SABA Math Course Program A

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SABA Math Course

Program A

Answerbook

Version 2013

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Math4all Foundation October 3, 2013 page 1

Content

1 Tables and graphs 3 1.1 Tables 3

1.2 Percentages 4 1.3 Graphs 5

1.4 Finding more values 6 1.5 Combine / Compare 7

2 Formulas 8

2.1 Using formulas 8 2.2 Plotting graphs 9 2.3 Equations 10 2.4 Inequalities 10 2.5 More variables 11

3 Linear relations 12

3.1 Directly proportional 12 3.2 Linear functions 13 3.3 Linear models 14 3.4 Linear equations 14

4 Exponential relations 16 4.1 Exponential growth 16 4.2 Calculations with powers 17 4.3 Real indices 18

4.4 Exponential functions 19

5 Change 20 5.1 In graphs 20 5.2 Per step 21

5.3 Average change 22

6 Counting 23 6.1 Possibilities 23

6.2 With or without repeating 24 6.3 Combinations 24

6.4 Pascal's triangle 25

7 Probability 26 7.1 Experiments 26 7.2 Reasoning 27 7.3 Tree diagrams 28

7.4 Probability distribution 28

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8 Normal distribution 30 8.1 Normal curve 30 8.2 Normal probabilities 31 8.3 Standardising 32 8.4 Normal or not 33

9 Probability models 34 9.1 Yes/No probabilities 34 9.2 Binomial distribution 35 9.3 Non-binomial distributions 35 9.4 Probability models 36

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1 Tables and graphs

1.1 Tables

a

1 The number of students in the educational system and the number of eductional institutions.

b 9

c P.e.: the number of students at ‘Spec. voortgezet onderwijs’ in 2002/’03 suddenly becomes 0, while the number of students in ‘Voortgezet onderwijs’ increases strongly.

And the number of schools in the categories ‘Speciale scholen’ and ‘Voortgezet onderwijs’ increases in 2002/’03.

d In 2002/’03: 913671 /692 ≈ 1320 students.

In 2007/’08: 941469 /658 ≈ 1431 students.

e 13 universities. An average of 166299 /13 ≈ 12792 students per university in 2000/’01.

An average of 212728 /13 ≈ 16364 students per university in 2007/’08.

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SABA PROGAM A > FORMULAS AND GRAPHS > TABLES AND GRAPHS

f The number of hbo-institutions decreased from 62 to 51.

On average there were³¹²⁶⁹⁸₆₂ ≈ 5044 studenten per hbo-instelling in 2007/’08.

On average there were³⁷⁴³⁷⁷₅₁ ≈ 7341 students per hbo-institution in 2000/’01.

a

2 Go Ahead Eagles (Deventer) leads with 39 points out of 18 matches.

b 12 ⋅ 3 + 3 ⋅ 1 + 3 ⋅ 0 = 39. A team gets 3 points per match won and 1 point for a draw.

c 597 goals (total number of goals scored)

d The goal difference is the differende between the number of goals scored and the number of goals against. De Graafschap has a goal difference of 36 − 14 = 22; SC Cambuur has a goal difference of 44 − 29 = 15.

e For De Graafschap (goal average 2,57) en Cambuur (goal average1,52) it would make no diﬀerence.

But FC Oss would be ranked above FC Omniworld. And FC Eindhoven above Fortuna Sittard.

a

3 Yes, it seems to be right. The number of parent birds remains 2, and they can only gather so much food per day.

b Apparently the parent birds try harder with more young.

With 12 young they even collect 12 ⋅ 0,70 = 8,40 gram per day.

c More young in one nest means that they can proﬁt from each other’s heat production.

d Make a table of the total heat production. With 12 young only 0,177 kcal of heat production per young is needed to stay warm. The rest can be used to grow.

a

4 See table.

b ₆₅₀₀⁶¹² c ₆₆₅⁵³

1.2 Percentages

a

1 € 37.49 b € 82.80

c 0.8 ⋅ 33.50 = 26.80; it is not correct.

a

2 € 1035.00 b € 1071.23

c No, the interest is compounded. The second next year you get interest over the original amount plus the interest from the previous year.

