pakuranga college€¦ · 12 2014 end of year l2 mat exam calculus you are advised to spend 60...

Name:__________________________________________ Teacher:___________________

PAKURANGA COLLEGE

12MAT Mathematics and Statistics Practice Exams, 2014

AS 91261 Apply algebraic methods in solving problems (4 Credits)

AS 91262 Apply calculus methods in solving problems (5 credits)

AS 91267 Apply probability methods in solving problems (4 credits)

Time allowed: 3 hours

You should answer ALL parts of ALL the questions in this booklet

You should show ALL your working for ALL questions.

The questions in this booklet are NOT in order of difficulty.

If you need more space for any answer, use the page provided at the back and clearly number the

question.

YOU MUST HAND THIS WORKBOOK TO THE SUPERVISOR AT THE END OF THE

EXAMINATION.

Achievement Achievement with Merit Achievement with Excellence Score Grade

Apply algebraic

methods in solving problems.

Apply algebraic methods,

using relational thinking, in solving problems.

Apply algebraic methods, using extended abstract thinking,

in solving problems.

Apply calculus


Apply calculus methods,


Apply calculus methods, using extended abstract thinking,


Apply probability


Apply probability methods,


Apply probability methods, using extended abstract thinking,


2

2 2014 End of year L2 MAT exam

AS 91261 Apply algebraic methods in solving problems Page 3

AS 91262 Apply calculus methods in solving problems Page 12

AS 91267 Apply probability methods in solving problems Page 19


Algebra You are advised to spend 60 minutes answering the questions in this booklet.

QUESTION ONE (a) (i) Factorise 5x2 – 6x – 8

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(ii) Solve 5x2 – 6x – 8 = 0

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(b) Write as a single fraction in its simplest form:

1

5

m

4

3

m

m

______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

(c) Solve 9

841642

2

x

xx = 5

______________________________________________________________________ ______________________________________________________________________

______________________________________________________________________ ______________________________________________________________________

______________________________________________________________________

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(d) (i) Find the value of c which satisfies this equation:

cxx

xx

283514

202

2

=

)32(7

5

x

x

______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

(ii) What does the value of c found in (i) above tell us about the nature of the roots of 14x2 – 35x +28c = 0 ?

______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

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(e) An elliptical running track has internal lengths of 3x and 2x as in the diagram. The width of the track is 5 metres. If the area inside the track is the same as the area of the track itself, find the inside dimensions of the track.

Note: The area of an ellipse = πab where a and b are the internal lengths.

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

2x

3x

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QUESTION TWO (a) Solve for x :

(i) logx125 = 3

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(ii) x = log264

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(b) The Department of Conservation begin a control programme on the number

of stoats on Karera Island. The number of stoats can be modelled by

S = 300 x (0.95)t

where S is the number of stoats and t is the time in months since the controls began. (i) After how long will the stoat population first drop below 200 ? _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________

_______________________________________________________________

_______________________________________________________________

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(ii) With less stoats the kiwi population on the island grows faster. It can be modelled by

K = 100 x (1.015)t

where K is the number of kiwis and t is the number of months since the stoat control programme was started. After how many months would the numbers in the kiwi and stoat populations be the same if the programme continued ?

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(iii) After three years the stoat control programme stopped. The rate of growth of the stoat population increased to 4% per month and the kiwi population began to decrease at a rate of 2% per month. How long after the programme was stopped will the numbers in the populations of kiwi and stoats be equal again ?

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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(c) The function is undefined for values of x less than 0. Find two solutions for m in the equation

( )

and explain why the negative solution is acceptable. _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

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QUESTION THREE (a) Simplify

(i)

2

3

4

m

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(ii) 3/26064.0 m

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

(b) Solve x

x

9

35 1 = 135

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

Assessor’s

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(c) Jonny is throwing a ball over the fence to his friend Ngaire.

