no

They can use up to 2 years (I think) of a previous term. LBJ could have had 2 full terms plus the remainder he had from JFK.

At least an associates

I think Witcher 3 falls into the same problems listed.

• Witcher Sense
• Busy HUD
• Fetch Quests
• Given direction to go w/ mini-map
• Talks to self

I mean, I loved the game, but it’s not minimalist. It’s like playing a movie.

close, they also have song of the year which is just for the songwriter. So record is the total sum of the recording with engineer and producer (according to the Grammy website).

yes/no. it is a single recording (song). Not sure it has to be a ‘single’. as maybe some bands/people may not put out a single.

I was just looking it up.

record goes to all involved (writer/produced/engineer)

Song of the year goes to the song writer.

edit.

maybe it should be pronounced like recording of the year.

Also, Not (A XOR B)

Don’t confuse this guy with ontological questions.

This is straight truth table level stuff.

B can still be true when a is false. iff means that b can only be true when a is true.

Also, the equivalent statement is.

vehicle if and only if car.

not

car only if vehicle

since a truck is a vehicle, the statement is false.

Somewhat wrong above:

A B a iff b

T T T

T F F

F T F

F F T

look online for truth tables.

Remnant 2 looks to be silver

Totally thought it was the one in MD

buffer overflow in reality

How is there always an xkcd for something?

https://pi-hole.net/

It’s a network level ad-blocker by blocking at the DNS level.

It was originally meant to run on a raspberry pi, but will run in docker or other Linux os as well. Very light weight and a great self-host project. Been running for years and support via patreon.

you aren’t talking about the Monty Hall problem then

There are more infinite real numbers between 0 and 1 than whole numbers.

https://en.wikipedia.org/wiki/Countable_set

deleted by creator

But the issue is that by switching doors, you have a 66% chance of winning, it doesn’t drop to 50% just because there are 2 doors, it’s still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won’t choose.

I found the easiest way to think about it as if there are 10 doors, you choose 1, then 8 other doors are opened. Do you stay with your first choice, or the other remaining door? Or scale up to 100. Then you really see the advantage of swapping doors. You have a higher probability when choosing the last remaining door than of having correctly choosen the correct door the first time.

Edit: More generically, it’s set theory, where the initial set of doors is 1 and (n-1). In the end you are shown n-2 doors out of the second set, but the probability of having selected the correct door initially is 1/n. You can think of it as switching your choice to all of the initial (n-1) doors for a probability of (n-1)/n.