Math - Question [ IMBA ]

[Dr. Boom]

Seas =)

Ok here we go - first off all the text was in German so I hope I will translate the following correct. ( This was from a test ( where I want work ( don't now the English word for this ) but I failed at this ^^ )

One child gets 1/2 - 100 €
One child gets 1/3 €
One child gets 1/4 + 50 €

How ( much or many <-- don't know ^^ ) money get they at all? ( I think it's to easy that I don't get it ^^ )

Please if you to this - write down the full way please - I need to know this!

 

[Darkness-4ever]

I don't understand your numbers and variables at all, theres no label for the 1/2 or the 1/4, and for the second one you have 1/3 € which completely throws me off as to the meaning of the fractioned unit-less numbers and the whole numbers with the €.

 

[DarkAngelAzazel]

Omg lol... this is easy as hell XD took me 1 minute of thinking
(I would put it into hidden tags but somehow they won't work for me )

First of all we calculate how much they get together in percent:

1. 1/2
2. 1/3
3. 1/4

Together = 13/12 and thats too much.

So the first child get - 100 and the last one +50. So all in all they get 13/12 - 50.

That means that the 50 which are all in all gone are 1/12. ( to make it fit. because all they get 1/1 = 12/12.)

That again means that if 1/12 = 50 -> 12/12 = 600.

Let's check:

first child: 1/2 - 100 -> 600 / 2 -100 = 200
second: 1/3 -> 600 / 3 = 200
third: 1/4 + 50 -> 600 / 4 + 50 = 200

All in all a gain 600

Aaaand that should be the right solution

(if you didn't get it I can give you a german explenation)

 

[Yixx]

The question was sentenced weird.. i didnt understand it at all..
Untill DarkAngelAzazal came with the answer which pretty much explained the question..

We need more riddles and maths questions! That would be fun!

 

[Darkness-4ever]

1/0 anyone?

 

[Dr. Boom]

Seas =)

@ DarkAngelAzazel: Wtf - now as I see the way - it's total simple - don't know why I get this way =O - thanks anyway =D [No need german I understand it well]

@ all: I said that I don't know how to write everything that you can understand it, because I'm German and this was a question in German ^^ sorry at all =)

[ Wtf: Opera want change "DarkAngelAzazel" to "Tranquilizes" =D ]

 

[Deleted member 157129]

Wait.. first child gets 1/2 of the money, which is -100 Euro. Third child gets 1/4 of the money which is 50 Euro? That just doesn't make sense. Can't see how this works either:

So the first child get - 100 and the last one +50. So all in all they get 13/12 - 50.

That means that the 50 which are all in all gone are 1/12. ( to make it fit. because all they get 1/1 = 12/12.)

Makes ANY sense at all.
If 13/12 is -50, then CERTAINLY 12/12 is not 600. 13/12 is the same as 12/12 + 1/12, and 1/12 is 50 by your calculation (as well as 12/12 being 600), thus 13/12 should be 650.


The way I see it, the first child gets 1/2 of the money, minus 100 Euro. The second child gets 1/3 of the money and the fourth child gets 1/4, plus 50 Euro. If x is the total amount of money (if is even sensible in this exercise), that makes for this statement:

(0.50x - 100) + 0.33x + (0.25x + 50)
1.08x - 50 is rounded total (you'd want to use the exact, which is 13/12, instead of 1.08).

 

[Dr. Boom]

Seas =)

650? Let's check this:

1) child: 650 / 2 - 100 = 225
2) child: 650 /4 + 50 = 212,50
3) child: 650 / 3 = 216,67 ( 216,66666....)

225 + 215,50 + 216,67 = 657,17

 

[Deleted member 157129]

I'm not saying the answer is 650, I'm saying, that by the numbers DAA has found, 13/12 would be 650, and not -50, if 12/12 is 600.

Though I'm rather confused by all the irregularities in notions, pluses and minuses are mixed with dashes and whatnot.

 

[Darkness-4ever]

I'm not saying the answer is 650, I'm saying, that by the numbers DAA has found, 13/12 would be 650, and not -50, if 12/12 is 600.

Though I'm rather confused by all the irregularities in notions, pluses and minuses are mixed with dashes and whatnot.

Don't forget the constants or -100 and +50, together = -50, so 13/12x - 50 = 600 since 13/12 = 650 and in short becomes 650 - 50 = 600.

And yeah I have no idea where the hell DAA jumped from 13/12 to 1/12 either.

