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Maths Driving test data: analysis and manipulation.
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Introduction The objective of this project is to successfully manipulate a large set of data in order to prove/ disprove a set of hypotheses. The data consists of results and statistics from a set of driving instruction and tests. The dependant variable in the data is the number of mistakes made during the test. The hypotheses will have been synthesized by myself and in response to influenced by the data that I have been given. The data will have to be subjected to sampling to reduce the vast amount of information. The data will then be processed so...

M 35 5 C Wed 14

M 40 4 C Thur 13

F 10 37 D Thur 10

F 17 31 D Wed 13

F 24 28 D Mon 10

F 31 24 D Mon 17

F 32 17 D Fri 14

F 40 19 D Fri 17

M 15 35 D Wed 14

M 25 28 D Thur 9

M 28 26 D Tue 12

M 29 27 D Thur 12

M 40 4 D Fri 14

M 40 20 D Wed 16

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Introduction In this investigation I...Introduction In this investigation I have been given the task of discovering a formula to tell us the value of n when 'n' is any given number. Also, in the second part of my investigation, look into whether there is a link between the phi of a number, and the product of its factors' phis. Such as: Is it true that 6 x 4 = 6 x 4? How to Find the Phi of a Number For any positive integer n, the phi function n is defined as the number of positive integers les than n which have no factor other than 1 in common are co-prime with n. So 10 = 4 because the positive integers less than 10 which have no factors other than 1, in common with 10 are 1, 3, 7 and 9 i.e. 4 of them. Also 16 = 8 because the integers less than 16 which have no factors other than 1, in common with 16 are 1, 3, 5, 7, 9, 11, 13 and 15 i.e. 8 of them. Part 1 i 3 = 1, 2 = 2 ii 8 = 1, 3, 5, 7 = 4 iii 24 = 1, 5, 7, 9, 11, 13, 17, 19, 24 = 9 iv 11 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10 To investigate this idea further, I will now find the phi functions for the numbers up to, and including 20, to see if I can find any patterns which will indicate to me a formula for n. ** In this investigation if the phi of any number has already been studied, only the phi will be given, without any working. ** ** Also only the numbers which are not factors of the phi in question will be stated. ** 1 = 0 2 = 1 = 1 3 = 1, 2 = 2 4 = 1, 3 = 2 5 = 1, 2, 3, 4 = 4 6 = 1, 5 = 2 7 = 1, 2, 3, 4, 5, 6 = 6 8 = 1, 3, 5, 7 = 4 9 = 1, 2, 4, 5, 7, 8 = 6 10 = 1, 3, 7, 9 = 4 11 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10 12 = 1, 5, 7, 11 = 4 13 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 = 12 14 = 1, 3, 5, 9, 11, 13 = 6 15 = 1, 2, 4, 7, 8, 11, 13, 14 = 8 16 = 1, 3, 5, 7, 9, 11, 13, 15 = 8 17 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 = 16 18 = 1, 5, 7, 11, 13, 17 = 6 19 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 = 18 20 = 1, 3, 7, 9, 11, 13, 17, 19 = 8 Sequence: 0, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8. Summary From studying the phi of numbers 1-20 there is clearly no outstanding link between the numbers, and their phi's. Although I can draw several points from this: Prime Numbers The prime numbers from the previous sequence were: 2, 3, 5, 7, 11, 13, 17, and 19. The phis of these were: 2 = 1 3 = 2 5 = 4 7 = 6 11 = 10 13 = 12 17 = 16 From this it can be seen that the phi of a prime number is always one less than the number in question e.g. 17 = 16. This is because there a no factors which go into a prime number, except for the number 1 and itself, and in the phi function the number in question is not considered, and the number 1 is always a factor, therefore all numbers except for the number in question, are all counted. We can now draw from this a simple formula where 'n' is any given PRIME number: n = n-1 Odd or Even? All of the results were even, except for the phi of 2 which is 1. Factors within a Phi Other than Prime numbers themselves and numbers which have factors which are prime numbers, have a phi of how many prime numbers exist in them, such as the phi of 8: 8 = 1, 3, 5, 7 In this phi only the numbers: 1, 3, 5 and 7 do not go into the number 8, and these numbers are all prime, except for 2 which is a factor of all even numbers therefore the phi of 8 has the same number of prime numbers in it, as the value of its phi, which is 4. This pattern also exists in the phis of: 3 4 8 already stated 10 14 16 The Phi's of Powers of Prime Number How to work out the Phi's of Powers of Prime Number The phi of a power of a prime number can still be found by identifying the number of positive integers less than n, which have no factor other than 1 in common are co-prime with n, but a simple amount of BIDMASS must be employed. This means that the phi of a power of a prime number can be found but multiplying the brackets out first, must occur, and then the phi of that number can be found. Below the first 3 prime numbers in the sequence of prime numbers 2, 3, 5, 7"¦.. and there first 5 power numbers and their phis have been found. Powers of Prime Numbers and their Phi's ** In this next part of my study the sign ^ will indicate a powered number, such as 2^2, 2^3, 2^4"¦"¦.. ** 2 = 1 2^2 = 2 2^3 = 4 2^4 = 8 2^5 = 16 3 = 2 3^2 = 6 3^3 = 18 3^4 = 54 3^5 = 162 5 = 4 5^2 = 20 5^3 = 100 5^4 = 500 5^5 = 2500 From the prime numbers and their powers phi's it can be seen that a clear pattern has become apparent. It can be seen that each time the power goes up by 1, the phi of that powered prime, is the previous powered prime's phi, multiplied by the prime number in question. I know this because it is evident in the previously stated phi's, for example from the powers of 2 phi functions, 1 x 2 is 2, 2 x 2 is 4, 4 x 2 is 8 and so on.   

