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state the reason. a B(20, 0.6) b B(30, 0.95) c B(40, 0.13) d B(50, 0.06) 212 2 Find the smallest possible value of n for which the following binomial distributions can be well-approximated by a normal distribution. a B(, 0.024) n b B(, 0.15) n c B(, 0.52) n d B(, 0.7) n 3 Describe the binomial distribution that can be... |
pack contains 2000 washers. Find the expectation and variance of the number of rubber washers in a retail pack. c Using a suitable approximation, find the probability that a retail pack contains at most 1620 rubber washers. 10 In a certain town, 63% of homes have an internet connection. a In a random sample of 20 home... |
build a new sports centre and 57% said they approved. Find approximately the probability that more than 75 out of 120 people said they approved, given that at least 60 said they approved. PS 16 A fair coin is tossed 400 times. Given that it shows a head on more than 205 occasions, find an approximate value for the pro... |
= 0.75, find P( Y > 6). 3 In Scotland, in November, on average 80% of days are cloudy. Assume that the weather on any one day is independent of the weather on other days. i Use a normal approximation to find the probability of there being fewer than 25 cloudy days in Scotland in November (30 days). ii Give a reason wh... |
122 8000) P( X > = ii Find the value of m. iii Find the probability that daily sales at this petrol station exceed 8000 litres on fewer than 2 of 6 randomly chosen days. b The random variable Y is normally distributed with mean µ and standard deviation σ. Given that σ = 2 3 µ, find the probability that a random value o... |
45 crates, on average, is underweight. A sample of 630 crates is selected at random. a Find the probability that more than 12 but fewer than 17 crates are underweight. b Given that more than 12 but fewer than 17 crates are underweight, find the probability that more than 14 crates are underweight. 216 PS 14 Once a wee... |
1 3 10 2 3 20 3 1 20 i Show that k 2=. ii Calculate XE( ) and XVar( ). iii Find the probability that two independent observations of X have a sum of less than 6. b The following table shows the probability distribution for the random variable.1 1 0.2 2 0.3 3 0.4 =. If one independent observation of each random variabl... |
over 15.5 knots. [3] [3] [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy ... |
have a driving license. 13 A fair eight-sided die is numbered 2, 2, 3, 3, 3, 4, 5 and 6. The die is rolled up to and including the roll on which the first 2 is obtained. Let X represent the number of times the die is rolled. a Find XE( ). b Show that P( X > 4) = 27 64. [4] [1] [1] Copyright Material - Review Only - No... |
′ ∩ = ) A B ′ ) ] = 0.364, as shown in the Venn diagram opposite., A B ∩ ′ = 0.196 P( P( ) P[( A ∪ B 0.286 and 1 A a Find the value of x and state what it represents. 0.196 x 0.286 b Explain how you know that events A and B are not mutually exclusive. c Show that events A and B are independent. B 0.364 3 Meng buys a p... |
- Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Practice exam-style paper 6 In each of a series of independent trials, a success occurs with a constant probability of 0.9. a The probability that none of the first n trials results in a failure is less than 0.3. Find the... |
, the table gives the value of ( )zΦ for each value of z, where Φ (z) Φ ( ) z = P( Z < z. ) Use Φ − ( z ) = − Φ for negative values of z.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.594... |
8554 0.8577 0.8599 0.8621 222 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 1.5 1.6 1.7 1.8 1.9 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.... |
2.9 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 0.9953 0.9955 0.9956 0.9957 0.9959 ... |
