Transcript for:
GCSE Maths Revision Summary

hi welcome to the chord maths ultimate cm4 revision video in this video we're going to go through the m4 topics from the ca gcse mavs course so the aim of this video is to spend about five or so minutes on each one of those topics to make sure you're familiar with all of those topics this video is really useful for you if you're advising for m4 i'd highly recommend watching the m3 and m2 videos as well if you're revising for m8 this video will be useful as well so this video will go through the m4 topics so in blue we've got the algebra topics such as the quadratic formula algebraic fractions perpendicular lines and so on in green we've got the geometry or the shape space and measure topics such as circle theorems volume of frustum and so on in orange we've got the statistics topics such as stratified sampling and histograms and in red we've got the number topic which is applying bounds so this is the revision checklist for m4 highly recommend you print this checklist it's in the description below so print it and add up to your notes beside each one of the topics you've got the video numbers on corporate math so if you do need a more detailed explanation in this video you've got much longer videos that you can watch in each one of these topics and make sure that you're really really confident in each of them but the end of this video is to spend about five or so minutes to make sure that you're familiar with each of the topics so this is the revision checklist it's in the description below and it's really useful also whenever you're watching this video you might want to as i'm going for each topic you might want to make notes as you're going through it but also there's an accompanying booklet of practice questions called the ultimate cm4 revision question booklet where i've put a question or two on each of the topics to make sure that you've got something to practice after you've watched that section in the video and you can scan the qr code to bring you to this video or you can scan the qr code for the answers and they'll be useful for you in this video i'll also borrow some of the corporate mouse revision cards so perhaps whenever i'm going through topics such as circle firms i'll be using the sections of the corporate revision cards and circle firms so the court matters revision cards might be very useful for you for m4 would highly recommend the higher sets also in terms of revision for m4 rather than leaving your revisions at the last minute and trying to cram it might be a good idea to use a little and often approach so the five a days in particular the higher five a days which are the green books and the higher plus five videos which the blue books might be useful for you okay so let's get started so our first topic is factorizing quadratics which is video 119 and corporate mavs so that's the video number that if you go to coopermaths.com forward slash contents and you scroll down that'll be the the video for this topic with factorizing quadratics there's two common approaches one's inspection which which is my preferred technique and also the split in the middle term i'll do the first few examples using both approaches but then as we go through the videos if there's any more quadratics to factorize i'll probably use inspection rather than the split in the middle technique just because it's my default approach but i will do the first few examples using both techniques so first question says to factorize 5x squared plus 13x plus 6. so it's a quadratic and we're going to factorize it so there's going to be two pairs of brackets like so and we're trying to factor out something that has a 5x squared at the beginning that means at the front of both brackets it won't just be an x and an x like it was often in m3 in this case it's going to be a 5x and an x because they will multiply together to give us our 5x squared if we had something such as 4x squared it's a bit harder because you're then having to figure out whether it's 4x and x or 2x and 2x but with this one being 5x squared that's quite nice so we've got 5x and nx now then if we were expanding our brackets we would do 5x times x which is our 5x squared then we would do 5x times this number and then x times that number and they would add together to give us our 13 x's in the middle and then finally the number here times the number here will give us six so a little times together to give us this number on the end so we know that we're looking for two numbers which will times together to be equal to six to go at the end of both brackets now they could be one and six they could be six and one it could be two and three or three and two and they it could also in theory be negative one and negative six and so on but they wouldn't work for this question because if we had 5x times negative 1 and then x times negative 6 we need to add them together it wouldn't give you a positive term in the middle so i know that because this is positive 13x these are going to be positive numbers here so we've got 1 and 6. so if we were to put that in the brackets whereas a one here and a six here so plus one and plus six and if we expanded our brackets we would get five x times x which is five x squared which is fantastic we would then have five x times six which is thirty x so that's a bit of a problem because that's quite large but then we've got one times x which is plus x and then finally we've got one times six is which is plus six so when we add the terms together in the middle we would get 5x squared plus 31x plus six but we didn't want that we wanted 5x squared plus 13x plus 6. so let's go back to our 5x and our x and we want to get 13x in the middle now using the options we've got here i'm looking at this bottom option here because i know if i use 2 i can do 5x times 2 which is 10x and then if i put my plus 3 here that will give me x times 3 which is 3x and if you add them together you get 13x so let's just check it 5x times x is 5x squared 5x times 2 is plus 10x 3 times x is plus 3x and 3 times 2 is just plus 6. and then if you add the terms in the middle you get 5x squared plus 13x plus 6. that's it and our answer would be 5x plus 3 x plus two so the main thing with factorizing quadratics is practice the more often you factorize quadratics the better you'll become out of it so you get to spot how to put those numbers in really quickly and easily okay so if you want to factorize this using the split in the middle technique i'd highly recommend you watch that video on cobra mouse but the technique is this first of all you take the number in front of the x squared this 5 the coefficient of x squared the number in front of the x squared the 5 and you multiply that by the number on the end of the quadratic the 6. so 5 times 6 is equal to 30 so you do 5 times 6 is equal to 30. now we're going to want to find numbers which will times to give it to be 30 but they will add together to give you the term in the middle which is 13. so in terms of numbers which will times together to be 30 and add together to be 13 they would be 3 and 10 because 3 times 10 is 30 and 3 plus 10 is 13. so what we're going to do is we're going to split this middle term up this 13x into a 3x and a 10x and if we do that we get 5 x squared now we could write plus 10x plus 3x or we could write plus 3x and plus 10x and it doesn't actually matter which way around you do that i'll show you that in a minute so let's do plus 10x and then plus 3x and then we've still got our plus 6 on the end so we've split this middle term up into 10x and a 3x so what we're now going to do is we're going to factorize the first two terms of this expression and we're going to factorize the second two terms of this expression so start off with the first two terms of this expression we've got 5x squared plus 10x now i can see that these two terms have 5x as a common factor so we'll write 5x and then we'll do a bracket and then factorizing this by taking out the 5x would leave us with an x to begin with because 5x squared divided by 5x is just x and then 10x divided by 5x is plus 2. now let's factorize the second two terms of this expression so we've got 3x plus 6 well we can divide both of these by 3 so we'll write plus three and then open up our brackets and dividing both of them by three will also give us x plus two and that's great because we've got the same bracket this x plus two twice we've got five x lots of x plus two and we've got three lots of x plus two now because we've got five x lots of it and we've got three lots of it all together that would be five x plus three lots of x plus two and that factorizes our five x squared plus thirteen x plus 6 because you can see that's what we got whenever we factorize it using inspection okay so that was one approach i just want to show you that if you had written it as plus 3x plus 10x you'll also get the same answer so this time let's write our 5 x squared plus 3x plus 10x so splitting up the 13x into 3x and 10x plus 6. now we're going to factorize the first two terms so we've got our 5x squared and our 3x well the common factor that would just be x so we'll take x out as a common factor so dividing both of them by x will give us five x plus three and then our second two terms we've got 10x plus six well if we had 10x plus six the common factor would be two so we'd write plus two and then bracket five x plus three and what's great is again there we've got the same bracket twice we've got x lots of five x plus three and we've got two lots of five x plus three so all together that would be x plus two lots of five x plus three so as you can see there are two different approaches to factorize this quadratic okay our next question is to factorize seven x squared plus twenty x minus three so if i was using inspection i would put my brackets down like so and put a 7x and an x now we know the numbers will times together to be negative 3 so it's going to be 1 and negative 3 or negative 1 and 3 or 3 and negative 1 or negative 3 and 1. so they would be my options but i know that i want to