Introduction In this investigation I have been given the task of discovering a formula to tell us the value of n when 'n' is any given number. Also, in the second part of my investigation, look into whether there is a link between the phi of a number,...

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This Database contains information about some...This Database contains information about some used cars, and many different makes of cars are included. I am going to include the following: "¢ Specifying clearly what I plan to do and why I am approaching this investigation in this way. "¢ Collating the data I need and representing it in a way which helps to develop my investigation. "¢ Interpreting my results and drawing a conclusion for them. Introduction In this project I am going to investigate Used Cars. I am going to look at the price of the cars and what kind of shape they are in. E.g. price when new, second hand prices, age, color etc. we will also see how, the younger the car the more expensive it will be because it will be a new car. And if the car is old the less expensive it will. I am going to collect the data I am going to use and try to further develop it to suit the investigation I am going to do. I am going to plot my results onto a scatter diagrams and interpret my results. Hypotheses 1 I am going to be testing Price and Mileage. There is a positive correlation between the old price of the car and the mileage. I.e. the more expensive a car is the less the mileage is. So I think that the higher the mileage the lower the price will be. Hypothesis 2 But it will all depend on the model of the car as well. E.g. if it is a brand new Lexus, the price is bound to be at around £35.000. Plan of Action I am going to use Second hand information. I am going to start with working with the price and mileage. The data is going to show a lot of use because the cheaper the price of the car the more mileage it will have on it meter gauge. And the more expensive the car the less it will have traveled. I am going to collect 40 samples of used cars, which will show a number of information that will be needed to know about the car. E.g. 92. Volkswagen at £8.710.00 that has traveled 50000 miles. In order for my graph too look professional and extremely accurate I have decided to use the computer program AUTOGRAPH. I hope to show that the more expensive the car is the less mileage it has. I plan to draw a cumulative frequency graph, histogram, box plot, whisker diagram along with a tally and pie chart. Data Collection have collected all of the 40 randomly selected samples out of 100. My tables have clear headings of type, model, old price, and mileage. Make Model Price when Price Age Color Engine Fuel MPG Mileage New Second Hand Size Ford Orion 16000 7999 1 black 1.8 unleaded 25-47 7000 Mercedes A140 Classic 14425 10999 1 red 1.4 unleaded 29-50 14000 Vauxhall Vectra 18580 7999 2 white 2.5 unleaded 26-44 20000 Vauxhall Astra 14325 6595 4 black 1.6 unleaded 32-52 30000 Nissan Micra 7995 3999 3 blue 1 unleaded 40-55 37000 Renault Megane 13610 4999 4 blue 1.6 unleaded 30-50 33000 Mitsubishi Carisma GDI 14875 5999 2 black 1.8 unleaded 34-56 24000 Rover 623 Gsi 22980 6999 4 green 2.3 unleaded 25-41 30000 Renault Megane 13175 6999 3 black 1.6 unleaded 30-50 41000 Vauxhall Tigra 13510 7499 4 marine 1.4 unleaded 33-57 27000 Fiat Bravo 10351 3495 5 red 1.4 unleaded 