Both children are 1.1 metres tall. Jonny is 2.5 metres away from the fence and Ngaire is 1.5 metres away on the other side. The ball reaches a peak on Jonny’s side of the fence at 4 metres, just beneath a tree.

(i) Form an equation to represent the path of the ball.

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________

(ii) Assuming that the ball just skims over the top of the fence,

what is the height of the fence?

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

Fence

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(d) Find the possible value(s) for k if the quadratic 3x2 – 2kx + 4k = 0

has two real roots.

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

_______________________________________________________________ _______________________________________________________________ _______________________________________________________________

Assessor’s

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Calculus

You are advised to spend 60 minutes answering the questions in this booklet.

QUESTION ONE:

(a) Find the gradient at the point where x = 1 on the curve 2736.0 25 xxxy .

(b) (i) Give the coordinates of the points on the curve 134 23 xxxy where the

gradient is 6.

(ii) Find the equation of the tangent to the curve 134 23 xxxy at the point (2, -1).

Assessor’s

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(b) (iii) The graph below shows the function )(xfy .

On the axes below, sketch the graph of the gradient function.

Assessor’s

use only

x

x


(c) The outline of an Easter Egg can be modelled by two functions:

the top curve is xxxxxf 43125.0)( 234 and the bottom curve is xxxg 4)( 2

x-4 -2 2 4

y

-4

-2

2

4

(i) The point (1 , 1.875) lies on the outline of the Easter Egg.

Find the gradient of the outline of the Easter Egg at the point (1 , 1.875) and give the other value

of x where the outline has the same gradient.

(ii) Explain why the outline is not smooth where the two curves meet. Justify your answer

using calculus concepts.

g(x)

f(x)

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QUESTION TWO

(a) Is the function 675)( 3 xxxf increasing, decreasing or stationary when x = 1. Use

calculus to justify your answer.

(b) The derivative of a function is 228)(' 3 xxxf .

Find f(x), if the graph of f(x) passes through the point (1, 0).

(c) Give the coordinates of the maximum turning point of the function

1185.1)( 23 xxxxf

and explain how this can be proved to be a maximum by investigating the second derivative.

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(d) Melting point of chocolate is about 45degrees Celsius; once it goes over 48 degrees the

chocolate burns and becomes lumpy.

The temperature T of chocolate when heated is given by the function

T = 4

3 t

2 + 20

where t is the time in minutes from when the chocolate is placed in the heat.

(i) Find the rate that the temperature of the chocolate is changing after 3 minutes.

(ii) Explain what your answer means in terms of heating or cooling of the chocolate

(iii) If the chocolate is cooled then heated again it is called “tempered chocolate”.

At the Cadberry factory, this temperature is controlled by the function

T = -0.01x4 + 0.4x

3– 5x

2+ 20.28x + 18

where T is the temperature of the chocolate and x is the time in minutes that the

chocolate has been heated.

Show that there are two peaks in temperature, first at 3 minutes and the next at around

17.14 minutes.

You must use calculus methods to show that these times give maximum values.

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QUESTION THREE

(a) (i) Find the equation for the curve, if the gradient function is given by dydx

= 12x2

+ 4x and

the local minimum point is (0,0).

(ii) Give the coordinates of any other point with the same gradient as the point (0,0) and

describe how this point differs from (0,0)

(b) (i) Find the gradient of the tangent to the curve y = 2x2

– 6x + 1 where the curve cuts the

y axis

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(ii) Find the minimum value of the curve y = 2x2

– 6x + 1

(c) A large helium balloon is released from the top of a building above the ground. The balloon’s

initial velocity is measured at 10ms-1

and rises at a constant acceleration of 2ms-2

for the first

minute after being released.

If the balloon is 620m above the ground after 20seconds, how high is the top of the building?

Assessor’s

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Probability

You are advised to spend 60 minutes answering the questions in this booklet.

At Hoopsdunk High School basketball is a major sport with both boys and girls teams.