get 20x whenever i expand so whenever i do 7x times that number and x times that number whenever i add them together they will simplify together to be 20x now straight away i know that if i want 20x it'd be really great a positive 20x if i put my plus 3 here i would have 7x times 3 which is 21. then if i had a minus 1 here i would then have x times -1 while bringing me down to 20x so that would be it so my answer would be seven x minus one and x plus three alternatively we could use a split in the middle technique and i'll just do that in blue ink now so seven multiplied by minus three so i'll be minus twenty one so we want two numbers which will times together to be minus 21 but we'll add together to be 20. and straight away i'm thinking 21 and minus 1. so 21 times -1 will be minus 21 but also minus 1 plus 21 would be 20. so there would be our two options so we've got our 21x and our x so let's split up the middle term so we could write 7x squared minus x plus 21x minus 3. so we've split up our middle term our 20x into a minus x and a 21x now let's factorize the first two terms so if we had 7x squared minus x we could divide both of those by x so that would be x bracket 7x and then taking x out here would be minus 1 whenever we factorize these first two terms so we get x bracket 7x minus one and our second two terms we can divide both of those by three so we would write plus three bracket and dividing both of these by three would give us seven x minus one and that's great because we've got the same bracket twice so putting our terms in front are x plus three in a bracket we would get x plus three seven x minus one and that's it okay let's have a look at our next question so next question is to factorize four x squared minus four x minus three for the rest of it i'm gonna stick to the inspection technique so i'm going to do bracket bracket bracket bracket so whenever i factorize this what i know is that the terms at the beginning of both brackets were multiplied together to give us 4x squared so that could be a 4x and an x or it could be a 2x and a 2x and our terms at the end will multiply together to give us negative three that's actually pretty good because i know that it's going to be either a 1 and a negative 3 or a negative 1 and 3 in either order so let's start off with something such as 2x and 2x i'm just going to use two x's at the beginning of both brackets and let's try minus three and plus one and let's see what we get so 2x times 2x that's 4x squared fantastic then we've got 2x times 1 that's 2x and then minus 3 times 2x that's minus 6x actually we've got the answer let's expand these brackets and see what we get so if we expanded these we would get so 2x times 2x is 4x squared 2x times 1 will be plus 2x minus 3 times 2x is minus 6x and then minus three times one which would be minus three and if we simplify these two middle terms we get four x squared minus four x minus three so that's it we've factorized it our next question our next question is to factorize fully 20x squared minus 125 y squared and this is video 120 now so this is the difference between two squareds and as you can see here we've got 20x squared and 125 y squared so we've got the x squared and the y squared but we don't have square numbers in front of them we've got 20 and 125 so let's see if we can actually factorize by taking out any numbers to begin with now because we've got 20x squared minus 125 y squared i can see that both of these the 20 and 125 have a factor of five so let's take five out as a common factor and let's divide both of them by five so that would be four x squared minus 25 y squared now that's great now this is difference between two squares because you can see you've got four x squared and you get 25y squared and the 4 and 25 are both square numbers so with b so put our 5 down and i remember the difference between two squareds you put your brackets down you square root both of them so square root the 4x squared that'd be 2x and that'll be the front of both brackets and square roots in 25 y squared that would be 5y and 5y and you put one with a plus sign and one with a minus sign and that's it okay let's have a look at our next topic so we've looked at factorizing quadratics now let's look at solving quadratic equations by using factorization and that's video 266 in corporate maths so we've been asked to solve 3x squared minus 10x minus 48 equals 0. so what i'm going to do is i'm going to factorize this left-hand side just like we would have in m3 but obviously it's just a harder factorization so let's put our brackets down bracket bracket bracket bracket equals zero and we've got three x squared so let's go for a free x and an x now we know the two numbers were multiplied together to be minus 48 and that means that one's going to be positive one's going to be negative and then whenever we expand our brackets we're going to get minus 10 x in the middle plus 8 minus 6. well 3x times x is 3x squared 3x times minus 6 would be minus 18 x x times 8 is plus 8x fantastic and then minus 48. so that would be 3 x squared minus 10x minus 48. so now we're factorized let's then solve it so because we know that these two brackets were multiplied together to give a zero that means that the brackets will be equal to zero so either this bracket is zero or this bracket zero now this bracket is quite easy to solve we know the x will equal six there because six minus six is zero this bracket whenever i have a three x e plus e equals zero i tend to write three x plus eight equals zero and then just minus c from both sides so three x equals minus eight and then divide by three x equals minus eight thirds like so so there are our two solutions the either x equals minus eight thirds or x equals six and that's it so we have solved the harder quadratic equation by factorizing okay let's have a look at our next topic so our next topic is algebraic fractions and this is video 21 on corporate maths and these algebraic fractions are a bit more complicated than the ones you'd get in m3 and we've been asked to express as a single fraction two over three x plus one plus 1 over x minus 1. now whenever we're adding together fractions we want the same denominator and to do that what we're going to do is we're going to multiply both the numerator and denominator of the first fraction by the denominator of the second fraction so we're going to multiply both of these by x minus one and for our second fraction we'll multiply both the numerator and the denominator by the denominator of the first fraction so we'll multiply both of these by three x plus one so whenever we do that we get well we'll write two and then we're going to multiply our numerator by the denominator of this fraction so two bracket x minus one like so and then our denominator would be well we would have three x plus one in a bracket and then we're gonna put another bracket beside it with x minus one like so so we've multiplied both the numerator and denominator both by x minus one now we're going to multiply both of these the numerator and denominator by three x plus one so when we do that we get we're multiplying the numerator by three x plus one because it was a one we could just write three x plus one because we know that we just multiplied it by one and then with our denominator of x minus one we're multiplying that by r three x plus one so we've got x minus one bracket three x plus one and that's it and as you notice we've got the same denominators for these fractions so we've got three x plus one times x minus one and here we've got three x plus one times x minus one so now that we've got the same denominators we can combine our fractions as a single fraction so we'd have our numerator so we've got two bracket x minus one and then we've got our second numerator so plus and then three x plus one and then we've got our denominator which is just three x plus one bracket x minus one like so now let's simplify our numerator so let's expand that bracket so 2 times x is 2x 2 times minus 1 that's minus 2 and then we've still got our plus 3x plus 1 and our denominator is still 3x plus 1 and x minus one and then finally simplifying our numerator well we've got two x plus three x that's going to be five x and then we've got minus two plus one well minus two plus one is minus one and our denominator was three x plus one x minus one like so and that's it so we've expressed two over three x plus one plus one over x minus one as a single fraction as five x minus one over three x plus one times x minus one like so okay let's have a look at our next topic so our next topic is called equations and that's video 111 and 111a in corporate maths so here's our example it says solve 2 over x minus 3 plus 1 over x minus 4 equals 2. so as you can see this left hand side we've got two algebraic fractions and we're going to add them together and combine them as a single fraction so we'll add them together to write them as a single fraction so let's do that so let's multiply both the numerator and denominator of this fraction by x minus 4 and then we'll multiply both the numerator and denominator of this fraction by x minus 3. so doing that would give us 2 bracket x minus 4 over x minus three times x minus four so multiplying both of these by x minus four and then plus and we're gonna multiply both of these by x minus three now because it's a one i'm just going to write x minus three on the numerator if there was a number there i would put it in a bracket and put the number in front then over and then i had x minus 4 and then i'm times that by x minus 3. and although the brackets are in a different order remember a times b and b times a is the same thing that equals 2. okay now we're going to combine them as a single fraction because they've got the same denominator i can write my x minus 3 x minus 4. my fraction i went up a bit there so i can write my x minus 3 and x minus 4 and on the numerator i have two bracket x minus four plus and then we've got our x minus three and that equals two so i've just written the two fractions as a single fraction