mathematics –bridging course for applied sciences · 2015-08-06 · mathematics –bridging...

Mathematics – Bridging course

for Applied Sciences

WS 2013/2014

1

Writing with decimal power and exponents

Writing with decimal power and exponents it is possible to write very small and very big numbers in a compact way. They build the base for scientific writing (SCI).

name decimal notation exponentation (scientific notation)

quintillion (trillion) 1 000 000 000 000 000 000 =1018

quadrillion 1 000 000 000 000 000 =1015

trillion (billion) 1 000 000 000 000 =1012

billion (Milliard) 1 000 000 000 =109

million 1 000 000 =10*10*10*10*10*10=106

hundred thousand 1000000 =10*10*10*10*10=105

ten thousand 10000 =10*10*10*10= 104

thousand 1000 =10*10*10= 103

hundred 100 =10*10=102

ten 10 =101

one 1 =100

tenth 0,1 =1/10=10-1

hundredth 0,01 =1/100=10-2

thousandth 0,001 =1/1000=10-3

Ten thousandth 0,0001 =1/10000=10-4

Scientific notation

presentation of numbers in the way:

A 10n

with 1 ≤ A < 10 and n integer number

e.g.: 0,000654 = 6,54*10-4

350010 = 3,50010*105

0,02800 = 2,800 *10-2

useful Link: http://science.widener.edu/svb/tutorial/scinot.html

Exercise `scientific notation´

1.) 25802=

2.) 0,0027=

3.) 87,9654=

5

4.) 818,5000=

5.) 12,85*102=

6.) 913,64*10-6=

Units /Prefixes

SI-units (french: Système international d’unités)

�International system for physical quantities

�The SI-system exists of 7 base units

�All other physical units derive from these base units � derived units

SI-units

7

measurement unit symbol

length metre m

mass kilogram kg

time second s

temperature kelvin K

amount of substance mole mol

electric current ampere A

luminous intensity candela cd

frequency hertz

force newton

pressure pascal

energy joule

power watt

voltage volt

sHz

1=

2s

kgmN

⋅=

22sm

kg

m

NPa

⋅==

s

kgmmNJ

⋅=⋅=

2

3

2

s

kgm

s

JW

⋅==

As

kgm

A

WV

⋅

⋅==

3

2

Derived SI-units, e.g.

Prefixes are added to unit names to produce multiple and sub-multiples of the original unit. All multiples are integer powers of ten to avoid a lot of positions after decimal point

e.g. 7000m = 7*103 m = 7 km

� unit is m (metre)

� k (standing for kilo) is the prefix and replaces the factor 1000 respectively 103

10

SI-Prefixesfactor prefix symbol/abbreviation

1015 peta P

1012 tera T

109 giga G

106 mega M

103 kilo k

102 hekto h

101 deka da

10-1 dezi d

10-2 centi c

10-3 milli m

10-6 micro µ

10-9 nano n

(10-10 Angström Å)

10-12 pico P

10-15 femto f

Exercise `use of prefixesWrite the following values with SI-prefix and vice versa:

e.g. 4,85* 10-9 g = 4,85 ng oder 2,58 mg= 2,58*10-3 g

1.) 3,16*10-3 m = 4.) 34,2 cL =

2.) 5,98*109 s = 5.) 2,50 ng =

3.) 58,89*103 g = 6.) 5µmol =

Conversion of units

example 1 159 km = ________cm

considerations:

km => prefix kilo = 103

cm => prefix centi = 10-2

conversion from a higher prefix to a lower prefix: number has to be multiplied by the factor (difference of power of ten)

� 159 km = 159 *105 cm

difference of thetwo prefixes:5 powers of ten(from 3 to -2)

Conversion of units

example 2 4 nm = ________cm

considerations :

nm => prefix nano = 10-9

cm => prefix centi = 10-2

conversion from lower prefix to a higher prefix: number has to be divided by the factor (difference of power of ten)

� 4 nm = 4/107 cm = 4*10-7cm

difference of the prefixes:7 powers of ten(from -9 to -2)

Exercise `conversion of units´

1. 7m (dm)

2. 6 km (m)

3. 5 dm (cm)

4. 5 dm (µm)

5. 32 nm (cm)

6. 560 cm (dm)

7. 940 mm (cm)

8. 5 µm (mm)

9. 5m (µm)

10. 72 dm (mm)

11. 37 m (mm)

12. 6300 mm (dm)

13. 6 µm (nm)

14. 700 m (km)

14

Convert the following quantities into the stated units.

