Phenol S.pdf
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N= sp3 F = p orbital of F overlap with sp3 hybrid.pdf
N= sp3 F = p orbital of F overlap with sp3 hybride orbital F is not central atom.
Solution
N= sp3 F = p orbital of F overlap with sp3 hybride orbital F is not central atom..
isotopes .pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. The document only contains the word "isotopes" repeated three times without any surrounding context.
I think you are missing an S species in the first.pdf
I think you are missing an S species in the first equation. Are you trying to from
H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN
HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion
[SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-.
S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 +
2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the
same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is
+5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 +
2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe
exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons)
Solution
I think you are missing an S species in the first equation. Are you trying to from
H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN
HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion
[SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-.
S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 +
2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the
same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is
+5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 +
2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe
exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons).
E=1 Solu.pdf
The document contains a simple equation stating that the value of E equals 1. No other information or context is provided.
Compound II and III only. .pdf
This one sentence document states that only Compound II and III should be used or considered. No other context or details are provided about what Compound II and III refer to or the purpose of only using those two specifically.
Until 2001, the 40-year-old Halamka also worked as an emergency room.pdf
Until 2001, the 40-year-old Halamka also worked as an emergency room physician, but he gave
that up to take on the additional responsibilities of being CIO of Harvard Medical School in
2002.
Two months later, Beth Israel Deaconess experienced one of the worst health-care IT disasters
ever. Over four days, Halamka’s network crashed repeatedly, forcing the hospital to revert to the
paper patient-records system that it had abandoned years ago. Lab reports that doctors normally
had in hand within 45 minutes took as long as five hours to process.
\"Everything’s the Web,\" Halamka says now. \"If you don’t have the Web, you’re down.\"
Hospitals come alive early. By 7 a.m., doctors and nurses started to send some of Beth Israel
Deaconess’s 100,000 daily e-mails. The pharmacy began filling prescriptions, transferring the
first bits of the 40 terabytes that traverse the network daily. Some of the 3,000 daily lab reports
were beginning to move.
By 8 a.m., the network again started acting as if it were flying into a headwind. Halamka realized
the network had settled down the night before only because hardly anyone was using it. When
the workday began in earnest, CPU usage spiked. The network started flapping. The problem
hadn’t been fixed.
Halamka’s team scrambled to find other possible sources of the trouble. One suspect was
CareGroup’s network of outlying hospitals in Cambridge, Needham, Ayer and elsewhere in
Massachusetts. They operated as a distinct network that plugged into Beth Israel Deaconess. The
community hospitals’ network was sluggish, and a billing application wasn’t working, according
to Jeanette Clough, CEO of Mount Auburn Hospital in Cambridge, which serves as the hub for
the outlying hospitals’ network.
To fix the problem, the CAP team decided to put a Cisco 6509 router between the core network
and PACS, eliminating spanning tree protocol and its seven-hop limitation. (The 6509 also has
switching capabilities, so the team decided to kill three switches inside PACS and use the 6509
for that too.)
At noon, Epstein came in to lend a hand...and walked into 1978. Epstein worked the copier, then
sorted a three-inch stack of microbiology reports and handed them to runners who took them to
patients’ rooms where they were left for doctors. (There were about 450 patients at the hospital.)
In time, the chaos gave way to a loosely defined routine, which was slower than normal and far
more harried. The pre-IT generation, Sands says, adapted quickly. For the IT generation, himself
included, it was an unnerving transition. He was reminded of a short story by the Victorian
author E.M. Forster, \"The Machine Stops,\" about a world that depends upon an Ÿber-computer
to sustain human life. Eventually, those who designed the computer die and no one is left who
knows how it works.
\"We depend upon the network, but we also take it for granted,\" Sands says. \"It’s a credit to
[Halamka] that we operate with a mind-set that the computers never go down. And that.
The Final Accounts of non-trading concerns consists of1. Receipts.pdf
The Final Accounts of non-trading concerns consists of:
1. Receipts and Payments Account
2. Income and Expenditure Account, and
3. Balance Sheet.
1. Receipts and Payments Account:
It is a Real Account. It is a consolidated summary of Cash Book. It is prepared at the end of the
accounting period. All cash receipts are recorded on the debit side and all cash payments are
recorded on the credit side.
Cash Book consisting of entries of receipts and payments in a chronological order while the
Receipts and payments is a summary of total cash receipts and cash payments.
It starts with opening balance of Cash and Bank and ends with closing balance of Cash and
Bank. It does not take into account outstanding amounts of receipts and payments. Receipts and
Payments may be of Capi tal or Revenue nature; they may relate to the current or previous year or
subsequent year; so long as they are actually received or paid, they must appear in this account.
Features of Receipts and Payment Account, In Brief:
1. It starts with opening balance and ends with closing balance
2. It is the summary of cash and bank transactions.
3. Actual cash transactions are entered.
4. It includes capital as well as revenue items.
5. It follows cash system of accounting
6. It shows cash position and excludes all non-cash items.
7. It is a real account.
8. It does not take any income/expense outstanding at the beginning or at the end.
2. Income and Expenditure Account:
It is a Nominal Account. It is in the form of Profit and Loss Account. It is concerned with only
revenue items—expenses and incomes. It records all losses and expenses on its debit side and all
incomes and gains on its credit side.
Of the incomes and expenses of revenue nature, only the portion pertaining to the current year is
shown in the Income and Expenditure Account i.e. amount relating to the previous year or future
year are excluded. Again, the incomes and expenses of current year, whether received or not,
must be shown.
In other words, incomes and expenses have to be adjusted for both out-standing and pre-
payments. All non-cash items, Depreciation, Bad Debts, Provision for Doubtful Debts etc. are
taken into account.
The difference between the debit side and the credit side is either surplus or deficit for the year
concerned and the difference will be transferred to the Capital Fund (also called General Fund or
Accumulated Fund) appearing in Balance Sheet.
Features of Income and Expenditure Account, In Brief:
1. It is prepared in lieu of Profit and Loss Account.
2. It is a nominal account.
3. It is based on mercantile system of accounting.
4. There is no opening balance.
5. It ends with Surplus or Deficit.
6. It excludes all capital income and capital expenses.
7. It includes only revenue items.
8. It records all expenses whether paid or not, and all incomes whether received or not.
3. Balance Sheet:
Balance Sheet in case of non-trading concern is prepared in the usual manner and consists of all
liabilities and .
Recommended
N= sp3 F = p orbital of F overlap with sp3 hybrid.pdf
N= sp3 F = p orbital of F overlap with sp3 hybride orbital F is not central atom.
Solution
N= sp3 F = p orbital of F overlap with sp3 hybride orbital F is not central atom..
isotopes .pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. The document only contains the word "isotopes" repeated three times without any surrounding context.
I think you are missing an S species in the first.pdf
I think you are missing an S species in the first equation. Are you trying to from
H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN
HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion
[SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-.
S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 +
2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the
same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is
+5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 +
2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe
exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons)
Solution
I think you are missing an S species in the first equation. Are you trying to from
H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN
HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion
[SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-.
S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 +
2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the
same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is
+5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 +
2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe
exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons).
E=1 Solu.pdf
The document contains a simple equation stating that the value of E equals 1. No other information or context is provided.
Compound II and III only. .pdf
This one sentence document states that only Compound II and III should be used or considered. No other context or details are provided about what Compound II and III refer to or the purpose of only using those two specifically.
Until 2001, the 40-year-old Halamka also worked as an emergency room.pdf
Until 2001, the 40-year-old Halamka also worked as an emergency room physician, but he gave
that up to take on the additional responsibilities of being CIO of Harvard Medical School in
2002.
Two months later, Beth Israel Deaconess experienced one of the worst health-care IT disasters
ever. Over four days, Halamka’s network crashed repeatedly, forcing the hospital to revert to the
paper patient-records system that it had abandoned years ago. Lab reports that doctors normally
had in hand within 45 minutes took as long as five hours to process.
