If you want any help with maths, ask me here

pinkbear

Account Number wrote: »

Can someone help me with part (ii) prove logbase5 24 = (3a + b) over (1-a)
when a = logbase10 2 and b = log base10 3?? I have up to logbase10 24 over logbase10 5 and am lost cos the change of base doesn't work?? Sorry about the way I laid it out in the post,have no clue where half the buttons are on a laptop keyboard.
https://scontent-fra3-1.xx.fbcdn.net/v/t34.0-0/p280x280/16468846_733399326814859_351788346_n.jpg?oh=143de0052eab1bb6fcc19e4c73f2f89b&oe=589523DD

It's not as difficult as it first looks.

log 24 = log 8 x 3 = log 8 + log 3 = log 2^3 + log 3 = 3log 2 + log 3
log 5 = log 10/2 = log 10 - log 2 = 1 - log 2

Done

Account Number

pinkbear wrote: »

It's not as difficult as it first looks.

log 24 = log 8 x 3 = log 8 + log 3 = log 2^3 + log 3 = 3log 2 + log 3
log 5 = log 10/2 = log 10 - log 2 = 1 - log 2

Done

Sorry,may have phrased my question badly ,but doesn't this just prove log10(24) over log10(5) equals all this? I'm trying to prove the algebra equals log5(24). This is what I have. https://scontent-amt2-1.xx.fbcdn.net/v/t34.0-0/p280x280/16506834_733815803439878_740835441_n.jpg?oh=10710046270c58f6b5d46039d2ec3f86&oe=5896C980

pinkbear

It does prove it (but sorry, I didn't go quite far enough).

Log (base 5) 25 = Log 24 / log 5 = (3log 2 + log 3)/(1 - log 2) = (3a + b)/(1 - a)

Liordi

How does the 'reverse chain rule' work?
Can't seem to make sense of it..

eg a question like this.

pinkbear

Liordi wrote: »

How does the 'reverse chain rule' work?
Can't seem to make sense of it..

eg a question like this.

The reverse chain rule is used when you notice that you have the integral of
(a function) times (something that looks like the derivative of that function).
e.g. Integral of (x^3)(3-2x^4)^6 dx
The function is (3-2x^4)^6
When you differentiate this you get 8x^3. This looks like x^3 so the reverse chain rule can be used in this situation.

I had a quick look on YouTube do see if your exact question is done somewhere, are it would take so many lines to write out, that you'd fall asleep before the end! Luckily Khan Academy have the exact question done here:
https://www.youtube.com/watch?v=7FQWBCeVIJM

Work through it, and I'm happy to answer anything you're confused about (quote the time on the video where you need further clarification).

Take Your Pants Off

Would someone explain to me the difference between the hypothesis testing. In 2 diff books they used diff methods and get diff answers ?!?
Are they used for diff situations or something.
Much appreciated

Take Your Pants Off

A similar situation arises here. I don't know which one to follow or use.

pinkbear

Hopefully this might clarify some of this really tricky topic for you…. but you may still have questions at the end.

Just to start with a few definitions…..

Hypothesis: An assumption made on the basis of limited evidence. It’s used as a starting point for further investigation.
Null hypothesis: The hypothesis that there is no significant difference between specified populations. This is the ‘status quo’, the thing being stated.
Hypothesis testing means testing a hypothesis by comparing it with the null hypothesis. The null hypothesis is only rejected if its probability falls below a predetermined significance level, usually 5%.
Significance level and confidence interval and p value: If your S.I. is 0.05 the corresponding C.I. is 95%. If your P value is less than 0.05 (the S.I.) the hypotheses test is statistically significant. If the C.I. does not contain the null hypothesis value, the results are statistically significant.

There are 3 formulas for hypothesis testing which have increasing complexity and increasing accuracy.

Formula (1)
The simplest one is 1/(square root of n)
This is all that is needed to be know for OL but wouldn’t get full marks at HL.

Formula (2)
The next one is 1.96(sigma)/(square root of n)

Formula (3)
For higher level, my understanding is that the more complex formulas should be known.
Margin of error = 1.96 x (the square root of (p(1-p)/n)

You may have to choose which one to use depending on what you’re given. E.g. In worked example 5.10 & 5.11, you were given information about p and about p(hat). You could be given information about the standard deviation, sigma, in which case you would use formula 2. I think avoid formula 1 if you want to get full marks. That example you show that uses it (the drugs company producing a new pain-relieving drug one)….. is that an ordinary level book? It makes sense if it is. I did it using formula 3, and I got .8+/-0.06272, so it doesn’t change the answer to the question, it just makes the confidence interval ‘tighter’.

