(x+3) = (x+1)(x+3)
So, we have (x+1)(x+3)/(x+3) = (x+1)
3) 4b^2 -4b -3 by 2b-1
4b^ -6b+2b -3 = 2b(2b-3) +1(2b-3) ( factorising)
= (2b+1)(2b-3)
(2b+1)(2b-3)/(2b-1) = 2b-3 (cancelling out common term )
4) (7x^2 +x^3)/(x^2 +5x -14)
We have to look for points where denominator is zero and function is undefined
x^2 +5x -14 = x^2 +7x -2x -14 = x(x+7)-2(x+7)
= (x-2)(x+7)
Denominator is zero at x =2 , -7
Domain : All real except x = 2 and x =-7
Solution
1) (-3xy^-3)^-4/(xy^-1)^-2 = (xy^-1)^2/(-3xy^-3)^4 ( revesring the num and denom.)
Now , (x^2*y^-2)/(81x^4*y^-12) = x^2-4*y^(-2+12)/81 = x^-2*y^10/81
2) x^2 +4x+3 by (x+3)
x^2 +4x +3 = x^2 +3x+x +3 ( Factorising)
x(x+3) +1(x+3) = (x+1)(x+3)
So, we have (x+1)(x+3)/(x+3) = (x+1)
3) 4b^2 -4b -3 by 2b-1
4b^ -6b+2b -3 = 2b(2b-3) +1(2b-3) ( factorising)
= (2b+1)(2b-3)
(2b+1)(2b-3)/(2b-1) = 2b-3 (cancelling out common term )
4) (7x^2 +x^3)/(x^2 +5x -14)
We have to look for points where denominator is zero and function is undefined
x^2 +5x -14 = x^2 +7x -2x -14 = x(x+7)-2(x+7)
= (x-2)(x+7)
Denominator is zero at x =2 , -7
Domain : All real except x = 2 and x =-7.
AnswerThe evolutionary history of humans since the most recent co.pdf
(a) Ka for C6H5NH3+ = Kw/Kb for C6H5NH2
= 10-14/4.27 x 10-10 = 2.342 x 10-5
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log([C6H5NH2]/[C6H5NH3+])
4.20 = -log(2.342 x 10-5) + log(0.50/[C6H5NH3+])
log(0.50/[C6H5NH3+]) = -0.4304
0.50/[C6H5NH3+] = 10-0.4304 = 0.3712
[C6H5NH3+] = 1.347 1.35 M
(b) Initial moles of C6H5NH2 = volume x concentration
= 1.00 x 0.50 = 0.50 mol
Initial moles of C6H5NH3+ = volume x concentration
= 1.00 x 1.35 = 1.35 mol
C6H5NH3+ + NaOH => C6H5NH2 + H2O + Na+
Moles of NaOH added = mass/molar mass
= 4.00/40.00 = 0.100 mol
Moles of C6H5NH2 = initial moles of C6H5NH2 + moles of NaOH added
= 0.50 + 0.100 = 0.600 mol
Moles of C6H5NH3+ = initial moles of C6H5NH3+ - moles of NaOH added
= 1.35 - 0.100 = 1.25 mol
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log(moles of C6H5NH2/moles of C6H5NH3+) since volume = 1.00 L for both
= -log(2.342 x 10-5) + log(0.600/1.25)
= 4.31
Solution
(a) Ka for C6H5NH3+ = Kw/Kb for C6H5NH2
= 10-14/4.27 x 10-10 = 2.342 x 10-5
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log([C6H5NH2]/[C6H5NH3+])
4.20 = -log(2.342 x 10-5) + log(0.50/[C6H5NH3+])
log(0.50/[C6H5NH3+]) = -0.4304
0.50/[C6H5NH3+] = 10-0.4304 = 0.3712
[C6H5NH3+] = 1.347 1.35 M
(b) Initial moles of C6H5NH2 = volume x concentration
= 1.00 x 0.50 = 0.50 mol
Initial moles of C6H5NH3+ = volume x concentration
= 1.00 x 1.35 = 1.35 mol
C6H5NH3+ + NaOH => C6H5NH2 + H2O + Na+
Moles of NaOH added = mass/molar mass
= 4.00/40.00 = 0.100 mol
Moles of C6H5NH2 = initial moles of C6H5NH2 + moles of NaOH added
= 0.50 + 0.100 = 0.600 mol
Moles of C6H5NH3+ = initial moles of C6H5NH3+ - moles of NaOH added
= 1.35 - 0.100 = 1.25 mol
Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH2]/[C6H5NH3+])
= -log Ka + log(moles of C6H5NH2/moles of C6H5NH3+) since volume = 1.00 L for both
= -log(2.342 x 10-5) + log(0.600/1.25)
= 4.31.
