[PDF] 2003 Applied Mathematics Advanced Higher - Maths 777

2003 Mathematics Higher Finalised Marking Instructions

additional pages have been added to the following 2003 Marking Instructions to show how an additional 20 marks could have been allocated to the 2003 examination Notes to the marking scheme for Higher Mathematics 2003 1 Illustrations where additional marks could be added to bring the overall total up to 130 are shown as follows:

2003 Applied Mathematics Advanced Higher - Maths 777

2003 Applied Mathematics Advanced Higher – Section B Finalised Marking Instructions Advanced Higher Applied 2003: Section B Solutions and marks

2003 Mathematics Advanced Higher Finalised - Madras Maths

Advanced Higher 2003: Section B Solutions and marks B1 Let x − 3 4 = y − 2 −1 = z + 1 2 = t then x = 3 + 4t y = 2 − t 2E1 z =−1 + 2t Thus 2(3 + 4t)+(2 − t)−(−1 + 2t)=4 1 9 + 5t = 4 t =−1 so the point is (−1, 3,−3) 1 B2 A2 = 4 − 3I A3 = 4A2 − 3A 1 = 16A − 12I − 3A = 13A − 12I 1 A4 = 13A2 − 12A 1 = 52A

CIE IGCSE Mathematics Paper 1 June 2003

Paper 1 June 2003 Page: 2 Q1(a) It is better to work out 7 1 + 2 9 first on the calculator but leave the answer on the calculator Then do √ ANS = 3 3 Beware of working out (√ 7 1 3) + 2 9 3 Q1(b) Cut off after the second figure from the decimal point (If the third figure is 5 or

CIE IGCSE Mathematics Paper 4 June 2003

4 Answer the whole of this question on a sheet of graph paper (a) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 4 units on the y-axis, draw axes for 04 ≤x ≤ 4 and 08 ≤y ≤ 8

CARIBBEAN EXAMINATIONS COUNCIL

MAY/JUNE 2003 INTRODUCTION This is the fifth year that Mathematics Unit I was examined on open syllabus and the fourth year for Unit II Twelve hundred and thirty-five candidates registered for the Unit I examinations Four hundred and four candidates registered for the Unit II

Playing with Mathematics: Play in Early Childhood as a

play is a special mode of thinking and doing” (McLane, 2003, p 11) In this sense, the process of play is characterised by a non-literal ‘what if’ approach to thinking, where multiple end points or outcomes are possible In other words, play generates situations where there is no one ‘right’ answer McLane (2003, p

Paper Reference(s) 5506/06 Edexcel GCSE

Friday 14 November 2003 −−−− Morning Time: 2 hours Materials required for examination Items included with question papers Ruler graduated in centimetres Formulae sheets and millimetres, protractor, compasses, pen, HB pencil, eraser Tracing paper may be used

The Anti-Anxiety Curriculum: Combating Math Anxiety in the

LaLonde, & Runk, 2003) In general, there is little empirical research about the causes of mathematics anxiety and even less on the effects and efficacy of timed testing as an instructional approach However, we do know that adding time requirements to tasks does increase anxiety, decrease accuracy and create a negative attitude toward the subject

JUNE 2007 CXC MATHEMATICS GENERAL PROFICIENCY (PAPER 2) Section I

Title: 06 CSEC Maths JUNE 2007 Author: Shereen Khan Created Date: 20190318015900Z

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2003 Applied Mathematics

Advanced Higher

Finalised Marking Instructions

2003 Applied Mathematics

Advanced Higher - Section A

Finalised Marking Instructions

Advanced Higher Applied 2003: Section A Solutions and marks A1.

P(Taxi Yellow | Witness states Yellow)

M1=

P(Taxi Yellow Witness states Yellow)

P(Witness states Yellow)

M1= P(Witness states Yellow | Taxi Yellow).P(Taxi Yellow)

P(WY | TY).P(TY) + P(WY | TG).P(TG)

1, 1=

0.8×0.15

0.8×0.15+0.2×0.85

1= 0.12

0.29=0.41

Alternative

Green

Yellow

0.8 0.8 0.2

0.20.68

0.17 0.12

0.030.85

0.15 0.12

0.17+0.12=12

29
[2][1] [1][1] A2. has mean 180 and s.d. 12T 1 has mean 20 and s.d. 5T 2

Total has mean 180 + 20 = 2001T=T

1 +T 2 variance 112 2 +5 2 =169 so s.d.1=13

Assuming that is normally distributed1T

the required value of is 3.091z

T=µ+z=200+3.09×13=240.17

1x=241

A3. (a) Assemble the lists into a single list of pupils numbered 1 to 180.1

Required sample size is 10% of 180 = 18.

Select a number at random from 1 to 10 e.g. 31

Take the 3rd pupil from the list and every 10th pupil thereafter.1 (b) Possible advantages: (just one needed)1 • systematic sampling is easily implemented • helps to provide a good spread

Possible disadvantages: (just one needed)1

• if there is a pattern in the population matched by the sequence an unrepresentative sample will result • in order to number the population, its size has to be known.

Page 2

A4.

