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#1 Re: Science HQ » electrostatics... » 2018-05-21 22:06:37

hi George,Y

Good to see you back here!!!

Bob

#2 Re: Help Me ! » Probability question to do with a tennis match » 2018-05-21 22:05:18

OK, here we go:

orTLfD1.gif

For one 'round' of advantage there are three outcomes: A WINS (P = 0.49), REPLAY FROM 40/40 (P = 0.42), B WINS (P = 0.09)

If the REPLAY occurs, there are  another three outcomes with the same probabilities.  Then again, then again and so on.

The REPLAYs could, in theory, go on forever so As chance of winning is the sum of an infinite number of probabilities.

If a GP has first term a and ratio r I will write it as GP(a,r)

Using the sum to infinity formula

Hope that helps,

Bob

#3 Re: Help Me ! » Probability question to do with a tennis match » 2018-05-21 21:44:03

hi mrpace

Give me a moment to make a 'tree diagram' for this and I'll show you how to do this.

Bob

#4 Re: Help Me ! » Find number in the series » 2018-05-18 01:07:48

That's very ingenious!  Why a quartic?

Bob

#5 Re: Help Me ! » Find number in the series » 2018-05-17 22:24:02

hi Siva882

Welcome to the forum.

At the moment I cannot see any obvious answer to this.  The differences started nicely as 1, 8 , 27 and then that didn't continue.  Sequence puzzles can be tough.  For any set of numbers you can always fit a formula that will generate the sequence but this is not usually what the puzzle requires.  Sometimes they use a single digit from each term to get the next; or some complicated sum using all the digits.  We have been asked before for next numbers in sequences and the best efforts of several members have not found an answer.  Check you've got the right numbers, please.

Bob

#6 Re: Help Me ! » Ask About Annuity » 2018-05-16 20:54:11

hi Monox D. I-Fly

I notice no one has answered this yet.  Maybe this post will 'bump it' to someone's attention.  In case not:

I don't know about using annuities for borrowing purposes so I cannot at the moment understand what you are doing here.  But it's still possible we can work together to a solution.  I have done this in the past.  The poster explains more about what they are doing and then I can understand enough to make a contribution.  Together we may get there smile

Here's what I do understand.

(1) If you borrow money you have to pay it back with interest. I cannot see what the rate for this would be in your question.

(2) If you invest money (usually for retirement) in an annuity, you pay regular amounts and the value accumulates because of this and interest earned. (at 10% ??)

(3) You want the ten year annuity value to  be equal to the ten year amount owed for the loan so it can be paid off.

Please let me know if I'm correct so far.

Bob

#7 Re: Help Me ! » Probability problem » 2018-05-15 20:55:56

Thanks!  I think I deserve an award for getting all those brackets right!  smile

Bob

#8 Re: Help Me ! » Probability problem » 2018-05-15 20:02:48

hi Grantingriver

The sum can be written as a series of GPs.  Simplifying, it becomes a single GP with a sum of 1.5

If GP(a,r) means a GP with first term a and ratio r:

If |r| < 1 then sum to infinity is a/(1-r)

 

Bob

#9 Re: Help Me ! » Probability problem » 2018-05-14 19:54:21

hi mrpace

Imagine there is a field of infinite daisy chains.

The probability that a given chain contains n daisies is given by the following.

P = 2 / (3^n)

What is the average length of the daisy chains in the field?

Is this the exact wording of the question?  If the field has infinite daisy chains, to me that means every chain is infinite in length.  So I guess that isn't what is intended.

If there are an infinite number of chains, each of which has some finite number of daisies, then the answer cannot be determined.  Say nearly every chain has 3 daisies and just a tiny number have a different number of daisies then the expected length would be close to 3.  Substitute x for 3 and you can see that any answer is possible because we don't know how many chains there are of each length.

The only way I can make sense of this is if the question reads "A field has an infinite number of daisy chains.  One chain has length 1, one chain has length 2, one chain has length 3 and so on.   This I can work out and I get the expected length = 1.5

If this interpretation is what you want then post back and I'll complete the proof.

