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hi larry04

Welcome to the forum.

I had a quick look and it just looked like any other calculator. Would you be able to post an example that shows it's superior features?

Bob

hi paulb203

CFGs are an example of a statistic: a 'thing' you can get by doing some calculations on the raw data. If you knew the actual heights of every plant, you could produce a step graph. I'll illustrate by making up some heights.

Let's say the first 9 were 31, 33, 34, 34, 36, 37, 38, 38, 40.

You could start with a point at (30, 0) meaning no plants at 30 (or lower). Then a point at (31,1) meaning one plant at 31 or lower. Then (33, 2) meaning 2 plants at 33 or lower. And so on up to (40,9).

Then you could join them with a series of straight lines looking a bit like a staircase.

But we cannot do that because we don't know the individual heights. The only points we can be confident about are (30,0) and (40,9). So the CFG is a compromise, intended to show something about how many plants there are at each height. You can estimate the median, for example, by looking for a point on the curve at (median, 21.5).

If you had another set of data for plants that had been growing in a new fertiliser you could compare the graphs to learn something about whether the new fertiliser helps growth. A fertiliser graph that starts rising more slowly and zooms up quickly at the end would suggest you get better growth from the fertiliser. If the graph rises slowly at first, then zooms up quickly in the middle, then slows up as the graph reaches 40, you could conclude that the fertiliser gives plants of more consistent height.

The point (30,0) means there are no plants with heights equal to or lower than 30. Note that the table shows 30 < h ≤ 40. It means heights above 30 up to and including 40.

Does that help?

Bob

hi Kate123

Welcome to the forum.

I work in finance!

Hhmm! That's interesting.

Subtract £10 from the bill and divide the remainder equally. Then add the £10 to the share of the 'generous extra' person.

Bob

Lynda,

This is a maths forum. We do not allow advertising and we also do not want out forum being used for non maths purposes.

If your only reason for joining is to advertise your product, then please desist. If you have a mathematical contribution to make, then great!

If I don't get a response from you I will delete your account and posts.

Bob, Administrator

I have changed my diagram in the earlier post to try and show what I mean. As S3 is more than S2 the missile can take a shorter path to hit the ship. So the direction changes at X from warship's direction to missile's direction.

Starting with XY, you can resolve S3 into a component parallel with the ship's direction and equal to S1, and a component at right angles to the ship's direction.

The distance XY = R, and angle XYS is INV COS (S1/S3) = θ

Time for missile = R/S3

Time for warship from W to X is WX/S2 but what is WX?

Draw a perpendicular from X to SY meeting SY at U. Then XU = R . SIN (θ)

Draw a perpendicular from X to WS meeting WS at V.

angle WXV = INV COS (S1/S2) =Δ

WV = D - XU. hence WX = WV / SIN (Δ)

That's enough to complete the calculation.

Note about my initial comment.

If an object (A) is moving and another object (B) must aim to travel straight towards A then B's path is continuously changing direction. This is illustrated by the Guardian's Four Dogs Puzzle:

https://www.theguardian.com/science/201 … in-pursuit

Bob

hi KerimF

I have drawn a diagram. The question says "(while moving on a straight line towards the side of the other ship)."

That seems contradictory. I'm marking the two vessel's positions with S for ship and W for warship. If W heads directly towards S while S is moving forwards then W must continuously change the direction of travel so that it is once more heading at S. If W doesn't do this it will end up where S was rather than where S is. But that means W's movement is not in a straight line.

So it seems that the best course for W is to try and close on S following a diagonal line. I'll need to check but I think that the optimal path is so that W's component of velocity in S's direction is S1 but using the extra velocity that S2 allows to move closer to S's path, reducing D until it reaches a certain value*. Then, since S3 is not infinite, the missile must similarly follow a diagonal track so that it closes on where S will be when the missile crosses S's path.

* Initially I though R here, but that won't work as the missile's diagonal path will be greater than R. So I need some more trig. to calculate the position of W at launch.

Sorry if that sounds complicated. I'll have a go at a making and posting a diagram for all the above.

Bob

ps. Now I look at the diagram, I've realised that WX and XY should not be part of the same straight line. As S3 is bigger than S2 the missile should be on a track that is less acute to the ship's direction. The angle between S's direction and the diagonal lines is determined by adjacent = S1, hypotenuse = S2 and then S3.

later edit: I have now changed the diagram.

