Answer 1 Basic Earning per shareBasic Earning per share = BEBS.pdf
•
0 likes•2 views
Answer 1 : Basic Earning per share Basic Earning per share = BE/BS = 25700/ 10600 = 2.42 Working Notes BE = Income available for common stock holders Net Income 50000 Less: Intt on Bonds (100000*6.5%) 6500 ------------- EBT 43500 Less Tax paid @ 30% 13050 ---------------- Income after tax 30450 Less Dividend paid to Prefered stock holder 4750 ------------------ Amount Available common 25700 Stock holder ----------- BS = Weighted avaerage no. of common stock = Share at begining 9600 May, 1 Share issued 1200 weight 1200 *8/12 800 SeP, 30 Share issued 800 Weight 800*3/12 200 ------------ Total no. of shares 10600 ----------------- Answer - 2 Dialute Earning Per share = Net Income/Avg shares+other convertible instuments = (50000-1950)/15100 = 48050/15100 = 3.18 Working notes Avg Share = 10600 Prefered stock (1000*2) = 2000 Bonds (100000/1000*25) = 2500 -------------------------- Total shares = 15100 --------------------------- Incremental EPS of Preferred Stock = 4750/2000 = 4750/2000 = 2.38 Incremental EPS of Bonds = interest/ incremental no. of shares due to conversion of bonds = 6500-1950/2500 = 4550/2500 = 1.82 Answer 3 Incremental Earning per share Bonds = $ 1.82 Preferred = $ 2.38 Solution Answer 1 : Basic Earning per share Basic Earning per share = BE/BS = 25700/ 10600 = 2.42 Working Notes BE = Income available for common stock holders Net Income 50000 Less: Intt on Bonds (100000*6.5%) 6500 ------------- EBT 43500 Less Tax paid @ 30% 13050 ---------------- Income after tax 30450 Less Dividend paid to Prefered stock holder 4750 ------------------ Amount Available common 25700 Stock holder ----------- BS = Weighted avaerage no. of common stock = Share at begining 9600 May, 1 Share issued 1200 weight 1200 *8/12 800 SeP, 30 Share issued 800 Weight 800*3/12 200 ------------ Total no. of shares 10600 ----------------- Answer - 2 Dialute Earning Per share = Net Income/Avg shares+other convertible instuments = (50000-1950)/15100 = 48050/15100 = 3.18 Working notes Avg Share = 10600 Prefered stock (1000*2) = 2000 Bonds (100000/1000*25) = 2500 -------------------------- Total shares = 15100 --------------------------- Incremental EPS of Preferred Stock = 4750/2000 = 4750/2000 = 2.38 Incremental EPS of Bonds = interest/ incremental no. of shares due to conversion of bonds = 6500-1950/2500 = 4550/2500 = 1.82 Answer 3 Incremental Earning per share Bonds = $ 1.82 Preferred = $ 2.38.
Report
Share
Report
Share
Download to read offline
Recommended
Step1 Error lies ; .1252 =.0625 Mole of O2 will .pdf
Step1 Error lies ; .125/2 =.0625 Mole of O2 will remain unreacted. Step2 We have
.125 mole NO2 and .0625 mole unreacted O2 Step3 P= nRT/V ; n=.1875; P=
.1875x.0821x300/4.5 =1.02625 atm=1.03 atm
Solution
Step1 Error lies ; .125/2 =.0625 Mole of O2 will remain unreacted. Step2 We have
.125 mole NO2 and .0625 mole unreacted O2 Step3 P= nRT/V ; n=.1875; P=
.1875x.0821x300/4.5 =1.02625 atm=1.03 atm.
a Happy Times Debt value 130 YTM 10.00 Mark.pdf
a Happy Times Debt value 130 YTM 10.00% Market Cap 350.00 Return on
equity 15% Joe\'s Debt 30.00 EBIT 17.00 Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value 233.98 Terminal
value is calculated using the formula - Year 5 Cash flow*(1+Growth rate)/(ROE-Growth rate)
Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34 Capex(12% of EBIT) 2.04 2.18
2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04 1.11 Profit before tax(EBIT-
Depreciation) 16.15 17.28 18.49 19.78 255.15 Tax 5.65 6.05 6.47 6.92 89.30 PAT 10.50
11.23 12.02 12.86 165.85 Net cash flow 8.29 8.87 9.49 10.15 162.95 PV of Cash flows
106.97 Market value of debt 30 EV of Joe 136.97 Shares outstanding 2.3
Maximum share price 59.55 (EV/Shares outstanding) b Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value (Using EV/EBIDTA of
9) 210.58 Terminal value is calculated using the formula - Year 5 Cash flow*(1+Growth
rate)/(ROE-Growth rate) Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34
Capex(12% of EBIT) 2.04 2.18 2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04
1.11 Profit before tax(EBIT-Depreciation) 16.15 17.28 18.49 19.78 231.75 Tax 5.65 6.05
6.47 6.92 81.11 PAT 10.50 11.23 12.02 12.86 150.64 Net cash flow 8.29 8.87 9.49 10.15
147.74 PV of Cash flows 99.41 Market value of debt 30 EV of Joe 129.41 Shares
outstanding 2.3 Maximum share price 56.26 (EV/Shares outstanding)
Solution
a Happy Times Debt value 130 YTM 10.00% Market Cap 350.00 Return on
equity 15% Joe\'s Debt 30.00 EBIT 17.00 Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value 233.98 Terminal
value is calculated using the formula - Year 5 Cash flow*(1+Growth rate)/(ROE-Growth rate)
Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34 Capex(12% of EBIT) 2.04 2.18
2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04 1.11 Profit before tax(EBIT-
Depreciation) 16.15 17.28 18.49 19.78 255.15 Tax 5.65 6.05 6.47 6.92 89.30 PAT 10.50
11.23 12.02 12.86 165.85 Net cash flow 8.29 8.87 9.49 10.15 162.95 PV of Cash flows
106.97 Market value of debt 30 EV of Joe 136.97 Shares outstanding 2.3
Maximum share price 59.55 (EV/Shares outstanding) b Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value (Using EV/EBIDTA of
9) 210.58 Terminal value is calculated using the formula - Year 5 Cash flow*(1+Growth
rate)/(ROE-Growth rate) Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34
Capex(12% of EBIT) 2.04 2.18 2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04
1.11 Profit before tax(EBIT-Depreciation) 16.15 17.28 18.49 19.78 231.75 Tax 5.65 6.05
6.47 6.92 81.11 PAT 10.50 11.23 12.02 12.86 150.64 Net cash flow 8.29 8.87 9.49 10.15
147.74 PV of Cash flows 99.41 Market value of debt 30 EV of Joe 129.41 Shares
outstanding 2.3 Maximum share price 56.26 (EV/Shares outstanding).
OH is a poor leaving group, so it first has toget.pdf
OH is a poor leaving group, so it first has toget hydrogenated (become OH2 +).
Also, it is highly encumbered withthe 2,2 dimethyl group.
Solution
OH is a poor leaving group, so it first has toget hydrogenated (become OH2 +).
Also, it is highly encumbered withthe 2,2 dimethyl group..
Molarity = MolesVolume = 22 =1 M .pdf
The document shows a calculation to determine the molarity of a solution, where the number of moles is 2 and the volume is also 2 liters, resulting in a molarity of 1M.
Hydrophobic molecules tend to be non-polar and, t.pdf
Hydrophobic molecules tend to be non-polar and, thus, prefer other neutral
molecules and non-polar solvents. Hydrophobic molecules in water often cluster
Solution
Hydrophobic molecules tend to be non-polar and, thus, prefer other neutral
molecules and non-polar solvents. Hydrophobic molecules in water often cluster.
There are two types of atomic bonds - ionic bonds and covalent bonds.pdf
There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their
structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and
bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 -
200 kcal/mol). Whether two atoms can form a covalent bond depends upon their
electronegativity i.e. the power of an atom in a molecule to attract electrons to itself. If two
atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the
atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and
negatively charged ion (anion). The bond between these two ions is called an ionic bond.
