Mathematics tertiary preparation 1 Open Access College (OAC) Tertiary Preparation Program Assignments Semester 1 2017 © University of Southern Queensland Published by University of Southern Queensland...

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Mathematics tertiary preparation 1 Open Access College (OAC) Tertiary Preparation Program Assignments Semester 1 2017 © University of Southern Queensland Published by University of Southern Queensland Toowoomba Queensland 4350 Australia http://www.usq.edu.au © University of Southern Queensland, 2017.1. Copyrighted materials reproduced herein are used under the provisions of the Copyright Act 1968 as amended, or as a result of application to the copyright owner. No part of this publication may be


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TPP7181 – Mathematics tertiary preparation 1 1 TPP7181 Mathematics tertiary preparation 1 Open Access College (OAC) Tertiary Preparation Program Assignments Semester 1 2017 © University of Southern QueenslandPublished by University of Southern Queensland Toowoomba Queensland 4350 Australia http://www.usq.edu.au © University of Southern Queensland, 2017.1. Copyrighted materials reproduced herein are used under the provisions of the Copyright Act 1968 as amended, or as a result of application to the copyright owner. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without prior permission. Produced by Learning Resources Development using the ICE Publishing System.Table of contents Page Assignment 1 1 Assignment 2 3 Assignment 3 9 Assignment 4 15 Assignment 5 23






TPP7181 – Mathematics tertiary preparation 1 TPP7181 – Mathematics tertiary preparation 1 1 TPP7181 Mathematics tertiary preparation 1 Open Access College (OAC) Tertiary Preparation Program Assignments Semester 1 2017 © University of Southern Queensland Published by University of Southern Queensland Toowoomba Queensland 4350 Australia http://www.usq.edu.au © University of Southern Queensland, 2017.1. Copyrighted materials reproduced herein are used under the provisions of the Copyright Act 1968 as amended, or as a result of application to the copyright owner. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without prior permission. Produced by Learning Resources Development using the ICE Publishing System. Table of contents Page Assignment 1 1 Assignment 2 3 Assignment 3 9 Assignment 4 15 Assignment 5 23 TPP7181 – Mathematics tertiary preparation 1 1 Assignment 1 Description Marks out of Wtg(%) Assignment 1 20 4 This is a Multiple Choice Quiz which will be opened on the 6th March and will close on 13th March at midnight. You will have 45 minutes to complete it and only one attempt will be allowed. Complete this assignment after you have studied module 1 and module 2. © University of Southern Queensland 2 TPP7181 – Mathematics Introductory Book © University of Southern Queensland TPP7181 – Mathematics tertiary preparation 1 3 Assignment 2 Description Marks out of Wtg(%) Assignment 2 45 10 Complete this assignment after you have completed module 3 and module 4. Attempt all questions that follow. Show sufficient working so that your tutor can follow your reasoning. Submit your assignment on A4 paper, not this sheet. Marks will be awarded for each question according to the marking criteria in the introductory book. You are advised to make a copy of this assignment before mailing. 1. Compare, using subtraction and division, the amount of sugar used in two recipes, if Recipe 1 used two and a half ( 1 22 ) cups and Recipe 2 used one and two third ( 2 31 ) cups. Show your calculations and then write your answers as sentences. (4 marks) 2. A house plan shows a scale of 1:40. If the rectangular lounge room of the house is to measure 4 metres by 6.4 metres, a. What are the dimensions of the room on the plan? b. What is the area of the lounge room on the plan? (3 marks) 3. A particular ‘street legal car’ is able to travel at 384 kmh–1. Express this speed in ms–1 (2 marks) 4. In a certain tertiary course, eight out of every 10 students that start the course finish it. a. If in this semester 294 students enrol in the course, how many are expected to finish? b. If the pass rate (for those completing the course) is 65%, how many of those sitting for the examination at the end of this semester should pass? (4 marks) 5. Consider a large water tank a. If 3900 litres occupies ¾ of the total capacity of the tank, how much water will the tank hold when it is full? b. If the tank is completely dry and water enters the tank at a rate of 7 litres per minute, how long will it take to half fill the tank? (3 marks) © University of Southern Queensland 4 TPP7181 – Mathematics Introductory Book 6. In the 2001-02 financial year Australian governments at all levels contributed a total of $37 546 000 000 towards education. Of this amount $21 283 000 000 was spent on the education of 3 238 900 students in primary and secondary schools; and $13 635 000 000 on the education of 1 456 500 students in tertiary education. a. How much government expenditure during this period was spent on education in areas other than those mentioned above? b. i. What percentage of all government education expenditure was spent on the tertiary education sector? ii. What was the total cost per student for tertiary education? (4 marks) 7. Evaluate the following on your calculator. Round your answers to two decimal places if necessary. a. 3 51024 b. 1 2256 − c. 3 45 5+ (3 marks) 8. Simplify the following, expressing your answer in index form. a. 8 2 412 12 12−÷ × b. 15 23 3−÷ (2 marks) 9. Complete the following. a. 230mg ______________ g= b. 36ML ______________ L= c. 23 km ______________ cm= d. 12.5cm ______________ m= e. 3.5 s ______________ ms=µ
Answered Same DayDec 26, 2021

Answer To: Mathematics tertiary preparation 1 Open Access College (OAC) Tertiary Preparation Program...