I have to own up here. When I did this I thought (5) was the answer. Then I compared with amnkb and read your comment. So I looked harder and put down both answers in my post.

And how many times did I tell students to read the question carefully?

Bob

Bob

ps. Aren't we all still students?

Bob

hi spmishra9

Welcome to the forum.

P=(A-b(q+r)^m)-c)q

Hmm. I've got several questions about this.

Firstly, that formula has 2 open brackets but 3 close brackets. It makes a difference to any calculation.

Next I'm unclear about what you are trying to maximise.

maximizing value of q

Do you mean vary q so that P is maximised?

It also depends on the sign of m and b. eg. if m is negative it changes the way P varies.

Thinking about it some more, I think the sign of every letter affects the outcome.

Please start by clearing up the bracket situation. I'll post back a formula in mathematical appearance so there no ambiguity and hopefully we can move to an answer.

later edit: I made up some numbers and tried getting a graph. Negative m doesn't seem to have an upper bound for P. When m is positive there's a clear maximum in positive q so I'm guessing that's what you're after. So I would probably differentiate to find the maximum. But I need a bracket consistent formula to do that properly please.

Bob

hi amnkb

Thanks, and thanks also for your patience.

We may be getting to the end of the line: http://www.mathisfunforum.com/viewtopic … 44#p436044

But maybe we will succeed in making a better mathematician and member too. I live in hope.

Bob

If it spends $40,000 on advertising, then 100,000 boxes of cereal will be sold and if it spends $60,000, then 200,000 boxes will be sold. Write a linear equation that relates the amount A spent on advertising to the number of

x boxesthe company aims to sell.

So x is for the boxes so y must be for the costs. => (100000 , 40000) and (200000 , 60000) are points on the line.

Bob

Sometimes we get posters who just want someone to do their homework for them. Our rules specifically say we don't do this. I'm a retired teacher. I expect students to do their homework themselves. But I will help if I can. My usual response is to push the thread back to the poster

eg wrote:

What formulas have you been given for this?

Mostlly that's the last we hear from the poster so we haven't wasted any time on them. If I do get a response then I've got something specific to work with.

Members: it's up to you, but I suggest you post nothing and await that response. If it doesn't happen then we haven't wasted any time or bandwidth.

Bob

See my important comment here: http://www.mathisfunforum.com/viewtopic … 44#p436044

Bob

sologuitar

Your behaviour is becomong less and less acceptable. Let me try to explain so there's no misunderstanding..

(1) Each day, when I log in I see a long string of posts from you. I find it very hard to remember which posts have been answered and which ones are still 'live'. If I've given you help but it hasn't been enough to get you to your own answer I'll try to give more help, but, at the moment I'm getting lost in all the posts.

(2) Every time you respond to a post you quote the previous post. That isn't helping either. It just increases the amount I have to wade through. It is possible to respond without quoting the previous post. What would help is a dialogue consisting of the following:

(i) post asking for help on a question.

(ii) (a) response from you saying that's it, I've got it now or (b) please offer more help about ......

(iii) if (a) then the thread is finished if (b) then wait for additional help and go back to step (ii)

(3) amnkb has been trying to help you as well as me. You were getting the whole solution from amnkb but you asked for this not to happen. I asked amnkb that whole solutions be hidden so you could still try to work it out for yourself. That is what is now happening! Each box is a hint and the 'contents' of the box is the answer you should have obtained for yourself. That takes a lot of time to set up but gives you maximum help. It's exactly what you asked for.

So it is very ungenerous of you to say

I am tired of clicking on rectangles to see the answer.

If you want help you have to accept it in whatever form it is offered. If you don't like it then say nothing! You are coming close to being banned for upsetting other members and I have to keep the peace.

So how should you go forward from here?

Look at the hints offered by me and by amnkb. If it is enough for you, then say "thank you", and move on. If it's not then say precisely what your problem is so someone can try to help. Stop being rude to the members who are giving up their time freely to help you.

Bob

hi mrpace,

What are we actually after here. Do you want a list of 72 matches (That's [18 x 8]/2 ) ?