Bob

#10 Re: Help Me ! » Vectors and Planes » 2018-05-11 20:09:04

hi Monox D. I-Fly

I thought 'who is this examiner to set the exam for a Saturday'? but then I noticed the post date.  smile Let's hope he did well.

This is what I did for number 4:

Note: It's easy to eliminate x and y to get an equation in z.

2 .... gives x = -z/k          and      3 ... gives y = -kz so substitute these into 1 to get an equation for z.

It has a quadratic denominator so the roots will lead to impossible to solve.

I used the function plotter https://www.mathsisfun.com/data/function-grapher.php? to see what values of z are possible.  There are three asymptotes, two vertical at the roots mentioned above and one horizontal.

As this is a valid function, any value of k that leads to a (unique) value of z means a single set of values for z, x and y.  That's most values of k.  There are two values at approximately - 0.6 and +0.8 (if my algebra is correct) where z tends to infinity so no solutions.  I cannot find any values of k where more than one solution exists.  At one point in my algebra I multiplied through by k to remove the fraction so I separately tested k=0 and that leads to a single solution too.

Bob

#11 Re: Maths Is Fun - Suggestions and Comments » AIM has been discontinued » 2018-05-10 19:45:16

hi Mathegocart

I don't see this field in either my own profile or yours.  Please explain what that acronym stands for.

Bob

#12 Re: Help Me ! » Algebra » 2018-05-09 19:27:56

hi Fish,

Welcome to the forum.

This problem seemed familiar so I searched using the word 'quartic'.

Here is the answer:

http://www.mathisfunforum.com/viewtopic.php?id=24336

Bob

#13 Re: Help Me ! » it should be m-1*n-1 right? » 2018-05-09 19:23:12

hi Pratham Kundur

Welcome to the forum.

I think this is from the MIF puzzles: https://www.mathsisfun.com/puzzles/brea … ution.html

Are you suggesting an alternative answer?

Your post title is unclear.

Did you mean

A simple test for any formula is to try some values.  Let's try m = n = 2

The above three give -1, +1, and +1 respectively.

None of these will give a correct number of breaks.

Bob

#14 Re: Help Me ! » Algebra, factor and remainder theorem » 2018-05-08 20:13:47

hi Grantingriver

I've tried this two ways (remainder theorem and 'long division' and both ways get 3q^3 = 12*.  I was expecting a simpler value for q.  Where am I going wrong?

* cubed not squared ???

Bob

#15 Re: Introductions » Differences in Differences of Cube Numbers » 2018-04-30 19:57:21

hi galoruce

cZ8LKaW.gif

I have chosen a 'random' cubic. The table above shows the values for f(x) = 5x^3 + 2x^2 - 7x + 9  for values of x from 1 to 8, with the first, second and third differences.  What your post concerns is the final column.  Yes, the values are all 30 and in the 6 times table.  Notice also that 30 = 5 *6.  This is no coincidence.  The cubic coefficient will always show up in the third differences column as that number times 6.  I'll try to show why:

The calculation for x is

5x^3 + 2x^2 - 7x + 9   ...............................(a)

For the next calculation x is replaced by x+1:

5(x+1)^3 + 2(x+1)^2 - 7(x+1) + 9 = 5x^3+ 3*5x^2 + 15x + 5 + 2x^2 + 4x + 2 - 7x - 7 + 9 .........(b)

Subtract (a) from (b)

3*5x^2 + 15x + 5  + 4x + 2  - 7 = 3x5x^2 + 19x   

Now calculate the next first difference by replacing x by x+1

3x5(x+1)^2 + 19(x+1)   = 3*5x^2 + 2*3*5x + 15 + 19x + 19

Subtracting to get the second difference:

= 2*3*5x + 16

Replacing x by x+1

2*3*5(x+1) + 16 = 2*3*5x + 2*3*5 + 16

Subtracting

2*3*5

I have tried to track the coefficient 5 throughout the algebra by highlighting it in red.  Hopefully I got them all.  You can see that the first difference 5 occurs as 3*5 in the calculation.  After another difference it occurs as 2*3*5.