15 a 70 2 b Boundaries at 4, 12, 24, 28 grams. Densities ∝ 28, 33, 17.5. c 310 We can be certain only that 0.1 0.4. 0.8 0.4<<a and that b< < 159, b = 636 b 23.5kg 10 11 = a a 4 hd 5 n 12 33cm 13 p = q29, = 94 Exercise 1C 1 a Points plotted at (1.5, 0), (3, 3), (4.5, 8), (6.5, 32), (8.5, 54), (11, 62), (13, 66). b i 23... |
a = 32, b = 45, c = 15, d = 33 a 65 b 24 a Ratio of under 155cm to over 155cm is 3:1 for boys and 1:3 for girls. b 81 or 82 c There are equal numbers of boys and girls below and above this height. d Polygon or curve through (140, 0), (155, 25), (175, 50). 9 a Points plotted at (18, 0), (20, 27), (22, 78), (25, 89), (2... |
ogram: Frequency density may be mistaken for frequency. Pie chart: does not show numbers of trees. b Pictogram: short, medium, tall; two, three and four symbols, each for six trees, plus a key. Shows 12, 18, 24 and a total of 54 trees. 6 a 1. Score (%) Frequency 30–39 40–49 50–59 60–69 70–79 80–89 90–99 3 5 6 15 5 4 2 ... |
continuous 4 a 6 b Five additional rows for classes 0–4, 20–24, 25–29, 30–34, 35–39. 5 6 a = b9, = 2 a 48 b 0.7 cm 7 a 120, 180 and 90 b 6.75cm c There is a class between them (not continuous). 8 a 30 days for region A, 31 days for region B. 225 b Bindu: unlikely to be true but we cannot tell, as the amount of sunshin... |
Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Answers 6 7 8 x = 12.4; b = –3 a = b4, = 5 a (–1.8, 2.8) b (26, –6) c TE(5.2, –1.2) → (19, –2) ET(5.2, –1.2) → (–9, 14) Location is dependent ... |
(They could consider using the greatest possible mean, 164.41.) ii Median safe current = median unsafe current 11 a First-half median is in 1–2; second-half median is in 4 – 5. b i 3 ii First half data are positively skewed (least possible mean is 100.8 s). 12 a Points plotted at (0, 0), (26, 15), (36, 35), (50, 60), ... |
, 14. ii 2.1h 11 16.4 12 81 13 14 15 a b Mode = 0, mean 1, median 0 = = Mean; others might suggest that none of the items are damaged. a 4006 − 2980 = $1026 b $3664 1.95 3 Measures of variation Prerequisite knowledge 1 16cm 2 a 4.5 b 27.3 Exercise 3A Box plots given by: smallest … … … … Item (units), as appropriate. Q2... |
0.095 Ω Ω 4.0... 25.8... 33.2... 38.8... 56.0 / Area (cm2). d Area < 6.3cm2 or area 58.3cm2. > Estimate ≈ 8 (any from 0 to 15) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge ... |
and (4, 2) iii (2, 2), (4, 4), (6, 6) b X Y, and Z are not mutually exclusive, b = 2, c = 6 3 5 ii 13 25 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Rev... |
108 b 0.48 ii 0.4 P( C ) ii 0.5 iii 0.06 D S 13 6 7 2 S ′S Totals 6 7 13 13 2 15 19 9 28 D ′D Totals Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review ... |
c 0.14 4 s =P( S s ) 5 a Proof b r 0 4 81 1 28 81 237 or 0.25 = 1 5 3 0.2 2 49 81 0 1 2 3 × 40 100 ) P( S × 75 100 ). P = P( and ) P( S b Yes; e.g. 30 100 P = a 312 or 531 441 b 310 or 59 049 c 37 or 2187 14 15 a 3 628 800 b 7 257 600 to show =P( R r ) 0.226 0.446 0.275 0.0527 c 0.774 6 a Proof b v =P( V v ) c 69 91 0... |
) = 0.896 17 a s =P( S s ) b 1 3 1 17 36 2 9 36 3 10 36 Exercise 6B E( 1 X ) = =p E( Y a b 0.2 2.1; Var( X ) = 0.93 2 3 4 5 6 7 ) 1.84; SD( = Y ) = 0.946 E( T ) = 5, Var( T ) 11.5 = m = 16; Var( V ) = 31.3956 831 ) =RVar( a 11; Var( = W ) = 79.8 a E(grade) profit. = 3.54; SD(grade) 1.20; A smallish = =SD 1.20; variabi... |
1 256 1 12 256 2 54 256 3 108 256 4 81 256 ; the probability of not obtaining B with each spin. End-of-chapter review exercise 6 1 2 X 9 14 ; E( = k ) 1 45 =XVar( 196 ) = 2 or 2.36 5 14 or 1.23 13 or =q 48 a =q b 34 3 a $6675 4 5 6 7 b 8 9 b 4.27 6 11 0.909 2 5 3 5 a i ii b j =P( J j ) a =S 1 0 0.3 1 2 0.6 0.1 ... |