 

[Paladon]

No, (13/12)*x = x + 50
ergo (x/12) = 50
hic et nunc 50*12 = x
in termini x = 600

 

[Deleted member 157129]

Oh, I see.. I thought he said 13/12 = 50 (or -50), not that the total, which I also mentioned in my solution, would be the result of 13/12 - 50.

 

[Darkness-4ever]

No, (13/12)*x = x + 50
ergo (x/12) = 50
hic et nunc 50*12 = x
in termini x = 600

Where did you get two x's from? I can only *see* one from the blanks left in his fractions, the 13/12*x, but not the x + 50. It may seem obvious to some but I honestly don't see it! Its killin me that I don't!

 

[Paladon]

Pfffff

x = searched number

x/2 - 100 + x/4 + 50 + x/3 = x
- aequi valere -
13x/12 = x + 50
x/12 = 50
x = 600
q.e.d.

 

[Deleted member 157129]

Where did you get two x's from? I can only *see* one from the blanks left in his fractions, the 13/12*x, but not the x + 50. It may seem obvious to some but I honestly don't see it! Its killin me that I don't!

It's from mine:

13/12x - 50 = x
Which is the equal to:
13/12x = x + 50

13/12 is the amount out of the total, which is x, and following the information provided; 13/12x - 50 is the total as well. Thus total = total.
You end up with:

13x = 12x + 600
x = 600

 

[Darkness-4ever]

Ooooooooh, and just like that I can leave this thread in peace.

 

[Paladon]

rest in peace thread

 

[The Reborn Devil]

Nono, no rest plez.
This must become the official maths thread of Teh Hive

Here a new question for you.
I know the answer to this one, but I want you to try


You're standing on flat land and looking at a mountain top "T". You want to measure the distance between you "P" and T and the height of the mountain. The angle between PT and the flat land is 36.2°. You then move 100 meters towards the mountain top to a point R. You measure the angle and it's now 37.9°.
Find the height of the mountain and the distance PT.
Enjoy

Edit: I thought a small picture could help a bit so I scanned the page from my book. The scan wasn't so good since the book was kinda hanging out of the scanner so I fixed it up a bit. It's at least readable.

 

[chilla_killa]

Nice exercise
Do you have to use Sinus, Cosinus and Tangens??
I could solve it, but I don't have time?

 

[Yixx]

one paper of 10x10 cm both sides filled with calcs..

Hope i have it right, and you can understand:

the bottom of the mountain is called A in here.. (rest is named the same)

RA = PA-100
Tan 36,2 = TA/PA
Tan 37,9 = TA/(PA-100)

TA = Tan 36,2 * PA
TA = Tan 37,9 * (PA-100)

Tan 36,2 * PA = Tan 37,9 * (PA-100)

Tan 36,2 / Tan 37,9 = (PA-100)/PA
Tan 36,2 / Tan 37,9 = PA/PA - 100/PA = 1 - 100/PA

1-(Tan 36,2 / Tan 37,9) = 100/PA
1 / (1-(Tan 36,2/Tan 37,9)) = PA/100

100 / (1-(Tan 36,2 / Tan 37,9)) = PA
PA = 1670,935

Tan 36,2 = TA/PA
Tan 36,2 * 1670,935 = TA
TA = 1222,940

PT² = TA² + PA²

PT = 2070,654 metres..

 

[The Reborn Devil]

You are correct, but it could've been easier

Here's the answer:

Here x = the distance between P - 100 (R) and the bottom of the mountain and h = the height of the mountain. Okay, here goes:

h = x * tan(37.9)
h = (100 + x) * tan(36.2)
x * tan(37.9) = 100 * tan(36.2) + x * tan(36.2)
x * tan(37.9) - x * tan(36.2) = 100 * tan(36.2)
x * (tan(37.9) - tan(36.2)) = 100 * tan(36.2)
x = (100 * tan(36.2)/(tan(37.9) - tan(36.2))

h = x * tan(37.9) = 1222.94
PT = x/cos(36.2) = 2070.66

The height of the mountain is approx. 1223 (or 1220 if you use less decimals throughout the calculation) and the distance between you and the top of the mountain (PT) is 2070.

