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# International Baccalaureate: Maths

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Meet our team of inspirational teachers Get help from 80+ teachers and hundreds of thousands of student written documents 1. ## Math Studies I.A

We cannot say what strong or weak correlation is because everyone describes it differently. Hence, we can only measure the degree of correlation in terms of a numerical value. We will then test the significance of the value based on our selected data. Background information Systemic Information/Measurement x Y xy Countries GDP per capita (PPP) Life expectancy Afghanistan 800 43.8 35040 640000 1918.44 Algeria 7,000 72.3 506100 49000000 5227.29 Andorra 42,500 82.67 3513475 1806250000 6834.3289 Anguilla 8,800 77.46 681648 77440000 6000.0516 Argentina 14,200 75.3 1069260 201640000 5670.09 Aruba 21,800 74.2 1617560 475240000 5505.64 Austria 39,200 79.21 3105032 1536640000 6274.2241 Bahamas,

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2. ## Math Portfolio: trigonometry investigation (circle trig)

By using theta, if it goes counter clock wise it will become positive while going in the direction of clockwise it becomes negative. Further more, based on our knowledge on Math Honor 1, we always know that hypotenuse in right triangles are the longest of among three sides. And thus since R is the hypotenuse, as explained before x and y can be any number as long as it is less than the radius. Therefore leading radius R always positive and real number.

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3. ## math

* The formula used to calculate the reducing balance depreciation is BVt= C ((1)-(rate/100)) ^n Note - n represents the number of years. Thus now I can use the present book value to calculate the rate at which the machine has depreciated. I will substitute the current book value in place of BVt in the equation. Then substitute C with the cost of the machine and substitute 2 in place of n as the current book value provided by the accountant is as of 2 years. I can then solve the equation in order to find the rate in a Graphic Display Calculator (GDC).

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4. ## IB Math Portfolio- Topic 1

Notice, the values of y never actually reach 0 because as the base increases, the smaller and closer to 1 the result must be in order to fit the expression. The same can be done with the other sequences: It is shown by this final sequence that the exponent of the base is increasing with direct correlation to the number of the term. The constant k, however, stays the same in all sequences. In looking back once more to the first sequence: Sequence 1 Term 1 Term 2 Term 3 Term 4 Term 5 The results show a pattern in relation to some of the different terms in the original statement: The top value, 3, represents the constant k.

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5. ## math portfolio

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6. ## infinite surds

One day Jimmy walked Kandee home from school and Kandee's dad saw them holding hands. He went crazy and ran after Jimmy. Luckily Jimmy was too fast and Kandee's dad didn't catch him. Kandee explained that the teacher told her to walk the poor blind penguin home. The funny thing is her dad actually believed her even though it looked like he knew what direction to run in. From then on Jimmy had to act blind in order to come to their house everyday.

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7. ## Logrithum bases

* Sequence 2: Log 3 81, Log 9 81, Log 27 81, Log 81 81, Log 243 81, Log 729 81... * Sequence 3: Log 5 25, Log 25 25, Log 125 25, Log 625 25, Log 3125 25, Log 15625 25... : : * Sequence 4: Log m mk, Log m2 mk, Log m3 mk, Log m4 mk, Log m5 mk, Log m6 mk ... Part II Now, I will write an expression for the nth term for each sequence.

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8. ## IB Math SL Type II Portfolio - BMI Index

First, let's take a look at the general cosine function: f(x) = a*cos(bx + c) + d. To find the amplitude (a), I will simply take the y-maximum (21.65) and subtract the y-minimum (15.20) and then divide by 2. This gives me 3.22 for the amplitude. Then to find the period I will have to see the length of one cycle, which is when the function goes from the maximum to the minimum and back to the maximum. By looking at the graph, I determine that one cycle is 30. I then take 2 and divide by 30 to get the period, since there are 2 radians in one cycle.

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9. ## IB Coursework Maths SL BMI

The independent variable for the above graph is the age in years, and this is shown on the x-axis. The dependent variable is the BMI, and this is shown on the y-axis. The highest BMI is 21.65, and the maximum age in years is 20. The parameters for this graph are: > Age in years: 2 - 20 > The BMI: 15.20 - 21.64 This graph does not show a line of best fit as I thought it would be clearer to see the exact points on the graph for each BMI value for the respective age. The function which I thought modelled the behaviour of the graph is the polynomial quadratic function, because it is clear that the graph has a similar shape to a parabola shape.

