Mathematics!!

[AugustaAdaByron]

Very good!  And that brings up the next question.  I always use 3.1415927 for Ï€.  So I guess that means when I was in high school, we had 8-digit calculators.  What does everybody else use?

 

 

log(x)/log(2)

 

OR ln(x)/ln(2)

 

 

Yes, that works, too.  I just decided to choose the highest power of 2 that is within the range.  Integer number.  Of course.

 

I don't even remember. I think I had (still have somewhere at home) a CASIO. I always liked Texas Instruments though. No idea why.

 

I like your attention to detail! 

[Artinum]

Since I didn't want to double post...

 

Here's another math thing.

 

One is more likely to win the grand prize for the lottery twice in a row buying only a single ticket (Megamillions) than he is to rebuild a solved rubix cube from the disassembled pieces. However, the odds of his rebuilt cube being solvable are 1/12

 

The way I'm interpreting that, you're suggesting that you would be more likely to win the lottery twice in a row with the same numbers than you would be to build a Rubix cube with all the pieces in the right places. I can only assume this is incorrect, as there are factories that assemble thousands of cubes a day without issue! I'm not familiar with the "Megamillions" draw - is it the same as the "choose six from 49" odds we have for the UK lottery? If so, you'd be looking at some pretty long odds (in the region of one in 196,000,000,000,000).

 

Very good!  And that brings up the next question.  I always use 3.1415927 for Ï€.  So I guess that means when I was in high school, we had 8-digit calculators.  What does everybody else use?

 

If I'm working out rough estimates, I just use 3.14. I seldom need to use a more accurate version of pi. In algebraic terms, I just keep pi as a constant variable (unless I want to come up with some rough bounds, in which case 3.14 is again pretty okay).

 

For angles, I prefer to think in degrees rather than radians. Computers seem to disagree. When I had a bash at coding an asteroids game, I created some lookup tables for angles (much faster than calculating them on the fly) and converted to degrees, because to hell with you, computer.

[goodbye_farewell]

The key is in the wording:

She did not say "by." She said "on." Very clever.

After all the explaining, this is the only bit I needed [emoji14]

[caeliter]

The way I'm interpreting that, you're suggesting that you would be more likely to win the lottery twice in a row with the same numbers than you would be to build a Rubix cube with all the pieces in the right places. I can only assume this is incorrect, as there are factories that assemble thousands of cubes a day without issue! I'm not familiar with the "Megamillions" draw - is it the same as the "choose six from 49" odds we have for the UK lottery? If so, you'd be looking at some pretty long odds (in the region of one in 196,000,000,000,000).

 

I meant to say, "assemble it blindfolded" but forgot to write the word blindfolded (went back and edited)

 

I was saying that, if I were to buy one megamillions ticket this week (odds are 1:258,890,850) and win, then buy another ticket next week (but only one ticket) and win again, that would be a more likely scenario (1:67,024,472,213,722,500) than a man correctly assembling a rubix cube without being able to see the pieces. (1:519,024,039,293,878,272,000)

 

Megamillions is draw 5 numbers from a pool of 75 and then draw a separate number from a pool of 15. Must match all 6 to win the top prize.

[Artinum]

Right, that makes more sense!

[Emerald_Dragonfly]

Ahhh, maths. How have I missed this thread? My real life super power is teaching Algebra I to 7th graders (loved every minute of the 7 years that I did it). :-)

Not a legit mathematician, though. I just really like it!

[Sam Ashen]

So are we going to start a Physics thread or are we going to dump problems over here because they are related?

[goodbye_farewell]

So are we going to start a Physics thread or are we going to dump problems over here because they are related?

Haha I was wondering the same thing...

[GoldLynx]

So are we going to start a Physics thread or are we going to dump problems over here because they are related?

 

I thought it would work fine to combine them into one thread, but there seems to be division in the troops. Feel free to start a physics thread.

[Estrix]

Seems fairly likely to me that it will be more or less the same individuals with an interest in both threads so they may as well be together. Although I suppose some people like the neat division of a thread each.

[Mark D]

If a chicken-and-a-half lays an egg-and-a-half in a day-and-a-half, how many eggs will one chicken lay in one day?

[Sam Ashen]

Is that related to:

 

The north side of a roof is sloped at 30 degrees and the south side of the same roof is sloped at 40 degrees.  If a rooster lays an egg at the top of the roof, which direction is it most likely to roll?

[AugustaAdaByron]

The answer to both is 42! (that's not factorial)

[Oramac]

OK Guys help me out here.  This ought to be a cakewalk for you.  

 

Say I have a product and I'm going to retail it.  My cost is $100.  I was taught that a 30% markup would be (cost*1.3) = $130.  I just got off the phone with a lady who told me that's wrong, and it should be (cost/.7) = $142.86.  

 

The second version doesn't make any damn sense to me.  

[Mark D]

Is that related to:

 

The north side of a roof is sloped at 30 degrees and the south side of the same roof is sloped at 40 degrees.  If a rooster lays an egg at the top of the roof, which direction is it most likely to roll?

 

 

Nope, there actually IS a numeric answer, and it's not 1.

 

 

OK Guys help me out here.  This ought to be a cakewalk for you.  

 

Say I have a product and I'm going to retail it.  My cost is $100.  I was taught that a 30% markup would be (cost*1.3) = $130.  I just got off the phone with a lady who told me that's wrong, and it should be (cost/.7) = $142.86.  

