How does the I.R.S. figures out age 59 and 1/2 for the I.R.A. distributions.?
459563
They get your birth date from the SSA, and they figure out whether you took the distribution after 59 years and 6 months after you were born.
1094661
Go to the website and look. They disclose that information.
972008
This is obviously a statistics problem.  Treat each mother as a sample and the data entry the age at which they had their first child.  If all you need is the "mean" point, it is a simple matter of knowing what a "normal distribution" is.  Half of the mothers will be above the mean date and half below (both 50%).  In a "formula" it would be (total of all the ages divided by the number of mothers)  More useful for the analyst is the shape of the distribution - the bell curve and the amount of data in the "tails".  To do this you need to perform a "t test" of the data set.  Since I assume that what you mean is the whole population of women, the distribution can be assumed to be "normal" and the "degrees of freedom" is infinite.  Look up "t test" in your statistics book, also known as "student's t".  The answer is there, in the "degrees of freedom" expressed in "standard deviations" from the mean.  Thus you can talk about "confidence intervals" from this.  The Wikipedia article is surprisingly hopeless for the non-mathematician.  My education is statistics came from "Biostatistical Analysis" by Zar, still in print and one of the only books from my former life I still keep on my shelf.
796185
a) 9¼% per year, compounded semiannually\n\nA = P(1 + r/100)^t\n= P*(1.04625^(2))^t\n≈ P*(1.09464)^t\nwhere t = number of years\n\nb) 9% pa, compounded continuously:\n\nA = P(lim n→∞ (1 + 0.09/n)^n)^t\n\nlim n→∞ (1 + 0.09/n)^n\n= lim n/0.09→∞ ((1 + 0.09/n)^n/0.09)^0.09\n= e^0.09\n\nSo A = Pe^(0.09t) (t in years)\n≈ P*1.09417^t which is slightly less than 9¼% per year, compounded semiannually
338128
#include<iostream.h>\n#define isleapyear(year) ((!(year % 4) && (year % 100)) || (!(year % 400) && (year % 1000)))\n\nclass Date\n{\nprivate:\nint  month, day, year;\n \npublic:\nDate(int month,int day,int year):month(month),day(day),year(year){};\nint isdatevalid();\nint weekday();\n\n};\n\n//\n//   return 1 if date is valid, 0 otherwise.\n//\nint Date::isdatevalid()\n{\n  if (day <= 0) return 0 ;\n  switch( month )\n{\n  case 1  :\n  case 3  :\n  case 5  :\n  case 7  :\n  case 8  :\n   case 10 :\n  case 12 : if (day > 31) return 0 ; else return 1 ;\n  case 4  :\n  case 6  :\n  case 9  :\n  case 11 : if (day > 30) return 0 ; else return 1 ;\n  case 2  : \n    if ( day > 29 ) return 0 ;\n      if ( day < 29 ) return 1 ;\n      if (isleapyear(year)) return 1 ;   // leap year\n    else return 0 ;\n}\n  return 0 ;\n}\n\n//\n// given month, day, year, returns day of week, eg. Monday = 0 etc.\n// tested for 1901 to 2099 (seems to work from 1800 on too)\n// \nint Date::weekday()\n{\n  int ix, tx, vx;\n \n  switch (month) {\n    case 2  :\ncase 6  : vx = 0; break;\ncase 8  : vx = 4; break;\ncase 10 : vx = 8; break;\ncase 9  :\ncase 12 : vx = 12; break;\ncase 3  :\ncase 11 : vx = 16; break;\ncase 1  :\ncase 5  : vx = 20; break;\ncase 4  :\ncase 7  : vx = 24; break;\n  }\n  if (year > 1900)  // 1900 was not a leap year\n    year -= 1900;\n  ix = ((year - 21) % 28) + vx + (month > 2);  // take care of February \n  tx = (ix + (ix / 4)) % 7 + day;              // take care of leap year\n  return (tx % 7);\n}\n \n \nint main()\n{\n  char week[7][10] = {\n  "Monday","Tuesday","Wednesday","Thursday",\n"Friday","Saturday","Sunday"\n};\n  Date *d;\n  int  month, day, year;\n  cout<<"Return the day of the week given the date."<<endl;\n  cout<<"Enter date in the form mm/dd/yyyy : ";\n  scanf("%d/%d/%d",&month,&day,&year);\n  d = new Date(month,day,year);\n  if (d->isdatevalid())\n  {\n    cout<<"The day of the week for this date is "<<week[d->weekday()]<<endl;\n  }\n  else\n    printf("%d/%d/%d not a valid date!",  month,day,year);\n  \n  getchar();   // wait \n  getchar();   // 2nd wait needed\n  return 0;\n}