 

[Dr. Boom]

Seas =)

What the hell dudes ( and girls don't know ) you eat mathbooks or what's going on =O

 

[zt2211]

Oceans =D

Find the curvature and components of acceleration of a particle if the following describes its position at time t:

• X(t) = sin(t)
• Y(t) = t · cos(t)
• Z(t) = 3^1/2 · t

Hints

• Try parametrizing the given equations into a single vector r(t)
• The components of acceleration are defined as a_N and a_T

Need more help?

• Parametrize X, Y, and Z into a vector r(t)
• a_N = (||v × a||) ÷ (||v||)
• a_T = (a · v) ÷ (||v||)
• Curvature = K = (||v × a||) ÷ (||v||)³

Parametrize? r(t)? What?

• Parametrizing, in this case, means converting given equations into a vector equation that effectively summarizes the information given in them
• It is common to name a vector with three components r(t)
• A vector consists of a magnitude and a parameter (often direction)
• A vector's dimension is determined by the number of magnitude/parameter pairs, in this case 3 {X,Y,Z}
• In three-dimensional vectors, such as in this case, parameters are often given as i, j, and k, which correspond to the X, Y, and Z coordinate system, respectively
• So, in this case, parametrization of the three equations would result in r(t) = sin(t)i + tcos(t)j + 3^1/2tk

What the hell are a and v?

• a and v are acceleration and velocity vectors, respectively
• v is the first derivative of the parametrized position vector r(t)
• v(t) = r'(t)
• a is the second derivative of the parametrized position vector r(t)
• a(t) = r''(t)

Derivative?

• A derivative of a function is itself a function that gives the slope of a line tangent to the original function at a given input value. It is one of the fundamental concepts of calculus, the other being integrals.
• Derivatives use the limiting process to calculate exact slopes at points on curves, but have many useful properties that allow for simple calculation
• One form of terminology for derivatives is the prefix d, where dy/dx means "the derivative of y in terms of x"
• A derivative taken in terms of a variable will produce results based on derivative rules, while all other variables have the prefix d attached
• The following rules assume the derivative is taken in terms of t and may prove useful in solving the problem given
• IMPORTANT: each rule may be applied to each separate term in a function, that is, each part wholly separated by a minus or plus symbol
• Let n and a = any constant, real number
• dtⁿ = (1/n)t^n-1 ; n>0 or n<0
• datⁿ = (a/n)t^n-1 ; n>0 or n<0
• dsin(at) = acos(t)
• dcos(at) = -asin(t)
• dn = 0 = da
• Chain rule: d(tsin(t)) = tcos(t) + sin(t)
• Notice the chain rule is the derivative of the product of two terms defined by previous rules; the term is split and the derivative of the first term is multiplied by the original second term then added to the derivative of the second term multiplied by the original first term
• Derivatives may be represented by the prime symbol, i.e. r'(t) is the derivative of the function r(t)

What the *♂☭♫ is the difference between · and ×?

• The first is a dot product (hence the dot symbol) and the second is a cross product (hence the "x" symbol)
• These are special multiplication symbols that refer to vector and matrix mathematics
• Dot products tell us to find the sum of the products of like parameters across vectors or matrices
• Dot products are easy to find; mutliply the corresponding vector parameters together and add them up (the parameters themselves cancel out when dot multiplied due to certain properties of matrices)
  
  Example

  • g₁ = ti + 3t²j - 3k
  • g₂ = cos(t)i + sin(t)j + 2cos(t)k
  • g₁ · g₂ = tcos(t)(i · i) + 3t²sin(t)(j · j) + -6cos(t)(k · k)
  • (i · i) = 1 = (j · j) = 1 = (k · k)
  • g₁ · g₂ = cos(t)(t - 6) + 3t²sin(t)
• Dot products produce scalar values (i.e., the value will be a number without a parameter)
• Cross products are a bit tougher and involve more advanced concepts such as matrix determinants. However, they can be simplified in a special case with three dimensions involved (which is convenient considering we live in 3D space, so the applications are limitless)
• To find a cross product in 3D space, multiply each term of the first vector to each term of the second vector, then drop the products which utilize the same parameter
  
