Striatus

Now, this is the model for the given data (i.e. at a surrounding temperature of 20 degrees C.) What we're each supposed to do is pick a different surrounding temperature, in 5 degree increments, and model out the curve of the D(t) as it decreases. As well, we're to calculate how long the water stays at or above 95 degrees C for each of our curves (in order to simulate a real scenario where we'd be cooking...something with bacteria in it, apparently. Not really sure what that'd be, other than raw meat. Maybe if you're boiling a caribou you've just slain. Who knows.)

The power is going to go out.

Everyone I know has left, either for Logging in Colorado or Rowing in Sacramento. The house is empty, and will be until tuesday.

And the power is going to be out from 2 pm Saturday until an estimated 12pm Sunday.

As a man who has been solely consumed by an Xbawks for the past two weeks
As the first human Spectre
As the Commanding Officer of the SSV Normandy SR-1
As an ELECTRONIC MAN

This is going to be a serious period of darkness in my life. I'm not going to sugar coat it here.
I plan on going off the deep end and swan diving into batshit territory. When day breaks on Sunday, if I haven't made myself a mayoral sash out of tortillas and had in-depth conversations with my sock-puppet chairman...

That would be great, actually.

Seditative: As for finding out how long the water stays above 95 degrees C, we just need to find all places where (since the temperature difference at 15 degrees C of 95+ C water would be 80) solve 80\geq 85e^{-0.0240465t} for t to get \frac{ln(\frac{80}{85})}{-0.0240465}=t=2.5211 hours of 95+ temperature, which is plenty long enough to boil some caribou.

HALSBASDLGBJDFA DEADMAU5 REMIXES