Probability & Statistics 1 4 a 0.121 b 0.000933 c 0.588 10 a 6.006 d 0.403 a 0.246 0.0146 0.254 a 0.140 0.177 a 0.599 0.349 a 0.291 a 0.330 a 0.15625 or 5 6 7 8 9 10 11 12 13 14 15 9 17 16 e 0.499 b 0.296 b 0.000684 b 0.257 b 0.648 b 0.878 16 6 18 23 5 32 b 0.578 19 a 0.0098 b a = 208, b = 3 c 68 20 a =p 0.5; the prob... |
91808 b 0.901 b 0.875 b 0.482 b 0.922 ii 0.0016 ii 0.484 c 0.4096 c 0.0672 c 0.0280 b 0.2016 iii 0.440 b Faults occur independently and at random. a 0.21 a 0.364 0.0433 b 0.21 b 0.547 c 0.21 a Not suitable; trials not identical (p not constant). b Not suitable; success dependent on previous two letters typed or X canno... |
0.40951 c 0.242 6 a 10 7 8 b 0.00772 or c 0.00162 a 2–12 b a 137 × 2–16 i 0.0706 ii 0.0494 iii 0.1176 b The students wear earphones independently and at random. 9 i 0.993 ii =n 22 8 The normal distribution Prerequisite knowledge 1 23.4 and 11.232 2 n = 32, p = 0.35 Exercise 8A a False 1 d False 2 a i σ σ> P Q b True e... |
k =k =k 1.035 0.003 =c =c 1.245 – 2.14 =c =c =c –1.90 1.96 Exercise 8C a 0.726 1 b 0.191 c 0.629 2 a 0.919 and 0.0808 b 0.613 and 0.387 c 0.964 and 0.0359 d 0.0467 and 0.953 f 0.954 h 0.319 j b d b d 0.0994 =b 15.5 = 23.6 d =g =j 42.7 17.5 e 0.285 and 0.715 g 0.423 0.231 i a c 35.0 18.5 =a =c e e a = 86.8 11.4 =f =h 0.... |
13 14 15 16 17 b 23 b =n 1000 b 0.0456 µ = 6.39 Exercise 8E a Yes; µ 1 b No; nq c Yes; µ d No; np = 2 12, σ= = <1.5 5 2 5.2, = σ = <3 5 = 4.524 2 3 4 5 6 a 209 =n =n 11 B(56, 0.25) c b d 34 =n =n 17 0.837 0.844 a p = 0.625; Var( H ) = 37.5 b 0.0432 7 a Proof b 0.292; np = 10 5 and > nq = 30 5 > b 4.45 ii 0.0118 8 9 a ... |
Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Practice exam-style paper 1 a 1.72 m b 16.75 2 a b =x 0.154; the value of A )∩ or A B ) ∩ ≠ 0 or equivalent. A B P( P( P( and ) B c Proof 3 a 36 b 5 9 a 0.35 or 0.693 b 95 137 a 11km 4 5 ... |
permutations Average: any of the measures of central tendency, including the mean, median and mode B Binomial distribution: a discrete probability distribution of the possible number of successful outcomes in a finite number of independent trials, where the probability of success in each trial is the same C Categorica... |
each other 245 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge Un... |
cumulative frequency graph 43–4 modal class 28–9 mode 27, 44 addition law 94–5 conditional 108–9, 111–15 dependent events 112–15 experiments, events and outcomes 91–3 of the geometric distribution 180 multiplication law for independent independent events 100–2, 111–12 multiplication law for independent events 100–2 ap... |
distribution 193 variation 55 see also measures of variation Venn diagrams 95–7 Wiles, Andrew 156 box-and-whisker diagrams 60 measures of central tendency 45–6 Z (standard normal variable) 195–9 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy... |
R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy R evie w C opy - C a m bridge U niversity Press - R evie w C opy Copyright Material - Review Only - Not for R... |
� (pa + pb)(pa pb) = a b ¡ ¡ (1.60) which is rational (since a and b are rational). If we have a fraction which has a denominator which looks like pa+pb, then pb achieving a rational we can simply multiply both top and bottom by pa denominator. ¡ or similarly c pa + pb c ¡ pb pa = = = = c pa + pb ¡ ¡ pa pa cpa a pb pb ... |
: = 7;8571 : : : = 7;4285 : : : = 7;4038 : : : ) x0 = (1.63) (1.64) (1.65) (1.66) ) x0 = 7;4285 : : : + 7;4038 : : : 2 = 7;4162 : : : (1.67) Using a calculator we flnd that p55 = 7;416198 : : :, which is very close to our approximation. 1.5 Accuracy The syllabus requires: write irrational (and rational) solutions round... |
has 5 signiflcant flgures. Estimating a number works by removing signiflcant flgures from your number (starting from the right) until you have the desired number of signiflcant flgures, rounding as you go. For example 6;827 has 4 signiflcant flgures, but if you wish to write it to 3 signiflcant flgures it would mean removing th... |