  Example

  • g₁ = ti + 3t²j - 3k
  • g₂ = cos(t)i + sin(t)j + 2cos(t)k
  • g₁ × g₂ = tcos(t)(i × i) + tsin(t)(i × j) + 2tcos(t)(i × k) + 3t²cos(t)(j × i) + 3t²sin(t)(j × j) + 6t²cos(t)(j × k) + -3cos(t)(k × i) + -3sin(t)(k × j) + -6cos(t)(k × k)
  • i × i = 0 = j × j = 0 = k × k
  • i × j = k = -(j × i)
  • j × k = i = -(k × j)
  • k × i = j = -(i × k)
  • g₁ × g₂ = tsin(t)(k) + 2tcos(t)(-j) + 3t²cos(t)(-k) + 6t²cos(t)(i) + -3cos(t)(j) + -3sin(t)(-i)
  • g₁ × g₂ = 3(2t²cos(t) + sin(t))i - cos(t)(2t + 3)j + t(sin(t) - 3tcos(t))k
• Cross products produce vector quantities (i.e., vector parameters will be preserved in the product)

**SPECIAL NOTICE: this post uses special characters as defined by Times New Roman ACSII values, so your browser may not display the characters correctly. I've therefore taken the liberty of replacing special characters and putting the entire post below in HIDDEN tags! Yay!**

No Special Characters

Find the curvature and components of acceleration of a particle if the following describes its position at time t:

• X(t) = sin(t)
• Y(t) = t * cos(t)
• Z(t) = 3^1/2 * t

Hints

• Try parametrizing the given equations into a single vector r(t)
• The components of acceleration are defined as a_N and a_T

Need more help?

• Parametrize X, Y, and Z into a vector r(t)
• a_N = (||v X a||) / (||v||)
• a_T = (a * v) / (||v||)
• Curvature = K = (||v X a||) / (||v||)³

Parametrize? r(t)? What?

• Parametrizing, in this case, means converting given equations into a vector equation that effectively summarizes the information given in them
• It is common to name a vector with three components r(t)
• A vector consists of a magnitude and a parameter (often direction)
• A vector's dimension is determined by the number of magnitude/parameter pairs, in this case 3 {X,Y,Z}
• In three-dimensional vectors, such as in this case, parameters are often given as i, j, and k, which correspond to the X, Y, and Z coordinate system, respectively
• So, in this case, parametrization of the three equations would result in r(t) = sin(t)i + tcos(t)j + 3^1/2tk

What the hell are a and v?

• a and v are acceleration and velocity vectors, respectively
• v is the first derivative of the parametrized position vector r(t)
• v(t) = r'(t)
• a is the second derivative of the parametrized position vector r(t)
• a(t) = r''(t)

Derivative?

• A derivative of a function is itself a function that gives the slope of a line tangent to the original function at a given input value. It is one of the fundamental concepts of calculus, the other being integrals.
• Derivatives use the limiting process to calculate exact slopes at points on curves, but have many useful properties that allow for simple calculation
• One form of terminology for derivatives is the prefix d, where dy/dx means "the derivative of y in terms of x"
• A derivative taken in terms of a variable will produce results based on derivative rules, while all other variables have the prefix d attached
• The following rules assume the derivative is taken in terms of t and may prove useful in solving the problem given
• IMPORTANT: each rule may be applied to each separate term in a function, that is, each part wholly separated by a minus or plus symbol
• Let n and a = any constant, real number
• dtⁿ = (1/n)t^n-1 ; n>0 or n<0
• datⁿ = (a/n)t^n-1 ; n>0 or n<0
• dsin(at) = acos(t)
• dcos(at) = -asin(t)
• dn = 0 = da
• Chain rule: d(tsin(t)) = tcos(t) + sin(t)
• Notice the chain rule is the derivative of the product of two terms defined by previous rules; the term is split and the derivative of the first term is multiplied by the original second term then added to the derivative of the second term multiplied by the original first term
• Derivatives may be represented by the prime symbol, i.e. r'(t) is the derivative of the function r(t)

What the **** is the difference between * and X?

• The first is a dot product (hence the dot symbol) and the second is a cross product (hence the "x" symbol)
• These are special multiplication symbols that refer to vector and matrix mathematics
• Dot products tell us to find the sum of the products of like parameters across vectors or matrices
• Dot products are easy to find; mutliply the corresponding vector parameters together and add them up (the parameters themselves cancel out when dot multiplied due to certain properties of matrices)
  