Useful links/interactive tasks:http://exercises.murov.info/ex1-2.htmhttp://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i17/bk8_17i2.htm

1. 5 t (kg) (note: the tonne (t) is no

SI-unit- but is gently used for masses)

2. 3 mg (kg)

3. 5 ng (mg)

4. 16 µg (g)

5. 36 ms (s)

6. 5 µs (s)

7. 25 L (nL)

8. 4 Gbit (Mbit)

9. 17 mol (µmol)

10. 25,5 mmol (mol)

15



16

1. 0,5 m2 (dm2)

2. 5 L (dm³)

3. 21 mL (dm³)

4. 1 m³ (mm³)

5. 36 cm³ (mm³)

6. 0,5 mm³ (cm³)

7. 0,6 L (cm³)

8. 12 mm³ (L)

examples: • 1 m2 = 1m*1m = 1*102 cm*1*102 cm = 1*104 cm2

• 1 m3 = 1m*1m*1m = (1*102 cm)*(1*102 cm)*(1*102 cm) = 1*106 cm3

• 1 m3 = 1000 L



17

1. 3 min (s)

2. 5 h (min)

3. 35 min (h)

4. 1 d (s)

5. 1,5 a (s)

6. 2,25 min (s)

7. 2 min 50s (s)

8. 273,15 K (°C)

9. 25 °C (K)

examples: • 2 a in min: 2a= 2a*365d/a*24h/d*60min/h= (2*365*24*60) min=1051200min • 20 C = (20+273,15) K



1. Write the value in scientific notation (using the writing of the power of ten with one pre decimal place)

2. Write the value in a suitable unit with a prefix.

18

Example:

mm

m

m

76,8

1076,8

00876,03

=

⋅= −step 1

step 2

µm

kg

L

mm

g

0024,0.5

0349,0.4

0098,0.3

00478,0.2

0048,0.1

19

²008463,0.10

²000574,0.9

000789,0.8

00145,0.7

0036,0.6

m

dm

cm

s

m

20

Exercise `calculating with units

][2325.5

][151.4

][1634.3

][5125.2

][125.1

mgµgmg

kgkgg

mmnmcm

mmµmcm

cmmmm

+

+

+

−

+

⋅

Convert into the stated units and calculate.

21


][236.9

][51.8

][367490.7

][78456.6

3

33

LcmL

dmdmmL

mgngµg

gmgg

−

+

+

+

⋅


22

Exercise `calculating with unit

][min162415.13

][min504.12

[min]5,345.11

[min]60min35.10

hss

hd

hs

s

++

+

+

+⋅


23

][164.3

³][51634.2

³][2265.1

2LdmmmV

mµmcmdmV

cmcmmmµmV

⋅=

⋅⋅=

⋅⋅=



24

a.) An ant is moving with 9 km/d. What is the velocity in km/h respectively cm/min?

b.) 25 mg/100mL = ________ mg/L

c.) 250 mmol/L = ________ mol/L

d.) 490 mg/L = ________ g/mL


25

Exercise `derived units

Ns

kgm

s

kgkm________________

50

2922

=⋅

=⋅

e)

Js

kgmgm______________

min5

500)25(2

2

2

2

=⋅

=⋅

f)

Ns

kgm

h

gs

mm

______________1

4)43(

2=

⋅=

⋅g)

26

Exercise `derived units

Pasm

kg

km

t______________

min)2(2,0

8,2822

=⋅

=⋅h)

Pasm

kg

µsnm

ng______________

2,02,0

222

=⋅

=⋅

i)

Summation notation (capital-sigma notation) ∑

In that example: the index k gets values from 1 (starting point) to 5 (stopping point): 1,2,3,4,5. The index is always incremented by 1

summation sign

Stopping pointupper limit of summation

function of index-representing each successive terme in the raw

Starting pointlower limit of summation

Index of summation(continous index)

Summation notation (capital-sigma notation) ∑

Example:

In that example the index k can only acceptvalues from 1 (starting point) to 4 (stoppingpoint) in entired steps:1,2,3 and 4, that will be added.

