TiMartin's Math Corner

to be fair i am in English atm

 

nerds

 

octo, ive got one for u, how many people do you need in a room (minimum) to have a better than even chance of any two of them sharing a birthday.

 

i've heard this one before and think i remember the answer but i'm going to give a shot at solving it out anyway just because it's a cool problem and i'm taking a probability and statistics class this semester anyway

the first place that this problem will confuse people is that it deals with pairs of people - for every n people there are (n! / (2!(n-2)!)), (or n choose 2 which has a normal notation but is impossible to do on this forum), pairs of people, which gets pretty big pretty fast, for example 10 choose 2 is 45, 15 choose 2 is 105

the probability that there will be at least 1 pair with the same birthday (let's call it p(B) -- B for birthday!) is equal to 1 minus the probability that there will be no pairs with the same birthday (let's call it p(B'))
so we have p(B) = 1 - p(B') which is true by definition and not really saying much but it helps us set up for the next part

let's define p(B') (the probability that there won't be any birthday pairs) for every n people. this will be the probability that a pair of people will not share the same birthday (364/365) multiplied by itself for every pair of people (n choose 2, or that more complicated thing up there), or (364/365)^(n choose 2)

now we have p(B) = 1 - (364/365)^(n choose 2) -- we're looking good here, this is solvable! let's set up the problem a little more.
we are looking for values of p(B) which are greater than or equal to 0.5, so we can say 0.5 <= 1 - (364/365)^(n choose 2)
...which we can further simplify to 0.5 >= (364/365)^(n choose 2)
set up a logarithm to isolate (n choose 2)
log base (364/365) of 0.5 <= n choose 2
... which becomes 252.652... <= n choose 2
252.652 <= (n! / (2!(n-2)!)) which equals (n! / (2 * (n-2)!)) which equals 1/2 * (n! / (n-2)!) which equals 1/2 * (n * (n-1)) which equals 1/2 * (n^2 - n), we can get rid of the 1/2 so...
252.652 * 2 <= n^2 - n
0 <= n^2 - n - (252.652 * 2)
solve for n fairly trivially and you get n = 22.984, since the derivative of n^2 - n - (252.652 * 2) is positive from n = 1/2 to infinity you can safely assume for any number of people n greater than 22.984 you will have an even or above chance of a pair sharing a birthday, nearest integer above that is 23 so there's your answer
since this is informal it would have been easier to cut out the last few steps and say "it's pretty easy to find out from here what it is" but i wanted to show the whole process since its a cool problem

TL;DR 23 why the hell did i waste so much god damn time writing this i have real homework i could be doing

 

Nice, way nicer to understand than anything i would do.

 

i need help. my friend tells me that 1+2=3. Then i say...if thats true then 2-1 must equal 3. He says no. Any help?

 

pi

 

yeah actually im kinda hungry. I heard about this kid who's working on memorizing pie

 

1.
You have 250 metres of fence.
The fence touches the wall of a building.
The area of the fenced area is 2500m^2.
How long are the sides of the fence?

Here's a pic to make it clearer.

The fence is green and the wall is black. No idea why I told you this, lol. If you are capable of solving this you obviously knew.


2.
A goalie kicks the ball. This parable shows the ball's flight: -0.08x^2+2.8x.
How far does the ball fly?

Who much further would the ball fly if he kicked it from 1.5 metres higher?

Draw the graph yourself.


3.
What's 9+10?

 

Do you go to school yet?

The whole thing? Wow!

 

Well I actually started to memorize it a little but I got about 150 digest before I couldn't do anymore but now I think I can do thirty

3.141592643589793236264502 I totally messed the end up but I forgot most of the digits I knew

 

The sad thing is that you will absolutely never need to know this.
Never.

edit: Wow! What a nice profile pic, tod! I can almost see your face behind that ipad or whatever!

 

The first one has 2 answers either
Single side: Other Sides:
22 m 114 m
or
~228.08 m 10.96 m

The second one is 35m off the ground and 35.53m from 1.5m up, so 53 cm further.

The third one is by far the hardest. I have no idea about how to solve it. (any clues?)

 

Can you show us how you did this?
Giving just the answer makes it seem like you used a calculator...

 

This is a very interesting question, i had to google it to find the answer but the method for finding it is very clever.
(48C9)/(52C13)*4
(Number of hands where person 1 has all 4 aces / the total number of possible hands.) x 4(cause there is 4 people who can have it)
= (1677106640/635013559600)*4
= ~0.01

 

I did but...
1. Let the sides be a and b where:
a+2b = 250 (1)
ab=2500 (2)

From (1)
a+2b = 250
a = 250-2b

Substitute a into (2)
(250-2b)b = 2500
250b-2b^2 = 2500
2b^2-250b+2500=0

Use the quadratic formula
(250(+-)sqrt((250^2)-4*2*2500))/4
From this you get 2 values of b:
~114m or ~10.96m

and from (1)
a+2b=250

so sub the values of b in and you get:
when b = 114m
a=22m
and when b = 10.96m
a = ~228.08m

2. This is a straightforward usage of the quadratic equation:
-0.08x^2+2.8x=0

(-2.8(+-)sqrt(2.8^2))/-0.16
(-2.8(+-)2.8)/-0.16
=0 or 35
so 35 m

From 1.5m up:
-0.08x^2+2.8x+1.5=0

(-2.8(+-)sqrt((2.8^2)-4*(-0.08)*1.5))/-0.16
= ~-0.53 or 35.53m
so 35.53m which is 0.53m or 53 cm further.

 

Lmao it was for a contest I still didn't win, wasn't even in top three but I came in 4th made me mad for wasting all that time just to be beat.

I've posted my face before, yeah that's an I pad. Thanks tho

 

yep it isn't that hard, but it looks like ur pretty good at math

 

On a flat surface three balls are put , each of which has a radius of 2 in such a way, that and are tangent at the point , balls and are tangent at the point , and balls and are tangent at the point . Subsequently on these balls another ball is put which has a radius of 3, and which is tangent to balls respectively at points .

• Prove that line segments and are parallel to each other.
• Calculate the perimeter of the trapezoid .