Covalent BondsIonic Bonds
Polarity:LowHigh
Formation:A covalent bond is formed between two non-metals that have similar
electronegativities. Neither atom is \"strong\" enough to attract electrons from the other. For
stabilization, they share their electrons from outer molecular orbit with othersAn ionic bond is
formed between a metal and a non-metal. Non-metals(-ve ion) are \"stronger\" than the
metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract
each other and form the ionic bond.
Shape:Definite shapeNo definite shape
What is it?:Covalent bonding is a form of chemical bonding between two non metallic atoms
which is characterized by the sharing of pairs of electrons between atoms and other covalent
bonds.Ionic bond, also known as electrovalent bond is a type of bond formed from the
electrostatic attraction between oppositely charged ions in a chemical compound. These kinds of
bonds occur mainly between a metallic and a non metallic atom.
Melting point:lowHigh
Examples:Methane (CH4), Hydro Chloric acid (HCl)Sodium chloride (NaCl), Sulphuric Acid
(H2SO4 )
Occurs between:Two non-metalsOne metal and one non-metal
Boiling point:LowHigh
State at room temperature:Liquid or gaseousSolid
Ability to conduct electricity in molten/aqueous state:
Yes
Ability to conduct electricity in solid state:
Solution
There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their
structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and
bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 -
200 kcal/mol). Whether two atoms can form a covalent bond depends upon their
electronegativity i.e. the power of an atom in a molecule to attract electrons to itself. If two
atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the
atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and
negatively charged ion (anion). The bond between these two ions is called an ionic bond.
Covalent BondsIonic Bonds
Polarity:LowHigh
Formation:A covalent bond is formed between two non-metals that have similar
electronegativities. .
T(n) = theta(nlog(2) n)T(n) = sqrt(2T(n2) + C1 where n1 so for.pdf
T(n) = theta(nlog(2) n)
T(n) = sqrt(2T(n/2) + C1 where n>1
so for n>1 T(n) = theta(nlog(2)n)
Solution
T(n) = theta(nlog(2) n)
T(n) = sqrt(2T(n/2) + C1 where n>1
so for n>1 T(n) = theta(nlog(2)n).
Showing a truth table that has the four motion sensors as inputs and.pdf
Showing a truth table that has the four motion sensors as inputs and an alarm trigger as the
output when at least two motion sensors are activated at the same time.
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Solution
Showing a truth table that has the four motion sensors as inputs and an alarm trigger as the
output when at least two motion sensors are activated at the same time.
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T.
Recommended
Step1 Error lies ; .1252 =.0625 Mole of O2 will .pdf
Step1 Error lies ; .125/2 =.0625 Mole of O2 will remain unreacted. Step2 We have
.125 mole NO2 and .0625 mole unreacted O2 Step3 P= nRT/V ; n=.1875; P=
.1875x.0821x300/4.5 =1.02625 atm=1.03 atm
Solution
Step1 Error lies ; .125/2 =.0625 Mole of O2 will remain unreacted. Step2 We have
.125 mole NO2 and .0625 mole unreacted O2 Step3 P= nRT/V ; n=.1875; P=
.1875x.0821x300/4.5 =1.02625 atm=1.03 atm.
a Happy Times Debt value 130 YTM 10.00 Mark.pdf
a Happy Times Debt value 130 YTM 10.00% Market Cap 350.00 Return on
equity 15% Joe\'s Debt 30.00 EBIT 17.00 Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value 233.98 Terminal
value is calculated using the formula - Year 5 Cash flow*(1+Growth rate)/(ROE-Growth rate)
Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34 Capex(12% of EBIT) 2.04 2.18
2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04 1.11 Profit before tax(EBIT-
Depreciation) 16.15 17.28 18.49 19.78 255.15 Tax 5.65 6.05 6.47 6.92 89.30 PAT 10.50
11.23 12.02 12.86 165.85 Net cash flow 8.29 8.87 9.49 10.15 162.95 PV of Cash flows
106.97 Market value of debt 30 EV of Joe 136.97 Shares outstanding 2.3
Maximum share price 59.55 (EV/Shares outstanding) b Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value (Using EV/EBIDTA of
9) 210.58 Terminal value is calculated using the formula - Year 5 Cash flow*(1+Growth
rate)/(ROE-Growth rate) Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34
Capex(12% of EBIT) 2.04 2.18 2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04
1.11 Profit before tax(EBIT-Depreciation) 16.15 17.28 18.49 19.78 231.75 Tax 5.65 6.05
6.47 6.92 81.11 PAT 10.50 11.23 12.02 12.86 150.64 Net cash flow 8.29 8.87 9.49 10.15
147.74 PV of Cash flows 99.41 Market value of debt 30 EV of Joe 129.41 Shares
outstanding 2.3 Maximum share price 56.26 (EV/Shares outstanding)
Solution
a Happy Times Debt value 130 YTM 10.00% Market Cap 350.00 Return on
equity 15% Joe\'s Debt 30.00 EBIT 17.00 Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value 233.98 Terminal
value is calculated using the formula - Year 5 Cash flow*(1+Growth rate)/(ROE-Growth rate)
Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34 Capex(12% of EBIT) 2.04 2.18
2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04 1.11 Profit before tax(EBIT-
Depreciation) 16.15 17.28 18.49 19.78 255.15 Tax 5.65 6.05 6.47 6.92 89.30 PAT 10.50
11.23 12.02 12.86 165.85 Net cash flow 8.29 8.87 9.49 10.15 162.95 PV of Cash flows
106.97 Market value of debt 30 EV of Joe 136.97 Shares outstanding 2.3
Maximum share price 59.55 (EV/Shares outstanding) b Year 1.00 2 3 4 5
EBIT(Growth rate 7%) 17.00 18.19 19.46 20.83 22.28 Terminal value (Using EV/EBIDTA of
9) 210.58 Terminal value is calculated using the formula - Year 5 Cash flow*(1+Growth
rate)/(ROE-Growth rate) Net working capital(6% of EBIT) 1.02 1.09 1.17 1.25 1.34
Capex(12% of EBIT) 2.04 2.18 2.34 2.50 2.67 Depreciation(5% of EBIT) 0.85 0.91 0.97 1.04
1.11 Profit before tax(EBIT-Depreciation) 16.15 17.28 18.49 19.78 231.75 Tax 5.65 6.05
6.47 6.92 81.11 PAT 10.50 11.23 12.02 12.86 150.64 Net cash flow 8.29 8.87 9.49 10.15
147.74 PV of Cash flows 99.41 Market value of debt 30 EV of Joe 129.41 Shares
outstanding 2.3 Maximum share price 56.26 (EV/Shares outstanding).
OH is a poor leaving group, so it first has toget.pdf
OH is a poor leaving group, so it first has toget hydrogenated (become OH2 +).
Also, it is highly encumbered withthe 2,2 dimethyl group.
Solution
OH is a poor leaving group, so it first has toget hydrogenated (become OH2 +).
Also, it is highly encumbered withthe 2,2 dimethyl group..
Molarity = MolesVolume = 22 =1 M .pdf
The document shows a calculation to determine the molarity of a solution, where the number of moles is 2 and the volume is also 2 liters, resulting in a molarity of 1M.
Hydrophobic molecules tend to be non-polar and, t.pdf
Hydrophobic molecules tend to be non-polar and, thus, prefer other neutral
molecules and non-polar solvents. Hydrophobic molecules in water often cluster
Solution
Hydrophobic molecules tend to be non-polar and, thus, prefer other neutral
molecules and non-polar solvents. Hydrophobic molecules in water often cluster.
There are two types of atomic bonds - ionic bonds and covalent bonds.pdf
There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their
structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and
bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 -
200 kcal/mol). Whether two atoms can form a covalent bond depends upon their
electronegativity i.e. the power of an atom in a molecule to attract electrons to itself. If two
atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the
atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and
negatively charged ion (anion). The bond between these two ions is called an ionic bond.