3 It makes no diﬀerence, for instance 100 × 0.6 × 1.21 = 100 × 1.21 × 0.6.

4 ₂₀₀⁹⁵ = 0.475, so 47.5%.

5 You get 300 grams for € 1.75.

If 250 grams costs € 1.75, 300 gram should cost ³⁰⁰₂₅₀⋅ 1.75 = 2.10 euro.

Your discount is 0.35 euro, approximately 16.7% of the price.

a

6 0.35 ⋅ 0.40 = 0.14, so 14%.

b 0.35 ⋅ 0.60 = 0.21, so 21%.

c _0.14⁶⁴⁰ ≈ 4571 kg.

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a

7 91.281 mln is 61.93%, so the total population was about ^91.281_0.6193≈ 147.394 mln.

b 132187 km²is 6.95%, so the total surface area was approximately ¹³²¹⁸⁷_0.0695 ≈ km². c Sulawesi: 188866 km²and 10.377 mln inhabitants.

The remaining islands: 142267 km²; 7.48%; 7.399 mln inhabitants; 5.02%.

d Java/Madura: approximately 691 inhabatants per km². e ^147394000= 77 inhabitants per km².

1.3 Graphs

a

1 A line graph is used.

b The number of victims.

c The number of victims decreases, except for the peak years 1976 en 1979. To study exact numbers you should use the table. The graph shows the trend better.

d You should compare the numbers of victims to the total size of the population for a better understand- ing.

e The number of burn victims was 5325. There is a total population of ⁵³²⁵_2.3 × 100000 = 231521739 in the USA.

a

2 ℎu�u�u�ℎu� and u�u�u�

b u�u�u�u�ℎu� and ℎu�u�u�ℎu�

c The graphs show the average development of height and weight of a child over time.

d In the bottom half you read out the height on the P₅₀line (approximately 170 cm). In the top graph you can read out the weight between the P₁₀and P₉₀lines correponding to <20 years (approximately 60 kg).

e Just do it.

f Marleen is slightly taller than average. Her weight is about average.

a

3 The water level changes continuously.

b Vase 1 belongs to graph 2.

Vase 2 belongs to graph 4.

c All graphs reach the level of 20 cm in half the time . a

4 De period is exactly 1 year.

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b Nature is blooming in this period and there is a lot of green foliage. So more CO₂is transformed into O₂.

c The minimum was 348 m³. And the maximum 360 m³.

d The total volume of air remains approximately equal. The plants determine whether there is a lot of carbondioxide (and less oxygen) or little carbondioxide (and more oxygen) in the air.

e Trend line ratio oxygen/nitrogen passes through −80 in jan.1989 and −150 in jan.1992.

f Trend line CO₂passes through 352 p.p.m.v. in jan.1989 and 356 p.p.m.v. in jan.1992. Yes, when less CO₂is changed into O₂the amount of CO₂increases and the amount of O₂decreases. This could be the result of deforestation of the tropical rainforests.

g Per 3 years there will be an increase of 4 p.p.m.v. CO₂ 8 356 +⁸₃⋅ 4 = 366²₃

1.4 Finding more values

a

1 Marleen’s measurements are between P₅₀and P₉₀: 173 cm b _{50 90}173 Tussen 40 en 50 km.

c After puberty, at about 16 years, girls hardly grow any more.

d 173 cm

7075 a

2 1944 and 1978.

b Not necessarily, the number of Americans may have increased.

c 1930-1935, 1947, 1955, 1959-1961 and 1970-1976.

d Approximately 2400 calls (an increase of 120 in 1983-1986, so 40 per year from 1986).

