Observation of people obeying temporary signs.

Psychology Coursework - Observation of people obeying temporary signs. Introduction: Milgram's work showed that when people are put in a situation where they have to obey orders given by an authority figure, they will do so, even if the task goes against their ethical beliefs. I wanted to observe the obedience of pupils when reading an order on a sign when written by either an authority figure or another pupil. Research aim: To find out if people will obey temporary signs on a drink vending machine in a Sixth Form common room, written by an authority figure and a school pupil. Hypothesis: Pupils will obey the sign written by an authority figure and disobey the sign when written a school pupil. Null Hypothesis: There will be no difference in obedience between the signs written by the authority figure and school pupil Variables: The independent variable was whom the signs were written by. The dependent variable was whether the pupil obeyed the signs or not. Population: Sixth Form pupils (16 and over), in UK. Sample: An opportunity sample of 40 Sixth Form pupils from a secondary school in Letchworth per lunch hour. They included male and female subjects, ranging from 16 to 18 years old. Method: Over the period of one lunch hour, it was observed how many Sixth Form pupils purchased drinks from the one drink vending machine in the Sixth Form common room. This was

  • Word count: 627
  • Level: AS and A Level
  • Subject: Maths
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Investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles.

Aim: To investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles. Task 1: a) The numbers 5, 12, 13 satisfy the condition. 5² + 12² = 13² Because 5² = 5x5 = 25 12² = 12x12 = 144 13² = 13x13 = 169 And so 5² + 12² = 25 + 144 = 169 = 132 b) The Numbers 7, 24, 25 also satisfy the condition. 7² + 24² =25² Because 7² = 7x7 = 49 24² = 24x24 = 576 25² = 25x25 = 525 And so 7² + 24² = 49+ 576 = 625 = 25² Task2: The perimeter and area of the triangle are: a) b) Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 2 6 5 2 3 30 30 7 24 25 84 84 Task3: Length of short side is going to be in fixed steps meaning that this is a linear sequence and the length of middle side and longest side is actually a quadratic sequence because they are not in fixed steps and in geometric sequence. 4 , 12 , 24 , 40 8 , 12 , 16 4 , 4 5 , 13 , 25 , 41 8 , 12 , 16 4 , 4 Length of shortest side: Term no 2 3 4 5 Sequence 3 5 7 9 1 Sequence 2n 4 6 8 0 Sequence 2n + 1 Length of the middle side: F (n) = an² + bn +c F= (1) = a x 1² + b x 1 + c = a + b + c = 4 - eqn1 F= (2) = a x 2² + b x 2 + c = 4a + 2b + c = 12 - eqn2 F= (3) = a

  • Word count: 1576
  • Level: AS and A Level
  • Subject: Maths
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Solving Equations by numerical methods.

Solving Equations by numerical methods. The fact that the vast majority of the numbers on the real number line are irrational (i.e. cannot be put in the form p/q, where p and q are integers) has an associated consequence that the vast majority of equations involving powers of x (i.e. polynomials) are insoluble by closed analytical techniques. This coursework's solution implies finding values of x say c1, c2-...c3 where f(c1)=0; f(c2)=0 and so on. Further transcendental equations and non- linear equations require numerical methods for their solution. It is to be noted that in no way are numerical methods inferior to analytical solutions, they are indeed the only practical solutions available, however on the down side no exact solution is possible and error bounds have to be placed on the solution given. In this coursework three main methods of solution from one-dimensional equations are given, each of the methods could be extended to the multi-dimensional case. The aim of this coursework is to show example of the methods in action, where these are successful and where they are not. A comparison between the methods will also be attempted. Method 1: Decimal Search SOLVING AN EQUATION WITH THIS METHOD Fig1. Shows the function f(x) = 12 ln(x) - x3/2. (Where ln(x) represents the natural logarithm of x.) To find the roots, the equation is: 2 ln(x) - x3/2 = 0 This equation

  • Word count: 2153
  • Level: AS and A Level
  • Subject: Maths
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Solving Equations Using Numerical Methods

Solving Equations Using Numerical Methods Numerical methods are used for solving equations which can't be solved with algebra. Quadratics would not need to use numerical methods as an accurate answer can be found using factorisation or by using the quadratic formula. I will be using numerical methods to solve cubic equations that have decimal answers. I will be using three methods in this coursework; 'Decimal Search', 'Newton-Raphson' and 'Graphical Convergence'. Decimal Search This method solves the equation by looking for a change of sign. We know that the root of the equation that we are looking for is in between the 2 numbers that the change of sign occurs in. We then repeat the same process as before but this time to a greater level of accuracy. When I have done this process to the level of accuracy required, I will be able to tell where the root of the angle is. I will now solve an equation using the decimal search method. The equation I will be using is y=x³-2x+0.5 I will be trying to find this root using decimal search. I will start with the integer immediately below the root (-2) and this will be the lower bound. I will find the values of f(x) for this and consecutive increments of 0.1 until a change of sign is found. Here is my table of values. From this table, I can see that the change in sign appears in between -1.6 and -1.5. To be certain that the root is

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  • Level: AS and A Level
  • Subject: Maths
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Examining, analysing and comparing three different ways in which to find the roots to an equation.