Covalent BondsIonic Bonds
Polarity:LowHigh
Formation:A covalent bond is formed between two non-metals that have similar
electronegativities. Neither atom is \"strong\" enough to attract electrons from the other. For
stabilization, they share their electrons from outer molecular orbit with othersAn ionic bond is
formed between a metal and a non-metal. Non-metals(-ve ion) are \"stronger\" than the
metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract
each other and form the ionic bond.
Shape:Definite shapeNo definite shape
What is it?:Covalent bonding is a form of chemical bonding between two non metallic atoms
which is characterized by the sharing of pairs of electrons between atoms and other covalent
bonds.Ionic bond, also known as electrovalent bond is a type of bond formed from the
electrostatic attraction between oppositely charged ions in a chemical compound. These kinds of
bonds occur mainly between a metallic and a non metallic atom.
Melting point:lowHigh
Examples:Methane (CH4), Hydro Chloric acid (HCl)Sodium chloride (NaCl), Sulphuric Acid
(H2SO4 )
Occurs between:Two non-metalsOne metal and one non-metal
Boiling point:LowHigh
State at room temperature:Liquid or gaseousSolid
Ability to conduct electricity in molten/aqueous state:
Yes
Ability to conduct electricity in solid state:
Solution
There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their
structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and
bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 -
200 kcal/mol). Whether two atoms can form a covalent bond depends upon their
electronegativity i.e. the power of an atom in a molecule to attract electrons to itself. If two
atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the
atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and
negatively charged ion (anion). The bond between these two ions is called an ionic bond.
Covalent BondsIonic Bonds
Polarity:LowHigh
Formation:A covalent bond is formed between two non-metals that have similar
electronegativities. .
T(n) = theta(nlog(2) n)T(n) = sqrt(2T(n2) + C1 where n1 so for.pdf
T(n) = theta(nlog(2) n)
T(n) = sqrt(2T(n/2) + C1 where n>1
so for n>1 T(n) = theta(nlog(2)n)
Solution
T(n) = theta(nlog(2) n)
T(n) = sqrt(2T(n/2) + C1 where n>1
so for n>1 T(n) = theta(nlog(2)n).
Showing a truth table that has the four motion sensors as inputs and.pdf
Showing a truth table that has the four motion sensors as inputs and an alarm trigger as the
output when at least two motion sensors are activated at the same time.
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Solution
Showing a truth table that has the four motion sensors as inputs and an alarm trigger as the
output when at least two motion sensors are activated at the same time.
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T
Sensor1
Sensor1
Sensor1
Sensor1
Output
1
F
F
F
F
F
2
F
F
F
T
F
3
F
F
T
F
F
4
F
F
T
T
T
5
F
T
F
F
F
6
F
T
F
T
T
7
F
T
T
F
T
8
F
T
T
T
T
9
T
F
F
F
F
10
T
F
F
T
T
11
T
F
T
F
T
12
T
F
T
T
T
13
T
T
F
F
T
14
T
T
F
T
T
15
T
T
T
F
T
16
T
T
T
T
T.
Due to persence of two large R groups which produ.pdf
Due to persence of two large R groups which produce steric hindence.
Solution
Due to persence of two large R groups which produce steric hindence..
Probability of obtaining six =p= 15Probability of not obtaining s.pdf
Probability of obtaining six =p= 1/5
Probability of not obtaining six=q=1- 1/5=4/5
number of throughs= n= 5
probability of obtaing at least one six =1-prbability of obtaing no six
[=1-(4/5)^5]
[=.67232]
probability of obtaing exactly three six= [^5C_3(1/5)^3(4/5)^2]
[=10xx(16/3125)=.0512]
Note use binomial dstribution t calculate probability.
Solution
Probability of obtaining six =p= 1/5
Probability of not obtaining six=q=1- 1/5=4/5
number of throughs= n= 5
probability of obtaing at least one six =1-prbability of obtaing no six
[=1-(4/5)^5]
[=.67232]
probability of obtaing exactly three six= [^5C_3(1/5)^3(4/5)^2]
[=10xx(16/3125)=.0512]
Note use binomial dstribution t calculate probability..
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced..pdf
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced.
Reaction 2 is also a redox reaction. In KClO3, Cl is in oxidation state Cl5+ and O is O2-. When
it decomposes, Cl is reduced to Cl- while O is oxidized to O2.
Reaction 3 is a neutralization reaction. An acid plus a base produces a salt plus water.
Solution
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced.
Reaction 2 is also a redox reaction. In KClO3, Cl is in oxidation state Cl5+ and O is O2-. When
it decomposes, Cl is reduced to Cl- while O is oxidized to O2.
Reaction 3 is a neutralization reaction. An acid plus a base produces a salt plus water..
Natalie is suffering form a disease called Cystic fibrosis. It is an.pdf
Natalie is suffering form a disease called Cystic fibrosis. It is an genetic condition in which the
lungs and the digestive system become clogged with sticky mucus. Symptoms usually starts in
early childhood. The main symptoms include:
Advice:
Solution
Natalie is suffering form a disease called Cystic fibrosis. It is an genetic condition in which the
lungs and the digestive system become clogged with sticky mucus. Symptoms usually starts in
early childhood. The main symptoms include:
Advice:.
I am sorry i could not see the problem. Please post again.Solu.pdf
I am sorry i could not see the problem.
Please post again.
Solution
I am sorry i could not see the problem.
Please post again..
b.false The alkanes are distinguished by having .pdf
b.false The alkanes are distinguished by having no functional group and being the
most chemically inert (non-reactive) of all organic families.
Solution
b.false The alkanes are distinguished by having no functional group and being the
most chemically inert (non-reactive) of all organic families..
In terms of structure A purine is a heterocyclic aromatic compound.pdf
In terms of structure : A purine is a heterocyclic aromatic compound that consists of a pyrimidine
ring fused to a imidazole ring.It consists of 2 carbon- nitrogen rings together with four nitrogen
atoms whereas pyrimidine is a heterocyclic aromatic compound consisting of nitrogen atoms that
are present on 1st and 3rd position of the ring.So,pyrimidine contains 1 carbon- nitrogen ring and
two nitrogen atoms.
In terms of molecular weight: Molecular weight of purine is 120.11 g/mol whereas molecular
weight of pyrimidine is 80.088 g/mol.
The pneumonic device that can be used to memorise which bases are purines vs pyrimidines are :
for purine nucleotides : It can be remembered as Pure as gold or pure as AG.So,purines are
Adenine and guanine.
for pyrimidines:It can be remembered as cut the Py because it sound like pie. Also, pie has one
ring and so does pyrimidine.So, pyrimidines are cytosine, uracil and thymine.
Solution
In terms of structure : A purine is a heterocyclic aromatic compound that consists of a pyrimidine
ring fused to a imidazole ring.It consists of 2 carbon- nitrogen rings together with four nitrogen
atoms whereas pyrimidine is a heterocyclic aromatic compound consisting of nitrogen atoms that
are present on 1st and 3rd position of the ring.So,pyrimidine contains 1 carbon- nitrogen ring and
two nitrogen atoms.
In terms of molecular weight: Molecular weight of purine is 120.11 g/mol whereas molecular
weight of pyrimidine is 80.088 g/mol.
The pneumonic device that can be used to memorise which bases are purines vs pyrimidines are :
for purine nucleotides : It can be remembered as Pure as gold or pure as AG.So,purines are
Adenine and guanine.
for pyrimidines:It can be remembered as cut the Py because it sound like pie. Also, pie has one
ring and so does pyrimidine.So, pyrimidines are cytosine, uracil and thymine..
In virtually every decision they make, executives today consider som.pdf
In virtually every decision they make, executives today consider some kind of forecast. Sound
predictions of demands and trends are no longer luxury items, but a necessity, if managers are to
cope with seasonality, sudden changes in demand levels, price-cutting maneuvers of the
competition, strikes, and large swings of the economy. Forecasting can help them deal with these
troubles; but it can help them more, the more they know about the general principles of
forecasting, what it can and cannot do for them currently, and which techniques are suited to
their needs of the moment. Here the authors try to explain the potential of forecasting to
managers, focusing special attention on sales forecasting for products of Corning Glass Works as
these have matured through the product life cycle. Also included is a rundown of forecasting
techniques.
