You need to create a mathematical model to 'describe' this situation. With only two parties the binomial distribution might work.

Probability in this 'constituency' of a voter supporting A is 0.65 (p) and for B is 0.35 (q).

Now you have to assume that these probabilities will remain for the rest of the country and that's where this comes unstuck.

Expected number of votes for A = np = 1200 000 x 0.65 = 780 000

Standard deviation is √ (npq) = 522.

You can use the normal distribution approximation for binomial results, which I have tried. I don't think it's worth showing the calculation because it makes it virtually certain that A will win overall.

The problem is with the underlined assumption. Think about it. 0.65 is pretty high and if that's the probability in every constituency then A will keep winning. In the UK that doesn't happen. There are big regional variations between support for the parties so the 'pundits' know they cannot assume general results from a single announcement. The pre-election opinion polls carefully select their samples by spreading across areas with so many people of each gender and age group.

To do what you want, you'd have to know the probabilities for each constituency; then you could calculate on an area by area basis.

The underlying theory is here: https://www.mathsisfun.com/data/standar … ution.html

The method works ok for bags of sugar if it's ok to assume consistency during the production run. Voters are more complicated.

In the UK they do an 'exit poll' asking a sample of people how they voted after they have voted. Providing the sample is across all areas like the opinion polls this does usually give a good prediction of the final outcome.

Bob

]]>I'll use the example from above.

We know there's a turning point at √(1/3) ≈ 0.57735

Investigate the gradient just left and just right of the turning point. I've chosen x = 0 and x = 1 to make the calculation easy.

The top line in the table shows the x values chosen and the second line shows the gradient at those points.

Below that I've sketched those gradients and the sketch shows it's a minimum.

Bob

]]>Welcome to the forum.

Thanks to zetafunc for the initial hint.

You might also find the following useful:

w^2 + w + 1 = 0 => w + 1 = - w^2

You can prove this equation either by substituting the complex values for each root (the working is identical) or like this:

Let w^2 + w + 1 = z ; then w^3 + w^2 + w = wz

As w^3 = 1 we end up with z = wz. So either z = 0 or w = 1. As we know that second isn't the case z must be 0.

Bob

]]>Welcome to the forum. Thanks for posting your answers to the earlier parts of the problem. It's a great help in knowing what you do understand about the problem.

I started with a diagram and marked on the height of the cliff and the initial velocity upwards. This is 'up' but the cliff is down and so is the direction of gravity. So you need to decide at the start which way you will take as the positive direction and then use + or - correctly for the info given.

You've got the right formula to use but the signs aren't consistent. If you take 'up' as positive then u = 24.5, g = - 9.8, and s = - 117.6 (that's taking the cliff top as zero).

So the quadratic comes out as

If you take 'down' as positive then it is

As you can see these amount to the same thing so it doesn't matter which you use; but be consistent.

I tried this and got 8 seconds but not with your quadratic. I've just tried to solve your version with a quadratic solver and the square root part comes out negative so it cannot be evaluated. So did you make another sign error which 'cancelled out' the first?

For the last part the ball is to be 49m above the ground. But we are both measuring from the cliff top so you need to work out the distance down from the top to the 49 position ie 117.6 - 49. Then you can use that as your new 's' value to once again get a quadratic.

Hope that helps,

Bob

]]>The problem is to estimate the Euclidean norm of the series

| SUM {k from 0 to infty} (I - A)^k A w_k |

where matrix A such that 0 <= A <= I

and vectors w_k such that | w_k | <= 1

]]>Welcome to the forum.

I'm happy to help students where they are having difficulty with a topic in maths but, if you read the page about what to do before you post you'll understand that we don't just do your homework for you.

This worksheet has been posted before. The company that makes these worksheets doesn't like it's copyright being abused by postings on the public forums.

There's no diagrams. What chance do I have helping with questions that are meaningless without the diagram.

When I am teaching a class, I know the past history so I know what they already know and where they might struggle, so I can target my teaching approach more usefully.

If you really cannot do any of this sheet then you need to get back to your teachers and ask for more lessons.

So, if you want my help, please say

(1) What can you do on this topic.

(2) Show what you have tried already.

(3) Be specific about what you are finding difficult.

Bob

]]>Welcome to the forum.

This picture shows

You'll find lots more help here:

https://www.mathsisfun.com/improper-fractions.html

Bob

]]>Welcome to the forum.

No diagrams? And no explanation of what this is about. Sorry but it'll be tough to help under the circumstances.

I'm going to assume this is an exercise on enlargements and scale factors. Here's an example of a quadrilateral that has been enlarged with a scale factor of x2.

AB is 4cm long and CD is 8cm long. The other lengths have doubled in size similarly.

If you don't know the scale factor you can work it out like this:

I not sure what 'prop' means in these questions. In Q8 one possible answer is not prop, so I'm guessing that some questions are deliberately not proper enlargements. How can you tell?

You have to check every enlarged length. If the sf is always the same then you can safely conclude that you have a proper enlargement with that sf. If some sf answers come out differently from others, then that question is not a proper enlargement.

Hope that helps,

Bob

]]>The thing is I'm better with discrete problems than geometry, and so I rarely gets the solution.

I have checked the source again. It seems to actually be: (BC + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2. (First term is BC not BA).

Sorry for the fault!

The simple proof of (BC+CA)(BC^2+CA^2-AB^2)=2*BC*CA^2:

1) AC/CB=AL/LB - angle bisector theorem

2) AL/LB*BD/DC*CH/AH - Cevas theorem for ABC and AD, BH, CL; BD=DC => AL/LB=AH/HC

3) CH=BC*cosC - because BHC=90

4) {1), 2), 3)} =>AC/CB=AL/LB=AH/HC=AH/(CB*cosC) => AC*cosC=AH => (AH+HC)*cosC=AH => (1+CH/AH)cosC=1.

cosC=(AC^2+BC^2-AB^2)/(2*AC*BC) and CH/AH=CB/AC (by {1), 2)}) => (1+CB/AC)*(AC^2+BC^2-AB^2)/(2*AC*BC)=1 =>

(BC+CA)(BC^2+CA^2-AB^2) = 2*BC*CA^2. Done!

phrontister wrote:

0<x<0

Looks like I messed up there.

I meant that x is non-zero, either positive or negative.

]]>

You are very welcome. It's been a pleasure to help someone who is so polite and has such a good grasp of the topic. I have always had a great regard for any pupil who really thinks it through and doesn't just go for the 'expected' answer. Hope the rest of your studies go well but do post again, either for help or just so we know how it's going.

Bob

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