Advanced Mathematics (C3) Coursework Introduction: For my investigation, I will be examining, analysing and comparing three different ways in which to find the roots to an equation. It will include the "Change of Sign Method", the "Newton-Raphson Method" and finally the method of rearranging "f(x) = 0" into the form "x = g(x)". I will be finding roots of equations using the methods, and hence compare the merits and flaws of the methods with each other. I will analyse which is the best in terms of factors such as a speed of convergence and ease of use with available software and hardware. Change of Sign Method: This method finds a root to an equation by looking at when values of f(x) change sign from positive to negative or vice versa. This works because when the value of f(x) changes its sign, it must have passed through the x axis, and thus the root is somewhere between the two values that changed. The technique of the method can be done in many ways such as by interval bisection or linear interpolation. However, I will be using the decimal search method. With decimal search, you first draw the graph of the curve that is to be investigated, and look for where the curve passes through the x axis, as these points are where the roots are. But as you cannot guess exactly where the roots are, you take intervals: the two points surrounding where the curve passes through the

  • Word count: 3405
  • Level: AS and A Level
  • Subject: Maths
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Methods for Advanced Mathematic

Methods for Advanced Mathematics (C3) Coursework Introduction: In this coursework, I am going to solve equations by using the Numerical Methods. The Numerical methods are used to solve equation that cannot be solved by using algebra. Here are the 3 Numerical methods: . Change of sign method 2. Newton-Raphson method 3. Rearranging f(x) = 0 in the form of x=g(x) Change of sign method This method is based on how an equation has a change of sign either from positive to negative or negative to positive on either side of a root when crossing the x-axis. It is best to sketch a diagram of the equation before starting with the method, so we would have a clear structure of what the equation looks like, the number of roots the equation has and the roots approximate position. To investigate this method I would use the following equation 0 = 3x3+11x2+2x-5 The graphs below show the function f(x) = 3x3+11x2+2x-5 x f(x) 6 051 5 655 4 371 3 81 2 67 1 0 -5 -1 -2 1 -3 7 -4 -29 -5 -115 -6 -269 x (1dp) f(x) -1 -0.9 -0.077 As we can see from the results table, there is a change of sign between [-1, -0.9]. This indicates the root is in the interval. To make sure the result is as accurate as possible, I will repeat the process until a sufficient number of decimal places are achieved. I will calculate this to 3 decimal places. A zoomed in of the root

  • Word count: 2354
  • Level: AS and A Level
  • Subject: Maths
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By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots.

.(a) By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots. (i) From the graph, we can see that the smallest positive root is 1 f (1) = 13 - 12 - 4(1) + 4 = 0 By the Remainder Theorem, 1 is a root and (x-1) is a factor of y = x3- x 2- 4x + 4 f (x) = x3- x 2- 4x + 4 = (x-1) (x2-4) = (x-1) (x-2) (x+2) Therefore, by the Factor Theorem, the equation y = x3- x 2- 4x + 4 has roots at x = -2, 1 and 2. Hence, the smallest positive root is 1 or (ii) From the graph, it seems that the smallest positive root is f () = 2() 3 - () 2 - 8() + 4 = 0 By the Remainder Theorem, is root and (2x-1) is a factor of y = 2x3-x 2-8x+4 f(x) = 2x3-x 2-8x+4 = (2x-1) (x2-4) = (2x-1) (x-2) (x+2) Therefore, by the Factor Theorem, the equation y = 2x3-x 2-8x+4 has roots at x = -2, and 2. Hence, the smallest positive root is . (iii) From the graph, it seems that the smallest positive root is f () = 3() 3 - () 2 - 12() + 4 = 0 By the Remainder Theorem, is a root and (3x-1) is a factor of y = 3x3-x 2-12x+4 f (x) = 3x3-x 2-12x+4 = (3x-1) (x2-4) = (3x-1) (x-2) (x+2) Therefore, by the Factor Theorem, the equation y = 3x3-x 2-12x+4 has roots at x = -2, and 2. Hence, the smallest positive root is . (iv) From the graph, it seems that the smallest positive root is f () = 4() 3 - () 2 - 16() +