Strategic planning focuses largely on managing interaction with environmental forces, which
include competitors, government, suppliers, customers, various interest groups and other factors
that affect your business and its prospects. Your ability as a small business owner-manager to
deal with these groups will vary widely depending on the group and on the timing. Also, you
may be able to get more of what you want from a supplier than from a competitor (although size,
distance, the percentage of the supplier\'s business you represent and your record of
dependability as a customer can affect this relationship). How you manage these and other
relationships is one of the decisions you will make during the strategic planning process.
Because of major changes in the business environment, your familiarity with strategic planning
and your ability to implement it is critical. At one time, business owner-managers assessed the
environment on a continuum that ran between very stable and very unstable. Businesses, such as
the producers of automobiles, furniture and other consumer goods, operated in a relatively stable
and predictable world. This also was true of many service firms, such as banks and savings and
loans. Typically, the environment included competition that was limited to a stable group of
competitors, loyal customers and a relatively slow transfer of information. Many small
businesses could thrive in this environment. Other small investors entered fields such as
xerography, computers and computer component production, software design and chemical
research. Some of these grew rapidly, becoming names with which we all are familiar: Xerox,
IBM, Apple and Microsoft. But many more failed.
Today, experts agree that more businesses face an unstable business environment. Improvements
in information processing and telecommunications have made major changes in most industries.
Along with this, improvements in transportation and the growth of foreign economies
(specifically in Europe and Asia) have created a global marketplace and redefined certain
industries. In addition, as consumers are exposed to more choices, loy.
Heat released by 0.025 mol of H3PO4 = 525 JHeat released by 2 mole.pdf
Heat released by 0.025 mol of H3PO4 = 525 J
Heat released by 2 moles of H3PO4 = 2/0.025 x 525
= 42000 J
Hrxn = - heat released by 2 moles of H3PO4
= -42000 J = -42 kJ
Solution
Heat released by 0.025 mol of H3PO4 = 525 J
Heat released by 2 moles of H3PO4 = 2/0.025 x 525
= 42000 J
Hrxn = - heat released by 2 moles of H3PO4
= -42000 J = -42 kJ.
An Arrhenius acid is a substance that when added .pdf
An Arrhenius acid is a substance that when added to water increases the
concentration of H1+ ions. An Arrhenius base is is a substance that when added to water
increases the concentration of OH1- ions. A Bronsted acid is a substance that can donate a
hydrogen ion. A Bronsted base is a substance that can accept a hydrogen ion. A Lewis acid
accepts lone pair electrons. A Lew base donates lone pair electrons.
Solution
An Arrhenius acid is a substance that when added to water increases the
concentration of H1+ ions. An Arrhenius base is is a substance that when added to water
increases the concentration of OH1- ions. A Bronsted acid is a substance that can donate a
hydrogen ion. A Bronsted base is a substance that can accept a hydrogen ion. A Lewis acid
accepts lone pair electrons. A Lew base donates lone pair electrons..
Five broad principles guide AARP’s evaluation of tax options. Propos.pdf
Five broad principles guide AARP’s evaluation of tax options. Proposals to reform the tax code,
stimulate the economy, raise additional revenue for any purpose, or modify specific tax
provisions should take into account these principles. Equity—Revenue-raising methods should
consider people’s ability to pay and should, to the extent possible, achieve vertical and horizontal
equity. Taxation should be progressive, and people with comparable incomes should be taxed at
comparable rates. Taxes may also be related to benefits derived from government services.
Economic neutrality—Taxes should be as neutral as possible in their treatment of economic
activity and should not unduly encourage behavior undertaken simply to avoid taxation. Often
this can be achieved by a combination of a broad taxable base and a low tax rate. In addition
taxes should not unduly hinder economic growth, induce inflation, or discourage savings. Taxes
may be used to mitigate economic costs imposed on society that were not fully captured through
market prices. Administrative efficiency—Taxes should be simple for taxpayers to understand
and comply with, and as easy as possible to administer. The need to collect data on taxpayers and
enforce tax laws should be balanced against the protection of individual liberties and privacy.
Revenue—The tax system must produce sufficient revenue to pay for important national, state,
and local priorities and maintain fiscal stability. Stable and reliable public policies and programs
require adequate and consistent sources of revenue. Social and other policy goals—A balance
must be struck between using the tax system to address social policy goals (through such
measures as tax expenditures or earmarking revenue) and raising adequate revenues simply and
equitably. Sometimes the above principles conflict with each other. In these cases AARP may
weigh the relative importance of each of the key principles. AARP’s position on tax policies take
into account the net fiscal impact of a given policy proposal, not only its tax component.
Economic neutrality—Typically taxes create some inefficiencies by distorting economic choices
and thus people’s behavior in the market. One major goal of most tax reform proposals is to
reduce these distortions and encourage stronger economic growth. Proponents of consumption
taxes note that income taxes encourage consumption over saving. Integrating the individual and
corporate income taxes would reduce distortions that influence how businesses are organized and
how income is transferred from corporations to shareholders. By eliminating certain tax
deductions and other preferences, fundamental restructuring of the current income tax would also
improve the neutrality and efficiency of the tax system. Administrative efficiency—The degree
of administrative complexity depends greatly on the form of the tax. A VAT can have high
administrative costs because it increases the number of taxpayers and requires detailed record
ke.
E(in ev) = 12400 ( in Ao) E = 12400 5400 = 2.296 ev E= 2.2.pdf
E(in ev) = 12400 / ( in Ao)
E = 12400 / 5400 = 2.296 ev
E= 2.296 * 1.6*10-19 J
power = energy / time
15 = n E /t
(n/t) = 15 /( 2.296 * 1.6 * 10-19) = 4.08 * 1019
Solution
E(in ev) = 12400 / ( in Ao)
E = 12400 / 5400 = 2.296 ev
E= 2.296 * 1.6*10-19 J
power = energy / time
15 = n E /t
(n/t) = 15 /( 2.296 * 1.6 * 10-19) = 4.08 * 1019.
cost = input(Enter the price of the dinner)rating - input(Ent.pdf
cost = input(\"Enter the price of the dinner:)
rating - input(\"Enter the rating of the dinner\")
tip = 0.0
if int(rating) == 1:
tip = 20.0
elif int(rating) == 2:
tip = 15.0
elif int(rating) == 3:
tip = 10.0
print float(cost)*(tip/100)
Solution
cost = input(\"Enter the price of the dinner:)
rating - input(\"Enter the rating of the dinner\")
tip = 0.0
if int(rating) == 1:
tip = 20.0
elif int(rating) == 2:
tip = 15.0
elif int(rating) == 3:
tip = 10.0
print float(cost)*(tip/100).
Conductance Conductance is the ability of a material to pass ele.pdf
Conductance : Conductance is the ability of a material to pass electrons
It is equal to reciprocal of the resistance.
Conductance C = 1/ R where R is resistance of the cell
Solution
Conductance : Conductance is the ability of a material to pass electrons
It is equal to reciprocal of the resistance.
Conductance C = 1/ R where R is resistance of the cell.
calcium chloride (CaCl2) is the ionic compoundCO2 , CS2 , SO2 and .pdf
calcium chloride (CaCl2) is the ionic compound
CO2 , CS2 , SO2 and OF2 are covalent compounds.
Solution
calcium chloride (CaCl2) is the ionic compound
CO2 , CS2 , SO2 and OF2 are covalent compounds..