\"Everything’s the Web,\" Halamka says now. \"If you don’t have the Web, you’re down.\"
Hospitals come alive early. By 7 a.m., doctors and nurses started to send some of Beth Israel
Deaconess’s 100,000 daily e-mails. The pharmacy began filling prescriptions, transferring the
first bits of the 40 terabytes that traverse the network daily. Some of the 3,000 daily lab reports
were beginning to move.
By 8 a.m., the network again started acting as if it were flying into a headwind. Halamka realized
the network had settled down the night before only because hardly anyone was using it. When
the workday began in earnest, CPU usage spiked. The network started flapping. The problem
hadn’t been fixed.
Halamka’s team scrambled to find other possible sources of the trouble. One suspect was
CareGroup’s network of outlying hospitals in Cambridge, Needham, Ayer and elsewhere in
Massachusetts. They operated as a distinct network that plugged into Beth Israel Deaconess. The
community hospitals’ network was sluggish, and a billing application wasn’t working, according
to Jeanette Clough, CEO of Mount Auburn Hospital in Cambridge, which serves as the hub for
the outlying hospitals’ network.
To fix the problem, the CAP team decided to put a Cisco 6509 router between the core network
and PACS, eliminating spanning tree protocol and its seven-hop limitation. (The 6509 also has
switching capabilities, so the team decided to kill three switches inside PACS and use the 6509
for that too.)
At noon, Epstein came in to lend a hand...and walked into 1978. Epstein worked the copier, then
sorted a three-inch stack of microbiology reports and handed them to runners who took them to
patients’ rooms where they were left for doctors. (There were about 450 patients at the hospital.)
In time, the chaos gave way to a loosely defined routine, which was slower than normal and far
more harried. The pre-IT generation, Sands says, adapted quickly. For the IT generation, himself
included, it was an unnerving transition. He was reminded of a short story by the Victorian
author E.M. Forster, \"The Machine Stops,\" about a world that depends upon an Ÿber-computer
to sustain human life. Eventually, those who designed the computer die and no one is left who
knows how it works.
\"We depend upon the network, but we also take it for granted,\" Sands says. \"It’s a credit to
[Halamka] that we operate with a mind-set that the computers never go down. And that.
The Final Accounts of non-trading concerns consists of1. Receipts.pdf
The Final Accounts of non-trading concerns consists of:
1. Receipts and Payments Account
2. Income and Expenditure Account, and
3. Balance Sheet.
1. Receipts and Payments Account:
It is a Real Account. It is a consolidated summary of Cash Book. It is prepared at the end of the
accounting period. All cash receipts are recorded on the debit side and all cash payments are
recorded on the credit side.
Cash Book consisting of entries of receipts and payments in a chronological order while the
Receipts and payments is a summary of total cash receipts and cash payments.
It starts with opening balance of Cash and Bank and ends with closing balance of Cash and
Bank. It does not take into account outstanding amounts of receipts and payments. Receipts and
Payments may be of Capi tal or Revenue nature; they may relate to the current or previous year or
subsequent year; so long as they are actually received or paid, they must appear in this account.
Features of Receipts and Payment Account, In Brief:
1. It starts with opening balance and ends with closing balance
2. It is the summary of cash and bank transactions.
3. Actual cash transactions are entered.
4. It includes capital as well as revenue items.
5. It follows cash system of accounting
6. It shows cash position and excludes all non-cash items.
7. It is a real account.
8. It does not take any income/expense outstanding at the beginning or at the end.
2. Income and Expenditure Account:
It is a Nominal Account. It is in the form of Profit and Loss Account. It is concerned with only
revenue items—expenses and incomes. It records all losses and expenses on its debit side and all
incomes and gains on its credit side.
Of the incomes and expenses of revenue nature, only the portion pertaining to the current year is
shown in the Income and Expenditure Account i.e. amount relating to the previous year or future
year are excluded. Again, the incomes and expenses of current year, whether received or not,
must be shown.
In other words, incomes and expenses have to be adjusted for both out-standing and pre-
payments. All non-cash items, Depreciation, Bad Debts, Provision for Doubtful Debts etc. are
taken into account.
The difference between the debit side and the credit side is either surplus or deficit for the year
concerned and the difference will be transferred to the Capital Fund (also called General Fund or
Accumulated Fund) appearing in Balance Sheet.
Features of Income and Expenditure Account, In Brief:
1. It is prepared in lieu of Profit and Loss Account.
2. It is a nominal account.
3. It is based on mercantile system of accounting.
4. There is no opening balance.
5. It ends with Surplus or Deficit.
6. It excludes all capital income and capital expenses.
7. It includes only revenue items.
8. It records all expenses whether paid or not, and all incomes whether received or not.
3. Balance Sheet:
Balance Sheet in case of non-trading concern is prepared in the usual manner and consists of all
liabilities and .
sickel cell anemia is caused by recessive allele. frequency of sicke.pdf
sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1.
according to Hardy-Weinberg law
frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2
q = square root of 0.1 = 0.316
then p = 1 - q = 1 - 0.316 = 0.684
p = frequency of healthy allele = 0.684
Frequency of healthy allele = 0.684
2.Ans.
how many people in a population of 10,000 will have sickel cell anemia.
sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia.
3.Ans.
how many people in a population of 10,000 will be sickel cell carriers.
sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers
4.Ans.
mutations occurs and new allele introduced into the population.
small population which causes genetic drift.
the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained.
3. Alkaptonuria caused by recessive allele.
survey revels that 5 in 50 people have Alkaptonuria
a. if a population of 50 people
if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1
q2 = 0.1 then q = square root of 0.1 = 0.316
q = 0.316
p = 1 - q =0.684
p = 0.684
b. Heterozygous to Alkaptonuria = 2pq
2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele
C. Homozygous Dominant = AA =p2
AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant
Solution
sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1.
according to Hardy-Weinberg law
frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2
q = square root of 0.1 = 0.316
then p = 1 - q = 1 - 0.316 = 0.684
p = frequency of healthy allele = 0.684
Frequency of healthy allele = 0.684
2.Ans.
how many people in a population of 10,000 will have sickel cell anemia.
sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia.
3.Ans.
how many people in a population of 10,000 will be sickel cell carriers.
sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers
4.Ans.
mutations occurs and new allele introduced into the population.
small population which causes genetic drift.
the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained.
3. Alkaptonuria caused by recessive allele.
survey revels that 5 in 50 people have Alkaptonuria
a. if a population of 50 people
if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1
q2 = 0.1 then q = square root of 0.1 = 0.316
q = 0.316
p = 1 - q =0.684
p = 0.684
b. Heterozygous to Alkaptonuria = 2pq
2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele
C. Homozygous Dominant = AA =p2
AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant.
I think that (e) have only non-singular generalized inversesSol.pdf
I think that (e) have only non-singular generalized inverses?
Solution
I think that (e) have only non-singular generalized inverses?.