Example summation notation:

In that example the index i can only acceptvalues from -1 (starting point) to 3 (stoppingpoint) in entired steps:-1, 0, 1, 2 and 3, that will be inserted in thefunction for i and than will be added.

Example summation notation:

In that example the index i can only accept valuesfrom 1 (starting point) to 7 (stopping point) in entired steps:1, 2, 3, 4, 5, 6 and 7, that will be inserted in thefunction for i and than will be added.

Exercises summaExercise `summation notation :Write in explicit/elaborated way

One more detailed video explanatory videoTo how to handle summation notations (in english): http://www.youtube.com/watch?v=hEPk36Yncxg

Exercise `summation notation :Write in explicit /elaborated way

Fractions

33

Expanding + Reducing

• Expanding means multiplying the numerator and denominator of a fraction by the same non-zero number

• Reducing means dividing the numerator and denominator of a fraction by the same non zero number

cb

ca

b

a

⋅

⋅=

cb

ca

b

a

:

:=

34http://www.wyzant.com/help/math/elementary_math/fractions/expanding_and_reducing_fractions

• Addition:

• Subtraction:

• Multiplication:

• Division:

db

bcda

d

c

b

a

⋅

⋅+⋅=+

db

bcda

d

c

b

a

⋅

⋅−⋅=−

db

ca

d

c

b

a

⋅

⋅=⋅

cb

da

c

d

b

a

d

c

b

a

⋅

⋅=⋅=:

35

Exercise `fractions 1´

db

bcda

d

c

b

a

⋅

⋅+⋅=+

db

bcda

d

c

b

a

⋅

⋅−⋅=−

=+5

2

4

3)a =+

7

8

6

5)b

=++3

1

11

5

9

8)c

=−2

5

3

1)a

=−5

2

6

1)b

=−−11

5

9

7

5

8)c

36

db

ca

d

c

b

a

⋅

⋅=⋅

=⋅

=⋅

2

5

6

3.2

9

4

8

7.1

=⋅

=⋅

4

52

8

56.6

4

33

16

3.5

37

=⋅

=⋅

42

4

7

35.4

27

39

14

12.3

Exercise `fractions 2

cb

da

c

d

b

a

d

c

b

a

⋅

⋅=⋅=:

=

=

2

5:

6

3.2

9

4:

8

7.1

=

=

4

52:

8

56.6

4

33:

16

3.5

38

=

=

42

4:

7

35.4

2

19:

14

11.3


=

−

+

=

16

12

8

34

5

3

2

.2

2

36

5

.1

=

+

⋅

=

+

−

28

25

16

35

7

21

19

.4

9

5

27

128

3

11

5

.3

39


Percentage calculations

40

Percentage

• % means: divide a number by 100

• Example: 32% = 32/100 = 0,32

• Converting in percentage: multiply a number with 100

• Example: 0,144 = 0,144*100 = 14,4%

41

Exercise `express as a percentage´

=

=

344,0.2

2

1.1

=

=

66,0.4

15

1.3

=

=

293,0.6

6

1.5

42

Exercise :convert the percentages into fractions.

=

=

=

=

%78.4

%4,44.3

%2,63.2

%5,81.1

=

=

=

%35,0.7

%94.24.6

%09,16.5

43

X% of Y (percentage of…)

Example:

25

10025,0

100%25

100%25

=

⋅=

⋅=

of

44

Exercise `…percentage of…´

1. 12% of 243

2. 33% of 148

3. 39% of 3290

4. 0,88% of 5

5. 40,2% of 23910

45

Systems of linear equations

• Aim: the simplest method for solving a system of linear equations is to repeatedly eliminate variables.