Approximately 530 packages ( an increase of 30 in 1984-1986, so 15 per year from 1986).

a

3 See the graph.

b From the wavelike behaviour of the graph.

c The trend is that unemployment increases approximately by a 1000 every year. In Jan.’95 there were 10700 unemployed. In Jan.’04 there will be (if the trend continues) approximately19700 unemployed.

d Mrt.’08: approximately 11900 + 13 ⋅ 1000 = 24900.

a

4 𝐵𝑀𝐼 ≈ 25

b Between 70 and 85 kg.

c -

d Puberty greatly inﬂuences people’s growth. After that length and weight remain more or less stable.

e Straight line through (0, 0) and (30; 97.2).

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1.5 Combine / Compare

a

1 The number of immigrants was equal to the number of emigrants at that time.

b At that time the number of Maroccan immigrants was equal to the number of Turkish immigrants.

c For the Turkish people between 1982 and 1985.

d -

e Possible answers: unemployment is large within these groups, a stricter immigration policy, an im- proved perspective in the country of origin.

f See the graph to the right.

g Decreasing most 1980.

a

2 The two cyclists have covered the same distance at the points of intersection.

b During the ﬁrst 95 minutes and between 115 and 160 minutes.

c At around 40 minutes, 95 minutes and 140 minutes.

a

3 2600 pieces; approximate proﬁt 200 kEuro.

b After 1000 pieces.

c After more than 2 months.

d See table.

4 Making a table by chaining the graphs:

age 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

length 97 104 112 118 124 130 136 142 148 157 161 164 167 168 168 weight 12.5 16.5 19 21.5 24 27 30 34 37 40 45 52 60 62 64 a

5 This point of intersection does not mean anything, because the number of horses and mules and the number of tractors have diﬀerently scaled axes.

b That should be between 1950 and 1960. Only then does the number of animals fall under the 10 mln.

A small table clariﬁes this : in 1952 both seem to be around 4 mln.

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2 ^Formulas

2.1 Using formulas

a 1 cm³

b 804.25 cm³ c 𝑉 = 16𝜋u�² d -

e for instance ℎ =⁽¹⁰⁰⁰⁾_(𝜋u�2). You could have made r the subject as well.

a

2 Relation between two variables. GR: Y1=X^3 b Relation between two variables. GR: Y1=400–5*X^2 c Not a relation between two variables.

a

3 u� = −0.5u� + 2.5

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SABA PROGAM A > FORMULAS AND GRAPHS > FORMULAS

b u� =²₃u� + 2 c u� = ±4 ⋅ √(u�) d u� = 6 /u�²

a

4 −2u�³− 12u�² b −u�²− 8u�

c u�²+ 15u� − 100 d 3u�³− 2u�²+ 3u� − 2

e u�²− 9

f 36u�²− 36u� + 9 a

5 3000 units b 3400 units c € 200

d 𝑅 = u� ⋅ u� = u� ⋅ (4000 − 20u�) = 4000u� − 20u�² e -

f u� = 100

2.2 Plotting graphs

a

1 𝑉 = 10𝜋u�²

b Enter Y1=10πX^2. Use the table to determine where u� gives a volume closest to 1000 cm³. You will ﬁnd u� ≈ 5.6 cm.

c 𝑉 = 2𝜋u�³

d The volume of the tin can is: 500 cm³. Make a table corresponding to: Y1=2πX^3. You will ﬁnd u� ≈ 4.3 cm.

a

2 𝐾 = 200 + 0.04u�

b 𝐼 = 0.10u�

c Use the table to find the first value of u� for which 𝐾 < 𝐼. You will find u� = 3334 a

3 Enter: Y1=250-0.5X^2 with window settings: 0 ≤ u� ≤ 500 and 0 ≤ u� ≤ 40000.

b Enter: Y1=0.04+200/X with window settings: 0 ≤ u� ≤ 100 and 0 ≤ u� ≤ 100.

c Enter: Y1=60/(30+0.5X) with window settings: 0 ≤ u� ≤ 500 and 0 ≤ u� ≤ 3.

a

4 12.83

b 𝐺𝑇𝐾 = ¹⁰⁰_u� + 0.1u�

c v.a.: u� = 0

d When the value of u� becomes very large, so does the value of 𝐺𝑇𝐾 become very large.

a

5 u� and u� represent the length and the width of the printed part of the poster, given in cm. 1 m²= 10000 cm². The size of the poster is (u� + 25) (u� + 20) = 10000.

b u� + 20 =_(u�+25)¹⁰⁰⁰⁰ and therefore u� = _(u�+25)¹⁰⁰⁰⁰ − 20.