For any of the questions that come up, the 3 formulas should give the same answer to the question, but they may have decimal points difference, as they get more accurate.

(Just as an aside…. where does the figure of 1.96 come from?
1.96 is read from the normal distribution table. It is the figure for z is less than or equal 0.9750, as this is the probability of the figure being within 95% of the mean. Look up your z-tables….. find 0.975 and you’ll see that it’s at 1.9 on the y and .06 on the x. 95% confidence is the only one that has ever come up I think, but perhaps other levels could come up. One of your pictures is of 90% confidence. In this case you’re looking for .95 on the z table which you’ll find at 1.6 and between .04 and .05, i.e. 1.645)

pinkbear

Here's a nice challenge for anyone interested.... from last year's mock paper..... I'll put up the answer in a day or two. It's very much based on the curriculum; there are no tricks or anything in it.

Tom has to retrieve a number of stones to put them in a bucket. It takes 5 seconds to retrieve the first stone and place it in the bucket. The time it takes him to place each subsequent stone in the bucket increases by 40%. How many stones can Tom retrieve and place in the bucket in 10 minutes?

Spanish Eyes

Hi folks,

I am not taking any exams anymore but did back in the day.

Was a C grade in maths at a struggle that is, and that was O level.

I have to say I cannot remember where the scenarios outlined above, or Pythagorus' Thereom etc, ever entered my fekkin head after LC.

Where is all that stuff relevant please!

At least with languages and literature there is something to be talked about. But maths. Feck sake. Calculators mean it's all a bit of a joke now surely.

Sorry to the maths nerds.

Take Your Pants Off

pinkbear wrote: »

Hopefully this might clarify some of this really tricky topic for you…. but you may still have questions at the end.

Just to start with a few definitions…..

Hypothesis: An assumption made on the basis of limited evidence. It’s used as a starting point for further investigation.
Null hypothesis: The hypothesis that there is no significant difference between specified populations. This is the ‘status quo’, the thing being stated.
Hypothesis testing means testing a hypothesis by comparing it with the null hypothesis. The null hypothesis is only rejected if its probability falls below a predetermined significance level, usually 5%.
Significance level and confidence interval and p value: If your S.I. is 0.05 the corresponding C.I. is 95%. If your P value is less than 0.05 (the S.I.) the hypotheses test is statistically significant. If the C.I. does not contain the null hypothesis value, the results are statistically significant.

There are 3 formulas for hypothesis testing which have increasing complexity and increasing accuracy.

Formula (1)
The simplest one is 1/(square root of n)
This is all that is needed to be know for OL but wouldn’t get full marks at HL.

Formula (2)
The next one is 1.96(sigma)/(square root of n)

Formula (3)
For higher level, my understanding is that the more complex formulas should be known.
Margin of error = 1.96 x (the square root of (p(1-p)/n)

You may have to choose which one to use depending on what you’re given. E.g. In worked example 5.10 & 5.11, you were given information about p and about p(hat). You could be given information about the standard deviation, sigma, in which case you would use formula 2. I think avoid formula 1 if you want to get full marks. That example you show that uses it (the drugs company producing a new pain-relieving drug one)….. is that an ordinary level book? It makes sense if it is. I did it using formula 3, and I got .8+/-0.06272, so it doesn’t change the answer to the question, it just makes the confidence interval ‘tighter’.

For any of the questions that come up, the 3 formulas should give the same answer to the question, but they may have decimal points difference, as they get more accurate.

(Just as an aside…. where does the figure of 1.96 come from?
1.96 is read from the normal distribution table. It is the figure for z is less than or equal 0.9750, as this is the probability of the figure being within 95% of the mean. Look up your z-tables….. find 0.975 and you’ll see that it’s at 1.9 on the y and .06 on the x. 95% confidence is the only one that has ever come up I think, but perhaps other levels could come up. One of your pictures is of 90% confidence. In this case you’re looking for .95 on the z table which you’ll find at 1.6 and between .04 and .05, i.e. 1.645)

Thanks helped a lot.
I know which one to use going forward now.
The drugs one was from text and test hl. The other one was from the active maths book.
Nevertheless thanks for the help.

pinkbear

Spanish Eyes, I know you’re just trolling, but on the off chance that you actually believe what you said or if anyone else reading does, I will respond.

Studying maths trains your mind in logic, and trains your memory. People who study maths are better at puzzles, and are better able to see the connections between words and concepts. The thought process in working out a maths problem is often similar to that involved in fine your car, setting up a home network, or writing code.