Question 1

(a ) ( i) The height of male basketball players at Hoopdunk High School is assumed to be normally

distributed with a mean of 185cm and a standard deviation of 8cm.

Calculate the probability that a player selected at random is between 185cm and 195cm tall.

(ii) Calculate the percentage of male basketball players at the school with heights between

175cm and 195cm.

(b) The height of female basketball players at the school is normally distributed with a mean of

175cm and a standard deviation of 5cm. There are 215 female basketball players at the

school. Calculate the expected number of female basketball players with heights less than

169cm or more than 181cm.

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(c) (i) Full size basketballs are used for the Premier Boys team. They are manufactured to

have a mean circumference of 765mm when correctly inflated. The manufacturer

will reject the smallest 5% (less than 755mm) and the largest 5% (more than

775mm).

Calculate the standard deviation of the balls assuming the circumference of

basketballs is normally distributed.

(ii) At the end of the season all basketballs from all teams, boys and girls, are weighed and

graphed as shown in the histogram.

Discuss the assumption that the weights of basketballs are normally distributed. In your

answer you should refer to key features such as the centre, shape and spread in relation to

the context.

0

10

20

30

40

50

60

Weight of basketball (in grams)

Freq

uen

cy

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Question 2

Carl and Tua, the Premier Boys team’s best shooters have a shoot off against each other. This

involves each player taking shots from the same distance. The highest score wins the game.

The probability of Carl winning the first game is 0.4.

If Carl wins a game then the probability of him winning the following game is 0.7.

If Tua wins a game then the probability of him winning the following game is 0.8 .

The boys decide to play three games against each other.

(a) (i) Calculate the probability that Tua will win all three games.

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

(ii) Calculate the probability that Carl will win at least two of the three games.

1st game

Carl

wins

Tua

wins

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(iii) If the boys play 40 matches of three games each over a season, find the expected

number of matches where Tua wins exactly two games (Carl wins one game and Tua

wins two games).

(iv) If Tua wins the first game, calculate the probability that Carl will win at least one of

the second and third games

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(b) A free throw is part of a game in basketball where a player takes two shots from a set

distance from the hoop and can score either zero, one or two shots.

Suppose, from each free throw, the probability of getting zero shots is ‘a’ and the

probability of getting one shot is ‘b’.

Show that the probability of getting either zero shots or four shots from two consecutive free

throws is

2a2 + b

2 – 2a - 2b + 2ab + 1

(Assume that the second turn at a free throw is not affected by the first)

Assessor’s

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Question 3

Hoopsdunk High School is thinking of charging people to watch the Premier Girls team play as part

of a fundraiser to improve the gym facilities. Over the last four years (80 games) the school has

recorded the number of spectators that have come to watch each of the Premier Girls games.

The results are shown in the graph.

(a) Describe a feature of the graph that shows the data collected over the four years does NOT

follow a normal distribution.

Assessor’s

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0

2

4

6

8

10

12

Number of spectators

Spectators at Hoopsdunk High School for Premier Girls team basketball games

Freq

uen

cy


(b) The school will pay for gym improvements earlier than planned if the chance of more than

150 people attending three games in a row is likely to happen.

Calculate the probability of this happening and explain whether the school is likely to pay

early for improvements. Assume that the attendance of one game is not affected by the

attendance of another.

(c) The coach of the Premier Girls team is interested to know if there is a link between the

number of spectators who turn up to watch the team play and the team’s results. The data

from the previous four years is summarised in the table.

Lose

Win

100 or less

spectators

22

12

34

101 or more

spectators

3

43

46

25

55

80

What is the probability that a game chosen at random

(i) had 100 or less spectators ?

(ii) was a game that the Premier Girls team won ?

(iii) was a game the team lost if it was known that 100 or less spectators attended ?

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(d) The school conducts a survey of students who are known to have attended at least one

Premier Girls game over the last four years, to find their likelihood of attending the next

game.

The results are summarised in the table.