31-53 51000 Vauxhall Vectra 18140 6499 4 blue 2.5 unleaded 26-44 49000 BMW 525i SE 28210 5995 8 white 2.5 unleaded 21-38 55000 Vauxhall Corsa 8900 4995 2 silver 1.6 unleaded 26-46 24000 Fiat Punto 8601 3995 4 gold 1.2 unleaded 38-51 31000 Rover 820 SLi 21586 3795 6 red 2 unleaded 22-38 51000 Mitsubishi Carisma 15800 5999 2 blue 1.8 unleaded 34-56 33000 Fiat Cinquecento 6009 1995 6 white 0.9 unleaded 43-60 20000 Rover 416i 13586 3795 6 silver 1.6 unleaded 30-55 49000 Nissan Micra 6295 1795 8 white 1.2 unleaded 40-58 47000 Daewoo Lanos 11225 5999 3 silver 1.6 unleaded 24-46 42000 Rover 114 Sli 8595 2495 6 red 1.4 unleaded 35-56 33000 Ford Escort 8785 1595 7 blue 1.3 unleaded 39-54 68000 Fiat Uno 6864 1495 8 red 1 unleaded 44-55 51000 Rover Metro 6645 895 7 nightfire 1.1 unleaded 40-58 43000 Vauxhall Nova 5599 1000 10 green 1.4 unleaded 31-50 75000 Toyota Corrolla 13800 7495 2 black 2 diesel 34-53 25000 Vauxhall Cavalier 10150 850 10 red 1.6 unleaded 33-53 73000 Volkswagen Golf 400 15 green 1.4 unleaded 33-49 Volkswagen Golf 9524 3695 7 red 1.4 unleaded 33-49 49000 Seat Ibiza 5995 795 7 silver 0.9 unleaded 32-53 45000 Rover 214i 9565 1700 8 blue 1.4 unleaded 32-50 55000 Ford Fiesta 7310 1050 8 blue 1.1 unleaded 45-62 90000 Fiat Tempra 10423 1295 6 silver 1.6 unleaded 31-50 81000 Ford Fiesta 7875 1495 11 red 1.8 unleaded 32-50 74000 Hyundai Sonnata 11598 1195 9 silver 2 unleaded 28-41 65000 Renault Clio 6795 1995 8 black 1.2 unleaded 41-61 47000 Citroen Debut 5715 1495 7 black 0.95 unleaded 50-62 50000 The scatter diagram of the old price and mileage shows me that there is a positive correlation between the price and mileage depreciation of a cars price. The equation of the equations line of best fit or the "Trend Line" is Y=0.04687x+1.304E+004. This equation shows the gradient and where the line crosses the axis. In order for me to construct my graph accurately I am going to use Autograph from the computer. Scatter Graph Cumulative Frequency Graph Mileage The cumulative frequency graph shows me the median, the range of the mileage and the interquartile ranges. Table of Values of Data Set 1: Class Int. Mid. Int. x Class Width Freq. Cum. Freq. 0 § x < 20000 1E+004 2E+004 6 6 20000 § x < 40000 3E+004 2E+004 8 14 40000 § x < 60000 5E+004 2E+004 20 34 60000 § x < 80000 7E+004 2E+004 3 37 80000 § x < 100000 9E+004 2E+004 2 39 100000 § x < 120000 1.1E+005 2E+004 1 40 120000 § x < 140000 1.3E+005 2E+004 0 40 140000 § x < 160000 1.5E+005 2E+004 0 40 160000 § x < 180000 1.7E+005 2E+004 0 40 180000 § x < 200000 1.9E+005 2E+004 0 40 Öf = 40 Öfx = 1.8E+006 Öfx² = 1.008E+011 Mean = 4.5E+004 Standard Deviation = 2.225E+004 Variance = 4.95E+008 Histogram The histogram of the graph was not as I expected because I found out that the cheaper the car the more mileage it has and the more expensive the car the less mileage it has. Table of Values of Histogram [Data Set 1]: Class Int. Mid. Int. x Class Width Freq. Cum. Freq. 0 § x < 20000 1E+004 2E+004 6 6 20000 § x < 40000 3E+004 2E+004 8 14 40000 § x < 60000 5E+004 2E+004 20 34 60000 § x < 80000 7E+004 2E+004 3 37 80000 § x < 100000 9E+004 2E+004 2 39 100000 § x < 120000 1.1E+005 2E+004 1 40 120000 § x < 140000 1.3E+005 2E+004 0 40 140000 § x < 160000 1.5E+005 2E+004 0 40 160000 § x < 180000 1.7E+005 2E+004 0 40 180000 § x < 200000 1.9E+005 2E+004 0 40 Öf = 40 Öfx = 1.8E+006 Öfx² = 1.008E+011 Mean = 4.5E+004 Standard