ConvertingSeconds.javaimport java.util.Scanner;public class Conv.pdf
ConvertingSeconds.java
import java.util.Scanner;
public class ConvertingSeconds {
public static void main(String[] args) {
//Declaring variables
int fromseconds,hours,minutes,seconds,toSeconds;
//Scanner class Object is used to read the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number of seconds entered by the user
System.out.print(\"Enter Total Number of Seconds :\");
fromseconds=sc.nextInt();
//calling the method secondTime() by passing the seconds as arguments
String str=secondTime(fromseconds);
//Converting the string to String array where \',\' is the delimeter
String arr[]=str.split(\",\");
//Converting the String into integer
hours=Integer.parseInt(arr[0]);
minutes=Integer.parseInt(arr[1]);
seconds=Integer.parseInt(arr[2]);
//calling the method inSeconds by passing the hours,minutes,seconds as arguments
toSeconds=inSeconds(hours,minutes,seconds);
//Displaying the hours,minutes,seconds to seconds
System.out.println(hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds corresponds to
\"+toSeconds+\" seconds\");
}
/*This method will converts the hours,minutes,seconds to seconds
* Params:hours,minutes,seconds
* Return:totalSeconds
*/
private static int inSeconds(int hours, int minutes, int seconds) {
//Converting the hours,minutes and seconds to totalseconds
int totSeconds=hours*3600+minutes*60+seconds;
return totSeconds;
}
/*This method will converts the seconds to hours,minutes,seconds
* Params:total Seconds
* Return:String
*/
private static String secondTime(int totseconds) {
//Declaring variable
int secs=totseconds;
//calculating the number of hours
int hours=totseconds/3600;
totseconds=totseconds-(hours*3600);
//calculating the number of minutes
int minutes=totseconds/60;
totseconds=totseconds-(minutes*60);
//Calculating the number of seconds
int seconds=totseconds;
//displaying the total seconds to hours,minutes,seconds
System.out.println(secs+\" Seconds corresponds to \"+hours+\" hours \"+minutes+\" minutes
\"+seconds+\" seconds\");
return hours+\",\"+minutes+\",\"+seconds;
}
}
_____________________________________________
Output:
Enter Total Number of Seconds :10000
10000 Seconds corresponds to 2 hours 46 minutes 40 seconds
2 hours 46 minutes 40 seconds corresponds to 10000 seconds
_____________________________________________
Output1:
Enter Total Number of Seconds :15000
15000 Seconds corresponds to 4 hours 10 minutes 0 seconds
4 hours 10 minutes 0 seconds corresponds to 15000 seconds
_________________________________________Thank You
Solution
ConvertingSeconds.java
import java.util.Scanner;
public class ConvertingSeconds {
public static void main(String[] args) {
//Declaring variables
int fromseconds,hours,minutes,seconds,toSeconds;
//Scanner class Object is used to read the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number of seconds entered by the user
System.out.print(\"Enter Total Number of Seconds :\");
fromseconds=sc.nextInt();
//calling the method seco.
Before Attempting to measure risk aversion, following two factors sh.pdf
Before Attempting to measure risk aversion, following two factors should be considered:
1 ) Estimating probabilities: It includes measuring the future events based on their possibility of
occurrence. generally, the Probabilities are assigned based on past experiences.
2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the
investors in their previous investment. By looking at their investment history and analyzing
behavior, this table can be prepared.
Solution
Before Attempting to measure risk aversion, following two factors should be considered:
1 ) Estimating probabilities: It includes measuring the future events based on their possibility of
occurrence. generally, the Probabilities are assigned based on past experiences.
2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the
investors in their previous investment. By looking at their investment history and analyzing
behavior, this table can be prepared..
Answer The general circumstance for retaining outside experts is wh.pdf
Answer: The general circumstance for retaining outside experts is where the auditor requires a
special skill or knowledge outside of accounting or auditing, the auditors feel they do not have
the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work
of a Specialist provides guidance and recognizes that auditors are not expected to be experts in
all areas, auditors may encounter material matters which are complex or subjective, and an
auditor may use the work of a specialist as evidential matter to evaluate financial statement
assertions (PCAOB AU 336, 2010).
In the case of auditing pensions, the auditors would most likely use an actuary as a specialist as
the auditors would not be expected to have that level of expertise. The actuary would be
knowledgeable of the entire pension development process and be able to more quickly
comprehend the nuances of the actuarial assumptions and mechanics of the calculations. For
example, some of the tasks that the actuary would be involved with the company on for pensions
includes the measurement of pension obligations, assignment of plan obligations to time periods,
development of a cost allocation procedure, development of a contribution allocation procedure,
determination of types and levels of benefits, projection of pension obligations, projection of
plan costs or contributions (AAA, 2010).
Much of the financial impact of pensions is based on going from current known data (number
employees, salaries, etc.) through a number of assumptions (how many years work, how many
years collect a pension, mortality rates, what rate able to invest funds at, etc.) to develop the most
likely funding needs over time to provide the retirement benefits as agreed to cover the pension
expenses. Outside of determining the current employees, salaries, etc., almost everything else is
an assumption. How do we make the most appropriate and best assumption to ensure having
dollars available in the future? How much do we need to put aside now that will grow and cover
the pension requirements? This is where the audit firm should employ an actuary to examine the
actuarial-type assumptions made by the company on its employees and current retirees to see if
they are reasonable assumptions given the company’s data. The actuary would be able to help
examine the pension plan assets, the investments, the effective earning rates from the
investments (in comparison to the discount rate being used), the projected pension expenses, and
the resultant pension plan funding situation. The actuary’s review should be able to assist the
auditor in identifying potential issue areas for the auditor’s further examination.
Solution
Answer: The general circumstance for retaining outside experts is where the auditor requires a
special skill or knowledge outside of accounting or auditing, the auditors feel they do not have
the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work
of a S.
AnsConsolidated financial statements are the aggregated or combin.pdf
Ans:
Consolidated financial statements are the aggregated or combined figure of statements between a
parent company and its subsidiaries.
Consolidated balance sheet
As on 31/12/2020
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,000
8,750,000
Liabilities and shareholders equity:
Current liabilities:
Accounts payable
630,000
280,000
910,000
Noncurrent liabilities:
Loan payable
420,000
700,000
1,120,000
1,050,000
980,000
2,030,000
Shareholders equity:
Share capital
1,680,000
420,000
2,100,000
Retained earnings
3,584,000
1,036,000
4,620,000
5,264,000
1,456,000
6,720,000
6,314,000
2,436,000
8,750,000
Consolidated income statement
For the year ended 31/12/2020
Particulars
P
S
Total
Revenue:
Sales
2,804,200
2,100,000
4,904,200
Royalties
210,000
--
210,000
Dividends
100,800
--
100,800
3,115,000
2,100,000
5,215,000
Expenses:
Cost of sales
1,680,000
1,260,000
2,940,000
Other
784,000
575,400
1,359,400
2,464,000
1,835,400
4,299,400
Net income
651,000
264,600
915,600
Retained earnings statement
For the year ended 31/12/2020
P
S
Total
Opening retained earnings
3,353,000
897,400
4,250,400
Net income
651,000
264,600
915,600
Dividends declared
(420,000)
(126,000)
(546,000)
Closing retained earnings
3,584,000
1,036,000
4,620,000
Figures are added. P + S = Total.
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,000
8,750,000
Liabilities and shareholders equity:
Current liabilities:
Accounts payable
630,000
280,000
910,000
Noncurrent liabilities:
Loan payable
420,000
700,000
1,120,000
1,050,000
980,000
2,030,000
Shareholders equity:
Share capital
1,680,000
420,000
2,100,000
Retained earnings
3,584,000
1,036,000
4,620,000
5,264,000
1,456,000
6,720,000
6,314,000
2,436,000
8,750,000
Solution
Ans:
Consolidated financial statements are the aggregated or combined figure of statements between a
parent company and its subsidiaries.
Consolidated balance sheet
As on 31/12/2020
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,00.
Answer b)S phaseS-phase is synthetic phase where semiconservative.pdf
Answer: b)S phase
S-phase is synthetic phase where semiconservative DNA replication takes place, which occurs
between G1 and G2 phase
Solution
Answer: b)S phase
S-phase is synthetic phase where semiconservative DNA replication takes place, which occurs
between G1 and G2 phase.
a)The NLS resides somewhere between amino acids 240 and 357. b)The.pdf
a)The NLS resides somewhere between amino acids 240 and 357.
b)The NES resides somewhere between amino acids 115 and 239.
Solution
a)The NLS resides somewhere between amino acids 240 and 357.
b)The NES resides somewhere between amino acids 115 and 239..
A database is generally used for storing related, structured data, w.pdf
A database is generally used for storing related, structured data, with well defined data formats,
in an efficient manner for insert, update and/or retrieval (depending on application).
On the other hand, a file system is a more unstructured data store for storing arbitrary, probably
unrelated data. The file system is more general, and databases are built on top of the general data
storage services provided by file systems.
A Data Base Management System is a system software for easy, efficient and reliable data
processing and management. It can be used for:
Creation of a database.
Retrieval of information from the database.
Updating the database.
Managing a database.
