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The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
Download (direct link): theatozofmath2002.pdf
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Solution. The trip falls into five stages, and since the speeds are constant, each stage is represented by a straight-line graph (see figure b):
♦ Stage 1. Harry travels 60 km in 3 h.
♦ Stage 2. He stops for lunch, and the graph is horizontal for 1 h.
♦ Stage 3. He travels 30 km in 2 h.
♦ Stage 4. Sightseeing for | h.
♦ Stage 5. From the point on the line at the end of stage 4, draw a straight line to
the time axis to meet it 3 h later.
The travel graph can now be drawn, where d is the distance from home in kilometers and t is the time taken in hours.
For the return journey Harry travels 90 km in 3 h:
Speed for return journey
90 km 3 h
30 km/h
Speed = distance -r time
To find the average speed for the whole trip we need the total distance traveled and the total time taken.
Total distance = 180 km Total time = 9.5h
180 km
Average speed
90 km out and 90 km back 3 + 1 + 2 + 0.5 + 3
Average speed = total distance T- total time
9.5 h = 18.9 (to 1 dp)
The average speed is 18.9 km/h.
References: Average Speed, Cartesian Coordinates, Gradient, Table of Values.
References: Königsberg Bridge Problem, Networks.
A tree diagram is a figure that resembles the branches of a tree, and is designed to show the events of an experiment and enable the probabilities of each event to be calculated. A tree diagram can be used in studying probability to list the sample space of an experiment, which is a list of all possible outcomes. This is explained in the entry
Event. It is also used to calculate probabilities. It is this latter use of tree diagrams that is described in this entry. The tree diagram is also known as a probability tree. The following examples explain how the tree diagram works in solving probability problems.
Example 1. Pat has three children, and none of them are twins. Each time a baby is bom the chances of giving birth to a boy or a girl are equal. Calculate the following probabilities:
(a) They are all girls.
(b) There are two boys and one girl.
(c) There are two boys and one girl, bom in that order.
(d) There are at least two girls.
Solution. The three births are represented by three sets of branches (see figure a; G = girl, B = boy) and at each birth the probability of having a boy or a girl is |. We have the following rules:
♦ To find the probability at the end of a branch we multiply the probabilities along the branch.
♦ If a solution is obtained from two or more sets of branches, we add together the probabilities at the end of each branch.
These two rules are put into practice as we solve the problem.
For (a),
Prob(3 G) = | Multiplying along the branch \ x \ x \
For (b),
Prob(2 B&1G) = | + | + | Adding ends of three branches
_ 3 — 8
For (c),
Prob(BBG in order) = | Multiplying along the branch | x | x |
For (d),
Prob(at least 2 G) = | Adding the ends of the four branches
with at least two girls in them
In the above problem the probability along each branch was the same value of | because previous births do not affect the probability of the next birth. In the next example the probabilities change along the branches.
Example 2. Jim bought his grandson Luke a bag containing nine lollipops. There are four red, three green, and two yellow lollipops. He lets Luke choose one without looking, and then another. You may assume that Luke does not put the first one back before choosing the second! Work out the following probabilities:
(a) They are both the same color.
(b) They are different colors.
(c) The first is green and the second is yellow.
Solution. At each choice the tree has three branches, which are R, G, Y (see figure b):
♦ First choice. There are four red, three green, and two yellow, making nine lollipops in the bag.
♦ Second choice. If one red has been taken, there are three red, three green, and two yellow left, making eight in all. On the other hand, if one green has been taken, there are four red, two green, and two yellow left, making eight in all, and so on.
The probabilities at the end of each of the nine branches are obtained by multiplying the probabilities along each of the branches.
For (a),
Prob(same color) = |§ + ^ ^
(b) To obtain the probability of the two different colors it is convenient to subtract the probability of the same color from 1:
Prob(different color) = 1 — ^ Using part (a)
- 13
— 18
For (c),
Prob(G & Y, in order) = ^ or ^
Note. The fractions representing probabilities may be simplified by canceling. References: Event, Experiment, Sample Space.
This is a process that involves guessing or estimating the solution to a problem and then improving on the guess to obtain a more accurate answer. This is sometimes called trial and improvement, which seems a better description of the process.
Example. The length of a rectangular garden is 20 meters greater than its width. If the area of the garden is 900 square meters, find the width of the garden.
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