Deviation = 2.225E+004 Variance = 4.95E+008 Bow and Whisk By the information I found in the Cumulative Frequency graph it has enabled me to draw the Bow and Whisker diagram for the mileage alone. PRICE My cumulative frequency graph of used price gives me information on the median the price range and the lower and upper quartiles. Table of Values of Data Set 1: Class Int. Mid. Int. x Class Width Freq. Cum. Freq. 0 § x < 10000 5000 1E+004 15 15 10000 § x < 20000 1.5E+004 1E+004 20 35 20000 § x < 30000 2.5E+004 1E+004 2 37 30000 § x < 40000 3.5E+004 1E+004 2 39 40000 § x < 50000 4.5E+004 1E+004 0 39 50000 § x < 60000 5.5E+004 1E+004 0 39 60000 § x < 70000 6.5E+004 1E+004 0 39 70000 § x < 80000 7.5E+004 1E+004 0 39 80000 § x < 90000 8.5E+004 1E+004 0 39 90000 § x < 100000 9.5E+004 1E+004 1 40 Öf = 40 Öfx = 5.9E+005 Öfx² = 1.76E+010 Mean = 1.475E+004 Standard Deviation = 1.491E+004 Variance = 2.224E+008 Grouped Data Statistics: Total Frequency, n: 40 Mean, x: 14750 Standard Deviation, x: 14914.3 Modal Class: 10000- Lower Quartile: 6666.67 Median: 12500 Upper Quartile: 17500 Semi I.Q. Range: 5416.67 ______________________________________ Histogram Table of Values of Histogram [Data Set 1]: Class Int. Mid. Int. x Class Width Freq. Cum. Freq. 0 § x < 10000 5000 1E+004 15 15 10000 § x < 20000 1.5E+004 1E+004 20 35 20000 § x < 30000 2.5E+004 1E+004 2 37 30000 § x < 40000 3.5E+004 1E+004 2 39 40000 § x < 50000 4.5E+004 1E+004 0 39 50000 § x < 60000 5.5E+004 1E+004 0 39 60000 § x < 70000 6.5E+004 1E+004 0 39 70000 § x < 80000 7.5E+004 1E+004 0 39 80000 § x < 90000 8.5E+004 1E+004 0 39 90000 § x < 100000 9.5E+004 1E+004 1 40 Öf = 40 Öfx = 5.9E+005 Öfx² = 1.76E+010 Mean = 1.475E+004 Standard Deviation = 1.491E+004 Variance = 2.224E+008 This histogram is very much surprising because it shows that more people who buy second hand cars would rather buy it in between 10000 and 20000 and there were no buyer around 40000 and 90000. Box and Whisker Table of Values of Data Set 1: Class Int. Mid. Int. x Class Width Freq. Cum. Freq. 0 § x < 10000 5000 1E+004 15 15 10000 § x < 20000 1.5E+004 1E+004 20 35 20000 § x < 30000 2.5E+004 1E+004 2 37 30000 § x < 40000 3.5E+004 1E+004 2 39 40000 § x < 50000 4.5E+004 1E+004 0 39 50000 § x < 60000 5.5E+004 1E+004 0 39 60000 § x < 70000 6.5E+004 1E+004 0 39 70000 § x < 80000 7.5E+004 1E+004 0 39 80000 § x < 90000 8.5E+004 1E+004 0 39 90000 § x < 100000 9.5E+004 1E+004 1 40 Öf = 40 Öfx = 5.9E+005 Öfx² = 1.76E+010 Mean = 1.475E+004 Standard Deviation = 1.491E+004 Variance = 2.224E+008 From this box and whisker, I can tell that the used car price is not symmetrical and not in equal distributions. There are more cars on the upper side of the market. Final Conclusion Overall, my investigation of the factors that effect the price of a second hand car as reached the following conclusion: There is evidence of positive correlation between the used price and the mileage. E.g. the higher the mileage the cheaper the car will be. There is also evidence of negative correlation on the histogram diagram. It has come to chow that only cheap cars are bought and not the high and expensive ones. My hypothesis in the introductions states that there should be a cars that have a higher price and a low mileage. And in my investigation it was proven true and not false. In conclusion I was able to prove my hypothesis to be correct in my investigation.   