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10. ## Mathematical Modeling

Million 554.8 609.0 657.5 729.2 830.7 927.8 998.9 1070.0 1155.3 1220.5 In order to plot a better understandable graph, I changed the number of years to intervals of five. That way, the graph will make more mathematical sense and show a more obvious trend. For this investigation, I use the software GeoGebra to plot all my graphs because it is high definition and clear, which shows the trend easily. The above graph is all the plotted data points given of the growth population of China.

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11. ## Investigating ratio of areas and volumes

Area A is now the area contained between the graph and the x-axis between the arbitrary points x = an and x = bn and area B is the area contained between the graph and the y-axis between the arbitrary points x = an and y = bn. By this the ratio area A: area B can more readily be investigated. Area A is the area contained in between the graph of y = x1/n and the x-axis between the two arbitrary points x = an and x = bn such that a<b.

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12. ## Crows Dropping Nuts

The following table shows the number of average drops it takes at a certain height to break the nut. Height of drop(m) 1.7 2.0 2.9 4.1 5.6 6.3 7.0 8.0 10.0 13.9 Number of drops 420 21.0 10.3 6.8 5.1 4.8 4.4 4.1 3.7 3.2 This is a graphical representation of the large nuts data using TI InterActive. Relationship of Height of the drop vs. Number of drops for the Large Nuts The variables in this data are that the height of the nut dropped affected the frequency, and this variable is put into an average.

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13. ## Maths Modelling. Crows love nuts but their beaks are not strong enough to break some nuts open. To crack open the shells, they will repeatedly drop the nut on a hard surface until it opens. So, through this portfolio I will attempt to create a function t

- large nuts. Parameters: After analyzing the data given I noticed some constraints in the data given. Firstly it is important to note that the data given is an average of the number of drops taken to break open a large nut. It is also important to note that the number of drops must be a whole number because it is impossible to drop a nut times. The domain of a function that models this graph would also have to be greater than zero because it is not possible to have a value for height that is negative.

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14. ## In this investigation I will be examining logarithms and their bases. The purpose of examining logarithmic bases is to find patterns between the bases, its exponents and the product.

The other sequences also follow a similar pattern. Sequence 2 can be written as a base of 3 to the power of n and sequence four also can be written as a base of m to the power of n. An expression to calculate the nth term of each of the four sequences are written below in the form , where p, q Z. 1. 2. 3. 4. The expressions above can also be proven using a graph. For each sequence, I can graph each term as well as the expression determined for the nth term of that sequence on the same axis.

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15. ## Logan's logo

X 0 1 2 3 4 5 6 7 8 9 10 Y 3.45 1.69 1.35 2.00 3.47 5.12 6.85 8.25 8.95 8.71 7.15 Table 2: observed data points for the curve g(x) The points of these curves are graphed on the following graph: Graph 1: Observed points of logo In the graph above the points were plotted in a coordinate system with a domain of and a range of, this was done according to the size of the diagram of the logo given.

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16. ## Matrix Binomials. In this Math Internal Assessment we will be dealing with matrices.

2 = 21 X3 has the element 4 repeated. 4 = 22 X4 has the element 8 repeated. 8 =23 X7 has the element 64 repeated. 64 = 26 X15 has the element 16384 repeated. 16384 = 214 X20 has the element 52488 repeated. 52488 = 219 X50 has the element repeated. =249 It should also be noted that whatever number X is raised to it is always 1 less than that of 2's exponent The same applies for Yn except the elements b and c, these elements are negative. This is if matrix Y= . The following general statement can be made for Xn and thus, Yn: Xn = therefore, Yn = To test the validity of this statement we will use different values for n: n X Y 5 --> =V --> =V ?

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17. ## IB Design Tech Design Cycle

For the bridge, I thought that tying 3 straws to form a bundle and using 4 bundles to form a square and 2 other bundles to cross inside the square will create a strong base. Every 5 inches I will place a base so that the building will hold up straight. My goal for the tower was to be 2 feet tall or even taller. For the joints there should be a 40 degrees triangle formed by 2 bundles cutting across the side of the building.

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18. ## Math Portfolio: Fishing Rod

the result is quadratic function From the graph it is visible that the first five points fits the quadratic function exactly however afterwards the points don't correspond the curve. This could be proved by calculating points from the function (see table below) x Guide number (from tip) 1 2 3 4 5 6 7 8 y Distance from tip (cm) 10 23 38 55 74 96 120 149 Distance from tip (cm) 10 23 38 55 74 95 118 143 Using graphing software the lack of correspondence to the curve can be narrowed (see graph below).