but with the same denominator now what we're to do is we're going to simplify this numerator so let's expand the brackets so 2 times x is 2x and 2 times -4 that's minus 8 plus x minus 3 over x minus 3 x minus 4 like so and that equals 2. so just expanding these brackets now let's simplify the numerator 2x plus x is 3x and then we've got minus 8 minus 3 that's minus 11 over x minus three x minus four and that equals two so now we've got three x minus eleven over x minus three times x minus four equals two now i don't want these brackets on the denominator here i want to actually get these off the denominator so what i'm going to do is multiply both sides of the equation by x minus 3 and x minus 4. so multiplying both sides by x minus 3 and x minus 4. so whenever we do that well we're timing this side by x minus three and x minus four to get rid of it on the denominator so we just be left with three x minus eleven and then on the right hand side where we're multiplying two by x minus three and x minus four so we would have two bracket x minus 3 and x minus 4. so now we're going to multiply this right hand side so we've got 3 x minus 11 and then let's expand these brackets so let's put our 2 down and put just a big bracket and then expand our brackets so x times x is x squared x times minus four is minus four x minus three times x is minus three x and minus three times minus four is plus twelve now let's simplify it so let's simplify these middle terms so we've got three x minus eleven equals two bracket x squared minus seven x plus twelve then expand our brackets so we've got three x minus eleven equals two x squared minus 14 x plus 24. and then finally whenever you're solving quadratics we wanted to equal zero so let's get rid of this 3x and minus 11. so let's minus 3x from both sides so that would give us minus 11 equals 2 x squared minus 17 x plus 24 and then we want to get rid of this minus 11 so let's add 11 to both sides so it'll be 0 equals 2x squared minus x plus 35. so we've now got a quadratic which we can hopefully factorize and solve so when we do that we get zero equals so we factorize our quadratic and we put 2x and x now we want two numbers that multiplied together to give us 35 and whenever we expand and add the terms together will give us minus 17x so i'm thinking it's gonna be a negative and a negative so let's put a five here so that's two x times minus five which would be minus ten x and then put a 7 here so it's minus 7 times x which is minus 17x fantastic and that's it factorized now let's solve it so we know that whenever two brackets multiplied together to give a zero that they have to be zero so that means that for our first bracket we've got two x minus seven equals zero so adding seven to both sides two x equals seven and dividing by two x equals seven over two so x can be equal to seven over two or three point five if you prefer decimals and then our right hand bracket well we know then that x equals five so our two answers are x equals seven over two or x equals five and that's it so there are two solutions okay so our next topic is the quadratic formula and that's video 267 on corporate mavs now sometimes we're given a quadratic that can't be factorized so we solve it by using the quadratic formula so if you've got a quadratic in the form ax squared plus bx plus c equals zero where a is a number in front of the x squared so a is the coefficient of x squared b is the coefficient of x the number in front of the x and c is the number on the end then you can substitute them into the quadratic formula and it will tell your solution straight away and the quadratic formula is x equals negative b plus or minus the square root of b squared minus 4ac all over 2a and this is a part of the chord mouse revision card and here's our example we've been asked to solve 2x squared minus x minus 9 equals 0. now sometimes in a quadratic formula question it will say solve 2x squared minus x minus 9 equals 0 and give your answers to one or two decimal places and that's a clear that you're going to be using the quadratic formula so let's label our a b and c to begin with so a equals b equals and c equals so a is a number in front of the x squared so a is going to be equal to 2. b is the number in front of the x so it says minus x that means minus 1x so that means that b is equal to minus one and finally c is a number on the end which is our minus nine so we have got a b and c now we're going to substitute those into the quadratic formula so x equals negative b well b is minus one so we've got negative b well if it's already negative that's going to make it positive so ask me one plus or minus the square root of b squared well minus 1 squared is going to be minus 1 squared minus 4 times a and i just put it in brackets that's 2 times c which is equal to minus 9 like so and that's all divisible by 2a and 2 times a is equal to 4. so we've got x equals 1 plus or minus the square root of minus 1 squared minus 4 times 2 times -9 all divided by 4. so now let's simplify what's under the square root sign so we've got x equals 1 plus or minus the square root of we've got minus 1 squared well negative 1 times negative 1 is 1 so it's going to be 1. and then we've got subtract and whenever i work out this 4ac part i always put my answer in a bracket so we've got 4 times 2 times -9 now 4 times 2 is equal to eight times minus nine is minus seventy two so we've already got a minus sign outside here so we've got four times two which is eight times minus nine which is minus seventy two so that goes there and then all divided by four next we've got 1 minus minus 72 that's going to be 1 plus 72 which is 73 so we're going to get x equals 1 plus or minus the square root of 73 all divided by 4. and now i'm actually just going to move this down here so i'm just pulling it down here now here we're going to have two solutions we're going to work out 1 plus the square root of 73 divided by 4 and we're going to do 1 minus the square root of 73 divided by 4. so let's write those out separately so we're going to have 1 plus the square root of 73 divided by 4 and we're going to have x equals 1 minus the square root of 73 divided by 4. and so let's just work those out in our calculator and see what we get so we get x equals 2.3 or x equals negative 1.886 and that's it so we've used the quadratic formula so x equals negative b plus or minus the square root of b squared minus 4ac all divided by 2a to solve a quadratic which couldn't be factorized okay so we've looked at the quadratic formula now sometimes in m4 you're given a question in a wordy context or a wordy situation and you have to create your own quadratic equation that you need to solve so let's have a look at a question like that now so we've been given a context a field has a width of x meters and a length of two x plus one meters and the area of the field is 600 meters squared find the width and the length of the field give your answers to one decimal place okay so first of all in this question well we've told a rectangular field remember we should write down the question has a width of x meters and a length of 2x plus 1 meters and the area is 600 so whenever we multiply the width and the length together we get the area so let's write that down x bracket 2x plus 1 equals 600. and then let's expand our brackets so that would be x times 2x is 2x squared plus x equals 600. and then we like our quadratics to equal zero so let's take away 600 from both sides that would give us 2x squared plus x and then we take away the 600 would be minus 600 and then on the right hand side we'd have equal to zero so we're going to solve this quadratic and we've been told to give our answers to one decimal place so that's a massive clue that we're not going to be factorizing because usually whenever you factorize and something with a 2x you'd either get whole numbers or maybe a 0.5 or something like that because we've been told to give answers to one decimal place implies we're going to be rounding it so let's use the quadratic formula here so let's use our a well a is gonna be equal to two b is going to be equal to one and c is going to be equal to minus six hundred or negative six hundred and our quadratic formula is x equals negative b plus or minus the square root of b squared minus 4 ac all divided by 2a so let's substitute our values in so we've got minus b well b is 1 so it's going to be x equals minus 1 plus or minus the square root of b squared well b is one one squared is one minus and then i like to put this in brackets four multiplied by a which is two multiplied by c which is minus six hundred and then our brackets and big squares in all divided by 2a well 2 times a is 4. now we've then got x equals minus 1 plus or minus the square root of and then we've got 1 minus and let's work out this bracket 4 times 2 times minus 600 equals minus 4 800 close brackets that's all divided by 4. and then let's simplify this so we've got x equals minus 1 plus or minus square root of now we've got one minus minus four thousand eight hundred so that's gonna be one plus four thousand eight hundred it's going to be four thousand eight hundred and one all divided by four and remember with this we've got our two solutions we've got either minus one plus the square root of four thousand eight hundred and one or divided by four or minus one minus the square root of four thousand eight hundred and one divided by four so let's write those out so x equals minus one plus the square root of four thousand eight hundred and one divided by four or x equals minus one minus the square root of four thousand eight hundred and one divided by four and whenever we work both of those out so typing this one into our calculator type in minus one plus the square root of 4801 divided by four gives us 17.0723 and so on or this one if we type this one into the calculator we get 17.5723 and so on so we've got our two values for x and if we go back up to the question we're told that x is the width for the field now if we go back down here again we've got one's positive one