120424
Try the following: 1)Get the actual amounts you are eligible for at ages 62, 65 and 70 from the SSA.  2) Assume your remaining life span. Since you do not appear to need the money to live, base your decision on the maximum SS $ you will receive based on your assumed age at death.  For example if you are entitled to $1,000/month at age 62, or $1,200 at age 65 or $1,700 at age 70 and you assume you will live until 85 then the math would suggest if you began to draw at 62 and lived until 85 you would draw a total of $276,000; $288,000 if beginning at age 65 and $306,000 if you waited until you were 70. Ofcourse these calculations do not include the added cash you would accrue if part or all of the money were invested as you aged.  Remember the RULE OF 72. What percent do I need to double my money???  Divide 72 by the number of years and the answer is the percent needed.  For example to double your money in 12 years you would need 6% compounded.  (12 times 6 = 72).  It also works in reverse - if I can invest at 9% APR and compound the interest it would take 8 years to double my money. (9 times 8 = 72).  Teach this rule to your children and grandchildren and they will easily see the value in saving....and they have the best advantage of all.....TIME
290570
I don't exactly know, but.. why not try going to their site, it probably states it somewhere within the site. Or even calling down there and asking would give you the information you need.
395363
Go to IRS.GOV.\n\nIn the search bar enter:\n\n"over 65 exemption"\n\nis should pop up a link for your answer.
980806
They should send you an annual statement.  You might want to try the web site:\nhttp://www.ssa.gov/planners/calculators.htm
649160
your tax professional can tell you. it depends on your age, income and whether or not each of you is a participant in a qualified plan through your employer. irs.gov is a great reference.
226966
take the date they were born and subtract it  from teh current date\n\n12/31/05 - 6/12/91 = 6 months, 19 days and 14 years
540631
just divide the given amount with the total amount and multiply it with 100. You will get the %age \nFor Example 63 out of 70 means\n(63/70)*100 = 90%
253208
I tried to calculate it for you on the website below, but I don't know how old you are, so you will have to do it yourself.
653803
There is a chart that explains all of it. Click on your level # and it will show you.
403150
you seem to only have trouble figuring out how to take in parameters. if you take some time to read from the link below perhaps it would become clearer to you\n\nyou can work out your logic like this\n\nsomeone is a teenager if:\n  age is greater or equal to 13\n   AND\n  age is less than or equal to 19\n\nsame as:\n  age >= 13 AND age <= 19\n\nhow to write age >= 13 in scheme?\n(>= age 13)\n\nhow to write age <= 19 in scheme?\n(<= age 19)\n\nand how do you AND these results together?\n(and (>= age 13) (<= age 19))\n\nso this tells you age IS a teenager.\n\nAll the operators we've used thus far {and, >=, <=} are binary operators. this means that they each take two parameters. go ahead, look back until this makes sense.\n\nto see if it is NOT a teenager, then we use the not operator. you've tried to use it already, but you passed to it two parameters. this is incorrect because 'not' is a unary operator, not binary. so you only pass it one argument:\n\n(not (and (>= age 13) (<= age 19)) )\n\nand that's your logic.
195991
Any quotient is a ratio, and by definition an average I.Q. will be 100.\nThe ratio, I.Q. = (mental age / chronological age)100.\n\nHave a look here:
290916
100...it's based on age