A. are aromatic because Benzene and Aromaticity.pdf
A. are aromatic because Benzene and Aromaticity Unusual stability of benzene,
heat of hydrogenation versus trienes Pi molecular orbitals of benzene Huckel’s rules: aromatic
anions and cations, aromaticity of heterocycles Cyclobutadiene, benzene, and Frost’s circle
Electrophilic aromatic substitution, mechanism and energy diagrams Br2, Br2/FeBr3, nitration,
sulfonation of benzene Friedel-Crafts alkylation and acylation, problems encountered
Electrophilic aromatic substitution Directing groups in detail, resonance versus inductive effects,
steric control Changing ortho / para directors to meta directors Addition and removal of
directing/blocking groups and retrosynthetic analyses of highly substituted benzenes
Electrophilic aromatic substitution of polycyclic aromatic compounds: napthlene and
phenanthrene Heterocycles Aliphatic heterocycles & reactivities: aziridines versus epoxides and
thioranes, synthesis and acidic activation of aziridines Aromatic heterocycles: acid-base
properties of pyridine, pyrimidine, pyrrole, imidazole, etc. Electrophilic aromatic substitution
reactivity and regioselectivity of pyrrole, furan, benzene, pyridine, etc. Synthesis of pyrroles,
furan, and thiophene from ?-dicarbonyl compounds Furan and thiophene in Diels-Alder
chemistry Fischer indole synthesis, regioselectivity of indoles in electrophilic aromatic
substitution Pyridine as a base, a nucleophile, and an acyl transfer catalyst (DMAP) Pyridine in
nucleophilic aromatic substitution: Chichibabin reaction and addition of PhLi Synthesis of
pyridines: Hantzsch condensation Synthesis of quinolines and isoquinolines from anilines
Solution
A. are aromatic because Benzene and Aromaticity Unusual stability of benzene,
heat of hydrogenation versus trienes Pi molecular orbitals of benzene Huckel’s rules: aromatic
anions and cations, aromaticity of heterocycles Cyclobutadiene, benzene, and Frost’s circle
Electrophilic aromatic substitution, mechanism and energy diagrams Br2, Br2/FeBr3, nitration,
sulfonation of benzene Friedel-Crafts alkylation and acylation, problems encountered
Electrophilic aromatic substitution Directing groups in detail, resonance versus inductive effects,
steric control Changing ortho / para directors to meta directors Addition and removal of
directing/blocking groups and retrosynthetic analyses of highly substituted benzenes
Electrophilic aromatic substitution of polycyclic aromatic compounds: napthlene and
phenanthrene Heterocycles Aliphatic heterocycles & reactivities: aziridines versus epoxides and
thioranes, synthesis and acidic activation of aziridines Aromatic heterocycles: acid-base
properties of pyridine, pyrimidine, pyrrole, imidazole, etc. Electrophilic aromatic substitution
reactivity and regioselectivity of pyrrole, furan, benzene, pyridine, etc. Synthesis of pyrroles,
furan, and thiophene from ?-dicarbonyl compounds Furan and thiophene in Diels-Alder
chemistry Fischer indole synthesis, regioselectivity of indoles in electrophilic aroma.
ASRASR is an expansive form of reaction between silica in aggrega.pdf
ASR:
ASR is an expansive form of reaction between silica in aggregates potassium and sodium alkalis
mostly from cement but also from aggregates, pozzolons admixtures and mixing water. The
reactivity is potentially harmful when it produces significant expansion. indication of presence of
alkali aggregate reactivity may be a network of closed and spall joints. or movement of portion
of structure.ASR can be controlled through a proper selection of aggregates.
Sulphate attack:
Excessive amount of sulphates in soil and water may attack concrete by reacting with hydrated
compounds in hydrant cement paste. these reactions can induce sufficient pressure to disintegrate
concrete. sulphate attck is more common where the concrete is subjected to both drying and
wetting for the best defence against the external sulfate attack.
DEF:
Delayed Ettringite formation is a special form of sulfate internal attack it occurs in concrete
which can be cured at delayed temperature for example where steam curing has been used. it was
originally identified in steam cured concrete sleepers. It causes Expansion of concrete due to
Etringite formation within the paste and can cause serious disintegration. DEF generally occurs
when the Ettringite crystals exert an expansive force. the total amount of ettingite formed during
hydration is decomposed.
Freezing/ Thawing:
The most potentially disrupting feature is freezing and thawing while concrete is wet particularly
in presence id decieng chemicals the detoriation is caused by freezing of water and subsequent
expansion of the paste during freezing water is displaced by ice formation in the paste so that is
accomodated and that is not disruptive. Microscopic air bubbles form the space for water to enter
and thus relieve the hydraulic pressure generated the concrete with low water cement ratio is
durable the concret with high cementatious ratio air entrained concrete with low water cement
ratio withstands great number of cycles to freezing and thawing.
Corrosion:
The concrete protects the embedded steel with its highly alkaline nature High PH Environment in
concrete causes the highly passive & non corroding protective oxide film to form on steel.
However the chlorides from sea water can destroy and penetrate the film once chloride corrosion
threshhold is reached the an electric cell is formed along steel between steel bars and electro
chemical process of carrions begins.
Solution
ASR:
ASR is an expansive form of reaction between silica in aggregates potassium and sodium alkalis
mostly from cement but also from aggregates, pozzolons admixtures and mixing water. The
reactivity is potentially harmful when it produces significant expansion. indication of presence of
alkali aggregate reactivity may be a network of closed and spall joints. or movement of portion
of structure.ASR can be controlled through a proper selection of aggregates.
Sulphate attack:
Excessive amount of sulphates in soil and water may attack concrete by reacting w.
anlSolutionanl.pdf
This document does not contain any substantive content to summarize. It consists of the letters "anl" repeated with no other words, sentences, or context provided. A meaningful summary cannot be generated from this sparse input.
Constructing Your Course Container for Effective Communication
Communicating effectively and consistently with students can help them feel at ease during their learning experience and provide the instructor with a communication trail to track the course's progress. This workshop will take you through constructing an engaging course container to facilitate effective communication.
More Related Content
More from aswrd
Due to persence of two large R groups which produ.pdf
Due to persence of two large R groups which produce steric hindence.
Solution
Due to persence of two large R groups which produce steric hindence..
Probability of obtaining six =p= 15Probability of not obtaining s.pdf
Probability of obtaining six =p= 1/5
Probability of not obtaining six=q=1- 1/5=4/5
number of throughs= n= 5
probability of obtaing at least one six =1-prbability of obtaing no six
[=1-(4/5)^5]
[=.67232]
probability of obtaing exactly three six= [^5C_3(1/5)^3(4/5)^2]
[=10xx(16/3125)=.0512]
Note use binomial dstribution t calculate probability.
Solution
Probability of obtaining six =p= 1/5
Probability of not obtaining six=q=1- 1/5=4/5
number of throughs= n= 5
probability of obtaing at least one six =1-prbability of obtaing no six
[=1-(4/5)^5]
[=.67232]
probability of obtaing exactly three six= [^5C_3(1/5)^3(4/5)^2]
[=10xx(16/3125)=.0512]
Note use binomial dstribution t calculate probability..
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced..pdf
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced.
Reaction 2 is also a redox reaction. In KClO3, Cl is in oxidation state Cl5+ and O is O2-. When
it decomposes, Cl is reduced to Cl- while O is oxidized to O2.
Reaction 3 is a neutralization reaction. An acid plus a base produces a salt plus water.
Solution
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced.
Reaction 2 is also a redox reaction. In KClO3, Cl is in oxidation state Cl5+ and O is O2-. When
it decomposes, Cl is reduced to Cl- while O is oxidized to O2.
Reaction 3 is a neutralization reaction. An acid plus a base produces a salt plus water..
Natalie is suffering form a disease called Cystic fibrosis. It is an.pdf
Natalie is suffering form a disease called Cystic fibrosis. It is an genetic condition in which the
lungs and the digestive system become clogged with sticky mucus. Symptoms usually starts in
early childhood. The main symptoms include:
Advice:
Solution
Natalie is suffering form a disease called Cystic fibrosis. It is an genetic condition in which the
lungs and the digestive system become clogged with sticky mucus. Symptoms usually starts in
early childhood. The main symptoms include:
Advice:.
I am sorry i could not see the problem. Please post again.Solu.pdf
I am sorry i could not see the problem.
Please post again.
Solution
I am sorry i could not see the problem.
Please post again..
b.false The alkanes are distinguished by having .pdf
b.false The alkanes are distinguished by having no functional group and being the
most chemically inert (non-reactive) of all organic families.
Solution
b.false The alkanes are distinguished by having no functional group and being the
most chemically inert (non-reactive) of all organic families..
