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International Baccalaureate: Maths
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Every four hours, the viral particles doubles = 200% (see figure 1-1).The viral particles replicate every four hours. So, for every one hour the viral particles will replicate with a value of (see figure 1-2) Therefore, the total amount of viral particles in the body for every one hour, would be equal to the start amount multiplied with 2 to the power of . So, this phase of illness can be calculated using the model By using the regression on the calculator, putting the values for every four hours the viral particles is replicating, I find the best fit line to be an exponential regression.
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Inmediatamente pude encontrar una relaci�n entre las respuestas de ambas matrices, con lo que hall� una f�rmula para las expresiones Xn e Yn. A continuaci�n muestro el proceso que llev� a cabo para hacerlo: Una vez hecho esto, hall� los valores de (X+Y)1, (X+Y)2, (X+Y)3 y (X+Y)4, para encontrar la f�rmula de la expresi�n (X+Y)n y as� luego poder relacionar las respuestas de la misma forma como lo hice en el proceso anterior. Luego, siendo las matrices A=aX y B=bY, tom� distintos valores constantes para a y b, y hall� con la calculadora A2, A3, A4; B2, B3 y B4.
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Math IA Type 1 In this task I will investigate the patterns in the intersection of parabolas and the lines y = x and y = 2x.
First I will the graph the functions then I will use my GDC to find the four intersections points as illustrated on the graph.[example 1] Graph showing the equations: [In blue] [In brown ] [In green] Example 1 ] I will find the values for x1, x2,x3,x4 for the function, using the 'calculate intersect' feature on a GDC[ TI - 84 plus] 2. Now I will look at different parabolas of the form with values of a greater than 1.
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Math IA type 2. In this task I will be investigating Probabilities and investigating models based on probabilities in a game of tennis.
Part 1 The ratio of the points won by Adam and Ben are 2:1 respectively. Therefore Adam wins twice as many points as Ben does. Therefore Adam wins of the points and subsequently Ben wins of the points. The distribution of X, the number of points won by Adam would be derived by using the binomial probability function and substituting variables and constants to arrive at an appropriate model for the distribution of X, the number of points won by Adam. The distribution chosen is the binomial probability distribution because in the case of the games of tennis, * There is a repetition of a number of independent trials in which there are two possible results, success [the event occurs ex.
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Infinite Surds. As we can see there are ten terms of this sequence where is the general term of the sequence when, is the first term of the sequence.
A graph has been plotted to show the relation between and. And it can be oticed that as long as gets larger, gets closer to a fixed value. To investigate more about this fixed value we take this equation into consideration as n gets bigger. n an-an+1 1 - 0.13956 2 -0.04428 3 -0.01380 4 -0.00427 5 -0.00132 6 -0.00040 7 -0.00013 8 -0.00004 9 -0.00001 We can figure out from the table above that when n gets larger, the term (an+an+1) gets closer to zero but it never reaches it So we can come to the conclusion: When n approaches infinity, lim (an-an+1)
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Graphical plot of the fish population Un versus the growth factor r of the logistic model un?1 ? run 2 IB Mathematics HL Type II Portfolio: Creating a logistic model International School of Helsingborg - Christian Jorgensen The slope of the trend line in figure 1 is determined graphically. The equation of the trend line is of the form y=mx + b: y ? ?1? 10?5 x ? 1.6 . Thus m= ?1? 10?5 . This can be verified algebraically: m ?
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5:41 1:07 1:35 5:54 2:28 2:16 2:54 1:21 3:36 3:52 4:36 2:05 2:58 1:50 1:01 1:02 1:28 1:58 0:32 3:46 5:46 2:00 3:08 3:39 1:30 2:26 1:24 1:37 4:36 0:44 1:17 1:52 1:25 6:02 2:09 3:08 4:31 6:54 1:49 1:06 1:17 2:50 2:02 4:32 0:25 2:39 3:30 9:10 1:14 2:46 2:32 4:07 1:13 5:46 2:30 1:38 2:16 4:22 1:16 1:19 4:59 2:42 2:07 2:06 c) Waiting Time for Each Client to Be Served 4:30 6:35 4:21 6:16 9:45 6:13 8:25 6:02 4:07 4:08 5:41 5:13 7:11 7:29 7:13 6:03 6:00 4:02 5:40 3:52 3:54 6:12 4:38 6:28 5:42 4:30 5:17
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In this equation A is the amplitude of the graph. Where max = maximum dependent variable value and min = minimum dependent variable value. The maximum value obtained from the data is 21.65 whereas the minimum value is 15.20. These values were then substituted into the equation, this is shown below. In the sin equation aforementioned B is the measure of how much the graph is stretched horizontally. B is calculated using the equation shown below. However, the period of this graph is unknown. To find the period of the graph the equation shown below must be used.
