This is a titration of a weak base w a strong ac.pdf
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This equation makes it able for you to plug-and-play the numbers. Let\'s classify the variables first. n is the number of people p is the profit acquired If you plug in p=0 (the situation when the profit is 0), you get 0 = 10(120-n)-50n Now all you have to do is solve for n. Solution This equation makes it able for you to plug-and-play the numbers. Let\'s classify the variables first. n is the number of people p is the profit acquired If you plug in p=0 (the situation when the profit is 0), you get 0 = 10(120-n)-50n Now all you have to do is solve for n..
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The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
There are two ways to solve this problem. I will .pdf
The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight
comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the
weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the
problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that
?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements
with atomic numbers less than roughly 24, which means that the element has one neutron for
every proton. Cesium, with an atomic number of 55, has considerably more neutrons than
protons in its nucleus. Therefore when measuring electron density, which is then used to
calculate bulk density, the algorithm should show an under-estimation of density due to this large
‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over-
estimated,” he continues. So what was the explanation? “When going through the data I noticed
an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long
story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher
photoelectric effect. This increases the absorption of gamma rays in the formation so that the
electron density appears higher than it is and physical density is therefore over estimated. This
far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now
StatoilHydro understood why the logging results were wrong when interpreted using
methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way
of interpreting the numbers accurately. The simplicity of the solution was stunning. After close
examination of the calibration data, a linear relationship was found between the photoelectric
factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk
density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is
that it relies only on the Pef measurements, which are logged simultaneously with bulk density,
and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model
The logged photoelectric factor isn’t only useful for correcting the density curve and estimating
saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the
photoelectric factor and its high vertical resolution with resistivity measurements from both the
drill pass and ream pass produces a very reliable and accurate net reservoir de?nition.
Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all
production wells high-quality resistivity image logs can be run. These logs provide important
information on structural dip, depositional environment, sedimentary features, facies and
geological correlations, all used by geomodelers to produce better reservoir models. “The results
match cor.
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
There are two ways to solve this problem. I will .pdf
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)..
Recommended
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
There are two ways to solve this problem. I will .pdf
The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight
comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the
weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the
problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that
?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements
with atomic numbers less than roughly 24, which means that the element has one neutron for
every proton. Cesium, with an atomic number of 55, has considerably more neutrons than
protons in its nucleus. Therefore when measuring electron density, which is then used to
calculate bulk density, the algorithm should show an under-estimation of density due to this large
‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over-
estimated,” he continues. So what was the explanation? “When going through the data I noticed
an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long
story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher
photoelectric effect. This increases the absorption of gamma rays in the formation so that the
electron density appears higher than it is and physical density is therefore over estimated. This
far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now
StatoilHydro understood why the logging results were wrong when interpreted using
methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way
of interpreting the numbers accurately. The simplicity of the solution was stunning. After close
examination of the calibration data, a linear relationship was found between the photoelectric
factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk
density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is
that it relies only on the Pef measurements, which are logged simultaneously with bulk density,
and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model
The logged photoelectric factor isn’t only useful for correcting the density curve and estimating
saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the
photoelectric factor and its high vertical resolution with resistivity measurements from both the
drill pass and ream pass produces a very reliable and accurate net reservoir de?nition.
Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all
production wells high-quality resistivity image logs can be run. These logs provide important
information on structural dip, depositional environment, sedimentary features, facies and
geological correlations, all used by geomodelers to produce better reservoir models. “The results
match cor.
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
There are two ways to solve this problem. I will .pdf
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)..
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
Advanced Java[Extra Concepts, Not Difficult].docx
This is part 2 of my Java Learning Journey. This contains Hashing, ArrayList, LinkedList, Date and Time Classes, Calendar Class and more.
Walmart Business+ and Spark Good for Nonprofits.pdf
"Learn about all the ways Walmart supports nonprofit organizations.
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The webinar may also give some examples on how nonprofits can best leverage Walmart Business+.
The event will cover the following::
Walmart Business + (https://business.walmart.com/plus) is a new shopping experience for nonprofits, schools, and local business customers that connects an exclusive online shopping experience to stores. Benefits include free delivery and shipping, a 'Spend Analytics” feature, special discounts, deals and tax-exempt shopping.
Special TechSoup offer for a free 180 days membership, and up to $150 in discounts on eligible orders.
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Answers about how you can do more with Walmart!"
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The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
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The empirical formula indicates the relative proportions of the elem.pdf
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If an oxide has the formula MO, then since O is normally found in th.pdf
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- 1. This is a titration of a weak base w/ a strong acid. So when you add the acid, it will dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form HA. So this will in affect become a buffered solution. So i believe you can use henderson hasselbach for this equation. Solution This is a titration of a weak base w/ a strong acid. So when you add the acid, it will dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form HA. So this will in affect become a buffered solution. So i believe you can use henderson hasselbach for this equation.