rule of addition for rational numbers (1.25) a b + c d = ad + bc bd 20 Step 2 : Fill in the values Fill in the values for a;b;c and d. Here you can read ofi that a = 7, b = 8, c = 5 and £ = 54 16 £ Step 3 : Minimise the denominator 54 16 is the correct answer, but it is not the simplest way to write it. We can see that... |
The syllabus requires: investigate number patterns, be able to conjecture a pattern and prove those conjectures recognise a linear pattern when there is a constant difference between consecutive terms recognise a quadratic pattern when there is a constant 2nd difference identify ‘‘not real’’ numbers and how they occur... |
ises a sequence is that it is a list which is ordered. In the alphabetised books example, someone who didn’t know the alphabet would not be able to work out how you had ordered the books. How would you be able to flnd your seats at the theatre or at a stadium if the seats were not ordered? Likewise if you are shown a se... |
help you study. Naturally you move another table and add it 23 Figure 2.1: Tables moved together to the existing one. Now six of you sit at the table. Another two of your friends join your table and you take a third table and add it to the existing tables. Now 8 of you can sit comfortably. Let assume this pattern cont... |
condition does not hold, the sequence is not an arithmetic sequence. It is also important to note the difierence between n and an. n can be compared to a place holder while an is the value at the place ’held’ by n. Like our study table above.Table 1 holds 4 people thus at place n=1 the value of a1 = 4. a1 = a3¡ n an 1 ... |
own quadratic sequences with a constant second ¡ ¡ 25 Figure 2.2: Tree diagram of series 2.1.3 Geometric Sequences Deflnition: A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by another constant number. This means that the ratio betwee... |
a1 = 2). The next day, each one then infects 2 of their friends. Now 4 people are infected. Each of them infects 2 people the third day and 8 people are infected etc. These events can be written as a geometric sequence: 2;4;8;16;32;::: Note the common factor between the events. Recall from the linear arithmetic sequenc... |
¡ for an in terms of an 1, which is called a recursive equation. So the recursive equation for a linear sequence of constant difierence d is an ¡ ¡ an ¡ an 1 = d ¡ (2.12) 27 We can do the same thing for quadratic sequences. There we noticed that 1 = Dn + d. Then the recursive equation for a quadratic sequence with an ¡... |
a(rn NOTE: equation 3 is not correct in the official syllabus. is a d missing.) there 28 When we sum terms in a sequence, we get what is called a series. If we only sum a flnite amount of terms, we get a flnite series. We use the symbol Sn to mean the sum of the flrst n terms of a sequence. For example, the sequence of n... |
1 to any integer n, we would write n Xi= times (2.19) 29 Since all the terms are equal to one, it means that if we sum to an integer n we will be adding n number of ones together, which is equal to n. n Xi=1 1 = n (2.20) Another simple arithmetic sequence is when a1 = 1 and d = 1, which is the sequence of positive int... |
shocked the teacher with the correct answer, 5050. 30 Writing out the sum of a sequence and then substituting in the general form for an arithmetic sequence gives us n n ai = Xi=1 Xi=1 d(i ¡ 1) + a1 (2.26) If there is a sum inside a sum, we can break it into two separate sums and calculate each part separately. n Xi=1... |
i ¡ 1) + 3 = 20 2 (2 £ 3 + 7 19) £ = 1390 (2.33) In this example, it is clear that using (2.31) is beneflcial. 31 2.2.2 Finite Squared Series When we sum a flnite number of terms in a quadratic sequence, we get a flnite quadratic series. The general form of a quadratic series is quite complicated, so we will only look at ... |