  Example

  • g₁ = ti + 3t²j - 3k
  • g₂ = cos(t)i + sin(t)j + 2cos(t)k
  • g₁ * g₂ = tcos(t)(i * i) + 3t²sin(t)(j * j) + -6cos(t)(k * k)
  • (i * i) = 1 = (j * j) = 1 = (k * k)
  • g₁ * g₂ = cos(t)(t - 6) + 3t²sin(t)
• Dot products produce scalar values (i.e., the value will be a number without a parameter)
• Cross products are a bit tougher and involve more advanced concepts such as matrix determinants. However, they can be simplified in a special case with three dimensions involved (which is convenient considering we live in 3D space, so the applications are limitless)
• To find a cross product in 3D space, multiply each term of the first vector to each term of the second vector, then drop the products which utilize the same parameter
  
  Example

  • g₁ = ti + 3t²j - 3k
  • g₂ = cos(t)i + sin(t)j + 2cos(t)k
  • g₁ X g₂ = tcos(t)(i X i) + tsin(t)(i X j) + 2tcos(t)(i X k) + 3t²cos(t)(j X i) + 3t²sin(t)(j X j) + 6t²cos(t)(j X k) + -3cos(t)(k X i) + -3sin(t)(k X j) + -6cos(t)(k X k)
  • i X i = 0 = j X j = 0 = k X k
  • i X j = k = -(j X i)
  • j X k = i = -(k X j)
  • k X i = j = -(i X k)
  • g₁ X g₂ = tsin(t)(k) + 2tcos(t)(-j) + 3t²cos(t)(-k) + 6t²cos(t)(i) + -3cos(t)(j) + -3sin(t)(-i)
  • g₁ X g₂ = 3(2t²cos(t) + sin(t))i - cos(t)(2t + 3)j + t(sin(t) - 3tcos(t))k
• Cross products produce vector quantities (i.e., vector parameters will be preserved in the product)

 

[The Reborn Devil]

That kind of maths is something I'll bump into next year
I could try though, just not in the correct way I suppose ^^
I'll define t myself and then find the different values and get the speed and acceleration.
Weeeee (I use radians here)

t = 1
X(1) = sin(1) = 0.841470985
Y(1) = 1 * cos(1) = 0.540302306
Z(1) = 3^1/3 * 1 = 1.44224957
Speed = Sqrt(X(1) * X(1) + Y(1) * Y(1) + Z(1) * Z(1) = 1.75501676
t = 2
X(2) = sin(2) = 0.909297427
Y(2) = 2 * cos(2) = -0.832293673
Z(2) = 3^1/3 * 2 = 2.88449914
Speed = Sqrt(X(2) * X(2) + Y(2) * Y(2) + Z(2) * Z(2) = 3.13685668
t = 3
X(3) = sin(3) = 0.141120008
Y(3) = 3 * cos(3) = -2.96997749
Z(3) = 3^1/3 * 3 = 4.32674871
Speed = Sqrt(X(3) * X(3) + Y(3) * Y(3) + Z(3) * Z(3) = 5.24989862

Lol, I fail. I knew actually from the moment I started calculating this wouldn't work at all
Guess I need those derivatives and cross/dot products.

 

[chilla_killa]

OMFG, I thought I was good in Maths, with my A, but you are much better then me and I want to study physics.
How old are you????

 

[The Reborn Devil]

He's probably older than us. I'm learning it next year

 

[chilla_killa]

Does anybody know about polynom division??

 

[Paladon]

Yup, why?

 

[Dr. Boom]

Seas =)

Poly. Div. is eeeaaaassssy =)

 

[Paladon]

I don't think it's easy at all but still handable.

 

[Deleted member 157129]

Didn't recognize the term, but a quick googling told me it has to do with reducing the degree of the upper expression in a fraction. If that's all there is to it, then it's quite easy, though potentially time-consuming.

 

[The Reborn Devil]

Polynom Division is fun
But I like the "solve()" command on my laptop better

 

[HINDYhat]

Find the curvature and components of acceleration of a particle if the following describes its position at time t:

• X(t) = sin(t)
• Y(t) = t · cos(t)
• Z(t) = 3^1/2 · t

Boooring. You could have at least chosen a difficult question. Let me open up my Calculus III notes and I'll solve it in a second:
r(t) = <sin(t), t*cos(t), sqrt(3)*t>
r'(t) = <cos(t), cos(t) - t*sin(t), sqrt(3)>
r''(t) = <-sin(t), -2sin(t) - t*cos(t), 0>

And since a_T = r'.r''/|r'|, we compute r'.r'' and |r'| for further use:
r'.r'' = -sin(t)cos(t) + [cos(t) - t*sin(t)]*[-2sin(t) - t*cos(t)]
|r'| = sqrt(cos²(t) + [cos(t) - t*sin(t)]² + 3)
Then:
a_T = (-sin(t)cos(t) + [cos(t) - t*sin(t)]*[-2sin(t) - t*cos(t)])/sqrt(cos²(t) + [cos(t) - t*sin(t)]² + 3)
Screw simplification because the components you gave suck.