If every player plays everyone else just once that's (18 x 17)/2 = 153 matches. So there's lots of pairs that never occur. There are loads of ways to come up with a list; I'd probably write a program to do it as it's somewhat tedious.

If in doubt choose a simpler example first.

Let's say there are just six players and each player must play just 3 matches.

AB, AD, AF, BC, BE, CD, CF, ED, EF is one solution but there are many others. To get this without lots of trial and error, I made a six by six table with A -F across the top and down the side. I put a cross in the leading diagonal as a player cannot play themselves. Then I chose opponents for A, mixed it about a bit for B and so on until I had 3 ticks in each row and column. By, for example, switching the As and Ds, you get another, different solution, which is why I say there are lots.

I'm not sure what else to say here.

Bob

hi KerimF

Apologies for not responding to this sooner. I had an idea when I read your first post but it has knocked around in my brain for some days before immerging as a method. It's got lots of steps and, if you wanted an actual formula, it could be done but it would be messy. Instead I'll outline an algorithm that will work.

If you just knew the three angles you could draw a triangle with those. But there are an infinite number of such triangles, all similar in shape, each scalable to scale up its size, but only one will have the right perimeter.

There's a formula, called Heron's formula after it's discoverer, that will calculate the area of a triangle if you know its perimeter. The semi perimeter = P/2 , sidea a, b, and c, and the formula is:

This Wiki page has an article about it https://en.wikipedia.org/wiki/Heron%27s_formula but there P already stands for the semi perimeter.

So how to make use of this.

Choose a random value for a; I'll call it p.

You can then use the sine rule to work out the other two sides, q and r.

Work out P' the perimeter of this triangle by P' = p + q + r

If P' = P then you're done because you had a really lucky choice for a. Unlikely though. But what you have got is the sides of a similar triangle that is either a scaled down or scaled up version of the correct triangle.

So you can work out the scale factor P/P' for converting your triangle to the correct one.

So now you know a, b, and c and so can proceed with Heron's formula.

Bob

hi sologuitar

If you know a mean and the number of values then you can work out what the total was by reversing the calculation.

total = mean times number of values

You can do this twice; once where n = 4 and then again where n = 5

The difference between these two totals must be the 5th test score.

Bob

hi amnkb

Well done for being able to do so many of sologuitar's questions. But he has made it very clear that he wants hints and help rather than the whole answer. See, for example:

http://www.mathisfunforum.com/viewtopic.php?id=30204

If you do this, you deny him the chance of improvement by doing the work himself. I have suggested that you hide your answer and just include a hint. That provides maximum help because he is getting just the hint towards his own solution, and he has the opportunity to compare his answer with yours. I would be grateful if you would do this in future.

Meanwhile, I am going to exercise my job as an administrator by hiding your full answer.

Bob

Ok. Here's a start:

The total in the H circle is 20. Don't write it inside the circle as the circle has 4 regions and they total 20 when you add up all four.

The overlap of all three (ie. those who do all three sports) is x.

We know that H and C is 7. H and C overlap in 2 regions that I'm calling H+C+F and H+C+F' (hockey, cricket but not football)

As H+C+F is x, that means that H+C+F' = 7-x

You have the information to similarly do H+C'+F (hockey and football but not cricket) and H'+C+F.

Finally you can do H+C'+F' (just hockey on its own) so that the four regions add up to 20 and the others similarly.

Take 18 from 50 to get the total of the 7 regions within the circles and make an eqaution using all those x expressions and solve for x. Then you can write in all the numbers for the regions. Check consistency with the initial information.

Bob

I meant times by 2. Every even number is double an integer eg 100 = 2 times 50

Sometimes I type x for multiply.

Bob

It says it is linear so y = mx + c again.

You can find enough information to get the coordinates of two points. Don't know the intercept this time. But you can plug in those coords and solve the simultaneous equations for m and c.

Or find m from the two points and then plug in that m with one set of coords to get c.

Bob

It's a matter of plugging in the information in y = mx + c.

The gradient is 2/25 and it's negative as the ramp is sloping down. It starts at 30 so that's the intercept.

Bob

If C is the across coordinate and F the up coordinate the 32F is the intercept with up axis.

(0,32) and (100, 212) are known points so substitute in y = mx + c ie. F = m.C + c where C is the centigrade and c is the intercept.

You've met this formula in another question for changing the subject of a formula.

Bob