• methods (3 possibilities) :

• by ´Equalizing´

• by ´Substitution´

• by ´Addition´

46

Generel formula:

47

System of linear equations with twovariables

222

111

)2(

)1(

cybxa

cybxa

=+

=+

´Equalizing´

3855102

822)2(

8)1(

=⇒+−=⇒=⇒=⇒

+−=−⇒−=

+−=

yyxx

xxxy

xy

Example

48

Exercise `system of linear euqations –solve by the method of `equalizing`

4933

24.)

17

82.)

=−

=−

−−=

−=−

xy

xyb

xy

xya

´Substitution´

6511

5

11

459

15444

11

154

11)2(

154)1(

=⇒

−=

−=⇒

+=

−=⇒

−=++⇒

+=

−=+⇒

=−

−=+

xx

y

yx

y

yy

yx

yx

yx

yx

Example

50

51

Exercise `system of linear euqations –solve by the method of `substitution`

64

90415.)

256

34116.)

=+

=+

+=

=+

yx

yxb

yx

yxa

´Addition´

Example

10

10

303

10

160166039

10037

203

10037

=

=⇒

=

=

=⇒

=−

=+⇒

=−

=+

y

x

y

x

xyx

yx

yx

yx

52

53

Exercise `system of linear euqations –solve by the method of `addition`

115338

27724.)

443

64

1

3

1

.)

=−

=+

=−

=+

yx

yxb

yx

yxa

Binominal theorem

222 2)( bababa ++=+

222 2)( bababa +−=−

)(*)(22bababa −+=−

54

Pascal`s triangle

55

Exercise Pascal`s triangle

56

Determine the following equations with Pascal`s triangle.

Exercise `binominal theorem´

calculate:

17²25.7

65536³32768²²6144³51216.6

8172²16.5

)365()365(.4

)35(.3

)³625(.2

)²52(.1

44

4

−

++++

++

+⋅−

−

+

−

x

mzmmzmzz

xx

xx

xy

x

x

57

Exponentsnmnm

aaa+=⋅)1

nm

n

m

aa

a −=)2

nmnmaa

⋅=)()3

n

n

aa

1)4 =−

74334xxxx ==⋅ +

25757 : xxxx == −

2,05

15 1 ==−

58

409622³)2( 12434 === ⋅

n

nn

b

a

b

a=

)5

nnnbaba ⋅=⋅ )()6

1)7 0 =a

aa =1)8

3222

5

5

55xxx

==

140 =

441 =

59

³³64³³³4)³4( babaab ==

Exercise `exponents´

)( nmnmaaa

+=⋅

=⋅ 34 33)a =⋅− 24) aad

=⋅ 04) bbb=⋅ 34) bae

=⋅ 44) bac

)( nm

n

m

aa

a −==

4

3

2

2)a =

−3

4

2

2)c

=4

0

)u

ub =

2

2

1

)x

xd

60

Roots

n

m

n maa =)1

aan n =)2

nn aa

1

)3 = 5125125 33

1

==

334 4 =

46444 22 32

3

===

61

nn

n

b

a

b

a=)5

nmnmm

nm n aaaa ⋅⋅ ===

11

)6

n

nx

x

11

)7−

=

nnn baba ⋅=⋅)4 5125255 333 ==⋅

5253

75

3

75===

2161616 422 === ⋅

5,044

12

1

==−

62

Exercise `roots´

=32 )2)(a =−32 )2)(b =42

1

))(xc =⋅ 520)d

=⋅ 32)3e =⋅ 22)3f =⋅ 3

2

3 22)h

=5

405)i =

4

3

81

27)j =

88

63)k

3 16) =g

63

Logarithmic functions

64

The equation exits of exactone real number.