In terms of structure A purine is a heterocyclic aromatic compound.pdf
In terms of structure : A purine is a heterocyclic aromatic compound that consists of a pyrimidine
ring fused to a imidazole ring.It consists of 2 carbon- nitrogen rings together with four nitrogen
atoms whereas pyrimidine is a heterocyclic aromatic compound consisting of nitrogen atoms that
are present on 1st and 3rd position of the ring.So,pyrimidine contains 1 carbon- nitrogen ring and
two nitrogen atoms.
In terms of molecular weight: Molecular weight of purine is 120.11 g/mol whereas molecular
weight of pyrimidine is 80.088 g/mol.
The pneumonic device that can be used to memorise which bases are purines vs pyrimidines are :
for purine nucleotides : It can be remembered as Pure as gold or pure as AG.So,purines are
Adenine and guanine.
for pyrimidines:It can be remembered as cut the Py because it sound like pie. Also, pie has one
ring and so does pyrimidine.So, pyrimidines are cytosine, uracil and thymine.
Solution
In terms of structure : A purine is a heterocyclic aromatic compound that consists of a pyrimidine
ring fused to a imidazole ring.It consists of 2 carbon- nitrogen rings together with four nitrogen
atoms whereas pyrimidine is a heterocyclic aromatic compound consisting of nitrogen atoms that
are present on 1st and 3rd position of the ring.So,pyrimidine contains 1 carbon- nitrogen ring and
two nitrogen atoms.
In terms of molecular weight: Molecular weight of purine is 120.11 g/mol whereas molecular
weight of pyrimidine is 80.088 g/mol.
The pneumonic device that can be used to memorise which bases are purines vs pyrimidines are :
for purine nucleotides : It can be remembered as Pure as gold or pure as AG.So,purines are
Adenine and guanine.
for pyrimidines:It can be remembered as cut the Py because it sound like pie. Also, pie has one
ring and so does pyrimidine.So, pyrimidines are cytosine, uracil and thymine..
In virtually every decision they make, executives today consider som.pdf
In virtually every decision they make, executives today consider some kind of forecast. Sound
predictions of demands and trends are no longer luxury items, but a necessity, if managers are to
cope with seasonality, sudden changes in demand levels, price-cutting maneuvers of the
competition, strikes, and large swings of the economy. Forecasting can help them deal with these
troubles; but it can help them more, the more they know about the general principles of
forecasting, what it can and cannot do for them currently, and which techniques are suited to
their needs of the moment. Here the authors try to explain the potential of forecasting to
managers, focusing special attention on sales forecasting for products of Corning Glass Works as
these have matured through the product life cycle. Also included is a rundown of forecasting
techniques.
Strategic planning focuses largely on managing interaction with environmental forces, which
include competitors, government, suppliers, customers, various interest groups and other factors
that affect your business and its prospects. Your ability as a small business owner-manager to
deal with these groups will vary widely depending on the group and on the timing. Also, you
may be able to get more of what you want from a supplier than from a competitor (although size,
distance, the percentage of the supplier\'s business you represent and your record of
dependability as a customer can affect this relationship). How you manage these and other
relationships is one of the decisions you will make during the strategic planning process.
Because of major changes in the business environment, your familiarity with strategic planning
and your ability to implement it is critical. At one time, business owner-managers assessed the
environment on a continuum that ran between very stable and very unstable. Businesses, such as
the producers of automobiles, furniture and other consumer goods, operated in a relatively stable
and predictable world. This also was true of many service firms, such as banks and savings and
loans. Typically, the environment included competition that was limited to a stable group of
competitors, loyal customers and a relatively slow transfer of information. Many small
businesses could thrive in this environment. Other small investors entered fields such as
xerography, computers and computer component production, software design and chemical
research. Some of these grew rapidly, becoming names with which we all are familiar: Xerox,
IBM, Apple and Microsoft. But many more failed.
Today, experts agree that more businesses face an unstable business environment. Improvements
in information processing and telecommunications have made major changes in most industries.
Along with this, improvements in transportation and the growth of foreign economies
(specifically in Europe and Asia) have created a global marketplace and redefined certain
industries. In addition, as consumers are exposed to more choices, loy.
Heat released by 0.025 mol of H3PO4 = 525 JHeat released by 2 mole.pdf
Heat released by 0.025 mol of H3PO4 = 525 J
Heat released by 2 moles of H3PO4 = 2/0.025 x 525
= 42000 J
Hrxn = - heat released by 2 moles of H3PO4
= -42000 J = -42 kJ
Solution
Heat released by 0.025 mol of H3PO4 = 525 J
Heat released by 2 moles of H3PO4 = 2/0.025 x 525
= 42000 J
Hrxn = - heat released by 2 moles of H3PO4
= -42000 J = -42 kJ.
An Arrhenius acid is a substance that when added .pdf
An Arrhenius acid is a substance that when added to water increases the
concentration of H1+ ions. An Arrhenius base is is a substance that when added to water
increases the concentration of OH1- ions. A Bronsted acid is a substance that can donate a
hydrogen ion. A Bronsted base is a substance that can accept a hydrogen ion. A Lewis acid
accepts lone pair electrons. A Lew base donates lone pair electrons.
Solution
An Arrhenius acid is a substance that when added to water increases the
concentration of H1+ ions. An Arrhenius base is is a substance that when added to water
increases the concentration of OH1- ions. A Bronsted acid is a substance that can donate a
hydrogen ion. A Bronsted base is a substance that can accept a hydrogen ion. A Lewis acid
accepts lone pair electrons. A Lew base donates lone pair electrons..
Five broad principles guide AARP’s evaluation of tax options. Propos.pdf
Five broad principles guide AARP’s evaluation of tax options. Proposals to reform the tax code,
stimulate the economy, raise additional revenue for any purpose, or modify specific tax
provisions should take into account these principles. Equity—Revenue-raising methods should
consider people’s ability to pay and should, to the extent possible, achieve vertical and horizontal
equity. Taxation should be progressive, and people with comparable incomes should be taxed at
comparable rates. Taxes may also be related to benefits derived from government services.
Economic neutrality—Taxes should be as neutral as possible in their treatment of economic
activity and should not unduly encourage behavior undertaken simply to avoid taxation. Often
this can be achieved by a combination of a broad taxable base and a low tax rate. In addition
taxes should not unduly hinder economic growth, induce inflation, or discourage savings. Taxes
may be used to mitigate economic costs imposed on society that were not fully captured through
market prices. Administrative efficiency—Taxes should be simple for taxpayers to understand
and comply with, and as easy as possible to administer. The need to collect data on taxpayers and
enforce tax laws should be balanced against the protection of individual liberties and privacy.
Revenue—The tax system must produce sufficient revenue to pay for important national, state,
and local priorities and maintain fiscal stability. Stable and reliable public policies and programs
require adequate and consistent sources of revenue. Social and other policy goals—A balance
must be struck between using the tax system to address social policy goals (through such
measures as tax expenditures or earmarking revenue) and raising adequate revenues simply and
equitably. Sometimes the above principles conflict with each other. In these cases AARP may
weigh the relative importance of each of the key principles. AARP’s position on tax policies take
into account the net fiscal impact of a given policy proposal, not only its tax component.
Economic neutrality—Typically taxes create some inefficiencies by distorting economic choices
and thus people’s behavior in the market. One major goal of most tax reform proposals is to
reduce these distortions and encourage stronger economic growth. Proponents of consumption
taxes note that income taxes encourage consumption over saving. Integrating the individual and
corporate income taxes would reduce distortions that influence how businesses are organized and
how income is transferred from corporations to shareholders. By eliminating certain tax
deductions and other preferences, fundamental restructuring of the current income tax would also
improve the neutrality and efficiency of the tax system. Administrative efficiency—The degree
of administrative complexity depends greatly on the form of the tax. A VAT can have high
administrative costs because it increases the number of taxpayers and requires detailed record
ke.