GR: Y1=10000/(X+25)-20 with window settings: 0 ≤ u� ≤ 500 and 0 ≤ u� ≤ 400.

c When u� = 0, it follows that u� = 380 and when u� = 0, it follows that u� = 475. In conclusion 0 ≤ u� ≤ 475 and 0 ≤ u� ≤ 380.

d In the table you can ﬁnd that u� = u� when u� ≈ 77.5 cm.

The dimensions of the poster will become 97.5 by 102.5 cm.

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SABA PROGAM A > FORMULAS AND GRAPHS > FORMULAS

2.3 Equations

a

1 u� = 37.5

b u� = ±√(37.5) ≈ ±6.12 c u� = 16

d u� = ±4 e u� = 7 of u� = 3

f u� = 7 g u� = 0.03

h there are no solutions because √(−4) is not a real number a

2 u� = 4

b u� = 1 ∨ u� ≈ 1.35 a

3 -

b 140 m³costs € 283 per year and 160 m³costs € 318, therefore between € 283 and € 318.

c 38 + 1.75u� = 250 gives u� ≈ 121.143 therefore a maximum of 121 m³. a

4 u� = 0 gives u� = −200 u� = 0 gives u� = 300 b u� = 0 gives 𝑊 = 0

𝑊 = 0 gives u� = 0 of u� = 200 c u� = 0 gives u� = ±√(96) ≈ ±9.80

u� = 0 gives u� = 8 of u� = −12 d u� = 0 gives u� = 20

u� = 0 does not lead to any value for u�

a

5 𝑉 = 200𝜋(1.5 + 0.5u�)²− 450𝜋

b Y1=200π(1.5+0.5a)^2-450π with window settings 0 ≤ u� ≤ 1000 and 0 ≤ u� ≤ 160000000 c Approximately 23 dips

2.4 Inequalities

a

1 u� < ¹₃

b u� ≤ −8 or u� ≥ 8 c u� > 8000

2 u� < −9 of u� > 10 a

3 11.5 ct/km b 3665 euros

c 1825 + 0.115u� < 4000 gives u� ≤ 18913

d 𝐾 (u�) = 2050 + 0.10u� when u� < 15000 𝐾 (u�) = 1825 + 0.115u� when u� ≥ 15000 a

4 u�𝐴(u�) = 110u� + 24 and u�𝐵(u�) = 120u�

b after 144 minutes

c between 2 hours and 2.8 hours

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2.5 More variables

a

1 -

b At u� = 8 on the u� -axis go straight up until you reach the graph corresponding to 𝐷 = 10; then read oﬀ the answer on the 𝑃 -axis.

c -

d 𝑃 = 0.00013 ⋅ 25²⋅ 20³= 650 kW.

a

2 The heating costs when there are no sun-hours and the outside temperature is 20°C.

b u� = 800 − 60 ⋅ 3.5 − 50 ⋅ (−4) = 790, so € 790 per day.

c When 60u� + 50u� = 800, for example when there are 5 sun-hours and there is an outside temperature of 30°C, or 10 sun-hour with an outside temperature of 24°C.

d - e 𝐾 = 340

f The costs are highest when u� = −2 and u� = 4, i.e. € 660. The costs are lowest when u� = 2 and u� = 10, i.e. € 100. Therefore, the cost can vary between € 100 and € 660.

a

3 u� = 0.5u�

b The number 3.6 arrives from the conversion of u� in km/h to m/s.

c u� = (14.4+1.8u�) u�

d 𝑁 = (14.4+1.8u�)^(60u�)

e The speed should be 70 km/h.

a

4 𝑉 = 50, so u� = 65, u� = 19.25 and u� = 0.39.