So you asked where this stuff is relevant? Well without maths there would be no technology, no engineering, no computers, no mobile phones, no printing machines, very primitive buildings, no planes, no ships no cars, no statistics, no stock market, no economies, no medicines….. I can provide a more comprehensive list if you with but I’m sure you get the gist.

Calculators don’t do anything other than speed up calculations that would otherwise be tedious, allowing us to focus on the more complicated concepts behind them.

Just as you have never used maths since leaving school, I use it on a daily basis in my job and in my life. I love it, and don’t consider myself a “nerd” in any kind of a derogatory sense…. I have friends and am sociable, and well able to discuss languages and literature as you mentioned, but my favourite subject has always been maths, just as others have a preference for music or business or art. I believe maths has helped greatly develop my logical reasoning, and memory, and I have been able to succeed at a great number of technical challenges over the years thanks to these skills. (Oh yes, not only do I love working with maths, but I earn good money at it too!)

sunny2004

I think pink bear is doing an amazing service to the community here.

That's coming from someone who is not doing any exams, I simply think she/he is very good for giving their time and knowledge.

BRAVO !!!

pinkbear

sunny2004 wrote: »

I think pink bear is doing an amazing service to the community here.

That's coming from someone who is not doing any exams, I simply think she/he is very good for giving their time and knowledge.

BRAVO !!!

Aw, thanks very much for the praise!

jinx257

Matrices...

I always found these easy and was able to do any problems and get full marks.... BUT what are they, what is their use and function? I never understood that...

pinkbear

jinx257 wrote: »

Matrices...

I always found these easy and was able to do any problems and get full marks.... BUT what are they, what is their use and function? I never understood that...

You do know that matrices are completely gone from the course now?

But to answer your question some uses of matrices are:
- in science and industry to track, record and display the results of research.
- In the study of electrical circuits (I've used them in Kirchoff's laws I think, and calculating battery power output)
- In quantum mechanics
- Encryption
- Displaying 3D images
- In robotics to calculate movements

I know this answer is a bit vague, as I'm not sure exactly how they are used in many of these applications, just that they are used and make the calculations much simpler.

evolving_doors

Think of each colour pixel on your TV/ phone screen as being represented by a number.
123
456
789
You can apply preset algorithms to manipulate the picture. A Basic manipulation would be to turn your image upside down or say to rotate from landscape to portrait would make it go to
369
258
147
The code can be found here, its a pretty basic interview question for graphics coders http://techieme.in/matrix-rotation/


Another use is algorithms for page ranking by Google (so you get to the 'top hit' website for your inputted word in the search bar).
Consider each number in the matrix as a probability of going to a certain page (transition matrix)... have a look at the first 4 pictures here for a better explanation http://www.math.cornell.edu/~mec/Winter2009/RalucaRemus/Lecture3/lecture3.html
Obviously these matrices would be MASSIVE as there's lots of sites out there with your 'key term' you want explored. But it's just pure numbers... messing with numbers is pretty easy stuff for a computer.

Another one is calculating probabilities as to an eventual outcome (as the number of iterations/ moves increases then a certain outcome is more likely)... markov chains

Strange it's not on the course....

ImaCabbage

Hello, I can't get my head around using "95% confidence intervals", i.e. 2015 P2 Q2/statistics. The notes provided in our book are very vague and the e-xamit solutions give very little guidance as well.

Any strong notes/advice on this specific area?

thetalker

Yeah thats in the supplementery material, I never really liked statistics but there's a formula you can just learn off.

Its 1.96 times the standard error(standard deviation over root n), you then add or subtract that from the mean to get the 95% confidence interval. same idea for a 5% margin of error.

Take Your Pants Off

Quite short pointing out needed, I get everything bar where the 4 came from in the (x^2+jx+4) is it the because you need 20 for the equation to match the 20 in the f(x). So you have 5 and you need to decide what by 5 will give you 20. Might be completely wrong but cannot get my head around it. The rest seems grand to me

TheBody

Take Your Pants Off wrote: »

Quite short pointing out needed, I get everything bar where the 4 came from in the (x^2+jx+4) is it the because you need 20 for the equation to match the 20 in the f(x). So you have 5 and you need to decide what by 5 will give you 20. Might be completely wrong but cannot get my head around it. The rest seems grand to me

That's correct. There are other ways to do the problem (if you are interested).

For example, given that x=5 is a root, you know that if you run 5 through f(x) you should get 0. That way you can find p.

If x=5 is a root, then x-5 must be a factor. So you could long divide x-5 into f(x) to get a quadratic which you can solve for the other roots.