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A2 U A3 U A4 U A5 7 21 A1, A2, A3, A4, A5, A6, A1 U A2, A2 U A3, A3 U A4, A4 U A5, A5 U A6, A1 U A2 U A3, A2 U A3 U A4, A3 U A4 U A5, A4 U A5 U A6, A1 U A2 U A3 U A4, A2 U A3 U A4 U A5, A3 U A4 U A5 U A6, A1 U A2 U A3 U A4 U A5, A2 U A3 U A4 U A5 U A6, A1 U A2 U A3 U A4 U A5 U A6
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1.892 log7 + log4 log28 1.447 log12 - log3 0.6021 log4 0.6021 log50 - log2 1.398 log25 1.398 log7 - log5 0.1461 log1.4 0.1461 log3 - log4 -0.1249 log0.75 -0.1249 log20 - log40 -0.3010 log0.5 -0.3010 This table to the left clearly shows that the log of 2 numbers subtracted from each other will equal the log of the numbers divided by each other.The table below clearly shows the log ()
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Therefore, my average was 0.80 meters. Finding the average between the through and the crest gave me the sinusoidal axis. The sinusoidal axis ended up being 6.475 meters. Calculations: Vertical Translation d= 6.475 12+12.3=24.3 ,24.3-2.=12.15 0.90+0.70=1.6 ,1.6-2.=1.6 0.8+12.15=12.95 ,12.95-2.=6.475 My next step was to find the graphs vertical stretch also known as the graphs amplitude; or the distance between the sinusoidal axis and the crest. In order to determine, it was necessary to subtract the sinusoidal axis from the average crest. My answer for the vertical stretch was than 5.675.
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Using the pattern I was able to find I multiplied 48 which was my answer for n=2 by 4. I got the answer that n=3 is equal to 192. In order to prove that my answer was accurate I had to actually count the number of sides in n=3. In order to do that I got in the internet a high resolution drawing of stage 3 this allowed me to precisely count the number of sides. The fact that I got 192 proved that a pattern had already been established for N. L= I was able to come up with the length of each sides by using the property of fractals.
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= 64 = 43 M4 = =, det (M2) = 256 = 44 M5 = =, det (M2) = 1024 = 45 M10 =, det (M2) = 1048576 = 410 M20 =, det (M2) = 1.099511628 x 1012 = 420 M50 =, det (M2) = 1.2676506 x 1030 = 450 By squaring each number in the matrix by the power of Mn you get the answer for each of the matrices. So if M2 =, you then square the matrix, so you multiply it by itself. . Multiplying the matrix by itself as this: gives you the new matrix which is, in this case.
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Make a conjecture about the value of D for these parabolas. * a= 2, b= -10, c =13 f(x) = 2x2-10x+13 X1: 1.40 X2: 1.70 X2-X1= 0.30 X3: 3.78 X4: 4.58 X4-X3= 0.80 D= 0.80-0.30 = 0.50 When a = 2, D = 0.5 * a=3, b= -9, c = 7 y = 3x2-9x+7 X1 : 0.82 X2: 1 X3 : 2.33 X4 : 2.85 X2- X1 = SL =1-0.82 = 0.18 X4-X3 = SR = 2.85-2.33 = 0.52 D= 0.52 - 0.18 =0.33 * a=4 , b=-15 , c= 14 X1= 0.75 X2= 1 X3= 1.5 X4 =2 SL = 0.25 SR= 0.50 D= 0.50 - 0.25 = 0.25 Evaluation
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The independent variable is the Age, which is the x axis on the graph while the dependent variable is the BMI, which is the y axis on the graph. The parameter is a constant in the equation of a curve that can be varied to yield a family of similar curves. 2. What type of function models the behaviour of the graph? Explain why you chose this function. Create an equation (a model) that fits the graph. I chose the polynomial, quadratic function because the graph looks similar to a parabola.