to negative we can't have a negative width of the field so this will have to be our solution so x will be equal to 17.0723 so that's going to be the width of our field and the length is 2x plus 1. so so the width is equal to x which is 17.0723 and so on and the length was two x plus one so to find the length we need to multiply this by two and add one so let's do that and notice i'm not rounding it yet so we're going to multiply this by two and then plus one and that gives us 35.1446 and so on now we've been told to give our answers to one decimal place so answers would be 17.1 meters and 35.1 meters and there are answers for the width from the length of the field that's it okay our next topic so our next topic is perpendicular lines so in m3 we looked at parallel lines and parallel lines were lines that had the same gradient so they were lines that they were you know they would never meet across each other and they would have the same gradient now we're going to look at perpendicular lines and these are lines that meet each other at 90 degrees they cross each other at a right angle now here we've got the chord mouse revision card and as you can see we've got an example here of a pair of perpendicular lines we've got the line y equals 2x which is going up in this direction and then going down in this direction we've got y equals minus a half x plus one and as you can see these two lines meet at 90 degrees they cross at 90 degrees now one thing to notice about the gradients is well here we've got two and here we've got minus a half now the reciprocal of two is a half the negative reciprocal of two would be minus a half so if you've know the gradient of one line if you find the negative reciprocal of it you'll find the gradient of the other line so here if you knew the gradient was equal to two if you find the negative reciprocal of it so take the reciprocal which is a half and then make it negative a half you will find the gradient of the line that's perpendicular to another interesting point is if you multiply the gradients of two perpendicular lines you will get negative one so if you have two times negative a half well two times a negative a half is negative one so the product of the gradients for perpendicular lines is negative one i tend to just use the negative reciprocal so if you know that the gradient of this line's a negative a half well the reciprocal of a half is two and then because it was negative it's not positive so the the gradient of the perpendicular line would be two okay let's have a look at a question based on that so here we've got a question it says a line is drawn that is perpendicular to y equals negative four x plus seven and it passes through the point eight three work out the equation of the line now we know that it's perpendicular to y equals negative four x plus seven so it's perpendicular to this line and this line has a gradient of negative four so if we find the negative reciprocal of negative four you will find the gradient of the line that's perpendicular to it so we first of all know this gradient was negative so the gradient of our line is going to be positive so we know it's going to be positive and the reciprocal of 4 well the reciprocal of 4 is equal to a quarter so we know that the gradient of our perpendicular line would be a quarter because the negative reciprocal of the gradient that we were given so we know the equation of our line will be y equals a quarter x plus c now we know that it passes through the 0.83 so if this is the x coordinate and that's the y coordinate we can substitute those into our equation and we can find our c so we know that our y is equal to three so we've got three equals a quarter x so it's a quarter of x and x is eight so it's a quarter of eight plus c so next we've got three equals a quarter of eight is two so we've got two plus c well c is going to have to equal one here so we've got with c is so that's it so we know that the equation of the line is perpendicular to y equals minus four x plus seven that passes through the point eight three would have to be y equals a quarter x plus one and that's it okay let's have a look at our next topic okay we're now going to look at our geometry our shape space and measure topics and we're going to start off by looking at circle theorems and these are videos 64 and 65 in corporate maps so here we've got our first circle of here so our first circle firm and this is part of the chord mouse revision card is the angle of the semicircle is 90 degrees so if you get the diameter of the circle and you've got two lines to go from the diameter up to a point on a circle or down the angle will always be 90 degrees so here's an example so if you were given this question and asked to find the value of x well here we've got our triangle we know that's a right angle so we need to find x well we know that that's 90 degrees and then we can just use the angles in a triangle to find x so 90 plus 34 190 plus 34 is equal to 124 degrees and if you take that away from 180 we can see what's left for x so 180 take away 124 would be 56. so that means that x is 56 degrees okay let's have a look at our next example now one thing to look out for whenever you're doing circle theorem questions is looking for isosceles triangles so from the center of the circle to the edges the radius so here we've got a radius and a radius there two radii that means they're the same length as each other so this would have to be an isosceles triangle so if we know that this angle is equal to 98 degrees we know this angle x and this angle over here would be the same so that's x as well so we can work out what x would be we could do 180 minus 98 which would be equal to 82 and then if you've done 82 divided by 2 that's equal to 41 degrees so x here would have to be 41 degrees so look out for our socialist triangles in circle firm questions where you've got particularly whenever radio i are involved okay next now here again you can see we've got from the center to the circumference of the radius so that's a radius that's a radius and that's a really so here we've actually got we've got two isosceles triangles this is an isosceles triangle and this is also an isosceles triangle and also another thing to note is the big triangle is a right angle triangle because we've got that diameter as well so that's a right angle at the top if you add these two angles together so let's look at our x y to begin with so we know that's the radius and that's there it is that means that y and 55 are the same as each other y equals 55. now i have 55 degrees now if y is equal to 55 you got 55 plus 55 or 55 plus 55 is equal to 110 degrees so 180 minus 110 is equal to 70 degrees that means x is equal to 70 degrees and finally because this is the diameter we know that x and z are in a straight line so we know that if we do 180 minus 70 we get x that's equal to 110 degrees so z is equal to 110 degrees and that's it so it is useful to look out for those slc's triangles okay next okay so our next circle here is that the angle of the circumference is half the angle at the center so here we've got our circle we've got the center of the circle you can see this angle is 120 degrees between the two radia and the angle at the top here between the two chords would be equal to 60 degrees so the angle of the circumference is half the angle of the center so let's have a look at some examples okay with our first example okay so we know the angle of the circumference is half the angle at the center well that means if we double the angle of the circumference we get the angle of the center so if we do 51 multiply by two that's equal to 102 degrees so that means the x here is equal to 102 degrees next we've got the angle at the center and we're trying to find the angle of the circumference so if we divide the 122 by 2 we'll get the angle of the circumference so that would be 61. so that means that x equals 61 degrees for this circle okay so the angle at the circumference is half the angle of the center or the angle of the center is double the angle of the circumference okay let's have a look at one more question so here we've got a circle this time here we've got a circle and we've got the angle of the circumference but we don't actually have the angle at the center we've got the angle inside of the shape here so what we're going to do is we're going to take the 234 away from 360 to find the angle at the bottom it has to be like this where you've got the angle of the circumference and the angle at the center so if we do 360 subtract 234 we get that's equal to 126 degrees so that's 126 degrees at the center now the angle of the circumference is half of that so if we do 126 divided by 2 we get that's equal to 63 degrees so the angle for x would be 63 degrees so the next circle firm is the angles in the same segment from a common quarter equal so if you look at this chord this blue line this dotted blue line you've got a pair of green lines coming up from either end of that chord and they're meeting at a point on the circumference of the circle and likewise we've got a pair of red lines that are coming off from either end of that chord meeting at a point at the circumference that means that this angle and this angle will be equal to each other now if you have a look at this example we've got a common chord here and we've got all these lines coming off from that chord either end of the chord to the circumference so that means that a b and 37 are all equal to each other so if a is equal to 37 degrees b is equal to 37 degrees as well that's it they're all lines coming off from that same chord okay let's have a look at our next example so we've got this circle we've got our 27 degrees our 88 degrees and our angle x and we've been asked to find this angle x as you can see we've got this chord at the bottom here and we've got our lines coming up here and here and also the lines coming off across here and here that means that this 27 will be the same as this angle here so that means that that angle there is 27 as well next thing to notice is we've