1XBin(100, 0.75)

is approximately 1XN and 1N(75, 4.33 2

1P(X70)=P

Z

70.5Š75

4.33

1=P(ZŠ1.04)

1=0.1492

A5. (a)t= r

1Šr

2 nŠ2 -0.655

1Š(Š0.655)

2 59
.1=Š6.66 The critical region for at the 5% level of significance with 59 d.f. 1 will be approx. .1t | t |>1.96 Since Ŧ6.66 lies in the critical region would be rejected1H 0 (b) there is evidence of a linear relationship between and 1xy A6. where is the Poi(1) distribution function.1P(X<10+a)=F(9+a)F(.)

1P(10Ša

We require F(9+a)ŠF(10Ša)0.99

gives a=7F(16)ŠF(3)=0.9626 gives M1a=8F(17)ŠF(2)=0.9830 gives a=9F(18)ŠF(1)=0.9923 Smallest integer is . [Method has to be clear.]1a=9 A7. (a)H 0 D =0 [Must infer differences]1H 1 D >0 t= d D S D n

3.83Š0

5.41 12

1=2.45

The critical region at the 5% signficance level with 11 df is .1t>1.796

Thus the null hypothesis would be rejected

at the 5% level of significance 1 so the data do provide evidence that the training course has been effective.1 (b) A sign test could have been used.1

Page 3

A8. (a) Assume that the journey time is normally distributed (with ).1=3 H 0 :µ=28 [Must be two-tailed]1H 1 :µ28 z= x n

25.125Š28

3 8

1=Š2.71

The critical region is or .1z<Š2.58z>2.58

Since the null hypothesis would be rejected1Š2.71<Š2.58 at the 1% level of significance i.e. there is evidence of a change.1 (b) p-value 1=2×(Š2.71)

1=2(1Š0.9966)=0.0068

The fact that the p-value is less than 0.01 confirms rejection of the null hypothesis at the 1% level of significance1 (c) The fact that 28 does not lie in the 99% confidence interval 1 confirms rejection of the null hypothesis at the 1% level. A9. (a) .p=78/ 100=0.78 A 95% C.I. for the proportion of ischaemic strokes in the population is

0.78±1.96

0.78×0.22

1001

0.78±0.081

or(0.70, 0.86) (b) The interval does not include 0.65 which means that 1 there is evidence of differing proportions.1 (c) Observed Expected

Died Survived Died Survived

Isch. 37 41 78 Isch. 40.56 37.44 78

Haem. 15 7 22 Haem. 11.44 10.56 22

52 48 100 52 48 100

1,1

Survival is independent of the type of stroke.H

0

Survival is dependent of the type of stroke.1H

1 X 2 (OŠE) 2 E x 2 =0.312+0.339+1.108+1.200 .1=2.959

Since 1

25%,1df

=3.841>2·959 is accepted at the 5% level1H 0 i.e. there is no evidence that survival depends on the type of stroke.1

Page 4

A10. (a) Since all the sample means plot within the chart limits there is no evidence of special cause variation.1

1µ=5018.86

Limits are given by 1µ±3

n as on chart].15018.86±3 288.3

5[=5018.86±386.794632.1, 5405.6

(b)P(45005500Š5018.86

288.3

4500Š5018.86

288.3
1 =(1.67)Š(Š1.80)1 =0.91761 (c) Adjust process so that mean becomes 5000.1

Reduce the variability in the process.1

[END OF MARKING INSTRUCTIONS]

Page 5

2003 Applied Mathematics

Advanced Higher - Section B

Finalised Marking Instructions

Advanced Higher Applied 2003: Section B Solutions and marks B1. f(x)=9Š4x, f(x)= -2 (9Š4x) 1/2 f(x)= -4 (9Š4x) 3/2 f(x)= -24 (9Š4x) 5/2

Taylor polynomial is

p(2+h)=1Š2hŠ4h 2 2Š 24h
3 6 =1Š2hŠ2h 2

Š4h

3 .3 Second degree approximation is 2p(2+0·03)=1Ŧ0·06Ŧ0·0018=0·9382 Principal truncation error term is .Ŧ4×0·03 3 =Ŧ0·0001 Hence second order estimate cannot be guaranteed accurate to 4 decimal places.2 B2.

L(x)=(xŠ0·2)(xŠ0·5)

(0·5)(0·3)0·741 (x 2

Ŧ0·7x+0·1)13·06Š(x

2

Ŧ0·5x)18·367+(x

2

Ŧ0·2x)4·490

=Ŧ0·367x 2

Ŧ0·947x+1·3064

B3. The first relation is linear since there is no term in of more than first degree.1a r Relation (i) is a second order relation. Its fixed point is found from , i.e. .2a

2a=3aŠ4a+9a=3

Sequence from (ii) is ; ; ; .2a

0 =1a 1 =1a 2 12 a 3 =Š3/8B4.

Let quadratic through , , be(x

0 ,f 0 )(x 1 ,f 1 )(x 2 ,f 2 .y=A 0 +A 1 (xŠx 0 )+A 2 (xŠx 0 )(xŠx 1

Then ; ; f

0 =A 0 f 1 =A 0 +A 1 hf 2=A 0 +2A 1 h+2A 2 h 2 and so A 1 f 1 Šf 0 h= ?f 0 h; A 2 f 2

Š2f

1 +f 0 2h 2 2 f 0 2h 2 Thus y=f 0 +xŠx 0 hf 0 (xŠx 0 )(xŠx 1 2h 2 2 f 0

Setting givesx=x

0 +ph y=f 0 +pf 0 12 p(pŠ1) 2 f 0 .5 (Can also be done by an operator expansion of .)(1+) pB5.quotesdbs_dbs5.pdfusesText_9