This Database contains information about some used cars, and many different makes of cars are included. I am going to include the following: • Specifying clearly what I plan to do and why I am approaching this investigation in this way. • Collating the data I need and representing it...

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Hidden Faces Coursework A cube...Hidden Faces Coursework A cube a total of 6 sides, when it is places on a surface only 5 of the 6 faces can be seen. However if you place 5 cubes side by side, there is a total of 30 faces, but out of this 30 only 17 can be seen. In this coursework I will be finding out the global formula for the total number of hidden faces for any number of cubes in any way positioned. To find this out I will be testing various numbers of cubes in different positions. This will enable me to find out several different formulae. Using the formulas found I will then be able to find out the global formula. I am generating only 3 formulae to get to the global formula. 1 row 6 faces 1 cube 1 hidden face 1 row 12 faces 2 cubes 4 hidden faces 1 row 18 faces 3 cubes 7 hidden faces 1 row 24 faces 4 cubes 10 hidden faces 1 row 30 faces 5 cubes 13 hidden faces 1 row 36 faces 6 cubes 16 hidden faces 1 row 42 faces 7 cubes 19 hidden faces 1 row 48 faces 8 cubes 22 hidden faces From the cubes drawn above I can see a pattern being formed. The number of hidden faces goes up by 3 every time a cube is added on the end. Cubes in a row Total faces Faces seen Faces unseen 1x1 6 5 1 1x2 12 8 4 1x3 18 11 7 1x4 24 14 10 1x5 30 17 13 1x6 36 20 20 1x7 42 23 19 1x8 48 26 22 The graph above show the number of hidden faces, the number of faces which can be seen and the total number of faces. Nth term 1 2 3 4 5 6 7 8 Total faces 6 12 18 24 30 36 42 48 difference + 6 + 6 + 6 + 6 + 6 + 6 + 6 The table above shows the total number of faces on an 'n' number of cubes. As we increase the number of cubes being added on the number of faces increases by 6. The formula to find out the total number of faces is: 6n E.g. 4 is the nth term so you multiply 4 by 6, which gives you a total of 24 which is the answer. Nth term 1 2 3 4 5 6 7 8 Seen faces 5 8 11 14 17 20 23 26 difference + 3 + 3 + 3 + 3 + 3 + 3 + 3 The table above shows the amount of faces seen on an 'n' number of cubes. The formula for working out the number of faces which can be seen is: 3n+2 E.g. 3 is the nth term so you have to multiply 3 by 3 33+2 Which gives a total of 9. you then add 2 which gives a final total of 11. Below shows the relationship between the cubes and the number of faces. Both hidden and seen. Nth term 1 2 3 4 5 6 7 8 Hidden faces 1 4 7 10 13 16 19 22 differences + 3 + 3 + 3 + 3 + 3 + 3 + 3 The graph above shows how many hidden faces there are related to the number of cubes. The graph and the table above shows the relationship between the number of cubes and the number of faces seen and unseen. Both the graph and the table above will now allow me to work out the formula for the number of hidden faces in one row. To find the global formula for the number of hidden faces in one row I have to refer to the table above. As you can see from the table it will be a linear equation because there is only 1 line of difference. The general linear equation is y=mx+c Therefore the linear rule is in the form of tn=an+c In the equation tn is the total number of hidden faces and n is the number of cubes. Therefore I need to find out the equation for a and c are. In the equation a is equal to the first difference. So I can replace the a with a 3, which makes tn=3n+c Now I need to find out the value of c so I can substitute it into the equation. To find c I will chose a number of cubes from the table and its results and place it into the equation. e.g. tn=an+c 13=35+c 13=15+c Now all I have to do is rearrange the formula so I can find out c. 13=35+c 13-15=c c=-2 tn=3n+-2 the equation is not in its simplest form so now I need to multiply out the brackets so I can get my final formula. tn=3n-2 To see whether or not the formula works I will test it using a number of cubes which Is not in my table above. I will use 10 cubes for this. tn=an+c tn=3n-2 tn=310-2 tn=30-2 tn=28 As I can see from the above equation I found out that my formula for hidden faces in one row has worked. For the next part of my coursework I need to generate another formula but for a different structure of cubes. I will be using 2 rows for this part of the coursework. 