� Sn = a1 + a1rn 1) = a1(1 + rn) (2.39) dividing by (r sequence since Sn = ¡ n i=1 a:ri ¡ 1 1) on both sides, we have the general form of a geometric 1) (2.40) a(rn r ¡ 1 ¡ P n Xi=1 a:ri ¡ 1 = 32 2.2.4 Inflnite Series Thus far we have been working only with flnite sums, meaning that whenever we determined the sum of a se... |
1 = 1 Xi=1 a1:ri ¡ 1 = a1 ¡ 1 r 1 < r < 1 ¡ (2.41) where a1 is the flrst term of the series, and r is the common ratio. (NOTE: the syllabus requires us to PROVE this series! how can we do that without a notion of a limit? again the syllabus talks nonsense.) We can see how this comes about by looking at (2.40), as. We c... |
say that 5;10;20; : : : is a geometric 3. Determine d and a9 for the following arithmetic sequence: 17;14;11; : : : It is given that 17;14;11; : : : is an arithmetic sequence, thus a2 ¡ 14 ¡ d = a2 = d 14 = 3 ¡ a1 = a3 ¡ 17 = 11 ¡ 3 ¡ To determine a9 we use an = a1 + d(n Thus: ¡ 1) with n = 9 1) ¡ 3)(9 an = a1 + d(n a... |
: a4 = 3 Thus we can use an = a1 + d(n 2, a8 = 1 2 and this is an arithmetic sequence. 1) ¡ a4 = 3 a8 = 1 2 = a1 + d(4 2 = a1 + d(8 ¡ ¡ 1) = a1 + 3d and 1) = a1 + 7d Subtract the one equation from the other to get rid of a1 and solve for da1 + 3d) 4d 1 4 (a1 + 7d) ¡ ¡ ¡ 1 4 )(7) 1 2 = a1 + ( ¡ 7 1 4 ) 2 = a1 ¡ a1 = 9 4... |
c (NOTE: page 25 of the syllabus lists some more situations, but they are corrupted. the syllabus and add them to this list) we need to get an uncorrupted version of identify the domain and range, axes intercepts, turning points (max/min), asymptotes, shape and symmetry, periodicity and amplitude, rates of change, inc... |
people can have the same height. 3.1.1 Variables, Constants and Relations A variable is a label which we allow to change and become any element of some set of numbers. For example, on a menu in a restaurant \price" is a variable on the set of real numbers, since for any menu item the manager can choose any price he or... |
ulae, we most often come across relations in words, tables and graphs. Instead of writing y = 5x, we could also say \y is always flve times as big as x". We could also give the following table: (NOTE: Working on a Latex-less machine, so table will come later) (I put in a table but not sure if it’s ok - Jothi) x 2 6 8 13... |
3.1. We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once. Therefore this is not a function. 40 Figure 3.1: Graph of y2 + x2 = 4 In a function y = f (x), y is called the dependent variable, because the value of y depends on what you choose as x. We s... |
�is less than’ and to’. Instead of saying that x is between 6 and 10, we often write 6 < x < 10. This directly means ’six is less than x which in turn is less than ten’. ‚ ‚ • ; 41 A set of certain x values has the following form: x : conditions, more conditions f g (3.1) We read this notation as \the set of all x valu... |
1 < x of all the possible x values is 3 or 6 • • [ (1;3] [ [6;10) (3.3) sign means the union(or combination) of the two intervals. We use where the the set and interval notation and the symbols described because it is easier than having to write everything out in words. [ 3.1.4 Example Functions In this section we wil... |
a very abstract way. Let’s say that we were investigating the function f (x) = 2x. We could then consider all the points (x; y) such that y = f (x), i.e. y = 2x. For example, (1; 2); (2:5; 5); and (3; 6) would all be such points, whereas (3; 5) 3. If we put a dot at each of those points, and then at would not since 5 ... |
