Timeline for Find your dog's age
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2020 at 12:52 | comment | added | Noodle9 | @JonathanAllan Nice one - thanks! :-) | |
| Aug 3, 2020 at 12:52 | history | edited | Noodle9 | CC BY-SA 4.0 |
deleted 61 characters in body
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| Aug 3, 2020 at 10:57 | comment | added | Jonathan Allan |
1.0644944 works too!
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| Aug 2, 2020 at 17:02 | comment | added | Noodle9 | @JonathanAllan Oh wow, that's great - thanks! :D | |
| Aug 2, 2020 at 17:01 | history | edited | Noodle9 | CC BY-SA 4.0 |
added 99 characters in body
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| Aug 2, 2020 at 16:48 | comment | added | Jonathan Allan |
...actually, lambda n:round(1.06449445**(n-31),2) would work for 36.
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| Aug 2, 2020 at 16:36 | comment | added | Jonathan Allan |
If we can't assume n is an integer we can't use ~; if we can assume n is an integer (in \$[1,122]\$) we can do lambda n:round(2.71828**((n-31)/16),2) for 38.
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| Aug 2, 2020 at 15:24 | comment | added | Jonathan Allan |
lambda n:round(math.exp(-~n/16-2),2) saves 4
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| Aug 2, 2020 at 14:52 | comment | added | Jonathan Allan |
@petStorm I disagree, "The final output in dog years must be a decimal, rounded to 2 digits." states nothing about necessitating trailing zeros (and that's assuming it means 2dp rather than 2sf). Furthermore, many answers will give no trailing zero, e.g. 0.5 when given 20.
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| Aug 2, 2020 at 14:15 | history | answered | Noodle9 | CC BY-SA 4.0 |