thanks in advance
更新1:
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4702 days= how many years?
2008-07-28 9:05 am
my cousin's is twelve years and 4702 days i want to know how old he would be in years but exact decimal wise like for example he's 12.514424 years
thanks in advance
thanks in advance
回答 (9)
2008-07-28 9:08 am
✔ 最佳答案
=> 4702/365=> 12.8821917808219178082191780821912 years
2008-07-28 10:13 am
ok take 365days(1 year) x 3 = 1095 then consider the leap year so on the forth year it is actually 366days so 1095 + 366 = 1461 is 4 years in days so take 1461 x 3 = 4383days or 12 years so if you added another 365days you would get 4748days which is more than 4702 so take 4702 - 4383 and you get 319days so... 4702days = 12 years and 319days
p.s. i figured you would want a mixed number and not a decimal
p.s. i figured you would want a mixed number and not a decimal
2008-07-28 9:45 am
4702 days/365 days
= 12.88219178082191780821917808219... years
= 12.88219178082191780821917808219... years
2008-07-28 9:24 am
12 years = 4383 days if you take account bissextile years
so 4702 days = 12 years +319 days
so 4702 days = 12 years +319 days
2008-07-28 9:18 am
4702/365 = 12.88 years
Exact Answer will be 11 years and 11 months approximately
Exact Answer will be 11 years and 11 months approximately
2008-07-28 9:14 am
4702/(365x3 + 366)
= 3.2183436
time by four = 12.8733744 years
^^^^^Did that to sorta account for leap years!
= 3.2183436
time by four = 12.8733744 years
^^^^^Did that to sorta account for leap years!
2008-07-28 9:09 am
4702/365=12.8876712328712328767123287671years
2008-07-28 9:08 am
4702 days = 12.8736494 years.
There are 365.242199 days in a year (averaged over many years). If it were exactly 365, there would be no need for leap years. However, the chances of the orbital period and rotational period of a planet being exact integer multiples of each other are essentially zero.
There are 365.242199 days in a year (averaged over many years). If it were exactly 365, there would be no need for leap years. However, the chances of the orbital period and rotational period of a planet being exact integer multiples of each other are essentially zero.
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