Anyway, continuing, a_N = |r' x r''|/|r'| so we need the r' x r'' vector. For that, I'll use a 3x3 determinant of the following matrix:
A = [i, j, k; cos(t), cos(t) - t*sin(t), sqrt(3); -sin(t), -2sin(t) - t*cos(t), 0]
det(A) = i*sqrt(3)*[2sin(t) + t*cos(t)] - j*sqrt(3)*sin(t) + k*[-cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t))]
and so:
r' x r'' = <sqrt(3)*[2sin(t) + t*cos(t)], sqrt(3)*sin(t), -cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t))>
|r' x r''| = sqrt(3*[2sin(t) + t*cos(t)]² + 3*sin²(t) + (-cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t)))²)
Then:
a_N = sqrt(3*[2sin(t) + t*cos(t)]² + 3*sin²(t) + (-cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t)))²)/sqrt(cos²(t) + [cos(t) - t*sin(t)]² + 3)

And finally, K = |r' x r''|/|r'|³
K = sqrt(3*[2sin(t) + t*cos(t)]² + 3*sin²(t) + (-cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t)))²)/sqrt(3*[2sin(t) + t*cos(t)]² + 3*sin²(t) + (-cos(t)*(2sin(t) + t*cos(t)) + sin(t)*(cos(t) - t*sin(t)))²)/sqrt(cos²(t) + [cos(t) - t*sin(t)]² + 3)³

Either I made a mistake somewhere in there, or you really gave horrible functions to start with.

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Anyway, participated in a math competition yesterday, and I think I did fairly well. Here's one of the questions that I'm pretty sure I got right. Anyone who gets it wins a cookie!

A piece of software generates random numbers between 0 and 1. It is designed so that for every x between 0 and 1, the probability that it will generate a number smaller than x is three times the probability that it will generate a number smaller than x/4. Moreover, the probability that it will generate a number higher than or equal to x is the same as the probability that it will generate a number smaller than 1 - x. Calculate the probability that this program will generate a number smaller than 1/21.

 

[Dr Super Good]

Dr. Boom, its a simple equation...
x/2 - 100 + x/3 + 1/4 + 50 = x
Solve that and bam, you have your answer.
x = 600

The_Reborn_Devil
mountain is 1222.93984961971m high with a hipotinuse of 2070.65376556243m from the orignal spot.

zt2211
Ah a physics wise guy I see. It is so obvious by the fact that those quantizations are of no use to man or beast outside of physics. Also I take it that your german by the fact that you are "." scalars instead of "x" for multiplying as we do in the UK. On computers we use * to represent multiplication as it is a prety universal symbol in all languages.

I did do some maths but I mistakenly read the formulas provided as velocity instead of positions so I decided to scrap it all (as it was logically wrong as V is first diervitave and A second). Overall the task is rather pointless as you just end up like hindy has show with god awful equations with tons of powers and trig. Yes it all represents something but it is just tedious to do.

 

[Yixx]

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Anyway, participated in a math competition yesterday, and I think I did fairly well. Here's one of the questions that I'm pretty sure I got right. Anyone who gets it wins a cookie!

A piece of software generates random numbers between 0 and 1. It is designed so that for every x between 0 and 1, the probability that it will generate a number smaller than x is three times the probability that it will generate a number smaller than x/4. Moreover, the probability that it will generate a number higher than or equal to x is the same as the probability that it will generate a number smaller than 1 - x. Calculate the probability that this program will generate a number smaller than 1/21.

I dont know the true mathemathical notation of chances, neither was I really excelling at chances, but we used P(X=Y) to write the probability that the Variable X is equal to Y.. im now using ==, <=, >=, < and > to represent respectively equal to, smaller than or equal to, greater than or equal to, smaller than and greater than..

In school, they taught me how to use binomial probability, i hope that is enough =\

we call our random number N, dont ask me why..
First, write down your story in maths..
3 * P(N<x) = P(N<x/4) (1)
P(N>=x) = P(N<1-x) (2)

P(N<1/21) is what we want to calculate.

