Selling Price = (CoGS * (1 + Tariff%)) / (1 - Margin)

I don’t mean like flapping their wings but - theoretically - could a wing suit skydiver find some perfect slope that works where they can gain altitude to a point of stall and land safely without a parachute? Alternatively could a glider do this?

Are we talking an extinction level event end of the dinosaurs type explosion? How much energy is released?

Firstly, please excuse the odd numbering system - it matches up to a spreadsheet that I am using!

I am going away on a golf trip in a week or so's time. There are ten golfers and we are playing three rounds. For each round, we need two groups of three, and one group of four.

I would like to organise the groups in such a way as to ensure that we can all play with as many different golfers as possible. My best current attempt is in the image. I was hoping that somebody might be able to tell me a) whether there is a better solution and b) if there is, either what the solution is, or how best to go about calculating it.

I have attempted to use websites that separate into groups using simpler parameters, but they don't give me anything better than what I have got thus far.

Thanks in advance!

I'm talking about big servers like things used for chat got, YouTube, Facebook, crypto mining, etc.

[Request] Ok no I didn’t, I’m 30, but I had a math thought. Let’s say I broke into my parents vodka, I think “I can drink 10% of the vodka and then fill the rest with water, mix it all up, and they’ll never know.” I do this, but the next time I think the same thing, “I’ll drink 10%, fill it up with water, and they’ll never know.” Of course, I’m a big dumb dumb and it will taste less and less of vodka. Assuming the water and vodka blended perfectly (aren’t immiscible) at water point does it all become water? Or, does it? Would this be logarithmic?

Moreover (although I guess this is sort of a chemistry question), let’s say they keep their 80 proof (40% alcohol) in the freezer, at what point in my taking of their vodka, does that liquid freeze?

The Pokémon TCG Pocket game has a mechanic where you are shown 5 cards face down, in which there is usually only one desirable card, and let you pick one at random to keep

That's all well-and-good, but recently they added an event where you get to "peak" at one of the cards before you make your choice.

Now - if the one you peak is correct, you can just pick it, so that math ends there. However, if the card you pick is incorrect, does the logic of the Monty Hall problem apply?

I, and I think most others, mentally decide a card to pick before it's even time (I always pick the bottom right). If I reveal a different card and it's incorrect, is it statistically probable for me to forsake my mental guess and pick one of the other 3 cards? This feels wrong, the game didn't know that was my choice, it feels like it should now be no less likely than the other 3 cards.

However, wouldn't the logic of the Monty Hall problem apply to this, and say it is incorrect? That, logically speaking, my initial probability doesn't change from 1/5 despite the fact another one was eliminate (this is, to be clear, under the theoretical that the 'peaked' card is wrong, as the peaked card being correct ends the scenario). If there was a 1/5 chance my initial guess was correct, there is a 4/5 chance it was wrong. If a card is revealed, there is still a 4/5 chance I was initially wrong, but if there are only 3 possible cards to switch to, they split that 4/5 3-ways, making them each 26.66% likely to be correct (as opposed to my 20%), no?

This is my way of understanding the Monty Hall problem but practically speaking I don't feel like that can be incorrect? The game doesn't even know my initial 'mental' pick, so how could there be a statistical difference if I choose it or swap it.

And if any of the 3 swaps are really 26.66% likely in that scenario, wouldn't that mean mentally envisioning one and then swapping it is actually (very slightly) more likely than peaking at one at random and then picking one of the remaining 4 (25%)? Again, in both these scenarios just not factoring in the possibility that the peak is correct, which should apply to both scenarios equally anyway and not change the end result.

I don't know what the flaw in my logic is but I can't imagine that 26.66% correct. If the 1/5 chance the "peaked" card was incorrect is distributed to all 4 cards then it's 25%, but wouldn't that mean the Monty Hall problem results in a 50% chance instead of a 66% chance?

Can anyone help me break down the probability in this scenario? Is there a flaw in my understanding of the probabiltiy or do you really increase your chances by mentally choosing one and then refusing to take it?

I'm reading The Sun Also Rises, and its mentioned that the whiskey and soda the characters but from a bar costs 12 francs. Would that have been a lot of money? Not quite a straightforward currency conversion as both France and Australia use different currencies than they did when the book was written.

A job I had awhile ago had a fun part of the commute. On the way home, once I merged onto the highway, if I got into a specific lane, I could follow that line for about 15 miles. It would merge me onto the interstate and drop me off on the right exit without ever changing lanes.

I took that route again tonight after a date night and my mind mused…what would be the longest stretch in the world where you could get into a lane and never change lanes?

Rules:

• No circles (e.g. multiple times around a roundabout)
• Assume no construction or lane blockages/closures
• A lane must be separated by painted lines, either dashed or solid. If a road ends to a non-painted street, that lane ends.
• A lane change requires driving over a dotted or straight line. If the lane exits to a different road, but it continues in a single lane without merging lanes, that is allowed.

I was reading a fiction book (One Piece), and a certain situation that happens WAY LATE into the story intrigued me, and thinking about what could have happened led me to make this weird question.

If there was a hole big enough for the mass and area of an entire island (average) to fall through to the center of the Earth, considering drag in either water or air or magma, how long would it take? Because whenever I search it up, the quickest answer always discounts any resistance from any fluid at all, and example results that do take it into account wouldn't translate well into this particular question.

People are speculating on the expected payments for a huge anti-trust suit against many schools, who colluded to cook the books on financial aid calculations.

So the question is. If there is 200M dollars available, and 200k POSSIBLE claimants, and the claim amount varies by the number of years attended for undergrad... what is the expected claim amount if every possible person submits a claim... for the classes of people who attended for 1, 2, 3, and 4 years of the relevant window.

I think it is a pretty interesting question.

So: There's this trivia game called that we use in my German (second language) class, and to finish the game, you have to collect 50,000 in-game dollars. To get money, you answer simple trivia questions. At first, a question is worth 1 dollar. The next is worth 2, then 3, etc until you break your answer streak by either being wrong or buying an upgrade.

I wrote the function f(X) = B((X/2)(2(C) + (X-1)(D))) to represent the total number of dollars earned in a given streak, where X is equal to the number of questions answered in a given streak, C is the number of dollars the first question in a streak is worth, and D is the number of additional dollars you earn for each successive question answered correctly in a streak. The initial function before buying any upgrades is f₁(x) = (X/2)(2+(X-1)).

With enough money, you can buy 4 upgrades from each of 3 categories that can change the output of the function. One is a "Multiplier" (B) upgrade, which will increase the B value. Another is the "Money Per Question" (C) upgrade, which will increase the C value. The last is the "Streak Bonus" (D) upgrade, which increases the D value.

B costs $50 for 1.5, $300 for 2, 2,000 for 3, and 12,000 for 5. C costs $10 for 5, $100 for 50, $1,000 for 100, and $10,000 for 500. D costs $20 for 3, $200 for 10, $2,000 for 50, and $20,000 for 250 Note that you do not have to buy the cheaper upgrades before the others, and they do not stack. The streak ends every time you buy an upgrade, but this only resets the streak bonus.

So, my question is: how can I optimize my purchase of upgrades in this silly little game so as to answer the least questions to earn $50,000, assuming I get none of them wrong?