Likely to attend

NOT likely to attend

Team won

previous game

153

92

245

Team lost

previous game

32

73

105

185

175

350

Using the idea of relative risk, use calculations to explain how much more likely a student is

to attend the next game based on whether the team lost the previous game.

Assessor’s

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Extra paper for continuing your answers, if required.

Clearly number the question.

Question number


Assessment Schedule: Mathematics and Statistics 91261: Algebra

Question

Evidence

Achievement (u)

Merit (r)

Excellence (t)

Apply algebraic

methods in solving

problems.

Apply algebraic

methods, using

relational thinking,

in solving

problems.

Apply algebraic

methods, using

extended abstract

thinking, in

solving problems.

TWO

(a)(i)

x3 = 125

x = 5

Correct

(a)(ii)

2x = 64

x = 6

Correct

(b)(i)

ln(0.95)t = ln(⅔)

t = 95.0ln

)ln( 32

= 7.9

Correct answer.

Accept 8 months.

(b)(ii)

300(0.95)t = 100(1.015)

t

ln

t

95.0

015.1= ln 3

t = 16.6

Correct answer.

Accept 17

months

(b)(iii)

S = 300(0.95)36

= 47

K = 100(1.015)36

= 171

47(1.04)t = 171(0.98)

t

t = 061.1ln

)638.3ln(

t = 21.8 See my notes

Finding and

equating the two

equations

t = 21.8 or 22

months.

Logical working

shown clearly.

(c)

log3(m2-14m + 49) = 4

m2-14m + 49 = 3

4 = 81

m2-14m-32 = 0

m = 16, -2

m = -2 is an acceptable

solution as when

substituted (m -7)2 will

be positive.

Correct

expansion and

equating to 81

Solutions found

Solutions found

and explanation

given.

N0 No response; no relevant evidence M5 1 of r

N1 attempt at one question M6 2 of r

N2 1 of u E7 1 of t

A3 2 of u E8 2 of t

A4 3 of u


Assessment Schedule: Mathematics and Statistics 91261: Algebra

ONE

(a)(i)

(5x + 4)(x – 2)

Correct

(a)(ii)

x = -0.8, 2

Correct

(b)

)4)(1(

)1(3)4(5

mm

mmm

)4)(1(

2032 2

mm

mm

One

denominator

Simplified

correctly

(c)

4x2 - 16x - 84 = 5x

2 – 45

x2 + 16x + 39 = 0

(x + 3)(x + 13) = 0

x = -13

x -3 as cannot divide by 0

Correct

factorisation

x = -13 only

solution

(d)(i)

)452(7

)4)(5(2 cxx

xx

=

)32(7

)5(

x

x

2x2 – 5x + 4c = (2x +3)(x – 4)

2x2 – 5x + 4c = 2x

2 – 5x – 12

c = -3

Factorisation

and

elimination of

common

factors

c = -3 found

(d)(ii)

If c = -3

14x2 – 35x – 84 = 0

2x2 – 5x – 12 = 0

a = 2, b = -5, c = -12

b2 – 4ac = 121 > 0

Roots are real and distinct

Discriminant

found

Nature of roots

described

(e)

Area track = π(2x + 5)(3x + 5)

-π.2x.3x

= 25πx + 25π

Inside area = 6πx2

Equating: 25πx + 25π = 6πx2

6x2 – 25x – 25 = 0

x = 5, -0.83 (eliminate)

Internal lengths are 10 and 15

Quadratic

equation

formed

Internal

lengths found



N2 1 of u E7 1 of t

A3 2 of u E8 2 of t

A4 3 of u


Assessment Schedule: Mathematics and Statistics 91261: Algebra, 2014

THREE

(a)

16

6m

Correct

(b)

(0.4m2)

2

= 0.16m4

Correct

(c)

x

x

2

1

3

3

= 27

3(x+1-2x)