Well, since a chance must always be smaller than or equal to 1 (as far as I know), P(N<x) must always be smaller than 1/3, for P(N<x/4) must at max be 1, and since 3 * P(N<x) is the same, P(N<x) is at max 1/3 (3)
P(N<x) + P(N>=x) = 1, since those are the exact opposite of each other, and thus must be 1 together (4)
Combining (3) and (4) tells us that P(N>=x) must always be greater than or equal to 2/3. (5)

Now we have something to start with, I am stuck..
All I know is that the probability is smaller than 1/3..

Well, please tell me because I'm dying to know! And please revive this thread, I want to learn stuff, especially about maths..

 

[HINDYhat]

Here's how I solved it. Also, you made a lot of mistakes. I'll use your notation:

(1) P(N < x) = 3*P(N < x/4)
(2) P(N >= x) = P(N < 1 - x)

So try to find P(N < 1/21):
P(N >= 1/21) = P(N < 1 - 1/21) = P(N < 20/21), by (2)
P(N < 20/21) = 3*P(N < 5/21), by (1)
Putting that all together, we have:
(1/3)*P(N >= 1/21) = P(N < 5/21)

Then, try to find P(N >= 5/21):
P(N >= 5/21) = P(N < 1 - 5/21) = P(N < 16/21), by (2))
P(N < 16/21) = 3*P(N < 4/21), by (1)
P(N < 4/21) = 3*P(N < 1/21), by (1)
Then, putting that last part all together, we have:
P(N >= 5/21) = 9*P(N < 1/21)

Now, consider the following probability: finding a number >= x or < x. Clearly, any number you choose will be either >=x or <x, so:
P(N >= 5/21) + P(N < 5/21) = 1
and
P(N >= 1/21) + P(N < 1/21) = 1

Replacing P(N >= 5/21) by 9*P(N >= 1/21), and
P(N < 5/21) by (1/3)*P(N < 1/21), we get the following system of equations:

9*P(N >= 1/21) + (1/3)*P(N < 1/21) = 1
P(N >= 1/21) + P(N < 1/21) = 1

Solving for P(N < 1/21), we get 1/13

 

[Yixx]

Wow, you really ARE smart!

Got more questions which we will probably not be able to solve?


Here is one from my maths book, probably really easy for you to solve, but still..

A special agent has been dropped in sea, 3 kilometers off coast.. The beach is a straight line. When the special agent swims 3 kilometers to the coast, he reaches point A, from there it is but 4 kilometers to run to point P, where is destination is.
The special agent can change his plans a little and swim for some more kilometers, then he has to run less.
His maximum speed in water is 4 km/h, where the water has no influence on his speed or direction, on the beach he can have a maximum speed of 8 km/h, and an infinite stamina.

Now the special agent is in the sea, deciding what to do, swim longer and run shorter?

1) give the equation for the time which the special agent has to use in order to get to point P, when the variable x is the distance from the point where he hits land, and point A.

2) What is the minimal time the special agent has to use to get to point P? (in minutes, so multiply the amount of hours by 60, duh)

3) What distance from point A to where the special agent hits land, with the time calculated above? (calculate x for the time found above)

 

[Dr Super Good]

HINDYhat, thats not ralley a maths questions but more like playing around with numbers.

Keep substututing the equations until you get something in the form you wanted. The only real maths was the last part which was suprisingly simple.

So how come I did not get this despite it needing only standard grade maths (substitutuion and solving dual simultanious equations)? I only ever get half of these stupid rearrange questions as I usually totally overlook it as an option and instead go down some obscure path which I only get to work half the time.

So what did I end up doing? Well I ended up getting a few measurements of chance by getting the overlaps of the equations and attempting to find the formula for less than x chance as a function of x. I managed to get that it envolved exponential elements but I could not align the constants correctly. Its like that damned formula for the block chance of blessed shield in Diablo II, I just cant get it...

The closest value I managed to obtain was for 1/20 or something but obviously using your method gets the correct result. Stupidly simple solution, so simple people will overlook it and try using university level maths to find a solution.

4 >= x >= 0
1. time(x) = (4-x)/8 + (((x^2)+9)^(1/2))/4 in hours

2. dtime(x)/dx = -1/8 + x/4(((x^2)+9)^(1/2))
-1/8 + x/4(((x^2)+9)^(1/2)) shows that the value is decreasing at x = 0 so will hit a minumum before starting to increase when
x/4(((x^2)+9)^(1/2)) = 1/8
2x = (((x^2)+9)^(1/2))
4(x^2) = (x^2)+9
3(x^2) = 9
(x^2) = 3
x = root(3) = about 1.7320508075688772935274463415059
plug this into the equation from 1 and you get 68.97114317 minutes.