Answer-Cultural relativism is a complex concept that has its inte.pdf
(1)
Q = 50 - 6.25P
When demand increases by 20%, the new demand Q2 = 1.2 x Q
So:
1.2 x Q = 1.2 x (50 - 6.25P)
or,
Q1 = 60 - 7.5P
(2)
Original Qd = 50 - 6.25P
Qs = - 10 + 12.5P
In equilibrium, Qd = Qs
50 - 6.25P = - 10 + 12.5P
18.75P = 60
P = 3.20
Q = - 10 + (12.5 x 3.2) = 30
(3)
New Qd = 60 - 7.5P
Equating with Qs:
60 - 7.5P = - 10 + 12.5P
20P = 70
P = 3.50, Q = - 10 + (12.5 x 3.5) = 33.75
(4)
(a) elasticity of demand = Change in quantity demanded / change in price
= 0.20 / [(3.5 - 3.2) / 3.2] **
= 2.13
** change in quantity demanded = 20% (0.20) given
(b) Elasticity of supply = change in quantity supplied / change in price
= [(33.75 - 30) / 30] / [(3.5 - 3.2) / 3.2]
= 0.125 / 0.0938 = 1.33
Solution
(1)
Q = 50 - 6.25P
When demand increases by 20%, the new demand Q2 = 1.2 x Q
So:
1.2 x Q = 1.2 x (50 - 6.25P)
or,
Q1 = 60 - 7.5P
(2)
Original Qd = 50 - 6.25P
Qs = - 10 + 12.5P
In equilibrium, Qd = Qs
50 - 6.25P = - 10 + 12.5P
18.75P = 60
P = 3.20
Q = - 10 + (12.5 x 3.2) = 30
(3)
New Qd = 60 - 7.5P
Equating with Qs:
60 - 7.5P = - 10 + 12.5P
20P = 70
P = 3.50, Q = - 10 + (12.5 x 3.5) = 33.75
(4)
(a) elasticity of demand = Change in quantity demanded / change in price
= 0.20 / [(3.5 - 3.2) / 3.2] **
= 2.13
** change in quantity demanded = 20% (0.20) given
(b) Elasticity of supply = change in quantity supplied / change in price
= [(33.75 - 30) / 30] / [(3.5 - 3.2) / 3.2]
= 0.125 / 0.0938 = 1.33.
Answer The answer is (C)DoubleSolutionAnswer The a.pdf
#include
using namespace std;
int main()
{
int n;
cout<<\"enter n between 1 and 10 : \";
cin>>n;
cout<<\"\ Roman numeral is : \";
switch(n)
{
case 1:
cout<<\"I\";
break;
case 2:
cout<<\"II\";
break;
case 3:
cout<<\"III\";
break;
case 4:
cout<<\"IV\";
break;
case 5:
cout<<\"V\";
break;
case 6:
cout<<\"VI\";
break;
case 7:
cout<<\"VII\";
break;
case 8:
cout<<\"VIII\";
break;
case 9:
cout<<\"IX\";
break;
case 10:
cout<<\"X\";
break;
}
return 0;
}
Solution
#include
using namespace std;
int main()
{
int n;
cout<<\"enter n between 1 and 10 : \";
cin>>n;
cout<<\"\ Roman numeral is : \";
switch(n)
{
case 1:
cout<<\"I\";
break;
case 2:
cout<<\"II\";
break;
case 3:
cout<<\"III\";
break;
case 4:
cout<<\"IV\";
break;
case 5:
cout<<\"V\";
break;
case 6:
cout<<\"VI\";
break;
case 7:
cout<<\"VII\";
break;
case 8:
cout<<\"VIII\";
break;
case 9:
cout<<\"IX\";
break;
case 10:
cout<<\"X\";
break;
}
return 0;
}.
Advantages1.Easy to share data over a network2.Better communica.pdf
a. The results are inconclusive because
the king size cigarettes are different sizes than the filtered cigarettes.
b. No, the filters do not appear to
make a difference because there is insufficient evidence to warrant the
rejection of the claim.
c. No, the filters do not appear to
make difference because there is sufficient evidence to warrant the rejection
of the claim.
d. Given the king size cigarettes have
the largest mean, it appears that the filters do make a difference(although
this conclusion is not justified by the results from analysis of variance)
Solution
a. The results are inconclusive because
the king size cigarettes are different sizes than the filtered cigarettes.
b. No, the filters do not appear to
make a difference because there is insufficient evidence to warrant the
rejection of the claim.
c. No, the filters do not appear to
make difference because there is sufficient evidence to warrant the rejection
of the claim.
d. Given the king size cigarettes have
the largest mean, it appears that the filters do make a difference(although
this conclusion is not justified by the results from analysis of variance).