occurs when y = 0, so it is always equal to b. So if asked to plot a straight line, there is no need to calculate lots of y values, you just need to flnd the x and y intercepts and draw a line through them. ¡ Parabolic Functions A parabola looks like a hill, either upside down (for a \positive" parabola) or right way u... |
minimum value of the function(that is the maximum or minimum value of y). At flrst it might seem di–cult to sketch the graph of a parabola but once a simple procedure is followed, then it becomes easier. When sketching the graph, we need to use some information about it. The only information we have are its shape, x an... |
-1 -8 0 -9 1 -8 2 -5 3 0 4 7 Hyperbolic Functions Hyperbolas look like 2 parabolas on their side which are mirror reections of each other around the diagonal (NOTE: sketch). Hyperbolic functions look like f (x) = a x + b (3.8) where a and b are constants. Just like for parabolas, a tells us how steep the curves are an... |
the algebraic inverse of exponents. When we say \inverse function" we mean that the answer becomes the question and the question becomes the answer. For example, in the expression ab = x the \question" is \what is a raised to the b power." The answer is \x." The inverse function would be logax = b or \by what power mu... |
log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base. To afiect a change of base, apply the change of base formula: logax = logbx logba (3.10) where b is any base you flnd convenient. Normally a and b are known, therefore logba i... |
of two straight lines with slopes turning point (b;c). + c = c. Therefore function a and a, which meet at the ¡ The function has the axis of symmetry x = b. In other words, the part of the function on one side of the vertical line x = b is the same as the reection about this line of the part of the function on the oth... |
not necessarily have xIt these do exist, then they will be the x-intercepts of the two intercepts. straight lines making up the absolute value function. 52 Chapter 4 Numerics 4.1 Optimisation The syllabus requires: Linear Programming (Grade 11) † 1. Solve linear programming problems by optimising a function in two var... |
these constraints translate into the inequalities x y and x 0 and y 0 and y ‚ ‚ ‚ ‚ • 53 mean that we can’t just take any x and y when looking for the x and y that optimise our objective function. If we think of the variables x and y as a point (x;y) in the xy plane then we call the set of all points in the xy plane t... |
is a maximum/minimum | in this case we have more than one solution. 54 6 6 4.1.1 Linear Programming The objective function is called linear if it looks like f (x;y) = ax + by where the coe–cients a and b are real numbers. For example, f (x;y) = 10x y is a linear objective function. If the objective function and all of... |
level line of the objective function (NOTE: Should I use this terminology?). Consider again the feasible region described in Figure 2y with this 4.1. Lets say that we have the objective function f (x;y) = x feasible region. 20 ¡ then we we get the level line y = 1 2 x + 10 which has been drawn in Figure 4.2. 2 + 5), f... |
� ¡ • ‚ Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. (NOTE: We could have inflnitely many solutions if the gradient of a constraint = the gradient of the level lines... should I mention this?). The fact that the value of our objective function along the line of... |
mention f to determine ‘the direction in which to move the ruler’. Even if I neglect the fact that students certainly know nothing about partial difierentiation, I’m still not sure whether I can mention and work with vectors in the plane...) r These points are su–cient to determine a method for solving any linear progr... |
20 flelds of available land on which he can plant crops. He must grow at least 5 flelds of maize. Also he cannot grow more than twice as much maize as wheat. Draw a graph to show the feasible region showing the possible number of flelds of wheat and maize the farmer can plant. What is the maximum number of flelds of wheat... |