If you replace every 5 with 'a' you get the general calculation.  So the third difference would then by 2*3*a.

Hope that helps,

Bob

#16 Re: Help Me ! » Algebra, factor and remainder theorem » 2018-04-27 18:56:59

hi Mercurus

Welcome to the forum.

Got to go out so this will be a quick hint.

Try the remainder theorem: http://www.mathsisfun.com/algebra/polyn … actor.html

I think that will do it.

Bob

#17 Re: Introductions » Differences in Differences of Cube Numbers » 2018-04-24 20:05:40

hi galoruce

Welcome to the forum.

Yes there is a reason and you can use some algebra to prove it.  I'll give you an outline of what to do and perhaps you can do it for yourself.  If not, post again and I'll fill in the gaps.

Start with a general expression for a cubic:

Then write out the expression if you increase x by 1:

If you expand those brackets and then create the difference expression by subtracting the first from the second you'll end up with a quadratic.

So first differences are always a quadratic.

Now repeat that by writing a second quadratic with x changed to (x+1) and subtract to get the second difference expression.

You should find it's a linear expression (ie. an x term and a constant)

The x term has a factor of 6.

And if you carry out one more difference calculation you'll end up with the third difference = 6a .

There's the proof.  smile

Bob

#19 Re: Introductions » Short intro » 2018-04-14 21:10:52

hi zahlenspieler

Welcome to the forum.

Bob

#20 Re: Help Me ! » a question of a dice in probabilty » 2018-04-14 21:09:49

hi Hannibal lecter

I'm sorry you didn't get what you wanted in time for your exam.  What you have to realise is that MIF is a world wide, and free to use, forum.  While you are studying, the person helping you may be sleeping, or at work, or just digging the garden.

When you know you have an exam approaching you need to allow plenty of time for other members to respond.  I'm surprised that you knew in advance what question would be asked.  This doesn't happen in the UK;  candidates can practise similar questions but the examiners set new questions for the exam itself.  Had I known this I would have been even more reluctant to provide the actual answer since it goes against what I believe is the purpose of the exam in the first place.  If you give an answer having no idea why it is correct but just copying someone else then you give the examiners a false impression of your understanding.

This is why I try to provide general help on the topic rather than just a model answer.  It says in the rules: "This is a Forum, it is not instant chat. Leave your message, and come back later (hours or days) to see what responses you got."  And in the advice when asking for help: "We are happy to help! But we don't do your homework for you."  This could equally have said: "We won't do your exam for you."

Bob

#22 Re: Help Me ! » [ASK] P to AC » 2018-04-09 11:19:25

This is to 'bump' my previous post as I have edited it.

Bob

#23 Re: Help Me ! » [ASK] P to AC » 2018-04-09 04:09:07

Thanks.  This is an edited version of the post taking into account my new thinking on this problem. 

I've set up 3D coordinates as follows:

Origin A.
x axis AD
y axis AB
z axis AE.

As P is on AH with AP = 3.PH this means that P is 3/4 of the way along AH.

Using this, P has coordinates (6,0,6)

Let Q be on AC so that PQ is perpendicular to AQ (ie. PQ is the shortest distance between P and AC)

Let Q be (j,j,0) for some number j.

Using row vectors:

PQ = ( j-6, j-0, 0-6 ) and AQ = ( j, j, 0 )

If these are perpendicular to each other then their dot product is zero:

So Q is the point ( 3, 3, 0 )  So

Bob

#24 Re: Help Me ! » [ASK] Parallelogram in Cuboid » 2018-04-08 19:43:44

Arhh.  That's better.

FP = PH = 2.5 by Pythag.  PO = 5 and angle OPH is 90 degrees so OH is given by

Angle PHO = PBO = alpha

Bob

#25 Re: Help Me ! » [ASK] G to BH » 2018-04-08 19:25:19

With that arrangement of the letters, BH is a diagonal and AG another.  The diagonals cross at 90 degrees and bisect each other, so the distance required is half a diagonal.

Bob

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