got two straight lines across each other that means the opposite angles are equal so if this angle is 88 degrees that means that that angle is 88 degrees also they're vertically opposite angles and finally you've got this little small triangle here with this x 88 and 27 that means that those three angles will add together to be 180 degrees so if we do 88 plus 27 that's equal to 115 degrees next if we do 188 subtract 115 degrees we get that's equal to 65 degrees that means that x is 65 degrees now you could have done this question in a slightly different way where we looked at the 88 and 27 and worked at this missing angle at the bottom and then looked at the chord and said well x is the same as that angle there and worked it out that way but that would be 65 as well anyway okay let's have a look at our next circle theorem so our next circle frame is if you've got a cyclic quadrilateral so that's a quadrilateral whether all the corners are on the circumference of a circle the opposite angles add up to 180 degrees as you can see here we've got 70 degrees and 110 and we've got 80 and 100 so the opposite angles add up to 180 degrees okay let's have a look at our example so here you can see we've got a cyclic quadrilateral this point this point this point and this point are all on the circumference of a circle that means the opposite angles will add up to 180 degrees this 114s outside but we can actually subtract that from 180 to find the angle inside so 180 subtract 114 degrees is equal to 66 degrees so that means this angle in here is 66 degrees now we can find our x and our y really quickly and easily so because the 66 is opposite the x and the cyclic quadrilateral we know if we take away the 66 away from 180 degrees we can find out what x is so 180 subtract 66 is equal to 114 degrees that means x is 114 degrees next our y well the opposite angles add up to 180 so if we do 180 subtract 121 that's equal to 59. so we know that y is equal to 59 degrees so we've worked out what x's and y's by using the fact that the opposite angles and a circular quadrilateral always add up to 180 degrees okay next circle theorem well the next circle firm is if you have the radius and a tangent they meet at 90 degrees so here you can see a b is the tangent so that means that the tangent and the radius meet at 90 degrees that's a 90 degree angle there now here we have got if this is 154 you can take that away from 180 that would be equal to 26 so it would be 26 degrees in there in a straight line next we've got a triangle we've got 90 degrees and 26 degrees if we add those together we get 116 degrees so 180 minus 116 degrees equals 64 degrees so that's 64 degrees there and then finally we have a straight line here so if we do 180 minus 64 we get that's equal to 116 degrees so that means x equals 116 degrees and that's it so the radius and the tangent will be to 90 degrees our next circle theorem it's got a name that i always remember that's alternate segment theorem and that is if you have a tangent the angle between the tangent and a chord equals the size of the angle opposite it inside the triangle so this 70 is the same as the angle opposite in the triangle and the 60 is equal to the 60 the angle opposite inside the triangle and that's called alternate segment theorem so let's have a look at our example so here we've got our tangent and the angle between the tangent and the chord is 55 so that 55 will be equal to the angle opposite it so x is equal to 55 degrees and here we've got our y that's equal to the angle opposite it's a y would be 60 degrees and that's it now what's interesting is if you go to the videos and worksheet section called mavs there's a video showing you the proof of this so if you want to see why they're equal to each other have a look at chord maps and you can see me explaining why okay next okay two more interesting circle theorems are if you've got a point the two tangents from that point to the circle will be the same length so as you can see here's his point c and you can draw two tangents to the circle the meter at d and e and the lengths of c d and c e will be equal to each other and then finally if you've got a chord the radius that cuts that chord in half will meet the chord at 90 degrees like so okay so that was circle theorem so we're now going to look at volume and surface area now in m4 we're asked to solve more complex problems for example involving frustums now whenever we look at frustums that will often involve cones and pyramids so the volume of a cone is found by a third pi r squared h which is a third the area of the base times the height and that also applies to pyramid so if you are given a pyramid the volume of that pyramid will be a third times the area of the base times its height okay so here we've got a comb and let's practice working out the volume of this cone because it would be useful for finding the volume of things such as frost rooms so here we've got a and you can see we've got the radius of the base is equal to 20 and we've got the height is equal to 21. so its volume is equal to a third times pi times by the radius of the base that's times by 20 squared times about its height 21. so when we work that out we get the volume would be equal to 2800 pi or if you want to write as a decimal 8796.46 centimeters cubed now that formula is given to you at the front of your m4 paper and this has been stepped from the chord mouse revision card so the volume of a cone is equal to a third pi r squared h or if it's a pair of massive volumes equal to third the air if the base the power squared is the base times the height and that'll be very useful for whenever you're finding the volume of a frustum and i'll talk about that in a moment also talking about volumes of pyramids here's a pyramid and if you've been given a pyramid and you're asked to find the volume of it again the volume is equal to third there at the base times the height so it's going to be the volume is equal to third times the area of the base that's going to be nine times six so nine times six is equal to 54 so times 54 times its perpendicular height so that's going to be from the base straight up to the top which is seven centimeters and whenever we work that out we get that's equal to 126 centimeters cubed and as i said the volume of a cone and a parameter will be very useful for finding the volume of a frustum very useful or essential to it okay next we've been found asked to find the volume of a frustum and the first first room sounds complicated but it's just a a cone or a pyramid with the top chopped off so here we've got this is video 360 and corporate mavs and it says shown as a frustum of a cone so as you can see the tops chopped off that had a perpendicular height of 20 centimeters so the it did go up to a top point just say there and down again and it did have a height of 20 centimeters in total now the height of this part is 10 so i would have had another 10 centimeters to have a total height of 20 centimeters and we've been asked to find the volume of the frustum so that's the bit that's left this part here so to do that what we're going to do is we're going to work out the volume of the current to begin with so the whole big cone we'll then work out the volume of the cone that's being chopped off and then take them away and see what's left so let's do the volume of the big cone so the volume is equal to a third times pi times the radius so for the big cone the radius is eight so eight squared times its height and we've been told this height was 20 centimeters okay whenever we work that out we can two either get eighty over three pi uh you'll learn about working exactly with pi in m seven i believe um alternatively you can write as a decimal number as one thousand three hundred and forty point four one two eight and so on centimeters cubed that's the volume of the whole big cone to begin with now the top was chopped off and that is a smaller cone and it's similar to the large coating you'll learn about similarity later which means enlargement but so we'll just work out the volume of this and so working out the volume of this smaller cone and it's got a volume of a third times pi times the radius 4 squared times its height 10 and whenever we work that out we get and that's equal to 160 over 3 pi or alternatively 167.5516082 centimeters cubed and then finally to work out the volume of the frustum well we're going to take that away so we're going to do 1340.412866 take away the 167.5516082 and when we subtract them we get 172.86 centimeters cubed and i've rounded that to two decimal places so if you're asked to find the volume of a frustum you just work out the volume of whatever shape there was to begin with whether it's a big cone or a big pyramid you work out the volume of what's being chopped off and then you take them away and that'll tell you the volume of what's left okay and our next topic now a m4 it doesn't actually specify if it's just the volume of frustums or the surface area of frustums so if i wanted to work at the surface area of this cone it would be pi times the radius which is 6 times the slant height which is 11 and that would be equal to 66 pi or 207.345 and so on centimeters squared but also remember this is the base of the cone then we need to work at the area of the base which would be pi times 6 squared and whenever we do pi times 6 squared we get 36 pi or 113.0973 and so on and when we add those two together we get 320.5 to two decimal places centimeters squared so to find the surface area of a cone you find the surface area of the curved surface so the top part of the cone here which should be found by pi r l so pi times the radius times this diagonal length and that gives you the curved surface area of the top but also remember to add on the air of the base so that circle and that will give you the total surface area of a cone if you're then asked to find the surface area of frustum you might find that the top part is being chopped off so you might want to then use that pi rl to work out the curved surface area that's been chopped