. We are usually told to maximise or minimise this function. Worked Example 2: Q: Consider the same situation as in worked example 1. The farmer can make a proflt of R100 on every fleld of wheat and R200 on every fleld of maize that he grows. How many flelds of wheat and maize must the farmer plant to maximise his proflt an... |
Q: A delivery company delivers wood to client A and bricks to client B. The company has a total of 5 trucks. A truck cannot travel more than 8 hours per day and it takes 4 hours make the trip to and back from client A and 2 hours for 59 20 15 10 5 A B 5 10 15 20 25 30 Figure 4.4: Graph of TODO client B. To honour an a... |
by the constraints is the shaded region, but since the variables x and y can only take on positive integer values, the feasible region actually consists of the collection of dots showing the integer values in the shaded region. Step 5: The objective function is the proflt (in Rands), which is P = 100x + 150y (4.21) Ste... |
(since St. Owen’s closed permanently at the end of my Grade 11 year). It was always one of my ambitions to attend the University of Cape Town (UCT) because it is a prestigious university. I applied to study medicine at UCT, but was not accepted, and so I enrolled for a science degree at UCT and have never regretted it... |
there are ten million adults paying R100 every month to a pension fund. That means that each month the pension fund receives R100x10 000 000 = R1b (i.e. one billion rand) in total each month. Now, each adult, like James, will want to receive a monthly pension which is greater than R100 when they retire. So, the pensio... |
studied previously. Calculus is more dynamic and less static. It is concerned with change and motion. It deals with quantities that approach other quantities. For that reason it may be useful to have an overview of the subject before beginning its In this section we give a glimpse of some of the main ideas intensive s... |
P. We write m = lim P! Q mP Q and we say that m is the limit of mP Q as Q approaches P along the curve. Since x approaches a as Q approaches P, we could also use Equation (5.1) to write ¡ ¡ The tangent problem has given rise to the branch of calculus called difierential m = lim a! (5.2) x f (x) x f (a) a calculus. 5.1.... |
, then some common alternative notations for the derivative are as follows: f 0(x) = y0 = dy dx = df dx = d dx f (x) = Df (x) = Dxf (x) The symbols D and d=dx are called difierential operators because they indicate the operation of difierentiation, which is the process of calculating a derivative. It is very important th... |
¡ 1. ¡ Rule 3: Linearity of Difierentiation If c is a constant and both f and g are difierentiable, then d dx (cf ) = c df dx df dx + d dx (f + g) = 5.2.1 Summary d dx c = 0 d dx (xn) = nxn ¡ 1 d dx (cf ) = c df dx dg dx (5.7) (5.8) d dx (f + g) = df dx + dg dx 5.3 Using Difierentiation with Graphs The syllabus requires: ... |
-intercepts by factorising ax3 + bx2 + cx + d = 0 and solving for x. First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary. 4. Find the turning points of the function by working out the deriva... |
�ll in f (x) f 0(x) = lim! h 0 • = lim h! 0 • f (x + h) h ¡ f (x) ‚ 4(x + h) + 2(x + h)2 h (4x + 2x2) ¡ ‚ Step 2 : Multiply out and simplify 4h + 4xh + 2h2 h ‚ [4 + 4x + 2h] h 0 • f 0(x) = lim! = lim 0 h! = 4 + 4x Step 3 : Substitute the value of x into f 0(x) Since f 0(x) = 4 + 4x then f 0(2) = 4 + 4(2) = 12 70 Worked... |
x5 + 4x3 d) 71 Step 1 : Write out the First Linearity Rule, equation (5.7) dx (cf ) = c df d dx Step 2 : Write out the Power Rule, equation (5.5) d dr (rn) = nrn ¡ 1 Step 3 : In this case n = 3 and c = 5 so... d 1 dr (5r3) = 5 3r3 ¡ £ Step 4 : Simplify d dr (5r3) = 15r2 e) Step 1 : Write out the Power Rule, equation (5... |