= 27

1 – x = 3

x = -2

Expressed as

powers of 3

x = -2 found

(d)(i)

Using midpoint as y-axis

y = - 4

9.2x

2 + 4

Equation found

(d)(ii)

Fence is at x = 0.5

y = - 4

9.2(0.5)

2 + 4

= 3.82 metres

Height found

from correct

equation

(e)

a = 3, b = -2k, c = 4k

b2 – 4ac > 0

4k2 – 4.3.4k > 0

4k(k – 12) > 0

k < 0 and k > 12

Discriminant

found and

set > 0

One value

found

Correct range

of values found



N2 1 of u E7 1 of t

A3 2 of u E8 2 of t

A4 3 of u

Grade point scoring (this can be varied for school requirements)

1 - 6 = Not Achieved

7 – 13 = Achieved

14 – 19 = Merit

20 – 24 = Excellence


Assessment Schedule: Mathematics and Statistics 91262: Calculus, 2014

Achievement Merit Excellence

Apply calculus methods in solving problems involves:

selecting and using methods

demonstrating knowledge of calculus concepts and terms

communicating using appropriate representations.

Apply calculus methods using relational thinking, in solving problems must involve one or more of:

selecting and carrying out a logical sequence of steps

connecting different concepts and representations

demonstrating understanding of concepts

forming and using a model

and relating findings to a context, or communicating thinking using appropriate mathematical statements.

Apply calculus methods using extended abstract thinking, in solving problems must involve one or more of:

devising a strategy to investigate a situation

demonstrating understanding of abstract concepts

developing a chain of logical reasoning, or proof

forming a generalisation

and using correct mathematical statements, or communicating mathematical insight.



Evidence Statement

One Expected Coverage Achievement(u) Merit(r) Excellence(t)

NØ = No response; no relevant evidence.

N1 = a valid attempt at ONE question.

N2 = ONE question demonstrating limited knowledge of calculus techniques.(1u)

A3 = TWO of u.

A4 = THREE of u.

M5 = ONE of r.

M6 = TWO of r.

E7 = ONE of t, with minor errors ignored.

E8 = 2 of t

(a)

dydx

= 3x4

+ 6x – 7

when x = 1, gradient = 2

Derivative found with one error and consistently used in finding the gradient at the point where x = 1.

derivative and

gradient at x = 1

found

(b) (i)

Provides graph of the gradient function

Graph of the gradient function sketched with x intercepts near x= 1 and x=-1, parabolic shape

(b) (ii)

dydx

= 3x2

– 8x + 3 = 6

x = 3 or – 13

points (3,1) (-0.33,-0.48)

Derivative found, equated to 6, partially solved

Coordinates of both points found

(b) (iii)

dydx

= 3x2

– 8x + 3

at x = 2, gradient = -1

tangent equation

y = – x + 1

derivative found evaluated with x=2

equation of the tangent

(c) (i)

f (x) = 0.5x3

+ 3x2

– 6x + 4

at x = 1 on f(x), gradient = 0.5

g(x) = 2x – 4 = 0.5, so

x = 2.25

the other x value is 2.25

finds the derivative for both functions

finds the gradient at x=1 for f(x)

finds the value of x where the gradients are the same

(c) (ii)

The gradients of the curves need to be the same at the points of intersection for the outline to be smooth.

At x=0 the gradient of f(x) is 4 and the gradient of g(x) is -4.

So the outline is not smooth

discusses the slope of the two curves needing to be equal

justifies the claim that the outline is not smooth by providing evidence the slopes are different where the curves meet



Two Expected Coverage Achievement Merit Excellence




A3 = TWO of u.

A4 = THREE of u.

M5 = ONE of r.

M6 = TWO of r.


E8 = 2 of t.