off to see what's left and then remember you'd have two circles so that information might be useful for you as well okay so now let's look at our statistics topics and our first statistics topic is called stratified sampling stratified sampling is really useful for whenever you want to find a suitable sample for groups where there's not the same number in each group so for instance here we've got year seven there's 120 year sevens there's only 80 year eight and there's a hundred year nines well it wouldn't be fair to take a sample of the same number of year sevens eights and nines because there's a lot more year sevens and year eights so you can use a stratified sample this keeps the sample in the same proportions as the population or close to so the formula for stratified sampling is the number in the category divided by the total times by the sample size and the question says ms holland wants to take a stratified sample of 15 students and it says how many year sevens should she serve us that's a bit of a tongue twister and so let's start off with year sevens so we want to find out how many years seven she was gonna want to survive so the number in that category well there's a hundred and twenty year seventh divided by the total well let's work out how many students are in total here so 120 plus 80 plus 100 is equal to 300. so we're going to divide that by 300 and then we're going to times by the sample size which is 15. so whenever we do 120 divided by 300 times by 15 we get that's equal to six so miss holland should sample six year sevens if you wanted to find out how many years you would sample you would do 80 divided by 300 times by 15 and for year 9 you would do 100 divided by 300 times by 15 and then that would be it now sometimes we'll stratify sample and you might get decimal numbers and then you would need to round them just make sure that whenever you've rounded them that you do get the total number of 15 so you know if you've rounded two up you might find that that will put you to 16 students so you might need to just round one of them up and so on okay let's have a look at our next topic our next topic is histograms so a histogram is a way of representing group data and here we've got our group data and it's a way of representing it and in a fair way whenever you've got categories that have got different sizes so as you can see here we've got time not the two two to four four to six six to ten ten to fourteen and fourteen to twenty so as you can see here we've got groups that are quite small so the first group's only got a width for class width of two seconds it's only two seconds wide from null to two but the last group goes from fourteen to twenty so it's actually got a width of six so it wouldn't be fair to draw a bar chart because some of the categories are much bigger or much wider than the other categories so you draw a thing called a histogram and the histogram looks at the frequency density and the frequency density is found by doing the frequency divided by the class width and just before i go on histograms is a very important topic so if you go to videos 157 to 159 and cobra mavs you'll see me talk about histograms in great detail and if you're studying for m4 i'd highly recommend you do that so what we're going to do is we're going to start off by drawing a histogram for this group frequency table so if we want to draw a histogram we're going to work out a thing called frequency density so let's put a column on for frequency density so i'm going to write f d standing for frequency density here so we've added on a column for fd or frequency density now what we're going to do is we're going to work out the frequency density for each one of these groups or classes so from null to two seconds well that's going to class with two it's too wide it's got a frequency of 10. so if we do 10 divided by 2 so the frequency divided by the clouds with 10 divided by 2 so 10 divided by 2 that gives us a frequency density of 5. so this category has a frequency density of 5. our next category well it's got a frequency of 13 and it's got a class width of 2. it's 2 wide going from 4 to 2 4 takeaway 2 is 2. so we're going to do 13 divided by 2 so 13 divided by 2 is equal to 6.5 our next group well again it's frequency is 18 and it's class width well it goes from 4 to it's class with again is two so we're gonna do eighteen divided by two and eighteen divided by two is nine our next group it's going from six to ten so it's going to classify for four because it's four wide and it's got a frequency of 16 so we're going to do 16 divided by 4 because we do the frequency divided by the class width that's 4. next the next category has got a frequency of 8 and we're going to divide that by the class width where that's equal to 4 because it's 4 wide so 8 divided by 4 is 2. and finally our last group has got a class width of 6 because it goes from 14 to 20 that's 6 wide and it's frequency 6 so we're going to do 6 divided by 6 which is 1. so we've got our frequency densities now what we're going to do is we're going to draw a histogram so we're going to draw bars for our each of our groups so our first group is from not to two so we're going to do a bar go from not to two and having a frequency density of five so it's going to go up to five so we've written a pencil you'd go draw a line from five across within a pencil you draw down and you draw your bar like so so that is your bar going from not to two and with a frequency density of five our next bar well from two to four it's going to go up to six point five so from two to four well we actually need to make our bar a little bit higher so we're gonna go now up to six point five so it's gonna be halfway between six and seven so it's there we're gonna go across to four and then down our next bar well from four to six goes up to nine so from four to six we'll go up to nine so again we need to make that bar a bit higher we're going to go up to nine now across and down so we've done from naught to two up to five we've done from two to four up to six point five and we've done from four to six up to nine now we're gonna do from six to ten and with a frequency density of four so from six which is here to ten we're gonna have a bar that goes up to four so we'll draw a cross to ten and then we're going to go down like so so that bar from 6 to 10 has got a height of 4. next we're going to go from 10 to 14 and let's get a height of 2. so from 10 to 14 we're going to go across from and down and finally the last group goes from 14 to 20 so we're going to go from 14 to 20 and it's got a height of one so we're going to go make sure it's heights one and we're going to go down like so and that's it we've drawn our histogram so we've drawn our histogram for our fruit frequency table now sometimes we're given a histogram and we may need to find out the frequency now if we go back frequency density was equal to frequency divided by class width if we wanted to make frequency the subject we would multiply both sides by class with so that would give us the frequency density times class width is equal to frequency so the frequency is equal to the frequency density times the class width so in other words if you multiply the height of each bar the frequency density by its clouds with how wide it is you will find the frequency so let's complete this group frequency table so we've got our categories not to 40. so as you can see our first bar went from 0 to 40. now if we look vertically we're going from 0 up to 1. there's 10 little squares that means that each square is not 0.1 so this bar has a height of 0.1 or frequency density of 0.1 and it's got a class width from not to 40 or 40. well if we do the frequency density 0.1 times by 40 that's equal to four so that means the frequency for this bar is equal to four so as you can see here the frequency would be four next on x bar and x bar goes from 40 to 60 and its frequency density well it goes up to 2.4 so the frequency density is 2.4 so we're going to do 2.4 multiplied by the class width which is well from 40 to 60 is 20 so multiplied by 20 is equal to 48 so that means that this category from 40 to 60 has a frequency of 48 there was 48 people and then our next category from 70 to 80 well the class with 10 is frequency density well it goes up to 3.9 so we're going to do 3.9 times 10 so 3.9 times 10 is 39 so it would have a frequency of 39 so to draw a histogram we use the formula frequency density as frequency divided by class width so if you've been given a histogram and you want to find the frequencies well we would do frequency density times class width okay let's have a look at a typical histograms question now so the histogram below shows information about the number of hours flown by some pilots and we're told that 45 pilots flew for more than 400 hours so we know that between 400 and 450 hours there's 45 pallets in here okay so our first question says the estimate how many pilots flew for under 100 hours so if you have a look at our histogram the first category goes from null to 200 so that's why the question says estimate because whenever we had this group frequency table it would have said not to 200 and i would have the frequency so because of screw frequency we don't actually know how many were between naught and 100 or 100 and 200 so this is going to be an estimate so if we have a look at this bar let's work out the frequency for this bar it's got a frequency density of 3.5 and the class width is 200 it goes from 0 to 200. so if we do 3.5 multiplied by 200 that would tell us the frequency and that's equal to 700 so that means there's 700 pilots in here now under 100 hours well that would be half of the pilots well our estimate our guess will be it's half of the pilots so we're going to do 700 divided by 2 and that's equal to 350. so that's our educated guess to how many pallets of flu under 100 we do know the 700 the flu between naught and 200 are and our estimate is that half of those 350 have flew for under 100. okay next question our next question says what fraction of pilots