This is the logarithm of the number b in respect to the base a

Logarithmic functions

)(log bx a=

bax =

baxb

short

xa =⇔=)(log

:

8³2

3)(log2

==

=

b

b

65

Special logarithms

• common logarithm:

(logarithm of x with respect to base 10)

• Binary logarithm:

(logarithm of x with respect to base 2)

• natural logarithm:

(logarithm of x with respect to base e) (mathematical constant e = 2,718…, called Euler's number )

)lg()(log10 rr =

)()(log2 rlbr =

)ln()(log rre =

66

Calculation of logarithm

)(log

)(log)(log

10

10

a

bb

a=

)4lg(

)256lg()256(log

4=

67

Exercises `logarithm 1´

( )

=

=

==

=

=

3

1log))100(log)

)1(log)27log)

25

1log))8(log)

310

53

52

fe

dc

ba

68

001,0log)10

g

Logarithmic rules

yxyx aaa loglog)(log)1 +=⋅

vuv

uaaa logloglog)2 −=

)15log()35log()3log()5log( =⋅=+

)625,0log(8

5log)8log()5log( =

=−

69

xkx a

k

a loglog)3 ⋅=

( ) )(log1

log)4 bn

b an

a ⋅=

)(log4)(log 4xx aa ⋅=

( ) ))(log)((log2

1)(log

2

1)(loglog 2

1

yxxyxyxy aaaaa +⋅=⋅==

70

Further rules

xx

aa log1

log)1 −=

01log)2 =a

)6log()6log()1log(6

1log −=−=

71

Exercise `logarithm 2´

Simplify the following terms:

2510

log4010

log) +a

nc 10log)

10

3log) af a

72

5010

log50010

log) −b

Exercise `logarithm 3´

=−+

=−+

=

=

)lg(*3²)²lg(*3

1.4

)lg(*3)lg(*5)lg(*3.3

7

³6lg.2

log.1

xyx

zyx

z

x

rs

xya

73

Calculation of xexample

)64,1()5log(

)14log(

)14log()5log(

)14log()5log(

145

≈=⇒

=⋅⇒

=⇒

=

x

x

x

x

74

Determine the solution

42

11

25

753.

107128.

943.

+

−−

−

=⋅

⋅=⋅

⋅=

xx

xx

xx

e

d

c

75

3

2

1 256log) xa =

2log) 2 −=xb

e-function/ln-function

76

Calculation of xexample

)2ln(26;)2ln(26

)2ln(26²

)2ln(26²

)2ln(3²5,0

202

21

3²5,03²5,0

+=+−=⇒

+−⇒

−=−⇒

=+−⇒

=⇒=− +−+−

xx

x

x

x

eexx

77

Exercise `e-function´

• Calculate:

10) =xea =⋅ )2exp()3exp()b

16)2ln() =xc =⋅+⋅ )43

4ln()

2

21

9

4ln()d

78

Determine x

3.4

0)1)(ln()1(.3

0)2()2(.2

0³)ln(³.1

1²

2

=

=−⋅−

=−⋅−

=⋅

+x

x

xx

e

xe

ee

xx

79

Trigonometric functions

80

radian b

Important angles in radian and degree

radian degree

0 0

π/6 30

π/4 45

π/3 60

π/2 90

π 180

2π 360

81

• Conversion of angles

� from degree to radian:

� from radian to degree:

• calculator:

deg = degree; rad = radian

• Sense of rotation:

clockwise rotating angles are negativ, counterclockwise rotating angles are positiv

απ

⋅°

=180

x

x⋅°

=π

α180

82

Exercise `converting angles´

1. 34° =

2. 126° =

3. -17° =

4. 0,94 =

5. -1,81 =

6. 5,97 =83

Trigonometric functions forright-angled triangles

84

The side opposite to the right angle (90 ) is called hypotenuse, here: c = hypotenuse

side b:vis-à-vis point B respectivelythe angle β.Side b is opposite the angle β � òppsite to the angle β.Side b is close to the angle α� àdjacent to the angle α

side a:vis-à-vis point A respectivelythe angle α.Side a is opposite the angle α � òppsite to the angle α.Side a is close to the angle β� àdjacent to the angle β