2; 4) (-3; 0) (0; 0) 73 5.5 Exercises 1. Draw the graph of y = x2 + x 6 for to this curve at x = 3, x = 1 and x = gradient of the curve at each of these points. ¡ x 6. Draw the tangents 5 ¡ 2, and hence flnd a value for the ¡ • • 2. Draw the graph of y = x2 4x ¡ 4 x for 0 hence flnd a value for the gradient of the curve ... |
a) y = x2 + 7x 7x2 (b) y = x (c) y = x3 + 7x2 (d) y = 3x2 + 7x ¡ (e) y = (x + 3)(x 1) (f) y = (2x + 3)(x + 2. Find the gradient of the following lines at the points indicated: ¡ (a) y = x2 + 4x at (0;0) x2 at (1;4) (b) y = 5x (c) y = 3x3 (d) y = 5x + x3 at ( (e) y = 3x + 1 (f) y = 2x2 1; ¡ x at (1;4) x + 4 ¡ 2x at (2;2... |
the regular polygons) can tell when polygons are similar. similar equilateral triangles are the line drawn parallel to one side of a triangle divides the other 2 sides proportionally A polygon is a shape or flgure with many straight sides. A polygon has interior angles. These are the angles that are inside the polygon.... |
The sum of the angles of a triangle is 180–. The exterior angle of any corner of a triangle is equal to the sum of the two opposite interior angles (NOTE: a diagram for these 2 rules). We have the following triangles: Equilateral All 3 sides are equal and each angle is 60–. (NOTE: need an example diagram) Isosceles Tw... |
other in half). There are not always lines of symmetry. (NOTE: what is a line of symmetry? we haven’t mentioned them before here. is this really a property?) (NOTE: need an example diagram) Rectangle This is a parallelogram with 90– angles. † Both pairs of opposite sides are parallel. Both pairs of opposite sides are ... |
Area of trapezium: 1 2 £ £ Area of parallelogram and rhombus: base Area of rectangle: length Area of square: length of side (NOTE: everything from here on in Extra is probably acceptable syllabus material, but it is here for now so i can see what needs to be brought back in to the main text. the theorems are not on th... |
equal in length. common sides (RHS) So 4 corresp sides in congruent triangles Hence diagonals are equal in (flgure 6 here) Proof: opposite sides equal 2. DC = DC all angles equal So BCD: 1. AD = BC 4 are equal 3. \D = \C AC = BD length. BCD ADC · 4 In 4 Rhombuses AOB and AOD: 1. AB = AD To prove that a quadrilateral is... |
good Similarly, the function y = f (x + a), is the result of shifting the function by a to the left (the function value at x + a is moved to x). f (x) f (x a) ¡ a f (x + a) f (x) a Figure 6.1: Graph of a function with the translations x x Note that this is just a simple shift either left or right of the entire graph. ... |
we can look at the function y = f (x). This is the same as ¡ saying y = f (x), which reects the function about the x-axis (every positive function value is changed to the corresponding negative function value and vice versa). ¡ f (x) f ( x) ¡ f (x) f (x) ¡ Figure 6.3: Graph of a function with the reections x! f (x). N... |
just a shrinking and stretching in the horizontal axis. ax and x!! x a. Note (NOTE: this is correct. vertical and horizontal rescalings are difierent... i think the author got confused and thought the same thing happened in each. this needs flxed.) Replacing y by by, where b is a positive constant and b > 1, gives by = ... |
the size of the angle subtended by the same arc at the circle (grade 12) the opposite angle of a cyclic quadrilateral are supplementary the tangent chord theorem (NOTE: really... word for word) thats what it says, 6.8.1 Circles & Semi-circles Circles: A circle centered at the origin with radius r is described by the r... |
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