It provides us with the many functionalities and is more advantageous than the traditional file
system in many ways listed below:
1) Processing Queries and Object Management:
In traditional file systems, we cannot store data in the form of objects. In practical-world
applications, data is stored in objects and not files. So in a file system, some application software
maps the data stored in files to objects so that can be used further.
We can directly store data in the form of objects in a database management system. Application
level code needs to be written to handle, store and scan through the data in a file system whereas
a DBMS gives us the ability to query the database.
2) Controlling redundancy and inconsistency:
Redundancy refers to repeated instances of the same data. A database system provides
redundancy control whereas in a file system, same data may be stored multiple times. For
example, if a student is studying two different educational programs in the same college, say
,Engineering and History, then his information such as the phone number and address may be
stored multiple times, once in Engineering dept and the other in History dept. Therefore, it
increases time taken to access and store data. This may also lead to inconsistent data states in
both places. A DBMS uses data normalization to avoid redundancy and duplicates.
3) Efficient memory management and indexing:
DBMS makes complex memory management easy to handle. In file systems, files are indexed in
place of objects so query operations require entire file scans whereas in a DBMS , object
indexing takes place efficiently through database schema based on any attribute of the data or a
data-property. This helps in fast retrieval of data based on the indexed attribute.
4) Concurrency control and transaction management:
Several applications allow user to simultaneously access data. This may lead to inconsistency in
data in case files are used. Consider two withdrawal transactions X and Y in which an amount of
100 and 200 is withdrawn from an account A initially containing 1000. Now since these
transactions are taking place simultaneously, different transactions may update the account
differently. X reads 1000, debits 100, updates the account A to 900, whereas X also reads 1000,
debits 200, updates A to 800. In bot.
9x^4y^10z^-2Solution9x^4y^10z^-2.pdf
This mathematical expression represents a single term with three variables being multiplied together with associated exponents. The term is 9x^4y^10z^-2, meaning 9 multiplied by x raised to the fourth power multiplied by y raised to the tenth power multiplied by z raised to the negative second power.
x= 38Solutionx= 38.pdf
x equals 3/8. The solution is provided that x equals 3/8. A short mathematical expression is given and the answer is shown to be x=3/8.
212+423 is same as 2 36+4 46 is same as 6 76 is same as 7 16...pdf
21/2+42/3 is same as 2 3/6+4 4/6 is same as 6 7/6 is same as 7 1/6.....keep it simple and reduce to
simplest answer to understand!!
Solution
21/2+42/3 is same as 2 3/6+4 4/6 is same as 6 7/6 is same as 7 1/6.....keep it simple and reduce to
simplest answer to understand!!.
1. Atoms Atoms are structured by 3 main components. Protons, ne.pdf
1. Atoms:
Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the
atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral.
Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively
charged.
2. Bonds:
You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may
have heard of include, NaCl and H2O. In order for these elements to form these compounds,
bonds need to form between them. When bonds are made energy is released.
3. Periodic Table:
The number of electrons that an atom of an element has is usually equal to it\'s atomic number.
More importantly there is a trend that you can determine what the number of valence (outer
most) electrons are. Group 1 has 1 valence electron, Group 2 has 2, Group 3 has 3, Group 8 has
0, Group 7 has 7, Group 6 has 6 (O, S, Se, Te), Group 5 has 5( N, P, As) and Group 4 has 4
(C,Si).
4. Electronegativity:
This means how reactive a certain element is. If an element has lots of valence electrons it will
have a high electronegativity because it wants that 1 electron, to fill its 8 compartments in its
shell. The trend for electronegativity is that electronegativity increases as you go across the
periodic table towards the right ->. As you go towards the right, you have more and more
valence electrons, therefore more \"want\" for electrons to fill up the compartments.
When you go (^) up the periodic table electronegativity also increases due to the distance the
electrons are away from the nucleus (core) of the atom.
5. Polarity:
Polarity just means you have 2 poles. Think of it like a magnet, you usually have a north pole
and a south pole OR a positive pole and a negative pole.
Different atoms have different charges. For example, H2O, consists of 2 hydrogens and 1
oxygen. Hydrogen is positively charged and oxygen is negatively charged. This is why water is
said to be polar. It consists of both a positive and a negative.
Water is also an example of hydrogen bonding, because Hydrogen is bonding with a large
molecule, oxygen.
Non-polar bonds are where there are no positives or negatives. An example would be a diatomic
molecule, such as oxygen gas: O2. Methane: CH4 is also considered to be non-polar because its
carbon atom covalently bonds with hydrogen atoms sharing almost completely equally.
Hope this helps! :)
Solution
1. Atoms:
Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the
atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral.
Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively
charged.
2. Bonds:
You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may
have heard of include, NaCl and H2O. In order for these elements to form these compounds,
bonds need to form between them. When bonds are made energy is released.
3. Perio.
1)importjavax.swing. = package contains JFrame and JButton class.pdf
1)
importjavax.swing.* => package contains JFrame and JButton classes
2)
javadoc comments begin with the /* symbol, and end with the */ symbol.
3)
If you want to read in values from the user in the console window you must use a
____Scanner______ object.
4)
float, int, long, and double are forms of ____C_____ data types.
5)
An easy way to access every element of a Two dimensional array is to use _____2_nested_____
for loops.
6)
To search the elements of an array, ___linear_____ search can always be used.
7)
There are two types of method calls, void return methods and ___data____ return methods.
8.)
import keyword is used to import built-in and user-defined packages into your java source file.
key word that is used to bring in java.utils is ____Import______
9. When an array name is passed to a method, it is passing the memory address, therefore it is
passing the array by ____reference_____.
because array name refers to the memory address of the array
Solution
1)
importjavax.swing.* => package contains JFrame and JButton classes
2)
javadoc comments begin with the /* symbol, and end with the */ symbol.
3)
If you want to read in values from the user in the console window you must use a
____Scanner______ object.
4)
float, int, long, and double are forms of ____C_____ data types.
5)
An easy way to access every element of a Two dimensional array is to use _____2_nested_____
for loops.
6)
To search the elements of an array, ___linear_____ search can always be used.
7)
There are two types of method calls, void return methods and ___data____ return methods.
8.)
import keyword is used to import built-in and user-defined packages into your java source file.
key word that is used to bring in java.utils is ____Import______
9. When an array name is passed to a method, it is passing the memory address, therefore it is
passing the array by ____reference_____.
because array name refers to the memory address of the array.
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site .pdf
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double
bond O.
2)This electrophile formed attacks the position of sulphur.
Because, the lone pair gets conjugated and forms Negative sites on both positions. But it
attacks -Carbon with less congestion(presence of less bulky groups).
3) It combines to form given product aong with release of HCl.
Solution
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double
bond O.
2)This electrophile formed attacks the position of sulphur.
Because, the lone pair gets conjugated and forms Negative sites on both positions. But it
attacks -Carbon with less congestion(presence of less bulky groups).
3) It combines to form given product aong with release of HCl..
(a) NaCl is an ionic compound. The intermolecular forces are electro.pdf
(a) NaCl is an ionic compound. The intermolecular forces are electrostatic interactions.
(b) CCl4 is a non-polar molecular compound. The intermolecular forces are London dispersion
forces.
(c) CHCl3 is a polar molecular compound. The intermolecular forces are dipole-dipole
interactions and London dispersion forces.
(d) C(diamond) is a covalent solid. The intermolecular forces are covalent bonds.
(e) Au is a metallic solid. The intemolecular interactions are metallic bonds.
Solution
(a) NaCl is an ionic compound. The intermolecular forces are electrostatic interactions.
(b) CCl4 is a non-polar molecular compound. The intermolecular forces are London dispersion
forces.
(c) CHCl3 is a polar molecular compound. The intermolecular forces are dipole-dipole
interactions and London dispersion forces.
(d) C(diamond) is a covalent solid. The intermolecular forces are covalent bonds.
(e) Au is a metallic solid. The intemolecular interactions are metallic bonds..