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19. ## Logarithm Bases - 3 sequences and their expression in the mth term has been given. All of these equations will be evaluated on a step-by-step basis in order to find an expression for the nth term.

Thus, to prove as the general expression for the first row of sequences: un = First term of the sequence = = log2n 8 (Taking the first term as u1 and so on,) L.H.S R.H.S u1 = log21 8 u1 = 3 log2 8 = 3 L.H.S = R.H.S. Hence, verified that is the general expression for the 1st row of the given sequence in the form . Similarly, other terms in the sequence were also verified by using the TI-83 where: U2: log48 (1.5)

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20. ## The aim of this particular portfolio investigation is to find the area under a certain curve by dividing it into trapeziums and adding the sum of the areas of these trapeziums to approximate the actual area. This is done instead of using the usual method

The height of each trapezium is found by subtracting the lower limit, which is zero, from the upper limit, which is one, and dividing it by the number of trapeziums that is two in this case. Hence, h will be 0.5.To find the parallel sides of the trapezium the x values of each trapezium are inserted into the function of . Using the trapeziums the area under the curve could be calculated as follows:- Area = area of first trapezium + area of second trapezium = (g (0)

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21. ## Math Portfolio type 1 infinite surd

Nonetheless, I will discuss the scope and limitations of my general statement. And finally I will explain how I arrived my general statement and the integer expression. II. BODY: The following expression is an example of an infinite surd: Consider this surd as a sequence of term an where: etc. The formula for an+1 in terms of an: Calculate the decimal values of the first ten terms of the sequence: Using MS Excel 2008 Terms Values 1 1.414213562 2 1.553773974 3 1.598053182 4 1.611847754 5 1.616121207 6 1.617442799 7 1.617851291 8 1.617977531 9 1.618016542 10 1.618028597 11 1.618032323 12 1.618033474 13 1.61803383 14 1.61803394 15 1.618033974 16 1.618033984 17 1.618033987 18 1.618033988

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22. ## Math Portfolio- Shay Areas

multiplied by the first base(b1) added to the second base(b2) divided by 2 represented by the formula (Ex 4): Ex. 4 As shown in (Ex 5) & (Ex 6) I will replace those variables for numbers that correspond to the two trapezoids seen in Ex 3. Ex. 5 As seen above I allowed: b2= 3.25 b1= 3 H = .5 Once you substitute these numbers into the equation simple arithmetic will allow one to come to an answer Thus, the area of the first trapezoid would be 1.5625. Ex. 6 Above I allowed: b2= 4 b1= 3.25 H = .5 After substituting the variables for values, the equation should look like: Thus, the Area of second trapezoid would be 1.81.

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23. ## Math Portfolio - SL type 1 - matrix binomials

For instance, the dimension of the following matrix is (2X2) since it has two rows and two columns:. For the addition of the matrices, corresponding elements need to be added. For example, If A and B were two different matrices but with the same dimensions, each elements are added according to its position of the matrix or the order of the elements. To add these two matrices, elements according to its order or position will be added; however, it needs to be remembered that this will only work when both matrices have same dimensions. Example, Scalar (real number)

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24. ## Investigation of the Effect of Different types of Background Music on

The independent variable was the genre of music played during memorization and recollection. The order in which the words were written did not affect their scores. All 26 participants were middle school band students aged 12 to 14, attending the Northwest Jackson Middle School. The results show that the null hypothesis must be retained for the tests conducted with the songs "Bad Mamma Jamma" and "Isn't She Lovely" (rap/hip-hop). Although there were slight decreases in the average number of words retained with these two songs, the decreases were not statistically significant. Introduction Research Question Does the type of background music played during memorization and recollection have a significant effect on students' ability to memorize and recall word lists?

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25. ## Matrix Binomials IA

Combining these terms would result in (X+Y)n-1. (X+Y)n-1: (X+Y)3-1 = (X+Y)2 = (X+Y)2 = X2+Y2 (X+Y)2 = which is the same as Xn-1+Yn-1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Now let A=aX and B=bY where a and b are constants or scalars. A=-1X A2= -1 --1 = A3= -1 -1 = A4= -1 -1 = When multiplied by -1, the matrix values become negative but remain the same. A=-1/2X A2= -1/2 --1/2 = A3= -1/2 -1/2 = A4= -1/2 -1/2 = When multiplied by -1/2, the matrix values become negative.

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