Also, there is a maths forum if you wish to use it.

http://www.boards.ie/vbulletin/forumdisplay.php?f=380

pinkbear

ImaCabbage:

Question is done out here: https://www.examinations.ie/tmp/1488550662_2696112.pdf but I’ve done it with a bit more detail below. If there's anything you don't get, let me know and I'll explain a bit more.

Firstly, to get your head around what exactly 'significance level', 'confidence interval' and 'p value' means....

Significance level and confidence interval and p value: If your significance level is 0.05 the corresponding confidence interval is 95%. If your P value is less than 0.05 (the significance level) the hypothesis test is statistically significant. If the confidence interval does not contain the null hypothesis value, the results are statistically significant.

Margin of error formulas. Choose one based on what you’re given to start.
1. Margin of Error = 1.96(sigma)/(square root of n)
OR
2. Margin of error = 1.96 x (the square root of (p(1-p)/n)

A survey of 100 shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was 90.45. The standard deviation was 20.73.

(a) Find a 95% confidence interval for the mean amount spend in a supermarket on that Saturday.

Answer
Margin of error = 1.96(sigma)/(square root of n) =
1.96(20.73)/(square root of 100)= 4.06

Confidence Interval = mean +/- margin of error = 90.45 +/- 4.06

We can be 95% confident that the mean amount spent was in the range from 86.39 to 94.51 (i.e. from 90.45-4.06 to 90.45+4.06)

(b) A supermarket has claimed that the mean amount spent by shoppers on a Saturday is 94. Based on the survey, test the supermarket’s claim using a 5% level of significance. Clearly state your null hypothesis, your alternative hypothesis, and your conclusion.

Answer
H0: Mean spend is 94 (Null hypothesis - nothing is changing, we are accepting the claim made)
H1: Mean spend is not 94 (we are rejecting the claim made)

Method 1: (Is the z-value of the value we are testing inside the z-value of the actual value?)
Mean, x bar = 90.45, sigma= 20.73, u=94, n-100

z = (x bar - u)/(sigma/square root of n) = (90.45-94)/(20.73/10) = -1.71
-1.71 > -1.96 —> Fail to reject null hypothesis

Method 2: (Is the value we are testing inside the actual mean plus or minus the margin of error?)
94 is inside the confidence interval from 86.39 to 94.51 worked out in part (a), so fail to reject null hypothesis.

Method 3: (Is the actual mean we know inside the range around the value we are testing?)
Confidence interval based on a sample of 100 based on the claim is:
94-Margin of error < x bar < 94+Margin of error
89.94 < x bar < 98.06
90.45 is inside this interval —> Fail to reject the null hypothesis

(c) Find the p-value of the test you performed in part (b) above and explain what this value represents in the context of the question

Answer
P(z<-1.71) = 1 - P(z<1.71) = 1 - 0.9564 = 0.0436

p-value = 0.0436 x 2 = 0.0872

Meaning: If the mean amount spent was 94, then the probability that the sample mean would be 90.45 by chance is 8.72%. As this is more than a 5% chance we do not reject the null hypothesis.

thetalker

So I've returned with less of a problem and more of a question

See there was a trig Q in which tan(3A)=-0.1, find all values of A between 0 and 360

So the teacher found inverse tan of both sides but instead of dividing by 3 keeps it as 3A, finds all the reference angles for up to 3 revolutions of 360 and then divides by 3.
I was wondering how the Q could be solved if we divide by 3 to begin with, I mean I know how to do it since you use multiples of 60 and subtract from that but I guess I'm not exactly sure why.

pinkbear

thetalker wrote: »

So I've returned with less of a problem and more of a question

See there was a trig Q in which tan(3A)=-0.1, find all values of A between 0 and 360

So the teacher found inverse tan of both sides but instead of dividing by 3 keeps it as 3A, finds all the reference angles for up to 3 revolutions of 360 and then divides by 3.
I was wondering how the Q could be solved if we divide by 3 to begin with, I mean I know how to do it since you use multiples of 60 and subtract from that but I guess I'm not exactly sure why.

Think of it two different ways….