  • Word count: 1351
  • Level: AS and A Level
  • Subject: Maths
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Sequence & Series

Sequence & Series Solutions . For the following sequences write down the next two terms of the sequence and an expression for the nth term. a. 5, 10, 15, 20, ... b. 1, 1/2, 1/3, 1/4, ... c. 0, 3, 8, 15, 24, ... d. 1/2, 1/6, 1/12, 1/20, ... a. The next two terms are 25, 30 and. b. The next two terms are 1/5, 1/6 and. c. The next two terms are 35, 48 and. d. The next two terms are 1/30, 1/42 and. 2. Write the following series in sigma notation a. 1+8+27+64+125 b. 2+4+6+8+...+20 c. d. e. -4-1+2+5+...+17 f. a. b. c. d. e. f. 3. Write down the first three terms, and where there is one, the last term of each of the following series a. b. c. d. e. a. b. 0+2+6+...+30 c. d. e. 4. Write down the nth term and the stated term of the following A.P's. a. 7+11+15+... (7th) b. -7-5-3-... (23rd) The nth term of an A.P. is a. b. 5. Find the sums of the following series a. 5+9+13+...+101 b. -17-12-7-...+33 c. 1+ 1 1/4 + 1 1/2 + ... + 9 3/4 The sum of an A.P. with n terms, where (this is a rearrangement of the formula for the last term) a. It can be seen that a=5, l=101 and d=4, therefore and so. b. Here a=-17, l=33 and d=5. So and . c. a=1, l=9 1/4 and d= 1/4 and so . The sum is then given by . 6. Evaluate a. b. a. b. 7. Find the sum of the A.P. -7-3+1+... from the 7th to the 13th term inclusive. From the AP it

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  • Level: AS and A Level
  • Subject: Maths
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Numerical Methods coursework

Numerical Methods Coursework Numerical Integration The Problem Integration means finding the area underneath a particular region of a function. At my current knowledge of maths I am not able to integrate various functions. Therefore I am going to use knowledge of numerical methods to produce an approximation to an area which does not have an analytic solution. In my coursework I will integrate the function: The graph below shows the graph of the function. The red arrow determines the region between 0.25 and 1.25, which then leads to the integral: I can not solve this problem using the knowledge of C1 and C2, because I am not able to integrate cos(x) yet. Due to this I suggest that this problem will be appropriate for numerical solution. The Approximation Rules To solve this problem, I am going to use knowledge of numerical integration studied in the "Numerical Methods" textbook. The approximate methods of definite integrals may be determined by numerical integration using: . The Trapezium Rule: The Trapezium Rule divides the area underneath the curve into trapeziums. We can then use the formula (where a, b are the bases and h is the height of the trapezium) to estimate the area. Dependent on the amount of trapezia (n) used the general formula is: 2. The Midpoint Rule: The Midpoint Rule divides the area underneath the curve into rectangles. We can then use the

  • Word count: 1047
  • Level: AS and A Level
  • Subject: Maths
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This coursework is about finding the roots of equations by numerical methods.

C3 Coursework - Numerical Solution to Equations This coursework is about finding the roots of equations by numerical methods. I am going to use three different methods to solve different equations. First of all, one root should be found successfully by using three different methods. Then, error bounds should be given and shown by graphic. A failure example is given and explained. Change of Sign Method I need to use an equation and use Autograph to get a rough interval and do a search to find intervals that show a change of sign. Here is an example x³-2x²-3x+4=0 Solve x³-2x²-3x+4=0 Here is the table of values x -3 -2 -1 0 2 3 4 F(x) -38 -6 4 4 0 -2 4 24 This shows that there are three intervals containing roots: (-2, -1), [1,2] and (2, 3) Then carry out a decimal search in one of the identified intervals to find that root to the desired level of accuracy. I use the interval (2, 3) x f(x) 2 -2 2.1 -1.859 2.2 -1.632 2.3 -1.313 2.4 -0.896 2.5 -0.375 2.6 0.256 2.7 .003 The change in sign tells us there is a root in the range (2.5, 2.6) Now use decimal search and Excel within the interval [2.5, 2.6] x f(x) 2.51 -0.31695 2.52 -0.25779 2.53 -0.19752 2.54 -0.13614 2.55 -0.07363 2.56 -0.00998 2.57 0.054793 2.58 0.120712 There is a root in the range (2.56, 2.57) Now use decimal search and Excel within the interval

  • Word count: 1797
  • Level: AS and A Level
  • Subject: Maths
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