Definitions in right-angled triangles:

Sine (sin) of an angle= opposite of an anglehypotenuse

cosine (cos) of an angle = adjacent of an anglehypotenuse

tangent (tan) of an angle = opposite of an angleadjacent of an angle

cotangent (cot) of an angle = adjacent of an angleopposite of an angle

Special angles

86

sin cos tan

0° 0 1 0

30° 0,5

45° 1

60° 0,5

90° 1 0 -

180° 0 -1 0

Determination of angles

87

Angles are determined by the inverse trigonometric functions:

Determinations in a right-angled triangles

88

examplel:

a = 7,6 cm; c = 15,5 cm; γ = 90

Exercise `trigonometric functions´

89

Calculate the missing values!

1. b = 2,4cm; c = 3,2cm; γ = 90

2. a = 5,2cm; α = 66,5 ; γ = 90

3. c = 21,5cm; β = 72,3 ; γ = 90

4. b = 12,6cm; α = 32,3 ; γ = 90

Function of sine and cosine

�1,5

�1

�0,5

0

0,5

1

1,5

�450 �360 �270 �180 �90 0 90 180 270 360 450

sin(x) bzw.cos(x)

angle in degree

function of sine and cosine

sin

cos

90

Important correlations:

+=

2sin)cos(

πxx

Additions theorem:

)sin()sin()cos()cos()cos(

)cos()sin()cos()sin()sin(

yxyxyx

xyyxyx

⋅⋅=±

⋅±⋅=±

m

−=

2cos)sin(

πxx

)cot(

1

)cos(

)sin()tan(

xx

xx ==

91

Solving equations

• Quadratic equations

• Root equations

• Fraction equations

92

Solving quadratic equations

qpp

x

qpxx

−±−=

=++

2

2;1

2

)2

(2

0

quadratic equation (p-q-formula)

7;13

)91(2

6

2

6

0916²

21

2

2,1

=−=

−−

±−=

=−+

xx

x

xx

93

About quadratic equations• quadratic equations have at most 2 solutions

• The solving formula:

gives:

− one solution if:

− two solutions if :

− no solution if:

qpp

−±− 2)2

(2

0)2

( 2 =− qp

0)2

( 2 >− qp

0)2

( 2 <− qp

94

Exercise `quadratic equations´

37²6.4

)415()53()97()27(.3

4421²15.2

03522²3.1

=+

−⋅−=−⋅−

=+

=+−

xx

xxxx

xx

xx

95

Root equations

example

4

95

352

=→

=+→

=+

x

x

x

proof

33

354

=→

=+

squaring is not a equivalent conversion-- > that means you have to proof

96

Exercise `root equations´

6323.3

2563.2

5453.1

=−++

+=+

=+−

xx

xx

x

97

Fraction equationsexample

4;5

204

²9

2

9

0209²

2²32010

)2(3)2(10

12

310

21

2/1

==⇒

−±=⇒

=+−⇒

−=−−⇒

−⋅=−−⋅⇒

⋅=−

−

xx

x

xx

xxxx

xxxx

HNxx

98

Exercise `fraction equations´

a

t

a

bb

a

t

b

b

a

bt

a

t

ta

a

22.3

6

32

23.2

12

1.1

2−=−

+=+

=−

−

99

determine t:

application

density (material constant)

101

)(

)()(

i

ii

Vvolume

mmassdensity =ρ

L

kg

mL

g

m

kg

dm

kg

cm

g

unitstypical

;;;³

:

33=

102

Fresh snowfall has a density of 0,20 g/cm³.

a. which weight has fresh snowfall of 30 cm thicknesson a flat roof of 20 cm length and 10 m width?

b. If this amount of snow melts. How many liter ofwater are formed?

103

A irregular formed piece of jewellery (trinket) weighs0,177N in air, at a thin fiber the lifting power in water is

0,017N .