1) tan (3A) = -0.1

But this means tan (3A +/- m180) = -0.1 as tan repeats every 180.
3A +/- m180 = inverse tan (-0.1) = -5.71 or 354.29

Divide everything by 3:
A +/- m60 = 118.1
A = 118.1 +/- m60
A = 58.1, 118.1, 178.1, 238.1m 298.1, 358.1…. (don’t exceed 360 which is the max value of A)

2) Let 3A = θ

tan (θ +/- m180) = -0.1
θ +/- m180 = inverse tan (-0.1) = -5.71 or 354.29
θ = 354.29 +/- m180
θ = 174.29, 354.29, 534.29, 714.29, 894.29, 1074.29…. (this time, don’t exceed 1081 which is the max value of 3A)

Divide everything by 3
θ/3 = A = 58.1, 118.1, 178.1, 238.1m 298.1, 358.1

So, you could have divided the 3A by 3, as long as you divided 180 by 3 also, and were adding and subtracting 60, up to a maximum of 360. Or left it as 3A and added and subtracted 180 up to a maximum of 3x360.

Does that answer your question?

thetalker

Yeah actually that clears everything up pretty well!

I was thinking it would repeat every 180 but wasn't sure how the division by 3 affected, but this shows me it very clearly, thanks

Ciarajohnstone

What do I need to know for co-ordinate geometry for ordinary level?

pinkbear

Ciarajohnstone: Main things you need to know are.....

The Line
Plotting points; drawing a line from an equation; applying formulas to find length of line, slope, midpoint; find the slope of a line given the equation; finding equation of a line; finding x and y intercepts for a graph; finding slope of perpendicular line; finding area of a triangle given 3 vertices; finding point of intersection of 2 lines; find if a point is on a line; find a point under a translation.

The Circle
Write down the equation of a circle given centre and radius; find the centre and radius from the equation; find if a point is on the circle; find the equation of a tangent to a circle; find where circle cuts the x and y axes; find common tangent to two circles;
Lots of the knowledge of the line is needed for the circle also.

If you want me to explain any of these further, let me know. I'm just doing one line question below here, as it's a bit different than other ones, and it would throw most people off.

Q. The line k passes through 2 points (0,2) and (4,0).
(i) Find the equation of k
(ii) Write down 3 inequalities which define the shaded area.

(i) Find slope using slope formula, then apply formula to get equation of k.
m=(y2-y1)/(x2-x1) = (0-2)/(4-0) = -1/2
y-y1 = m(x-x1)
y-0 = -1/2 (x-4)
y = (-1/2)(x) + 2

(i) The shaded area is a right angled triangle with vertices at (0,0),(0,2) and (4,0)
So it is 1. Above the x axis, 2. To the right of the y axis, 3. Below the line from part (i).
1. y>=0
2. x>=0
3. y<=(-1/2)(x)+2

thetalker

Hi again I have a question from my mocks that stumped me, my teacher handed out corrections but I don't understand the logic behind the steps, probability especially the choose and pick parts have always been something I only really half learnt.

Anyway this is the q;

A bag contains 5 yellow, 3 green, 6 blue and 2 black discs, if 4 discs are chosen at random find the probability that.

(i); the discs are all the same colour (I got this one but didn't use the 'choose' method my teacher did

(ii): the discs are all different colours

(iii); two of the discs are black

Any solution or insight would be greatly appreciated!

pinkbear

My answers are below. I don't usually use choose and pick.... can't remember them, and prefer to just work from logic. Let me know if they are the same as the teacher's answers.

thetalker wrote: »

Hi again I have a question from my mocks that stumped me, my teacher handed out corrections but I don't understand the logic behind the steps, probability especially the choose and pick parts have always been something I only really half learnt.

Anyway this is the q;

A bag contains 5 yellow, 3 green, 6 blue and 2 black discs, if 4 discs are chosen at random find the probability that.

(i); the discs are all the same colour (I got this one but didn't use the 'choose' method my teacher did
The discs can't be green or black as there are less than 4. So there could be 4 yellow or 4 blue. So it's the P(all yellow) + P(all blue)
P(all yellow) = P(1 yellow) x P(2nd yellow) x P(3rd yellow) x P(4th yellow)
= 5/16 x 4/15 x 3/14 x 2/ 13 = 1/364
P(all blue) = P(1 blue) x P(2nd blue) x P(3rd blue) x P(4th blue)
= 6/16 x 5/15 x 4/14 x 3/ 13 = 3/364
1/364+3/364 = 1/91

(ii): the discs are all different colours
P(a yellow) x P(a green) x P(a blue) x P(a black) =
5/16 x 3/15 x 6/14 x 2/13 = 3/728

(iii); two of the discs are black
P(black) x P(black) x P(not black) x P(not black) =
2/16 x 1/15 x 1 x 1 = 1/120

Any solution or insight would be greatly appreciated!

thetalker

Your first one is right and uses the same method as me but the second and third are wrong. Or at least different to the answers I was given.

ii gives the answer as 9/91

iii gives the answer as 1/20

If you can see how these answers could be found using your method I'd still be interested in knowing.