flew between 200 and 400 hours so we want to look at the fraction of pilots so we're going to need to know how many pallets are all together so let's get rid of that dotted line to begin with and we know the 700 in this bar we've been told there's 45 in this bar because 50 multiplied by 0.9 is 45 so we know there's 45 in there let's work on how many pellets are in this bar and this bar so for this bar here it has a frequency density of 4.7 so a frequency density of 4.7 multiplied by a class width of 100 would be equal to 470. so there's 470 pilots in here and this bar has a height frequency density of 2.5 and again the class move is 100 2.5 times 100 is 250. so we know the 700 pilots between norton 200 470 pallets between 200 and 300 250 pallets between 300 and 400 and 45 pallets between 400 and 450. we've been asked to find out what fraction of pilots flew between 200 and 400 so that's going to be all together so 470 plus 250 that's 720 out of all the pilots all together so it's going to be 700 plus 470 plus 250 plus 45 that's equal to 1465 and if i type that into my calculator it'll simplify nicely for me to 144 over 293 okay our next question okay let's have a look at our next question so our next question says the histogram shows the heights of two thousand four hundred trees so we've got their heights in centimeters from not all the way up to two hundred and and we've been asked to find an estimate for the mean height we've been asked to find an estimate for the median height so if i was working this out the first thing i want to know is the frequencies so we've got the frequency densities of 8 16 20 and 5 and i would want to work out the frequencies for each one of those bars so we've got here this bar 75 this bar would be 125 and yeah okay so for the frequency for this bar from 0 to 75 what i would do the frequency density 8 times the class width which is 75 so it has a frequency of 600 next from 75 up to 125 it's going to class for 50 times by 16 is 800 next from 125 to 150 that's a class width for 25 times by 20 is equal to 500 and finally from 150 up to 250 that's 100 as a class width times by 5 is equal to 500 and let's just add our frequencies up 600 plus 800 plus 500 plus 500 is equal to 2400 which we're told in the question which is fantastic so the question asks us to work out an estimate of the mean height and an estimate of the median height let's start off with the mean height now i'm actually just going to draw a frequency table like so and our frequencies were 600 800 500 and 500 and we've got our height so it's not to 75 75 to 125 125 to 150 and 150 up to 250. now we want to find an estimate of the mean so remember that's where we add on our fx column and we do the midpoint times the frequency midpoint times the frequency and so on so the midpoint for our first category from not the 75 well well that's going to be 37.5 so we're going to do 37.5 times 600 that's equal to 22 500. next we'll do the midpoint where the midpoint of 75 125 is 100 times that by the frequency is 80 000. next we do the midpoint of 125 150 and that's 137.5 remember we can find the midpoint by just adding the numbers and dividing by two times by 500 is equal to 68 750 and finally we're going to do the midpoint of those numbers which is 200 times about 500 which is equal to 100 000. next to find the estimated mean we add up our f x column so we find the grand total and when we do that we get two hundred and seventy one thousand two hundred and fifty and then we divide that by the total frequency but we're told in the question that's two thousand four hundred so we could add up the frequencies or we could just use the fact that we're told in the question so when we divide 271 thousand of 225 by 2400 the grand total divided by how many there are that would tell us that our estimate for the mean height remember they're estimates because we're using the midpoints and we don't actually know the actual values that's equal to 113.02 centimeters and that's it so that's the r estimate of the mean so that's the two decimal places and that's it okay next we will find us then to find an estimate of the median okay so our next question says find an estimate of the median height so if we go back to our histogram and we look at each bar its area represents its frequency because the frequency density the height of the rectangle multiplied by the class width the width for the rectangle that gives us our frequency or the area of the rectangle so that means that the frequency is represented by the air of the rectangle so that means that we can find the vertical line that we draw down so the half the area of the histogram's on the left and half the area of the histogram's on the right that means that we can read off and find out what the medium would be so let's find the area for this histogram now i could count the little tiny squares if i wanted to but i'm actually going to count the bigger blocks because i can actually see in this question that there's lots of these whole blocks and each of the bars sort of is split up into these larger blocks apart from obviously this one where i've then got half blocks so if i count them so there's 48 big blocks so if i can find out where the halfway the 24th block is isn't divided so there's 24 blocks on the left and 24 blocks on the right we can find out where the median would be so let's find out where the 24th block would be so we've got 1 2 3 4 5 6 7 8 9 10 11 12. so we've got 12 blocks in that one so we need number 12 we've got one two three four five six seven eight well twelve plus eight is twenty now we need another four but we've got if we look across here we've got one two three four five six seven eight so we actually only need four blocks that's half of those so if we took our learner pencil and we went halfway down here that would mean that we would have 24 blocks worth on the left and 24 blocks worth on the right and that value would be was in between 100 and 125 and exactly in the middle so the median would be 112.5 centimeters and that's it so that's one way found in the median now you may have been able to guess that that's not actually the approach that i use whenever i work out the median just because i don't like counting the little blocks and trying to figure out where that line is i use approach called linear interpolation so it's video 52 in corporate valves and the formula for it is the lower bound plus number into category divided by number in category times class width it sounds quite complicated but it's not releases lower bound plus number into category divided by number in category times cloud swift let me explain what i mean by that so we're looking for the median value so if there's 2400 we divide that by two because remember for group data it's just an estimate anyway so we can do 2400 divided by 2 is equal to the 1200 value and if i was looking for the 1200 value i would look at my frequency table and i would say well the 1200 value is not in the first category because there's only 600 in there if i look at the next category there's 800 in there so all together if i work at the cumulative frequency there's one thousand four hundred trees altogether so the one thousand two hundredth tree is in this category here so we're going to look at this category and we're gonna say what's this lower bound that's 75 plus the number into the category we're looking for the 1200th tree so we've gone 600 we need to go another 600 so we need to go in over 600 into this category so we'll write 600 it's 600 into the category divided by the number in the category well that's 800 and then times by the class with the class width is 50. so if we do 75 plus 600 over 800 times 50 we should get that median that we got before and the answer is 112.5 centimeters which is the same as before the reason i like that formula is you can then work at the lower quartile upper quartile interquartile range really quickly and easily from histograms and the formula is not actually that complicated okay let's have a look at our last histogram question this is the hardest histogram question we've got here and it says 24 cars traveled faster than 40 miles an hour so we know 24 cars traveled faster than 40 miles an hour and it says estimate the number of cars traveling between 25 and 35 miles per hour so here we've got our histogram and we've got the frequency density going up vertically with no numbers that's not going to be very helpful and we've got our speed going across horizontally and we've been told that 24 cars so we know there's 24 cars in this bar here and it says work out an estimate for the total number of cars traveling between 25 and 35 miles per hour so we want to work out an estimate for how many cars would be in here now first of all what makes this question a lot harder than the previous ones is we don't know the frequency densities so there's two different approaches you could use for this question the first one because we know that 24 cars travel more than 40 miles per hour we know that the frequency for this bar is 24. so if you do 24 the frequency divided by the class width we can get the frequency density 24 divided by 10 is 2.4 so we know the frequency density there will be 2.4 so we know the frequency density is 2.4 now we can figure out our scale so we've got two point fours here well there's one two three lines going up to that so if we divide that by three that's no point here so no point eight one point six two point four three point two four four point 4.8 and 5.6 so now that we've got our scale for our frequency density we can now work out our estimate for how many cars are traveling between 25 and 35 miles per hour so let's work out the frequency for this 20 to 30 so we would do the class width which is 10 times 4. so that means there's 40 all together in this bar so that means that if we divided it by 2 this should be 20 in here and likewise for this bar this one going from 30 to 40 miles per hour so we would do 10 times 5.6 which is 56 and if we divided that by two there'd be 28 and 20 plus 20 is equal to 20 plus 28 