 

The second version doesn't make any damn sense to me.  

 

You're calculating price as 30% higher than cost.  She's calculating cost as 70% of final price (obviously that makes final price = cost+42.86%).

 

Sounds like she does a lot of government contracts.......

 

"The first rule of government contracts is you never build one when you can build two for twice the price". - From the movie Contact.

[Sam Ashen]

Hmm.  I am looking at....

2615 Volts Line to Ground

32.31 Amps

7300 Watts

(per phase)

 

What's wrong with imaginary numbers?

 

14500 Volts Line to Ground

190 Amps

68000 Watts per Phase, 3 Phases

 

at the difference a 0.32 degree phase angle shift.

 

 

 

^^This is exactly how I was feeling about trying to express electrical loss in horsepower.

[Oramac]

^^This is exactly how I was feeling about trying to express electrical loss in horsepower.

 

Meh.  I don't like cats.  

[caeliter]

Nope, there actually IS a numeric answer, and it's not 1.

 

Can a chicken lay 2/3 of an egg? (Did I do it wrong?)

[Mark D]

Can a chicken lay 2/3 of an egg? (Did I do it wrong?)

Nope exactly right.

Question two:

If a chicken and a half lays an egg and a half in a day and a half, how long will it take a centipede with one wooden leg to kick the seeds out of a dill pickle?

Answer: blue, because aliens don't surf.

Here endeth the lesson.

[Sam Ashen]

Facebook newb here.  I think I am at 90 friends now.

 

What is the probability none of them share a birthday?

[GoldLynx]

If a chicken-and-a-half lays an egg-and-a-half in a day-and-a-half, how many eggs will one chicken lay in one day?

 

This means that three chickens (twice the number) will lay three eggs in those one and a half days. In three days, therefore, three chickens will lay six eggs. So in three days, one chicken will lay a third that number, or two eggs. Two eggs in three days =2/3 egg/day*chicken.

 

[AugustaAdaByron]

Facebook newb here.  I think I am at 90 friends now.

 

What is the probability none of them share a birthday?

 

Very very small. Less than 0.1%. (Assuming we're just thinking of day and month and not year.)

 

To make it easier to describe let's assume that none of them was born on the 29th of February and that they could have been born on any day of each year with equal probability. Consider any arbitrary list of your friends.

 

Your first friend could have been born on any day of the year. Thus, 365 possible days.

Your second friend will have to have been born on a different day of the year than the first one. Thus, 364 possible days.

The third friend will have to have been born on different days than the previous two. Thus, 363 possible days.

...

Your 90th friend will have to have been born on a different day of the year than the previous 89. Thus, 276 possible days.

 

However, since each one of them could have been born on any day with equal probability. We get 365^{90} different events that could have occurred, each one of them with equal probability.

This means that the probability that non of them share birthday is,

 

(365 x 364 x 363 x ... x 276)/365^{90}.

 

 

Nice problem!

 

btw, I have 3 Facebook friends who have birthday today.

 

P.S.: It's very possible that I've made some mistake on the calculations. Some + or - 1, for example. (But you get the idea.)

 

P.S2: When you reach 366-367 friends then the probability will be 0 and it will be certain that two of them share birthday.

[Sam Ashen]

 

Very very small. Less than 0.1%. (Assuming we're just thinking of day and month and not year.)

 

To make it easier to describe let's assume that none of them was born on the 29th of February and that they could have been born on any day of each year with equal probability. Consider any arbitrary list of your friends.

 

Your first friend could have been born on any day of the year. Thus, 365 possible days.

Your second friend will have to have been born on a different day of the year than the first one. Thus, 364 possible days.

The third friend will have to have been born on different days than the previous two. Thus, 363 possible days.

...

Your 90th friend will have to have been born on a different day of the year than the previous 89. Thus, 276 possible days.

 

However, since each one of them could have been born on any day with equal probability. We get 365^{90} different events that could have occurred, each one of them with equal probability.

This means that the probability that non of them share birthday is,

 

(365 x 364 x 363 x ... x 276)/365^{90}.

 

Well done!  I get something like 1 in 162,500 or so.

 

At about 22 or 23 or so it is about even, if I remember correctly.  Once you hit 30 or so you have a very good bet.

 

Now if I want to consider the number 22, I might approximate as follows:

 

P = (354/365)²² = 51% probability of 22 unique birthdays.

 

Putting it into Excel I get 52.4%.  The difference is in one of the assumptions I named.  Number 23 puts us less than even.

 

Nice problem!

 

Thanks!  

[AugustaAdaByron]

Yup! Around 23 it is even. I'm too lazy to actually do the calculations though. After 100 it's practically certain.

[AugustaAdaByron]

OK, that's a bit more logic-based than just mathematics and it's fairly well-known but let me say it.

 

Let us assume that there is an island where of all the inhabitants there are exactly k of them with blue eyes. No one however knows the colour of their eyes, no one can tell them and they have no reflective surfaces to see the colour of their eyes (no, not even the water). They can only see the colour of the eyes of all other inhabitants. The population is not very big and thus they all meet together early every night. However, for some completely crazy and random law whoever has blue eyes has to leave the island at midnight. One day a new person visits and right after midnight makes the comment "At least one of you has blue eyes". The person is honest so they're telling the truth. The inhabitants are all reasonable and law-abiding so as soon as someone finds out that they have blue eyes they leave the island at midnight. 

 

What's going to happen after they learn that at least one person has blue eyes?