2 rows 2 cubes 12 faces 4 hidden Faces 2 rows 4 cubes 24 faces 12 hidden faces 2 rows 6 cubes 36 faces 20 hidden faces Cubes In A Row Total Faces Seen Faces Hidden Faces 2x1 12 8 4 2x2 24 12 12 2x3 36 16 20 2x4 48 20 28 2x5 60 24 36 2x6 72 28 44 2x7 84 32 52 2x8 96 36 60 12n 4n+4 8n-4 The graph above shows us the number of cubes, faces a hidden faces. It also shows the formulae for finding out the number of hidden faces, total faces and number of faces seen. Above shows the relationship between the number of cubes and the number of faces. By looking at the graph and the chart I have generated my second equation to find the number of hidden faces in 2 rows. The table below shows the relationship between the number of cubes and the number of hidden faces. I did not draw the other two tables because they were not relevant in finding out the global formula. nth term 2 4 6 8 10 12 14 16 tn 4 12 20 28 36 44 52 60 1st diff + 8 + 8 + 8 + 8 + 8 + 8 + 8 Formula for working out the total number of hidden faces: This pattern also has only the 1st difference so I could see that this was going to be a linear equation as well. So again I would have to follow the rule: tn = an + c As you know tn is total number of hidden faces and n is the number of cubes, so the two unknowns are once again a and c. a has been replaced with 8 as that represents the 1st difference, therefore the equation now looks like: tn = 8n+c. Again two methods can be used to work out the c term. The first way, which can be used, is to use the zero term: so as the 1st difference is +3 the working out would be 4 "“ 8 = -4. Another method that could be used is, you pick any number from the table above and it's result, and work it out as an equation: tn = 8n+c 44 = 86+c 44 = 48+c You need to rearrange the formula so c is on it's own, and change the + to -. c = 44 "“ 48 c = -4 So the second general formula would look like: tn = 8n - 4 To make sure my formula is right I will test it but I will use a value of cubes which is not in my table. I will use 20 cubes. Now all I have to do is substitute the information found into the equation. The final part of my investigation is to generate another formula, all these three formulae should help me find out the global formula. To get my third and final formula I will investigate using 3 rows of cubes. This should hopefully help me get the global formula. Cubes In A Row Total Faces Seen Faces Hidden Faces 3x1 18 11 7 3x2 36 16 20 3x3 54 21 33 3x4 72 26 46 3x5 90 31 59 3x6 108 36 72 3x7 126 41 85 3x8 144 46 98 18n 5n+6 13n-6 The table above shows the formulae for total faces, hidden faces and seen faces. The graph above show the relationship between the faces and the number of cubes. nth term 3 6 9 12 15 18 21 24 tn 7 20 33 46 59 72 85 98 1st diff + 13 + 13 + 13 + 13 + 13 + 13 + 13 The table above shows the difference between the number of hidden faces and the number of cubes. I have to do all the things listed above in the previous investigations, but using the information gained in this investigation. I have to substitute the letters with numbers. To make sure my formula works I will have to test it out first. I will use a number which is not part of my table. I will use 30 cubes. tn=an+c tn=13n-6 tn=1330-6 tn=390-6 tn=384 The final formula is 13n-6 The final section of my coursework is to find out the global formula for finding out the total number of hidden faces in any given number of cubes and rows. I know that the formulae for finding out the different surfaces areas are: length multiplied by width, length multiplied by height and height multiplied by width. I also know the 6LWH will give me the volume of the cube, which is the total number of faces in a group of cubes. This information will help me find out the global formula. As 6LWH is the total number of faces, this can be used as the first part of my formula. 6LWH Now I need to subtract the faces which are not inside of the cube, to do this I will use the formulae for finding out the surface area. Firstly I will have to minus the top are of the cubes as these are all visible, the formula for the surface area is length multiplied by width. This makes our formula 6LWH-LW Now I have to minus the four surface areas that are on the side of the cuboids. two of these will always be equal, as will the other two sides. To find out the surface area of the two different sized sides, I will have to do length multiplied by width for one and width multiplied by height for the other. But because these surfaces are in pairs I will have to multiply them both by two. This will give me the final formula. 6LWH-2HW+2HL-LW The formula shown above is used to work out hidden faces in any number of cubes in any formation. This concludes my coursework and I reached my target, which was to find the global formula by only using three formulae.   

Hidden Faces Coursework A cube a total of 6 sides, when it is places on a surface only 5 of the 6 faces can be seen. However if you place 5 cubes side by side, there is a total of 30 faces, but out of this 30 only 17...

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