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That is why continued fractions are used to solve quadratic irrational numbers. Example: Questions A 1. Let be 2 Let be 3 Let be 4 Conclusion: Therefore whereby 2. n tn tn+1 tn-tn+1 1 2 1.5 0.5000000000 2 1.5 1.666666667 -0.1666666667 3 1.666666667 1.6 0.0666666667 4 1.6 1.625 -0.0250000000 5 1.625 1.615384615 0.0096153846 6 1.615384615 1.619047619 -0.0036630037 7 1.619047619 1.617647059 0.0014005602 8 1.617647059 1.618181818 -0.0005347594 9 1.618181818 1.617977528 0.0002042901 10 1.617977528 1.618055556 -0.0000780275 11 1.618055556 1.618025751 0.0000298045 12 1.618025751 1.618037135 -0.0000113842 13 1.618037135 1.618032787 0.0000043484 14 1.618032787 1.618034448 -0.0000016609 15 1.618034448 1.618033813 0.0000006344 16 1.618033813 1.618034056 -0.0000002423 17 1.618034056 1.618033963
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For example, the nut cannot break on 1.4 drops; it would either break on the first or second attempt. However, the data presented here is an average over a large number of trials, and because of this n can be any positive number, either an integer or fractional. Now another point to note about the value of n is that it represents the 'average' number of drops over a number of trials for different size nuts. This is assumed to imply that n is the mean of the data set for each value of h. However if the data set was not large enough the mean may not be a precise measure, and error bars should be associated with each value of n because of this the functional fit, which may represent the actual value of n, f(h)
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D = I SL - SR I D = I 0.618 - 1.618 I D = I -1 I D = 1 Therefore the difference between the SL and SR is one. 2. Find values of D for other parabolas of the form y = ax2 + bx + c, a>0, with vertices in quadrant 1, intersected by the lines y=x and y = 2x. Consider various values of a, beginning with a = 1. Make a conjecture about the value of D for these parabolas.
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+ (1/2 (3.015+3.062) x 0.125) + (1/2 (3.062+3.14) x 0.125) + (1/2 (3.14+3.25) x 0.125) + (1/2 (3.25+3.390) x 0.125) + (1/2 (3.390+3.562) x 0.125) + (1/2 (3.56+3.765) x 0.125) + (1/2 (3.765+3.4) x 0.125) = = 0.375 + 0.379 + 0.389 + 0.399 + 0.415 + 0.434 + 0.457 + 0.485 = 3.333 A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8 +A9 + A10 = (1/2 (3+3.01) x 0.1) + (1/2 (3.01+3.04) x 0.1) + (1/2 (3.04+3.09) x 0.1) + (1/2 (3.09+3.16) x 0.1) + (1/2 (3.16+3.25) x 0.1) + (1/2 (3.25+3.36)
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The midpoint in the data presented below is the maximum value plus the minimum value divided by 2. The midpoint for a standard sine curve is zero. So the first step is to change from the standard f(x) = sin (x), to f(x) = sin (x) + d where'd' is the vertical shift. So at this stage the model for the data is f(x) = sin (x) + 18.425 However this does not accurately representing the data, there are further changes that take place. The next step is to increase the amplitude to the same amplitude of the data.
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For the Number of sides i am basically going to count the number of sides. The length of each side can be calculated because each side gets split up into three parts. The perimeter will be measured using the simple formula, Perimeter = length of one side x number of sides. The area of the shape will be manually calculated for each of the smaller triangles and multiplied to the number of triangles. Formula = Method Taking the initial side length (length of one side of the triangle at Stage 0)
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_Vertical and horizontal translation (the scales are represented by c and d) Once we have indentified the type of the function that models the behaviour of the graph, it is possible to use algebraic methods to create an equation that fits the graph BMI. _Consider the graph of the function y=sinx The distance from a point where y= -1 (the minimum value) to the nearest point where y=1 (the maximum value) is. According to the graph BMI, we can see that the point (5; 15.2) represents the minimum value of the graph if it is to resemble the shape of the graph y=sinx.
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This can show by; This in one way or the other can be proved by; Considering, if 0 is put into the function instead of Will result in 3 which is the interval of when is 0. And also considering intervals, 0.5 and 1 respectively instead for, in the function, the results are 3.5 and 4 respectively, which proves my intervals. Having found all the required lengths, it's now possible to calculate an approximation of an area under the curve of the given function above using 2 trapeziums Therefore, areas for the two trapeziums; In order to estimate the total
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Gradient of f(x) =sinx.at every /4 unit d) make a conjecture for the derived function f �(x) conjecture: f �(x) = cosx e) Use your GC to test your conjecture graphically. Explain your method and your findings. Modify your conjecture if necessary 1. The graph joined scatter plots with curve of f�(x) =cosx is shown in figure 2 The derivative of f(x) =sinx found by calculator with the curve in figure 2is shown in figure 3 Figure 2 scatter plots and f�(x) =cosx Figure 3 derivative of sine function found by calculator Figure 2reveals that the curve of function f�(x)
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Therefore: And: Assuming that a = 2 Assuming that a = 3 Assuming that a = 6 Assuming that a = -4 Assuming that b = -1 Assuming that b = 4 Assuming that b = 5 Assuming that b = 7 To find the expression for I observed that the final results achieved were always the value chosen for a multiplied by the matrix X and the product to the power of n. For example (assuming a = 2): As a result we can see that: And since we know Hence: To find the expression for I did the same as for changing the value a to b and X to Y.
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