3. x = root(3) = about 1.7320508075688772935274463415059

Really easy question, what is the point of 3 as you already found it out to get 2.

 

[HINDYhat]

DSG, that's the point. It's deceivingly simple. If you don't realize how simple it is, you get caught up in ultra formulas. But I can assure you that it's not random number bashing either: the goal was to get a number divisible by four such that the first rule would apply.

Also, participated in the AMC today. It was hard :<

Also, Yixx, where is point P with respect to point A? Is it aligned on the beach, or is it further off?

 

[Dr Super Good]

Its obvious it is a triangle.

on the beach he can have a maximum speed of 8 km/h

Knowing how simple the question looks, it has to be a triangle and most likly a L shaped one where his desternation is at pi/2 from his straight course to land. Also the coast must be perpendicular to the line from him to A where it is a direct cut off and his speed changes instantly.

Any other meaning of the question is impossible to solve due to missing data or would be beyond this highschool maths book he must be getting it from.

 

[Yixx]

DSG got it right, and about that last thing, i dont know why they asked it, i took the 3 last questions from the whole assignment, because the first 5 were waaaay too easy to solve, it was just calculating the time he needed to reach point A first and then P, and calculating time for going to P immediately.. stuff like that..

HINDYhat, DSG is also right here, it was a simple triangle, but i should have made a drawing or something to explain the situation better, indeed.

Any more questions? We are currently having sine and cosine in school, everybody has huge problems with it, while I'm sleeping -___-, seriously they just do the simplest stuff there, not even funny anymore..

Anyway, a vector question:

In 3D space 2 points are given: A(3,-4,5) and B(5,2,1).
Give the equation of plane P which has the property of: distance(P,A) = distance(P,B), or |PA| = |PB| for every x,y and z.

wow, i only now realize that this one is reaaaly easy xD

 

[Dr Super Good]

I have no clue what you are after as what does the distance function do between a plane and a position vector, but as a rough guess.
x+3y-2z+5=0

The plane whos normal vector is AB and is located at the midpoint.

 

[Yixx]

They want us to get familiar with working with planes, lines, equations of them and stuff..

Saw the property? |PA| = |PB|
meaning:
root((x-3)² + (y+4)² + (z-5)²) = root((x-5)² + (y-2)² + (z-1)²)

answer following option 1

x²-6x+9 +y²+8y+16 +z²-10z+25 = x²-10x+25 +y²-4y+4 +z²-2z+1
4x+12y-8z=-20

that is basically the answer..

Another way of doing it is using the fact that the line AB is a normal vector of P, then you will get an equation in the form of:
ax + by + cz = d, where you know a,b and c, but not d. Then you would have to know a point on the plane.. To get that, you can use the fact that |PA|=|PB|, using parametric description of the line AB with starting point A and another with starting point B..
that would leave it to something like this:
(x,y,z) = A + a*AB
(x,y,z) = B + b*AB
since the x,y and z are the same, that would leave you with this:

x_A + a* x_AB = x_B + b*x_AB
and so on for the y and z..
using simple maths gives you values for a and b..
using them in the parametric description gives you values for x,y and z..

Now put those x,y and z in the equation of the plane, which gives you a value for d.
Et voila, you have the equation of plane P.

answer following option 2

A(3,-4,5) B(5,2,1)
AB(2,6,-4)
(x,y,z) = (3,-4,5) + a*(2,6,-4)
(x,y,z) = (5,2,1) + b*(2,6,-4)
1) 3+2a = 5+2b
2) -4+6a = 2+6b
3) 5-4a = 1-4b

2a-2b=2 : a-b=1
6a-6b=6 : a-b=1
4b-4a=-4 : a-b=1
this means that we can just pick 2 values which work..
a=2,b=1..
(x,y,z) = (7,8,-3) according to the first parametric description
(x,y,z) = (7,8,-3) according to the second parametric description.. so this works..

(7,8,-3) is a point on P.
so d = 2x+6y-4z = 2*7+6*8-4*-3 = 14+48+12 = 74

equation:
x+3y-2z=37

So now we got 2 equations with different values for d, which are both not the same as the answer my book says..
is that possible? equations with different d values??