Runs Newtons method to find roots of f(z) . It col function [a.pdf
% Runs Newton\'s method to find roots of f(z) . It col
function [a,n,TOL,N]=P3(a,n,TOL,N)
A=zeros(n,n,\'uint8\');
f = inline(\'z^3-1\'); % Define f and f\'.
fprime = inline(\'3*z^2\');
rts=roots([1 0 0 -1])
h=2*a/n;
for j=1:n+1, % Try initial values with imaginary parts between
y = -a + (j-1)*h; % -a and a
for i=1:n+1, % and with real parts between
x = -a + (i-1)*h; % -a and a.
z = x + 1i*y;
zk = z;
kount = 0; % kount is the total number of iterations.
conv1 = 0; % conv1,2,3 count iterations when approx soln is within
conv2 = 0; % TOL of root1,2,3.
conv3 = 0;
while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5,
kount = kount + 1;
zk = zk - f(zk)/fprime(zk); % This is the Newton step.
if abs(zk-root1) < TOL, % Check for convergence to root1.
conv1 = conv1 + 1;
else
conv1 = 0;
end;
if abs(zk-root2) < TOL, % Check for convergence to root2.
conv2 = conv2 + 1;
else
conv2 = 0;
end;
if abs(zk-root3) < TOL, % Check for convergence to root3.
conv3 = conv3 + 1;
else
conv3 = 0;
end;
end;
if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green.
if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red.
if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue.
end;
end;
imshow(A)
Solution
% Runs Newton\'s method to find roots of f(z) . It col
function [a,n,TOL,N]=P3(a,n,TOL,N)
A=zeros(n,n,\'uint8\');
f = inline(\'z^3-1\'); % Define f and f\'.
fprime = inline(\'3*z^2\');
rts=roots([1 0 0 -1])
h=2*a/n;
for j=1:n+1, % Try initial values with imaginary parts between
y = -a + (j-1)*h; % -a and a
for i=1:n+1, % and with real parts between
x = -a + (i-1)*h; % -a and a.
z = x + 1i*y;
zk = z;
kount = 0; % kount is the total number of iterations.
conv1 = 0; % conv1,2,3 count iterations when approx soln is within
conv2 = 0; % TOL of root1,2,3.
conv3 = 0;
while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5,
kount = kount + 1;
zk = zk - f(zk)/fprime(zk); % This is the Newton step.
if abs(zk-root1) < TOL, % Check for convergence to root1.
conv1 = conv1 + 1;
else
conv1 = 0;
end;
if abs(zk-root2) < TOL, % Check for convergence to root2.
conv2 = conv2 + 1;
else
conv2 = 0;
end;
if abs(zk-root3) < TOL, % Check for convergence to root3.
conv3 = conv3 + 1;
else
conv3 = 0;
end;
end;
if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green.
if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red.
if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue.
end;
end;
imshow(A).
The java Payroll that prompts user to enter hourly rate .pdf
/**
* The java Payroll that prompts user to enter
* hourly rate of pay and number of hours worked.
* Then calculates the gross pay and net pay
* and print to console.
* */
//Payroll.java
import java.util.Scanner;
public class Payroll
{
public static void main(String[] args)
{
//declare variables for hourly rate and hours worked
double hourlyRate;
int hoursWorked;
//Set tax rate as 0.15 (15 percent )
final double WITH_HOLD_TAX=0.15;
//Set grossPay=0
double grossPay=0;
//Set tax =0
double tax=0;
//Set netPay=0
double netPay=0;
//Create an instance of Scanner class
Scanner inputScanner =new Scanner(System.in);
System.out.println(\"Enter hourly rate of pay\");
//prompt for hourly rate
hourlyRate=Integer.parseInt(inputScanner.nextLine());
System.out.println(\"Enter number of hours worked\");
//prompt for number of hours
hoursWorked=Integer.parseInt(inputScanner.nextLine());
//Calculate grossPay
//multiply hoursWorked by hourlyRate
grossPay=hoursWorked*hourlyRate;
//calculate with hold tax
tax=grossPay*WITH_HOLD_TAX;
//calculate netPay
//subtract tax from grossPay
netPay=grossPay-tax;
//print gross pay and net pay to console
System.out.println(\"Gross Pay : \"+grossPay);
System.out.println(\"Net pay : \"+netPay);
}
}
---------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------
Sample output:
Enter hourly rate of pay
10
Enter number of hours worked
50
Gross Pay : 500.0
Net pay : 425.0
Solution
/**
* The java Payroll that prompts user to enter
* hourly rate of pay and number of hours worked.
* Then calculates the gross pay and net pay
* and print to console.
* */
//Payroll.java
import java.util.Scanner;
public class Payroll
{
public static void main(String[] args)
{
//declare variables for hourly rate and hours worked
double hourlyRate;
int hoursWorked;
//Set tax rate as 0.15 (15 percent )
final double WITH_HOLD_TAX=0.15;
//Set grossPay=0
double grossPay=0;
//Set tax =0
double tax=0;
//Set netPay=0
double netPay=0;
//Create an instance of Scanner class
Scanner inputScanner =new Scanner(System.in);
System.out.println(\"Enter hourly rate of pay\");
//prompt for hourly rate
hourlyRate=Integer.parseInt(inputScanner.nextLine());
System.out.println(\"Enter number of hours worked\");
//prompt for number of hours
hoursWorked=Integer.parseInt(inputScanner.nextLine());
//Calculate grossPay
//multiply hoursWorked by hourlyRate
grossPay=hoursWorked*hourlyRate;
//calculate with hold tax
tax=grossPay*WITH_HOLD_TAX;
//calculate netPay
//subtract tax from grossPay
netPay=grossPay-tax;
//print gross pay and net pay to console
System.out.println(\"Gross Pay : \"+grossPay);
System.out.println(\"Net pay : \"+netPay);
}
}
---------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------
Sample output:
E.
They represent the position of the other function.pdf
They represent the position of the other functional groups in the compound with
respect to most senior functional group. Hope you know the seniority order of functional groups
in nomenclature
Solution
They represent the position of the other functional groups in the compound with
respect to most senior functional group. Hope you know the seniority order of functional groups
in nomenclature.
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
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sickel cell anemia is caused by recessive allele. frequency of sicke.pdf
sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1.
according to Hardy-Weinberg law
frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2
q = square root of 0.1 = 0.316
then p = 1 - q = 1 - 0.316 = 0.684
p = frequency of healthy allele = 0.684
Frequency of healthy allele = 0.684
2.Ans.
how many people in a population of 10,000 will have sickel cell anemia.
sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia.
3.Ans.
how many people in a population of 10,000 will be sickel cell carriers.
sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers
4.Ans.
mutations occurs and new allele introduced into the population.
small population which causes genetic drift.
the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained.
3. Alkaptonuria caused by recessive allele.
survey revels that 5 in 50 people have Alkaptonuria
a. if a population of 50 people
if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1
q2 = 0.1 then q = square root of 0.1 = 0.316
q = 0.316
p = 1 - q =0.684
p = 0.684
b. Heterozygous to Alkaptonuria = 2pq
2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele
C. Homozygous Dominant = AA =p2
AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant
Solution
sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1.
according to Hardy-Weinberg law
frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2
q = square root of 0.1 = 0.316
then p = 1 - q = 1 - 0.316 = 0.684
p = frequency of healthy allele = 0.684
Frequency of healthy allele = 0.684
2.Ans.
how many people in a population of 10,000 will have sickel cell anemia.
sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia.
3.Ans.
how many people in a population of 10,000 will be sickel cell carriers.
sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers
4.Ans.
mutations occurs and new allele introduced into the population.
small population which causes genetic drift.
the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained.
3. Alkaptonuria caused by recessive allele.
survey revels that 5 in 50 people have Alkaptonuria
a. if a population of 50 people
if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1
q2 = 0.1 then q = square root of 0.1 = 0.316
q = 0.316
p = 1 - q =0.684
p = 0.684
b. Heterozygous to Alkaptonuria = 2pq
2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele
C. Homozygous Dominant = AA =p2
AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant.