E(in ev) = 12400 ( in Ao) E = 12400 5400 = 2.296 ev E= 2.2.pdf
E(in ev) = 12400 / ( in Ao)
E = 12400 / 5400 = 2.296 ev
E= 2.296 * 1.6*10-19 J
power = energy / time
15 = n E /t
(n/t) = 15 /( 2.296 * 1.6 * 10-19) = 4.08 * 1019
Solution
E(in ev) = 12400 / ( in Ao)
E = 12400 / 5400 = 2.296 ev
E= 2.296 * 1.6*10-19 J
power = energy / time
15 = n E /t
(n/t) = 15 /( 2.296 * 1.6 * 10-19) = 4.08 * 1019.
cost = input(Enter the price of the dinner)rating - input(Ent.pdf
cost = input(\"Enter the price of the dinner:)
rating - input(\"Enter the rating of the dinner\")
tip = 0.0
if int(rating) == 1:
tip = 20.0
elif int(rating) == 2:
tip = 15.0
elif int(rating) == 3:
tip = 10.0
print float(cost)*(tip/100)
Solution
cost = input(\"Enter the price of the dinner:)
rating - input(\"Enter the rating of the dinner\")
tip = 0.0
if int(rating) == 1:
tip = 20.0
elif int(rating) == 2:
tip = 15.0
elif int(rating) == 3:
tip = 10.0
print float(cost)*(tip/100).
Conductance Conductance is the ability of a material to pass ele.pdf
Conductance : Conductance is the ability of a material to pass electrons
It is equal to reciprocal of the resistance.
Conductance C = 1/ R where R is resistance of the cell
Solution
Conductance : Conductance is the ability of a material to pass electrons
It is equal to reciprocal of the resistance.
Conductance C = 1/ R where R is resistance of the cell.
calcium chloride (CaCl2) is the ionic compoundCO2 , CS2 , SO2 and .pdf
calcium chloride (CaCl2) is the ionic compound
CO2 , CS2 , SO2 and OF2 are covalent compounds.
Solution
calcium chloride (CaCl2) is the ionic compound
CO2 , CS2 , SO2 and OF2 are covalent compounds..
A. are aromatic because Benzene and Aromaticity.pdf
A. are aromatic because Benzene and Aromaticity Unusual stability of benzene,
heat of hydrogenation versus trienes Pi molecular orbitals of benzene Huckel’s rules: aromatic
anions and cations, aromaticity of heterocycles Cyclobutadiene, benzene, and Frost’s circle
Electrophilic aromatic substitution, mechanism and energy diagrams Br2, Br2/FeBr3, nitration,
sulfonation of benzene Friedel-Crafts alkylation and acylation, problems encountered
Electrophilic aromatic substitution Directing groups in detail, resonance versus inductive effects,
steric control Changing ortho / para directors to meta directors Addition and removal of
directing/blocking groups and retrosynthetic analyses of highly substituted benzenes
Electrophilic aromatic substitution of polycyclic aromatic compounds: napthlene and
phenanthrene Heterocycles Aliphatic heterocycles & reactivities: aziridines versus epoxides and
thioranes, synthesis and acidic activation of aziridines Aromatic heterocycles: acid-base
properties of pyridine, pyrimidine, pyrrole, imidazole, etc. Electrophilic aromatic substitution
reactivity and regioselectivity of pyrrole, furan, benzene, pyridine, etc. Synthesis of pyrroles,
furan, and thiophene from ?-dicarbonyl compounds Furan and thiophene in Diels-Alder
chemistry Fischer indole synthesis, regioselectivity of indoles in electrophilic aromatic
substitution Pyridine as a base, a nucleophile, and an acyl transfer catalyst (DMAP) Pyridine in
nucleophilic aromatic substitution: Chichibabin reaction and addition of PhLi Synthesis of
pyridines: Hantzsch condensation Synthesis of quinolines and isoquinolines from anilines
Solution
A. are aromatic because Benzene and Aromaticity Unusual stability of benzene,
heat of hydrogenation versus trienes Pi molecular orbitals of benzene Huckel’s rules: aromatic
anions and cations, aromaticity of heterocycles Cyclobutadiene, benzene, and Frost’s circle
Electrophilic aromatic substitution, mechanism and energy diagrams Br2, Br2/FeBr3, nitration,
sulfonation of benzene Friedel-Crafts alkylation and acylation, problems encountered
Electrophilic aromatic substitution Directing groups in detail, resonance versus inductive effects,
steric control Changing ortho / para directors to meta directors Addition and removal of
directing/blocking groups and retrosynthetic analyses of highly substituted benzenes
Electrophilic aromatic substitution of polycyclic aromatic compounds: napthlene and
phenanthrene Heterocycles Aliphatic heterocycles & reactivities: aziridines versus epoxides and
thioranes, synthesis and acidic activation of aziridines Aromatic heterocycles: acid-base
properties of pyridine, pyrimidine, pyrrole, imidazole, etc. Electrophilic aromatic substitution
reactivity and regioselectivity of pyrrole, furan, benzene, pyridine, etc. Synthesis of pyrroles,
furan, and thiophene from ?-dicarbonyl compounds Furan and thiophene in Diels-Alder
chemistry Fischer indole synthesis, regioselectivity of indoles in electrophilic aroma.
ASRASR is an expansive form of reaction between silica in aggrega.pdf
ASR:
ASR is an expansive form of reaction between silica in aggregates potassium and sodium alkalis
mostly from cement but also from aggregates, pozzolons admixtures and mixing water. The
reactivity is potentially harmful when it produces significant expansion. indication of presence of
alkali aggregate reactivity may be a network of closed and spall joints. or movement of portion
of structure.ASR can be controlled through a proper selection of aggregates.
Sulphate attack:
Excessive amount of sulphates in soil and water may attack concrete by reacting with hydrated
compounds in hydrant cement paste. these reactions can induce sufficient pressure to disintegrate
concrete. sulphate attck is more common where the concrete is subjected to both drying and
wetting for the best defence against the external sulfate attack.
DEF:
Delayed Ettringite formation is a special form of sulfate internal attack it occurs in concrete
which can be cured at delayed temperature for example where steam curing has been used. it was
originally identified in steam cured concrete sleepers. It causes Expansion of concrete due to
Etringite formation within the paste and can cause serious disintegration. DEF generally occurs
when the Ettringite crystals exert an expansive force. the total amount of ettingite formed during
hydration is decomposed.
Freezing/ Thawing:
The most potentially disrupting feature is freezing and thawing while concrete is wet particularly
in presence id decieng chemicals the detoriation is caused by freezing of water and subsequent
expansion of the paste during freezing water is displaced by ice formation in the paste so that is
accomodated and that is not disruptive. Microscopic air bubbles form the space for water to enter
and thus relieve the hydraulic pressure generated the concrete with low water cement ratio is
durable the concret with high cementatious ratio air entrained concrete with low water cement
ratio withstands great number of cycles to freezing and thawing.
Corrosion:
The concrete protects the embedded steel with its highly alkaline nature High PH Environment in
concrete causes the highly passive & non corroding protective oxide film to form on steel.
However the chlorides from sea water can destroy and penetrate the film once chloride corrosion
threshhold is reached the an electric cell is formed along steel between steel bars and electro
chemical process of carrions begins.
Solution
ASR:
ASR is an expansive form of reaction between silica in aggregates potassium and sodium alkalis
mostly from cement but also from aggregates, pozzolons admixtures and mixing water. The
reactivity is potentially harmful when it produces significant expansion. indication of presence of
alkali aggregate reactivity may be a network of closed and spall joints. or movement of portion
of structure.ASR can be controlled through a proper selection of aggregates.
Sulphate attack:
Excessive amount of sulphates in soil and water may attack concrete by reacting w.
anlSolutionanl.pdf
This document does not contain any substantive content to summarize. It consists of the letters "anl" repeated with no other words, sentences, or context provided. A meaningful summary cannot be generated from this sparse input.