(a)

f (x) = 15x2

– 7

when x = 1, gradient = 8

This means that f(x) is

increasing

derivative found and evaluated at x=1

derivative found and states that f(x) is increasing

(b)

f(x) = 2x4

– x2

– 2x + c

0 = 2(1)4

– (1)2

– 2(1) + c

c = 1

so f(x) = 2x4

– x2

– 2x + 1

function integrated and c evaluated

(c)

f (x) = 3x2

– 3x – 18 = 0

x = 3 or – 2

f’’(-2)=-15

f’’(3)=15

so (-2,21) is a local maximum.

derivative found and both x values found

Finds the second derivative

gives the coordinates for the maximum TP and explains the second derivative being positive means that

(-2,21) is a maximum.

(d) (i)

(d) (ii)

T ' = 83

t

At t=3 , T ‘ = 8°C per min. The chocolate is heating up

derives T evaluates T when t=3 (units not required) and explains it is increasing in temperature (heating)

(d) (iii)

T ' = -0.04x3

+ 1.2x2

– 10x + 20.3

shows T '= 0,when x=3 and17.1

T '' = -0.12x2

+ 2.4x – 10

at x=3 and 17.1 shows T''<0

or otherwise using Calculus

derives T, makes T’ =0

shows T’=0 for both x values

justifies both points are maximums by calculus methods

35 2014 End of year L2 exam


Three Expected Coverage Achievement Merit Excellence




A3 = TWO of u.

A4 = THREE of u.

M5 = ONE of r.

M6 = TWO of r.


E8 = 2 of t.minor error.

E8 = Excellence correct.

(a) (i)

y = 4x3 + 2x

2 + c

at (0,0)

c = 0

so y = 4x3 + 2x

2

0 = 12x2

+ 4x

x = 0 or – 13

– 1

3 , 2

27

anti-differentiates to give an expression for y, includes some evidence of considering the constant of integration, and that c=0.

finds both coordinates of the other point where the gradient is 0

(a) (ii)

(b) (i) y ' = 4x – 6

at x=0, gradient = -6

derivative and gradient found

(b) (ii)

0 = 4x – 6

x = 1.5

minimum value y= -3.5

equates derivative to 0 and solves for x

finds minimum value

(c)

v = 2t + c

at t=0, v=10 so c=10

v = 2t + 10

S = t2

+ 10t + c

when t=20, s= 620, so c=20

S = t2

+ 10t + 20

when t=0, s=20

the height of the building

is 20m

Finds the velocity equation, finds the velocity equations and integrates a second time to find an expression for distance, with minor errors

integrates twice and interprets the distance at t=0 as the height of the building

t= with minor errors or omissions

2t completely correct


Assessment Schedule: Mathematics and Statistics 91267: Probability, 2014 Evidence Statement

Question

Expected Coverage

Answers will vary depending on

whether the candidate uses the tables

or a graphing calculator.

Achievement (u) Merit (r) Excellence (t)

ONE

(a) i

P(185< X < 195) = 0.394 Probability found.

(a) ii P(175 < X < 190) = 0.789

= 78.9%

Probability found but must

be expressed as a

percentage. Rounded to

79% is acceptable.

(b) P (X < 169 or X > 181)

= 0.115 x 2

= 0.23

0.23 x 215 = 49 or 50 girls

Probability of 0.115

found.

Correct number of

girls found

(c)

z-score = 1.645

so

so

so σ = 6.08 mm

z-score found

Equation set up Equation solved to

at least 1 d.p.

(d) Comments should be in context and

for t students should have linked the

previous context to realise there are

two separate normal distributions

(slightly overlapping), one for girls

basketball team and one for boys

basketball team OR other valid

reason discussed and validated in

context.

Evidence to support this, in context,

could include

- Each distribution is symmetrical

- Mean and median of each

distribution are centrally located

to that distribution

- Unimodal

- Spread from the middle of each

distribution fits the shape of a

Normal distribution

- Idea of overlapping distributions

does not mean that basketballs

are not normally distributed, it

just means that the lightest of

the ‘boys’ set are lighter than

the heaviest of the ‘girls’ set.