Is the trinket made of gold?

obtain: F=m*g

g= 9,81 N/kg; density (gold)= 19,3 kg/dm³; density (water)= 0,998 kg/dm³

Dilutions

104

Dilutions

dilute: the concentration of a solved substance in a solution is reduced.

105

100

ml

8 P

kt/1

00 m

L

+100 mLwater

shake10

0 m

l8 Pkt/200 mL=4 Pkt/100 mL

Add water

106

Initial volume + water = final volume

Dilution factor F

volumeinitial

volumefinal

ionconcentratfinal

ionconcentratinitialF

==

example: Create 50 mL of a physiologic salt solution 0,9% out of 10% salt solution:

initial concentration: 10%

final concentration : 0,9%

final volume: 50 mL

Needed initial volume ? Needed volume of water?

107

mLmLmLwaterV

mLmL

volume initial

lumeinitial vo

mL

%,

%

lumeinitial vo

mefinal volu

entrationfinal conc

nncentratioinitial coF

5,455,450)(

5,4%10

%9,050

50

90

10

=−=⇔

=⋅

=⇔

=⇔

==

Exercise `dilutions´

1. You have a 10 times concentrated buffer which should be diluted to one times. Create 500 mL of the buffer. How many mL buffer and water are needed?

109

Exercise `dilutuions´

2. For determining the concentration of extracted DNA you dilute 100µL of the DNA solution with 400µL water. The diluted solution has a concentration of 50 mg DNA per mL. What is the concentration of the initial solution? Calculate the dilution factor! How many DNA is isolated if the volume of the initial solution was 5mL?

110

Diagrams

111

112

informative diagram title

Especially if more than twographs are presented in thediagram a legend is necesarry.

axis labeling including units,x-axis: axis of abscissae

(independent (preset) dimension)

axis

lab

elin

gin

clu

din

gu

nit

s,y-

axis

: axi

so

fo

rdin

ates

(dep

end

ent

dim

ensi

on

)

Axis classification in a well spent way ofusing the given place

Single data – sign points or crosses

Best-fit-curve(=regression line, = trend line)

Diagram - example

113

Exercise `diagrams/linear equations´

114

1.) Measurements of the solubility L of a salt in water depending on the temperature T give following data:

a.) Draw the corresponding diagram.b.) Determine the equation of the regression line by the diagramc.) Determine the solubility of the salt by 36,5 C

c1.) graphically (in /by diagram)c2.) calculative (by the linear equation)

i 1 2 3 4 5 6Ti [°C] 0 20 40 60 80 100

Li [g/100mL] 70,7 88,3 104,9 124,7 148,0 176,0

Exercise `diagrams/linear equations´

115

2.) The concentration of an apple juice sample is determined. Therefore standards

(solutions of juices with known concentrations) are prepared. The sample and the

standards are measured by a photometer. That means which concentration causes

which „colour intensity“ (extinction) and in the other way around which „colour

intensity“ (extinction) is which concentration?

a.) Draw the diagram on millimetre

paper.

b.) Determine the equation of the

regression line.

c.) Calculate the concentration of

the sample.

Solution Concentration in % Extinction

Sample

Exercise `linear equations´

3.) Two points of a straight line are known (pair of variates): P(2/3) and Q (-1/-3).Determine the linear equation.

4.) A layer of lipids is 100 nm and grows 5 nm per day .a.) Give an equation to calculate the lipid layer at any

time.b.) How thick is the lipid layer after one week?

Arithmetic mean

a) 5,0,8,6,9,5

b.) 22,94; 22,90; 22,92; 22,76; 22,80; 22,85; 22,84; 22,86; 22,83; 22,87

c.) 6m, 7m, 4m, 4m, 5m, 3,m 4m, 7m, 0m, 5m, 5m, 6m, 2m 117

Definition of the arithmetic mean:

∑=

=+++⋅=n

i

in xn

xxxn

x1

21

1)...(

1

mathematics –bridging course for applied sciences · 2015-08-06 · mathematics –bridging...

Documents