Use xkcd's formula to approximate the world population

Pyth, 21 20 bytes

-1 byte by Dennis

c/J-*3.d3C\ᙹ4T+33J

These have the same byte count but are ASCII-only:

c/J%*3.d3 523 4T+33J
c/-J%*3.d3*44T33 4TJ

I don't know Pyth, so still probably possibly golfable. Using the same algorithm:

TI-BASIC, 23 bytes

max(3getDate-5753
{.1int(Ans/4),Ans+33

getDate returns a list of three floats {YYYY,MM,DD} in some order depending on the date format setting (TI-84s don't have a true int datatype); the max( will be the year. Multiplying and subtracting inside the max( saves a close-paren.

Javascript (ES6), 55 54 48 bytes

-~((n=Date().substr(13,2)*3+280)/4-9.1)/10+' '+n

Works in Firefox 33; theoretically supports all years from 2000 to 2099. If programs that dump the result onto the console are not allowed, use this 51-byte function:

(n=Date().substr(13,2)*3+280)=>-~(n/4-9.1)/10+' '+n

Full program, 55 bytes:

n=Date().substr(13,2)*3+280,alert(-~(n/4-9.1)/10+' '+n)

Getting the year was quite expensive, but after using the deprecated getYear() instead of getFullYear(), all of the numbers in the equation became smaller, saving a lot of bytes. EDIT: Thanks to an eeevil trick, I skipped new and getYear() altogether. >:D

Suggestions welcome!

Python 2, 80 bytes

from datetime import*
y=date.today().year%40
print int(61.55+.75*y)/10.,y*3+280

Now rounds!