I think that (e) have only non-singular generalized inversesSol.pdf
I think that (e) have only non-singular generalized inverses?
Solution
I think that (e) have only non-singular generalized inverses?.
ConvertingSeconds.javaimport java.util.Scanner;public class Conv.pdf
ConvertingSeconds.java
import java.util.Scanner;
public class ConvertingSeconds {
public static void main(String[] args) {
//Declaring variables
int fromseconds,hours,minutes,seconds,toSeconds;
//Scanner class Object is used to read the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number of seconds entered by the user
System.out.print(\"Enter Total Number of Seconds :\");
fromseconds=sc.nextInt();
//calling the method secondTime() by passing the seconds as arguments
String str=secondTime(fromseconds);
//Converting the string to String array where \',\' is the delimeter
String arr[]=str.split(\",\");
//Converting the String into integer
hours=Integer.parseInt(arr[0]);
minutes=Integer.parseInt(arr[1]);
seconds=Integer.parseInt(arr[2]);
//calling the method inSeconds by passing the hours,minutes,seconds as arguments
toSeconds=inSeconds(hours,minutes,seconds);
//Displaying the hours,minutes,seconds to seconds
System.out.println(hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds corresponds to
\"+toSeconds+\" seconds\");
}
/*This method will converts the hours,minutes,seconds to seconds
* Params:hours,minutes,seconds
* Return:totalSeconds
*/
private static int inSeconds(int hours, int minutes, int seconds) {
//Converting the hours,minutes and seconds to totalseconds
int totSeconds=hours*3600+minutes*60+seconds;
return totSeconds;
}
/*This method will converts the seconds to hours,minutes,seconds
* Params:total Seconds
* Return:String
*/
private static String secondTime(int totseconds) {
//Declaring variable
int secs=totseconds;
//calculating the number of hours
int hours=totseconds/3600;
totseconds=totseconds-(hours*3600);
//calculating the number of minutes
int minutes=totseconds/60;
totseconds=totseconds-(minutes*60);
//Calculating the number of seconds
int seconds=totseconds;
//displaying the total seconds to hours,minutes,seconds
System.out.println(secs+\" Seconds corresponds to \"+hours+\" hours \"+minutes+\" minutes
\"+seconds+\" seconds\");
return hours+\",\"+minutes+\",\"+seconds;
}
}
_____________________________________________
Output:
Enter Total Number of Seconds :10000
10000 Seconds corresponds to 2 hours 46 minutes 40 seconds
2 hours 46 minutes 40 seconds corresponds to 10000 seconds
_____________________________________________
Output1:
Enter Total Number of Seconds :15000
15000 Seconds corresponds to 4 hours 10 minutes 0 seconds
4 hours 10 minutes 0 seconds corresponds to 15000 seconds
_________________________________________Thank You
Solution
ConvertingSeconds.java
import java.util.Scanner;
public class ConvertingSeconds {
public static void main(String[] args) {
//Declaring variables
int fromseconds,hours,minutes,seconds,toSeconds;
//Scanner class Object is used to read the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number of seconds entered by the user
System.out.print(\"Enter Total Number of Seconds :\");
fromseconds=sc.nextInt();
//calling the method seco.
Before Attempting to measure risk aversion, following two factors sh.pdf
Before Attempting to measure risk aversion, following two factors should be considered:
1 ) Estimating probabilities: It includes measuring the future events based on their possibility of
occurrence. generally, the Probabilities are assigned based on past experiences.
2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the
investors in their previous investment. By looking at their investment history and analyzing
behavior, this table can be prepared.
Solution
Before Attempting to measure risk aversion, following two factors should be considered:
1 ) Estimating probabilities: It includes measuring the future events based on their possibility of
occurrence. generally, the Probabilities are assigned based on past experiences.
2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the
investors in their previous investment. By looking at their investment history and analyzing
behavior, this table can be prepared..
Answer The general circumstance for retaining outside experts is wh.pdf
Answer: The general circumstance for retaining outside experts is where the auditor requires a
special skill or knowledge outside of accounting or auditing, the auditors feel they do not have
the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work
of a Specialist provides guidance and recognizes that auditors are not expected to be experts in
all areas, auditors may encounter material matters which are complex or subjective, and an
auditor may use the work of a specialist as evidential matter to evaluate financial statement
assertions (PCAOB AU 336, 2010).
In the case of auditing pensions, the auditors would most likely use an actuary as a specialist as
the auditors would not be expected to have that level of expertise. The actuary would be
knowledgeable of the entire pension development process and be able to more quickly
comprehend the nuances of the actuarial assumptions and mechanics of the calculations. For
example, some of the tasks that the actuary would be involved with the company on for pensions
includes the measurement of pension obligations, assignment of plan obligations to time periods,
development of a cost allocation procedure, development of a contribution allocation procedure,
determination of types and levels of benefits, projection of pension obligations, projection of
plan costs or contributions (AAA, 2010).
Much of the financial impact of pensions is based on going from current known data (number
employees, salaries, etc.) through a number of assumptions (how many years work, how many
years collect a pension, mortality rates, what rate able to invest funds at, etc.) to develop the most
likely funding needs over time to provide the retirement benefits as agreed to cover the pension
expenses. Outside of determining the current employees, salaries, etc., almost everything else is
an assumption. How do we make the most appropriate and best assumption to ensure having
dollars available in the future? How much do we need to put aside now that will grow and cover
the pension requirements? This is where the audit firm should employ an actuary to examine the
actuarial-type assumptions made by the company on its employees and current retirees to see if
they are reasonable assumptions given the company’s data. The actuary would be able to help
examine the pension plan assets, the investments, the effective earning rates from the
investments (in comparison to the discount rate being used), the projected pension expenses, and
the resultant pension plan funding situation. The actuary’s review should be able to assist the
auditor in identifying potential issue areas for the auditor’s further examination.
Solution
Answer: The general circumstance for retaining outside experts is where the auditor requires a
special skill or knowledge outside of accounting or auditing, the auditors feel they do not have
the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work
of a S.
AnsConsolidated financial statements are the aggregated or combin.pdf
Ans:
Consolidated financial statements are the aggregated or combined figure of statements between a
parent company and its subsidiaries.
Consolidated balance sheet
As on 31/12/2020
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,000
8,750,000
Liabilities and shareholders equity:
Current liabilities:
Accounts payable
630,000
280,000
910,000
Noncurrent liabilities:
Loan payable
420,000
700,000
1,120,000
1,050,000
980,000
2,030,000
Shareholders equity:
Share capital
1,680,000
420,000
2,100,000
Retained earnings
3,584,000
1,036,000
4,620,000
5,264,000
1,456,000
6,720,000
6,314,000
2,436,000
8,750,000
Consolidated income statement
For the year ended 31/12/2020
Particulars
P
S
Total
Revenue:
Sales
2,804,200
2,100,000
4,904,200
Royalties
210,000
--
210,000
Dividends
100,800
--
100,800
3,115,000
2,100,000
5,215,000
Expenses:
Cost of sales
1,680,000
1,260,000
2,940,000
Other
784,000
575,400
1,359,400
2,464,000
1,835,400
4,299,400
Net income
651,000
264,600
915,600
Retained earnings statement
For the year ended 31/12/2020
P
S
Total
Opening retained earnings
3,353,000
897,400
4,250,400
Net income
651,000
264,600
915,600
Dividends declared
(420,000)
(126,000)
(546,000)
Closing retained earnings
3,584,000
1,036,000
4,620,000
Figures are added. P + S = Total.
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,000
8,750,000
Liabilities and shareholders equity:
Current liabilities:
Accounts payable
630,000
280,000
910,000
Noncurrent liabilities:
Loan payable
420,000
700,000
1,120,000
1,050,000
980,000
2,030,000
Shareholders equity:
Share capital
1,680,000
420,000
2,100,000
Retained earnings
3,584,000
1,036,000
4,620,000
5,264,000
1,456,000
6,720,000
6,314,000
2,436,000
8,750,000
Solution
Ans:
Consolidated financial statements are the aggregated or combined figure of statements between a
parent company and its subsidiaries.