is equal to 48 cars so that's our estimate for how many cars should be traveling between 25 and 35 miles per hour so that's one approach now that question was quite nice because the information we were told is there's 24 cars traveling more than 40 miles per hour so we were told the frequency for one of the bars sometimes in the question you might be given sort of you know so many cars are traveling more than 35 miles per hour it's a bit more complicated or another approach is to consider the frequency and to look at the area of that bar that represents that frequency so as you can see there's one there's one two three four five six smaller squares inside that bar and that represents 24 cars traveling more than 40. so if you do 24 divided by six you'll find that there's four cars in each one of these four four four four four four each one of those squares represents four cars so if you want to find an estimate for how many cars are traveling between 25 and 35 we can consider how many of those squares are in that region so as you can see there's 1 2 3 4 5 6 7 8 9 10 11 12. so you could do 12 times 4 is equal to 48 and that's the same estimate again so that is another approach alternatively sometimes you can find that it doesn't work out even as nicely as that because it might not be a whole number of blocks so what you could do is actually say well all together there's one two three four five six seven eight nine ten little squares going across and there's five ten fifteen of those so there's a hundred and fifty little squares so 150 squares equals 24 cars you could then divide 150 by 24 and say well 150 divided by 24 is equal to 6.25 so 6.25 squares equals one car then you could look at the 25 to 35 region and see how many little squares there are in those so as you can see we've got 5 10 15 20 25 times five so there's 125 little squares there and from 30 to 35 there's 5 10 15 20 25 30 35 x 5 is equal to 170 so all together there would be 125 plus 175 which is 300 squares and if we divide that by 6.25 we can see how many cars would be so 300 divided by 6.25 is also equal to 48 cars and that's it okay let's have a look at our last topic so our last topic is a plan bounds and that's video 184 on corporate maps so we've looked at an m3 appliance on bonds already but we're going to look at some harder situations now so we've got the lengths of time taken by four people to complete a puzzle are listed below and it says each time it's given to one decimal place so the four people have taken seconds to one decimal place 8.4 seconds to one decimal place 35.5 seconds to one decimal place and 19.8 seconds to one decimal place and we've been asked to work out the greatest possible range so remember the range is equal to the largest subtract the smallest so the largest subtract the smallest so here our largest value would be this one 35.5 and our smallest value would be 8.4 but each of these times have been rounded to one decimal place okay so let's find the upper bound and lower bound for these values now before we do that i'd highly recommend you watch the m3 video just to make sure that you're confident finding the upper and lower bounds so the upper bound so this number has been rounded this 8.4 has been rounded to one decimal place so the upper bound would be 8.45 seconds because it could go up to 8.45 seconds but not including and the lower bound would be well it could go down as far as 8.35 seconds because that's the lowest possible value that would round up to 8.4 seconds likewise for 35.5 the upper bound would be 35.55 seconds and the lower bound would be equal to 35.45 seconds so we find the upper bound and lower bound for each of those values and we've chosen those ones because we're looking at the range which is the largest subtract the smallest now we want to work out the greatest possible range now that's going to be found by using the greatest possible value for the largest value and it's going to be found by using the smallest possible value for the smallest value because when we take them away we get the biggest possible answer so we're going to want to use the upper bound for the largest value and we're going to want to use the lower bound for the smallest value because when we take them away we'll get the biggest possible answer so we're going to do 35.55 seconds and we're going to take away 8.35 seconds and whenever you do that you get 27.2 seconds now you can actually try over values to see if you can get a bigger answer you could try 35.45 and 8.45 and take those away and so on but you're going to find that's the biggest possible answer you want to find that you want to make the biggest difference between them so you're going to want to use the biggest biggest value and the smallest smallest value so when you take them away you get the greatest possible range if they wanted the smallest possible range you'd want them to be as close together as possible so you use the lower bound for the largest value and the upper bound for the lower value so that they're closer together and then give you the smallest possible range okay next question says natalie runs 100 meters to the nearest 10 meters so it's 100 meters to the nearest 10 meters that means it could be the lower bound would be 95 meters and the upper bound would be equal to 105 meters there the upper and lower bounds for how far she's run and it takes her 14 seconds to the nearest second so the lower boundary so she could do it in 13.5 seconds or the upper bound the greatest possible amount of time it would take would be 14.5 seconds there the lower bound and upper bound for 14 seconds at the near second so the question says to work out the greatest possible speed so we're working at speed so the speed is found by the distance divided by the time now but we've been asked to find her greatest possible speed so to do that what you want to do is you want to run the largest distance in the smallest possible time because the largest distance in the smallest possible time will give you the greatest speed so the greatest possible distance would be 105 meters that's the greatest possible distance she could run and in smallest amount of time would be 13.5 seconds because if she ran a large distance in a smaller amount of time that means she's running even faster so if you done 105 the upper bound divided by the lower bound that's going to give you the greatest possible speed so 105 divided by 13.5 would be equal to seven point seven seven eight meters per second to three decimal places that would be her greatest possible speed if you wanted to find her slowest possible speed you should do the smallest distance in the greatest amount of time and that means she's traveling slower and so on okay let's have a look at one last question okay last question on the plan binds we've been told that w is equal to the square root of c divided by x minus one and we've been told that c is equal to 9.5 rounded to one decimal place that means the lower bound for c would be equal to 9.45 and the upper bound for c would be equal to 9.55 and we've been told that x is equal to 4.81 to 2 decimal places so the lower bound for x would be equal to 4.805 and the upper bound for x would be equal to 4.815 and we've been asked to find the lower bound of w so if we want to find the lowest possible value for w we want to divide the lowest possible value for the square root of c by the biggest possible value whenever you want to find the lowest possible something you want to divide the smallest amount by the biggest demand that will give you the smallest answer so whenever we do this lower bound for c well the slower bound for c would be the square root of 9.45 because that will be smaller than the square root of 9.55 and then we're going to divide that by x well we want to divide it by the biggest possible thing to make give us a small answer so we're going to do 4.815 minus 1 because the upper bound 4.815 and then we take away one and when we do that we get 0.805789 and so on okay so that's the lower bound for w so whenever you want to find the lowest possible answer you would want to do the lowest divided by the biggest and if you wanted to find the upper bound for w you'd want to do the largest divided by the smallest so you'd use the upper bound for c and you'd use the lower bound for x and that's it so i really hope you found this video useful in this video we've spent about five or so minutes maybe for histograms a bit more than five minutes but in this video i spent around five or so minutes on each one of these topics i really hope you found it useful this revision checklist can be found in the description below so remember to print it add it to your notes stick it on your wall whatever as long as you go through these c m4 topics and make sure you're really confident with them also if you need a more detailed explanation you've got all the video numbers there you can go to cover mavs and watch those also as well as this video remember there's that bumper pack of questions so as you're going through the video it might have been useful to try some questions on each of them if you haven't done that feel free to click on that now and you'll have some questions on every single topic in the video and there's the answers for them as well also in this video we've used quite a lot of the code mouse revision cards the higher revision cards will be really really useful for you if you're studying for m4 and in terms of the five videos with that little and often approach to study can be really useful i can really help you retain information that you've learned in class for m4 highly recommend looking at the green higher books and the blue higher plus books so that'll be really useful in terms of using a little and often approach invest in five or ten minutes every single day will really make a massive difference to your results i really really hope you found this video useful this has been the ultimate cm4 revision video and please like the video please subscribe to code mavs and just all the very best of luck with your studies good luck