Robert answered on Dec 26 2021
120 Votes
Assignment 3

1) The total number of handshakes at a party of n guests, where everyone is shaking
everyone’s hands only once: s = n(n-1)/2
a) Explaining the formula: let’s try to imagine the situation if we went to a party and
everyone shakes everyone else’s hands only once! However, no one shakes their own
hand so every person ends up having n-1 handshakes, so there would be total n(n-1)
handshakes, but it includes each handshake twice because if A is handshaking with B,
then we have also considered the same handshake when B is handshaking with A. we

have counted twice the handshakes as it should be, Therefore, the total number of
handshakes = (1/2)*n*(n-1)
b) In the question, it’s given; n =12
So, total number of handshakes would be = 12*(12-1)/2 = 12*11/2 =132/2 = 66
2) Arranging the catering for wedding feast:
a) Formula for total catering cost: Let’s assume that number of guests = n (including the
family and ourselves)
And, we know that it costs $25 per person, so we got 25*n for guests, now we are
going to have to add the booking cost which is $120.
Therefore the total catering cost for the wedding = $(120 + 25*n)
b) To fill the table we just have to use the formula, we define above: cost = $(120 +
25*n); n= number of guests.
No. of Guests 20 40 60 80 100
Catering Cost($) 620 1120 1620 2120 2620
c) Plotting the Graph:
d) Look at the graph below that we have plotted, and the trend line we have got, for
every next 20 guests, cost of catering increases by 500, and hence it’s a linear graph,
for every next 10 guests, cost of catering would increase by 250. So, for 110 guest,
cost of catering = 2620 + 250 = 2870
e) Given the cost of catering = $2000
We will simply apply the formula we derived in the first part, cost = $(120 + 25*n)
Therefore, 2000 = 120 + 25*n
And, 1880/25 = n = 75.2
Because people can’t be expressed in decimal point (obviously); total number of
guests = 75 (including the family and us)
3) Looking at the graph given which is how income per person has changed in north and
south Korea from 1960 to 2013
a) Income per person in South Korea in 1960 = $1190
In 2013 = $32,000
620
1120
1620
2120
2620
y = 500x + 120
0
500
1000
1500
2000
2500
3000
3500
4000
20 40 60 80 100
C
at
e
ri
n
g
C
o
st

Number of Guests
No. of Guests Vs. Catering Cost
b) Percentage change in income from south Korea from 1960 to 2013 = 100*(Income
per person in South Korea in 2013 - Income per person in South Korea in 1960)/
Income per person in South Korea in 1960
= 100*(32,000 – 1190)/1190 = 2589.08 %
c) Comparison of 2013 income per person for north and south Korea by division:
2013 income per person for South Korea/2013 income per person for North Korea
= 32,000/1390 = 168.42
That means compared to income per person for North Korea, South Korean income
per person is 168.42 times higher.
Comparison of 2013 income per person for north and South Korea by subtraction:
2013 income per person for South Korea - 2013 income per person for North Korea =
32,000 – 1390 = $30,610
That means compared to income per person for North Korea, South Korean income
per person is $30,610 more.
4)
a) Let’s assume the person gets paid $100 every fortnight and let’s assume further that
he pays $40 each month towards the repayment of that loan.
So, we can create an imaginary bank balance:
Period(Date) Balance($) Loan Payment($) Balance Remained($)
15th Jan 100 0 100
30th Jan 200 40 160
15th Feb 260 0 260
29th Feb 360 40 320
15th March 420 0 420
30th March 520 40 480
15th April 580 0 580
b) We will try to understand how the balance changes from looking at the graph we
just drew.
There are continuous dips in the graph at the end of each month, which shows some
money has been taken out to towards the repayment of loan, it goes up again and
then again a dip at the end of the month. So, if the person has a balance of $100 at
15th Jan, it will be $200 on 30th jab but he will have to give $40 towards loan
repayment, so it will be a dip in the graph.
5)
a) From the given graph, we can see, the x-intercept = 4
The y-intercept = -2
b) We can see from the graph that the line passes through (0, -2) and (4, 0). So, the
slope of the line would be: m = The equation of the line would be = [0-(-2)]/[4-0] =
1/2
Therefore, the equation of the line would be: y-0 = (1/2) * (x-4)
Or, 2y = x - 4
Or, x - 2y - 4 = 0
c) If the another line is parallel to this one, then slope of that line would also be 1/2
And, it’s passing from (0, 1). Therefore the equation of that line would be:
y - 1 = (1/2) * (x - 0)
0
100
200
300
400
500
600
700
15th jan 30th jan 15th feb 29th feb 15th march 30th march 15th april
B
al
an
ce
R
e
m
ai
n
in
g
Date
Or, 2y-1 = x
Or, x-2y-1 = 0
6) y=4- 3x
It fits in the format of y= mx +c, where m is the gradient and c is the y-intercept
So in this given equation, we have gradient = -3, intercept = 4
7)
a) It’s given that the gradient of the line is -1, i.e. it’s making a 135 degrees angle with...
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