Consolidated balance sheet
As on 31/12/2020
Assets
P
S
Total
Current assets:
Cash
70,000
28,000
98,000
Accounts receivable
210,000
224,000
434,000
Inventory
252,000
140,000
392,000
532,000
392,000
924,000
Noncurrent assets:
Land
140,000
--
140,000
Equipment
7,000,000
3,780,000
10,780,000
Amortization, equipment
(2,478,000)
(1,736,000)
(4,214,000)
Investment in S
1,120,000
--
1,120,000
5,782,000
2,044,000
7,826,000
Total assets
6,314,000
2,436,00.
Answer b)S phaseS-phase is synthetic phase where semiconservative.pdf
Answer: b)S phase
S-phase is synthetic phase where semiconservative DNA replication takes place, which occurs
between G1 and G2 phase
Solution
Answer: b)S phase
S-phase is synthetic phase where semiconservative DNA replication takes place, which occurs
between G1 and G2 phase.
a)The NLS resides somewhere between amino acids 240 and 357. b)The.pdf
a)The NLS resides somewhere between amino acids 240 and 357.
b)The NES resides somewhere between amino acids 115 and 239.
Solution
a)The NLS resides somewhere between amino acids 240 and 357.
b)The NES resides somewhere between amino acids 115 and 239..
A database is generally used for storing related, structured data, w.pdf
A database is generally used for storing related, structured data, with well defined data formats,
in an efficient manner for insert, update and/or retrieval (depending on application).
On the other hand, a file system is a more unstructured data store for storing arbitrary, probably
unrelated data. The file system is more general, and databases are built on top of the general data
storage services provided by file systems.
A Data Base Management System is a system software for easy, efficient and reliable data
processing and management. It can be used for:
Creation of a database.
Retrieval of information from the database.
Updating the database.
Managing a database.
It provides us with the many functionalities and is more advantageous than the traditional file
system in many ways listed below:
1) Processing Queries and Object Management:
In traditional file systems, we cannot store data in the form of objects. In practical-world
applications, data is stored in objects and not files. So in a file system, some application software
maps the data stored in files to objects so that can be used further.
We can directly store data in the form of objects in a database management system. Application
level code needs to be written to handle, store and scan through the data in a file system whereas
a DBMS gives us the ability to query the database.
2) Controlling redundancy and inconsistency:
Redundancy refers to repeated instances of the same data. A database system provides
redundancy control whereas in a file system, same data may be stored multiple times. For
example, if a student is studying two different educational programs in the same college, say
,Engineering and History, then his information such as the phone number and address may be
stored multiple times, once in Engineering dept and the other in History dept. Therefore, it
increases time taken to access and store data. This may also lead to inconsistent data states in
both places. A DBMS uses data normalization to avoid redundancy and duplicates.
3) Efficient memory management and indexing:
DBMS makes complex memory management easy to handle. In file systems, files are indexed in
place of objects so query operations require entire file scans whereas in a DBMS , object
indexing takes place efficiently through database schema based on any attribute of the data or a
data-property. This helps in fast retrieval of data based on the indexed attribute.
4) Concurrency control and transaction management:
Several applications allow user to simultaneously access data. This may lead to inconsistency in
data in case files are used. Consider two withdrawal transactions X and Y in which an amount of
100 and 200 is withdrawn from an account A initially containing 1000. Now since these
transactions are taking place simultaneously, different transactions may update the account
differently. X reads 1000, debits 100, updates the account A to 900, whereas X also reads 1000,
debits 200, updates A to 800. In bot.
9x^4y^10z^-2Solution9x^4y^10z^-2.pdf
This mathematical expression represents a single term with three variables being multiplied together with associated exponents. The term is 9x^4y^10z^-2, meaning 9 multiplied by x raised to the fourth power multiplied by y raised to the tenth power multiplied by z raised to the negative second power.
x= 38Solutionx= 38.pdf
x equals 3/8. The solution is provided that x equals 3/8. A short mathematical expression is given and the answer is shown to be x=3/8.
212+423 is same as 2 36+4 46 is same as 6 76 is same as 7 16...pdf
21/2+42/3 is same as 2 3/6+4 4/6 is same as 6 7/6 is same as 7 1/6.....keep it simple and reduce to
simplest answer to understand!!
Solution
21/2+42/3 is same as 2 3/6+4 4/6 is same as 6 7/6 is same as 7 1/6.....keep it simple and reduce to
simplest answer to understand!!.
1. Atoms Atoms are structured by 3 main components. Protons, ne.pdf
1. Atoms:
Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the
atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral.
Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively
charged.
2. Bonds:
You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may
have heard of include, NaCl and H2O. In order for these elements to form these compounds,
bonds need to form between them. When bonds are made energy is released.
3. Periodic Table:
The number of electrons that an atom of an element has is usually equal to it\'s atomic number.
More importantly there is a trend that you can determine what the number of valence (outer
most) electrons are. Group 1 has 1 valence electron, Group 2 has 2, Group 3 has 3, Group 8 has
0, Group 7 has 7, Group 6 has 6 (O, S, Se, Te), Group 5 has 5( N, P, As) and Group 4 has 4
(C,Si).
4. Electronegativity:
This means how reactive a certain element is. If an element has lots of valence electrons it will
have a high electronegativity because it wants that 1 electron, to fill its 8 compartments in its
shell. The trend for electronegativity is that electronegativity increases as you go across the
periodic table towards the right ->. As you go towards the right, you have more and more
valence electrons, therefore more \"want\" for electrons to fill up the compartments.
When you go (^) up the periodic table electronegativity also increases due to the distance the
electrons are away from the nucleus (core) of the atom.
5. Polarity:
Polarity just means you have 2 poles. Think of it like a magnet, you usually have a north pole
and a south pole OR a positive pole and a negative pole.
Different atoms have different charges. For example, H2O, consists of 2 hydrogens and 1
oxygen. Hydrogen is positively charged and oxygen is negatively charged. This is why water is
said to be polar. It consists of both a positive and a negative.
Water is also an example of hydrogen bonding, because Hydrogen is bonding with a large
molecule, oxygen.
Non-polar bonds are where there are no positives or negatives. An example would be a diatomic
molecule, such as oxygen gas: O2. Methane: CH4 is also considered to be non-polar because its
carbon atom covalently bonds with hydrogen atoms sharing almost completely equally.
Hope this helps! :)
Solution
1. Atoms:
Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the
atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral.
Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively
charged.
2. Bonds:
You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may
have heard of include, NaCl and H2O. In order for these elements to form these compounds,
bonds need to form between them. When bonds are made energy is released.
3. Perio.
1)importjavax.swing. = package contains JFrame and JButton class.pdf
1)
importjavax.swing.* => package contains JFrame and JButton classes
2)
javadoc comments begin with the /* symbol, and end with the */ symbol.
3)
If you want to read in values from the user in the console window you must use a
____Scanner______ object.
4)
float, int, long, and double are forms of ____C_____ data types.
5)
An easy way to access every element of a Two dimensional array is to use _____2_nested_____
for loops.
6)
To search the elements of an array, ___linear_____ search can always be used.
7)
There are two types of method calls, void return methods and ___data____ return methods.
8.)
import keyword is used to import built-in and user-defined packages into your java source file.
key word that is used to bring in java.utils is ____Import______
9. When an array name is passed to a method, it is passing the memory address, therefore it is
passing the array by ____reference_____.
because array name refers to the memory address of the array
Solution
1)
importjavax.swing.* => package contains JFrame and JButton classes
2)
javadoc comments begin with the /* symbol, and end with the */ symbol.
3)
If you want to read in values from the user in the console window you must use a
____Scanner______ object.
4)
float, int, long, and double are forms of ____C_____ data types.
5)
An easy way to access every element of a Two dimensional array is to use _____2_nested_____
for loops.
6)
To search the elements of an array, ___linear_____ search can always be used.