An S corporation is formed by filing an Articles of Incorporation wi.pdf
let be A=x1i+y1j
B=x2i+y2j
A+B=(x1+x2)i+(y1+y2)j
lA+Bl^2=(x1+x2)^2+(y1+y2)^2
now lAl^2=x1^2+y1^2
lBl^2=x2^2+y2^2
and AB=x1x2+y1y2
now
lAl^2+2AB+lBl^2=x1^2+x2^2+2x1x2+2y1y2+y1^2+y2^2=(x1+x2)^2+(y1+y2)^2=lA+Bl^2
there fore lA+Bl^2=lAl^2+2AB+lBl^2
or
lA+Bl^2=lAl^2+lBl^2+2lAllBl=lAl^2+2AB+lBl^2
Solution
let be A=x1i+y1j
B=x2i+y2j
A+B=(x1+x2)i+(y1+y2)j
lA+Bl^2=(x1+x2)^2+(y1+y2)^2
now lAl^2=x1^2+y1^2
lBl^2=x2^2+y2^2
and AB=x1x2+y1y2
now
lAl^2+2AB+lBl^2=x1^2+x2^2+2x1x2+2y1y2+y1^2+y2^2=(x1+x2)^2+(y1+y2)^2=lA+Bl^2
there fore lA+Bl^2=lAl^2+2AB+lBl^2
or
lA+Bl^2=lAl^2+lBl^2+2lAllBl=lAl^2+2AB+lBl^2.
A)depository institutionsB) contractual savings institutionsC)in.pdf
Income statement Net Sales / Income from Operations Rs m 8,741 Other
Operating Income Rs m 58 Total Income Rs m 8,799 Expenditure
(Increase) / Decrease in Finished / Trading Goods Rs m (74) Consumption of
Raw Materials and Movement in Work-in-Progress Rs m 5,686 Purchase of Trading
Goods Rs m 223 Employees Cost Rs m 532 Depreciation Rs m
207 Other Expenditure Rs m 1,370 Total Rs m 7,944
Profit / (Loss) from Operations Before Other Income, Financial Expenses and Exceptional Item
Rs m 855 Other Income Rs m - Profit Before Financial
Expenses and Exceptional Item Rs m 855 Interest Rs m 147
Profit / (Loss) After Financial Expenses But Before Exceptional Item Rs m 708
Exceptional Items Rs m - Profit from Ordinary Activities Before Tax Rs m
708 Tax Expense - Income Tax Rs m 241 Net Profit for the
Period Rs m 468 Balance Sheet ( Rs m) Sources of Funds Shareholders\'
Fund Share Capital 383 Equity Share Warrants - Reserves and
Surplus 5,956 Total 6,340 Minority - Loan Fund
Secured Loans 3,810 Unsecured Loans 3,690 Total 7,500
Deferred Tax Liability Net 1,052 Total 14,892 Application of
Funds Fixed Assets Gross Block 13,106 Less: Depreciation 4,699
Net Block 8,407 Capital Work-in-Progress 779 Share in Joint
Venture - Goodwill on Consolidation - Investments
4 Current Assets, Loans and Advances Inventories 4,194
Sundry Debtors 1,752 Cash and Bank Balances 2,315 Other Current
Assets 2 Loans and Advances 1,844 Total Current Assets, Loans
and Advances 10,106 Less: Current Liabilities and Provisions Current
Liabilities 3,886 Provisions 520 Total Current Liabilities and
Provisions 4,407 Net Current Assets 5,699 Deferred Revenue
Expenditure 3 Total 14,892 Change in Retained earning
Opening Balance Rs m 1,844 Add Transfer from Profit and Loss Account Rs m
500 Less Transfer to Profit & Loss Account Rs m (436) Common
Stock Repurchased and Cancelled Rs m - Closing Balance Rs m
1,908
Solution
Income statement Net Sales / Income from Operations Rs m 8,741 Other
Operating Income Rs m 58 Total Income Rs m 8,799 Expenditure
(Increase) / Decrease in Finished / Trading Goods Rs m (74) Consumption of
Raw Materials and Movement in Work-in-Progress Rs m 5,686 Purchase of Trading
Goods Rs m 223 Employees Cost Rs m 532 Depreciation Rs m
207 Other Expenditure Rs m 1,370 Total Rs m 7,944
Profit / (Loss) from Operations Before Other Income, Financial Expenses and Exceptional Item
Rs m 855 Other Income Rs m - Profit Before Financial
Expenses and Exceptional Item Rs m 855 Interest Rs m 147
Profit / (Loss) After Financial Expenses But Before Exceptional Item Rs m 708
Exceptional Items Rs m - Profit from Ordinary Activities Before Tax Rs m
708 Tax Expense - Income Tax Rs m 241 Net Profit for the
Period Rs m 468 Balance Sheet ( Rs m) Sources of Funds Shareholders\'
Fund Share Capital 383 Equity Share Warrants - Reserves and
Surplus 5,956 Total 6,340 Minority - Loan Fund
Secured Loans 3,810 Unsecured Loans 3,690 Total 7,500
Deferred Tax Liability Net 1,052 Total 14,892 Application of
F.
A is correct. History holds ten commands in the buffer by default..pdf
whats the right answer ??
Solution
whats the right answer ??.
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4.pdf
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4:5. moles of O2 = (0.221 mol/4) x
5 = 0.276 mol
Solution
4P + 5O2 = 2P2O5 the molar ratio of P to O2 is 4:5. moles of O2 = (0.221 mol/4) x
5 = 0.276 mol.
1. Corporations, limited liability companies (LLCs), general and lim.pdf
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- 1. Can't see the question Solution Can't see the question