More from aswrd (20)
Due to persence of two large R groups which produ.pdf
Due to persence of two large R groups which produ.pdf
Probability of obtaining six =p= 15Probability of not obtaining s.pdf
Probability of obtaining six =p= 15Probability of not obtaining s.pdf
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced..pdf
Reaction 1 is a redox reaction. Na is oxidized while Cl is reduced..pdf
Natalie is suffering form a disease called Cystic fibrosis. It is an.pdf
Natalie is suffering form a disease called Cystic fibrosis. It is an.pdf
I am sorry i could not see the problem. Please post again.Solu.pdf
I am sorry i could not see the problem. Please post again.Solu.pdf
b.false The alkanes are distinguished by having .pdf
b.false The alkanes are distinguished by having .pdf
In terms of structure A purine is a heterocyclic aromatic compound.pdf
In terms of structure A purine is a heterocyclic aromatic compound.pdf
In virtually every decision they make, executives today consider som.pdf
In virtually every decision they make, executives today consider som.pdf
Heat released by 0.025 mol of H3PO4 = 525 JHeat released by 2 mole.pdf
Heat released by 0.025 mol of H3PO4 = 525 JHeat released by 2 mole.pdf
An Arrhenius acid is a substance that when added .pdf
An Arrhenius acid is a substance that when added .pdf
Five broad principles guide AARP’s evaluation of tax options. Propos.pdf
Five broad principles guide AARP’s evaluation of tax options. Propos.pdf
E(in ev) = 12400 ( in Ao) E = 12400 5400 = 2.296 ev E= 2.2.pdf
E(in ev) = 12400 ( in Ao) E = 12400 5400 = 2.296 ev E= 2.2.pdf
cost = input(Enter the price of the dinner)rating - input(Ent.pdf
cost = input(Enter the price of the dinner)rating - input(Ent.pdf
Conductance Conductance is the ability of a material to pass ele.pdf
Conductance Conductance is the ability of a material to pass ele.pdf
calcium chloride (CaCl2) is the ionic compoundCO2 , CS2 , SO2 and .pdf
calcium chloride (CaCl2) is the ionic compoundCO2 , CS2 , SO2 and .pdf
A. are aromatic because Benzene and Aromaticity.pdf
A. are aromatic because Benzene and Aromaticity.pdf
ASRASR is an expansive form of reaction between silica in aggrega.pdf
ASRASR is an expansive form of reaction between silica in aggrega.pdf
Recently uploaded
Constructing Your Course Container for Effective Communication
Communicating effectively and consistently with students can help them feel at ease during their learning experience and provide the instructor with a communication trail to track the course's progress. This workshop will take you through constructing an engaging course container to facilitate effective communication.
How to Create a More Engaging and Human Online Learning Experience
How to Create a More Engaging and Human Online Learning Experience Wahiba Chair Training & Consulting
Wahiba Chair's Talk at the 2024 Learning Ideas Conference. BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...Nguyen Thanh Tu Collection
https://app.box.com/s/y977uz6bpd3af4qsebv7r9b7s21935vdBÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...Nguyen Thanh Tu Collection
https://app.box.com/s/qhtvq32h4ybf9t49ku85x0n3xl4jhr15ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
-------------------------------------------------------------------------------
Find out more about ISO training and certification services
Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
-------------------------------------------------------------------------------
For more information about PECB:
Website: https://pecb.com/
LinkedIn: https://www.linkedin.com/company/pecb/
Facebook: https://www.facebook.com/PECBInternational/
Slideshare: http://www.slideshare.net/PECBCERTIFICATION
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective UpskillingExcellence Foundation for South Sudan
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.Wound healing PPT
This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications.
A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function.
Healing is the body’s response to injury in an attempt to restore normal structure and functions.
Healing can occur in two ways: Regeneration and Repair
There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc.
Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 𝟏)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐄𝐏𝐏 𝐂𝐮𝐫𝐫𝐢𝐜𝐮𝐥𝐮𝐦 𝐢𝐧 𝐭𝐡𝐞 𝐏𝐡𝐢𝐥𝐢𝐩𝐩𝐢𝐧𝐞𝐬:
- Understand the goals and objectives of the Edukasyong Pantahanan at Pangkabuhayan (EPP) curriculum, recognizing its importance in fostering practical life skills and values among students. Students will also be able to identify the key components and subjects covered, such as agriculture, home economics, industrial arts, and information and communication technology.
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐍𝐚𝐭𝐮𝐫𝐞 𝐚𝐧𝐝 𝐒𝐜𝐨𝐩𝐞 𝐨𝐟 𝐚𝐧 𝐄𝐧𝐭𝐫𝐞𝐩𝐫𝐞𝐧𝐞𝐮𝐫:
-Define entrepreneurship, distinguishing it from general business activities by emphasizing its focus on innovation, risk-taking, and value creation. Students will describe the characteristics and traits of successful entrepreneurs, including their roles and responsibilities, and discuss the broader economic and social impacts of entrepreneurial activities on both local and global scales.
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Recently uploaded (20)
B. Ed Syllabus for babasaheb ambedkar education university.pdf
B. Ed Syllabus for babasaheb ambedkar education university.pdf
Constructing Your Course Container for Effective Communication
Constructing Your Course Container for Effective Communication
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
How to Create a More Engaging and Human Online Learning Experience
How to Create a More Engaging and Human Online Learning Experience
IGCSE Biology Chapter 14- Reproduction in Plants.pdf
IGCSE Biology Chapter 14- Reproduction in Plants.pdf
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Answer 1 Basic Earning per shareBasic Earning per share = BEBS.pdf
- 1. Answer 1 : Basic Earning per share Basic Earning per share = BE/BS = 25700/ 10600 = 2.42 Working Notes BE = Income available for common stock holders Net Income 50000 Less: Intt on Bonds (100000*6.5%) 6500 ------------- EBT 43500 Less Tax paid @ 30% 13050 ---------------- Income after tax 30450 Less Dividend paid to Prefered stock holder 4750 ------------------ Amount Available common 25700 Stock holder ----------- BS = Weighted avaerage no. of common stock = Share at begining 9600 May, 1 Share issued 1200 weight 1200 *8/12 800 SeP, 30 Share issued 800 Weight 800*3/12 200 ------------ Total no. of shares 10600 ----------------- Answer - 2 Dialute Earning Per share = Net Income/Avg shares+other convertible instuments = (50000-1950)/15100 = 48050/15100 = 3.18 Working notes Avg Share = 10600 Prefered stock (1000*2) = 2000 Bonds (100000/1000*25) = 2500
- 2. -------------------------- Total shares = 15100 --------------------------- Incremental EPS of Preferred Stock = 4750/2000 = 4750/2000 = 2.38 Incremental EPS of Bonds = interest/ incremental no. of shares due to conversion of bonds = 6500-1950/2500 = 4550/2500 = 1.82 Answer 3 Incremental Earning per share Bonds = $ 1.82 Preferred = $ 2.38 Solution Answer 1 : Basic Earning per share Basic Earning per share = BE/BS = 25700/ 10600 = 2.42 Working Notes BE = Income available for common stock holders Net Income 50000 Less: Intt on Bonds (100000*6.5%) 6500 ------------- EBT 43500 Less Tax paid @ 30% 13050 ---------------- Income after tax 30450 Less Dividend paid to Prefered stock holder 4750 ------------------ Amount Available common 25700 Stock holder ----------- BS = Weighted avaerage no. of common stock
- 3. = Share at begining 9600 May, 1 Share issued 1200 weight 1200 *8/12 800 SeP, 30 Share issued 800 Weight 800*3/12 200 ------------ Total no. of shares 10600 ----------------- Answer - 2 Dialute Earning Per share = Net Income/Avg shares+other convertible instuments = (50000-1950)/15100 = 48050/15100 = 3.18 Working notes Avg Share = 10600 Prefered stock (1000*2) = 2000 Bonds (100000/1000*25) = 2500 -------------------------- Total shares = 15100 --------------------------- Incremental EPS of Preferred Stock = 4750/2000 = 4750/2000 = 2.38 Incremental EPS of Bonds = interest/ incremental no. of shares due to conversion of bonds = 6500-1950/2500 = 4550/2500 = 1.82 Answer 3 Incremental Earning per share Bonds = $ 1.82 Preferred = $ 2.38