Some specific numerical evidence

needs to be mentioned e.g. one ND

(girls) has a mean of around

535grams

. One appropriate

comment whether

arguing for or

against Normal

Distribution.

Two relevant

comments

recognising weight

of basketballs are

normally

distributed = t

This must include

some numerical

evidence and also a

suggested reason

e.g. boys and girls

basketballs …

some inflated and

some left less

inflated …etc….

N0 no relevant comments

N1 attempt at 1 question A3 two of u M5 one of r E7 one of t

N2 one of u A4 three of u M6 two of r E8 two of t



Question Expected Coverage Achievement (u) Merit (r) Excellence (t)

TWO

(a) - i

P(RRR) = 0.6 x 0.82 = 0.384

Correct probability

(a) - ii P( C ≥ 2games)

= CCC + CCR + CRC + RCC

= 0.4 x 0.72 + (0.4 x 0.7 x 0.3) +

(0.4 x 0.3 x 0.2) + (0.6 x 0.2 x 0.7)

= 0.388

Probability calculated

using at least three of

the four combinations

possible.

Correct

probability

(a) - iii P( Rua 2 games of 3)

= RRC + RCR + CRR

= (0.6 x 0.8 x 0.2) + (0.6 x 0.2 x 0.3)

+ (0.4 x 0.3 x 0.8)

=0.228

40 x 0.228= 9 or 10 matches

At least one probability

of the three

combinations

calculated

Correct expected

number – allow

answer of 9.12

as long as

workings seen

(a) - iv So would need CC + CR + RC

Or 1 – RR = 1 – 0.82

= 0.36

Correct

probability

calculated

(d) Three branch tree diagram with branches

labelled a, b and 1 – a – b

Or a, b and 1 –(a + b)

P (0 or four points from two free shots)

can only happen by two 0’s or two 2’s

So a2 + [(1 – a –b)(1 – a – b)]

= a2 + (1 – a – b –a + a

2 + ab – b + ab + b

2)

= 2a2 + b

2 +2ab – 2a – 2b +1 as required.

e.g. Tree diagram set

up with branches

labelled or evidence at

least that the three

branches will add

up to 1.

.. or other similar

evidence.

Attempt to set

up expression

and multiply out

but no errors.

Logical reasoning

and workings lead

to correct final

expression. Must

see at least some

evidence of

multiplying out

and collection of

like terms.






Question Expected Coverage Achievement (u) Merit (r) Excellence (t)

THREE

(a)

Not unimodal

No tailing off

Not bell shaped

…or other relevant evidence

Correct and relevant

evidence stated.

(b) P(spec > 150) = 25/80

3 games in a row (25/80)3

= 0.031

Very unlikely BoT will take the

view of paying early for gym

improvements as that level of

attendance 3 games in a row is

only 3.1%. They can hold onto

their money!!

Probability for more

than 150 spectators

attending calculated

Probability for three

games in a row

calculated.

Correct and

reasonable

conclusion reached

based on their

probability

calculation from.

CON allowed here.

(c)- i P( spec ≤ 100) =34/80

=17/40

Correct probability or

equivalent.

.

(c)- ii P(win) = 55/80

= 11/16

Correct probability or

equivalent.

(c)- iii P (lost | spec ≤ 100) = 22/34

=

11/17

Correct probability

or equivalent.

(d) P(attend | won previous game) =

153/245

P(attend | lost previous game) =

32/105

153/245 ÷ 32/105 = 2.05

Just over twice as likely to attend

the next game if the team had

won the previous game.

One of either probability

calculated

Relative risk

calculation done

correctly

Making meaningful

conclusion to

explain the

difference.

.




Judgement Statement

Not Achieved Achievement Achievement

with Merit

Achievement

with Excellence

Score range 0 – 8 9 – 14 15 – 19 20 – 24

pakuranga college€¦ · 12 2014 end of year l2 mat exam calculus you are advised to spend 60...

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