7)
There are two types of method calls, void return methods and ___data____ return methods.
8.)
import keyword is used to import built-in and user-defined packages into your java source file.
key word that is used to bring in java.utils is ____Import______
9. When an array name is passed to a method, it is passing the memory address, therefore it is
passing the array by ____reference_____.
because array name refers to the memory address of the array.
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site .pdf
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double
bond O.
2)This electrophile formed attacks the position of sulphur.
Because, the lone pair gets conjugated and forms Negative sites on both positions. But it
attacks -Carbon with less congestion(presence of less bulky groups).
3) It combines to form given product aong with release of HCl.
Solution
1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double
bond O.
2)This electrophile formed attacks the position of sulphur.
Because, the lone pair gets conjugated and forms Negative sites on both positions. But it
attacks -Carbon with less congestion(presence of less bulky groups).
3) It combines to form given product aong with release of HCl..
(a) NaCl is an ionic compound. The intermolecular forces are electro.pdf
(a) NaCl is an ionic compound. The intermolecular forces are electrostatic interactions.
(b) CCl4 is a non-polar molecular compound. The intermolecular forces are London dispersion
forces.
(c) CHCl3 is a polar molecular compound. The intermolecular forces are dipole-dipole
interactions and London dispersion forces.
(d) C(diamond) is a covalent solid. The intermolecular forces are covalent bonds.
(e) Au is a metallic solid. The intemolecular interactions are metallic bonds.
Solution
(a) NaCl is an ionic compound. The intermolecular forces are electrostatic interactions.
(b) CCl4 is a non-polar molecular compound. The intermolecular forces are London dispersion
forces.
(c) CHCl3 is a polar molecular compound. The intermolecular forces are dipole-dipole
interactions and London dispersion forces.
(d) C(diamond) is a covalent solid. The intermolecular forces are covalent bonds.
(e) Au is a metallic solid. The intemolecular interactions are metallic bonds..
Runs Newtons method to find roots of f(z) . It col function [a.pdf
% Runs Newton\'s method to find roots of f(z) . It col
function [a,n,TOL,N]=P3(a,n,TOL,N)
A=zeros(n,n,\'uint8\');
f = inline(\'z^3-1\'); % Define f and f\'.
fprime = inline(\'3*z^2\');
rts=roots([1 0 0 -1])
h=2*a/n;
for j=1:n+1, % Try initial values with imaginary parts between
y = -a + (j-1)*h; % -a and a
for i=1:n+1, % and with real parts between
x = -a + (i-1)*h; % -a and a.
z = x + 1i*y;
zk = z;
kount = 0; % kount is the total number of iterations.
conv1 = 0; % conv1,2,3 count iterations when approx soln is within
conv2 = 0; % TOL of root1,2,3.
conv3 = 0;
while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5,
kount = kount + 1;
zk = zk - f(zk)/fprime(zk); % This is the Newton step.
if abs(zk-root1) < TOL, % Check for convergence to root1.
conv1 = conv1 + 1;
else
conv1 = 0;
end;
if abs(zk-root2) < TOL, % Check for convergence to root2.
conv2 = conv2 + 1;
else
conv2 = 0;
end;
if abs(zk-root3) < TOL, % Check for convergence to root3.
conv3 = conv3 + 1;
else
conv3 = 0;
end;
end;
if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green.
if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red.
if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue.
end;
end;
imshow(A)
Solution
% Runs Newton\'s method to find roots of f(z) . It col
function [a,n,TOL,N]=P3(a,n,TOL,N)
A=zeros(n,n,\'uint8\');
f = inline(\'z^3-1\'); % Define f and f\'.
fprime = inline(\'3*z^2\');
rts=roots([1 0 0 -1])
h=2*a/n;
for j=1:n+1, % Try initial values with imaginary parts between
y = -a + (j-1)*h; % -a and a
for i=1:n+1, % and with real parts between
x = -a + (i-1)*h; % -a and a.
z = x + 1i*y;
zk = z;
kount = 0; % kount is the total number of iterations.
conv1 = 0; % conv1,2,3 count iterations when approx soln is within
conv2 = 0; % TOL of root1,2,3.
conv3 = 0;
while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5,
kount = kount + 1;
zk = zk - f(zk)/fprime(zk); % This is the Newton step.
if abs(zk-root1) < TOL, % Check for convergence to root1.
conv1 = conv1 + 1;
else
conv1 = 0;
end;
if abs(zk-root2) < TOL, % Check for convergence to root2.
conv2 = conv2 + 1;
else
conv2 = 0;
end;
if abs(zk-root3) < TOL, % Check for convergence to root3.
conv3 = conv3 + 1;
else
conv3 = 0;
end;
end;
if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green.
if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red.
if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue.
end;
end;
imshow(A).
The java Payroll that prompts user to enter hourly rate .pdf
/**
* The java Payroll that prompts user to enter
* hourly rate of pay and number of hours worked.
* Then calculates the gross pay and net pay
* and print to console.
* */
//Payroll.java
import java.util.Scanner;
public class Payroll
{
public static void main(String[] args)
{
//declare variables for hourly rate and hours worked
double hourlyRate;
int hoursWorked;
//Set tax rate as 0.15 (15 percent )
final double WITH_HOLD_TAX=0.15;
//Set grossPay=0
double grossPay=0;
//Set tax =0
double tax=0;
//Set netPay=0
double netPay=0;
//Create an instance of Scanner class
Scanner inputScanner =new Scanner(System.in);
System.out.println(\"Enter hourly rate of pay\");
//prompt for hourly rate
hourlyRate=Integer.parseInt(inputScanner.nextLine());
System.out.println(\"Enter number of hours worked\");
//prompt for number of hours
hoursWorked=Integer.parseInt(inputScanner.nextLine());
//Calculate grossPay
//multiply hoursWorked by hourlyRate
grossPay=hoursWorked*hourlyRate;
//calculate with hold tax
tax=grossPay*WITH_HOLD_TAX;
//calculate netPay
//subtract tax from grossPay
netPay=grossPay-tax;
//print gross pay and net pay to console
System.out.println(\"Gross Pay : \"+grossPay);
System.out.println(\"Net pay : \"+netPay);
}
}
---------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------
Sample output:
Enter hourly rate of pay
10
Enter number of hours worked
50
Gross Pay : 500.0
Net pay : 425.0
Solution
/**
* The java Payroll that prompts user to enter
* hourly rate of pay and number of hours worked.
* Then calculates the gross pay and net pay
* and print to console.
* */
//Payroll.java
import java.util.Scanner;
public class Payroll
{
public static void main(String[] args)
{
//declare variables for hourly rate and hours worked
double hourlyRate;
int hoursWorked;
//Set tax rate as 0.15 (15 percent )
final double WITH_HOLD_TAX=0.15;
//Set grossPay=0
double grossPay=0;
//Set tax =0
double tax=0;
//Set netPay=0
double netPay=0;
//Create an instance of Scanner class
Scanner inputScanner =new Scanner(System.in);
System.out.println(\"Enter hourly rate of pay\");
//prompt for hourly rate
hourlyRate=Integer.parseInt(inputScanner.nextLine());
System.out.println(\"Enter number of hours worked\");
//prompt for number of hours
hoursWorked=Integer.parseInt(inputScanner.nextLine());
//Calculate grossPay
//multiply hoursWorked by hourlyRate
grossPay=hoursWorked*hourlyRate;
//calculate with hold tax
tax=grossPay*WITH_HOLD_TAX;
//calculate netPay
//subtract tax from grossPay
netPay=grossPay-tax;
//print gross pay and net pay to console
System.out.println(\"Gross Pay : \"+grossPay);
System.out.println(\"Net pay : \"+netPay);
}
}
---------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------
Sample output:
E.
They represent the position of the other function.pdf
They represent the position of the other functional groups in the compound with
respect to most senior functional group. Hope you know the seniority order of functional groups
in nomenclature
Solution
They represent the position of the other functional groups in the compound with
respect to most senior functional group. Hope you know the seniority order of functional groups
in nomenclature.
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