𝐿 = 1, 𝑆 = 2 and 𝐷 = 40 gives 𝐵 = 65 ⋅ 1 + 19.25 ⋅ 2 + 0.39 ⋅ 40 = 119.1 mL.

𝐵_{(u�⊤u�)}= 19.25 ⋅ 2 = 38.5 mL, so 32.2%. 𝐵_{(u�u�u�u�)}= 0.39 ⋅ 40 = 15.6 mL, so 13,1%.

b Car 1 drives at 50 km/h = 13.9 m/s, so for 600 m the ﬁrst car takes 43.2 s.

Car 2 drives at 70 km/h = 19.4 m/s, so for 600 m this car will take 30.9 s.

Therefore the second car must wait for 12.3 s.

c Car 1: 𝑉 = 50, so u� = 65, u� = 19.25 and u� = 0.39. 𝐿 = 0.9, 𝑆 = 0 and 𝐷 = 0, so 𝐵 = 58.5 mL.

Car 2: 𝑉 = 70, so u� = 91.6, u� = 37.73 and u� = 0.39. 𝐿 = 0.9, 𝑆 = 1 and 𝐷 = 12, so 𝐵 = 124.85 mL.

Car 2 needs more than twice the amount of petrol.

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3 Linear relations

3.1 Directly proportional

a

1 Own answer b Cyclist A: u� = 20u�

Cyclist B: u� = 150 − 25u�

c Cyclist A.

d u� = 3¹₃ hours a

2 Not directly proportional

b Directly proportional; constant of proportionality is 3.

c Not directly proportional

d Directly proportional; constant of proportionality = ¹₃. e Directly proportional; constant of proportionality = ¹₃.

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SABA PROGAM A > FORMULAS AND GRAPHS > LINEAR RELATIONS

f Directly proportional; constant of proportionality = −¹₃. a

3 Yes.

b No.

c Not necessarily. If you had covered no distance at all at u� = 0, it is.

d No, the speed increases all the while, so the distance travelled per second changes.

e Yes, in similar rectangles the ration of length and width remains the same.

f No.

a

4 𝐾 = ₆₀₀⁷¹u�

b 𝐾 = ₂₀₀¹¹u�

c From 22106 km per year. Use your graphic calculator to find the first value of a where more than 1400 euros difference between both values for 𝐾 occurs.

a

5 De call-out charge is not included in the wages.

b The call-out charge is now included.

c 𝑇𝐾 = 22.50u� + 30 a

6 26 km in 45 minutes converts to 34²₃ km/h.

b u� = 34²₃u�

3.2 Linear functions

a

1 Own answer b Cyclist 1: u�1= 20u�

Cyclist 2: u�2= −25u� + 150

c u�1= u�2leads to u� = 3¹₃(using the graphic calculator).

They meet after 3 hours and 20 minutes.

a

2 3u� − 5 = 0 so 3u� = 5. It follows that: u� = ⁵₃. The points of intersection with the axes are: (⁵₃, 0) and (0, −5).

b u� − 4 = 0 so u� = 4. The points of intersection with the axes are: (4, 0) and (0, −4).

c −0.5u� + 4 = 0 so 0.5u� = 4 giving u� = 8. The points of intersection with the axes are: (8, 0) and (0, 4).

d −2 (u� + 3) = 0 so u� + 3 = 0 giving u� = −3. The points of intersection with the axes are: (−3, 0) and (0, −6).

a

3 75 euros b 0.09 euros

c u� = 87.50 + 0.10u� where u� is given in € and u� in km a

4 For each hour a similar increase of 35 euros can be seen;

compare 5 hours worked with 10 hours worked, the costs do not double as well b u� = 35u� + 65

c 𝐾 = 35 ⋅ 6 + 65 = 275 d 𝐾 = 35 ⋅ 2⁵₆+ 65 ≈ 164.17

a

5 u� = 200 gives u� = 2.5 ⋅ 200 − 300 = 200. The proﬁt is € 200.

b The expenses made to organise the party.

c The takings per visitor (the entry fee).

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SABA PROGAM A > FORMULAS AND GRAPHS > LINEAR RELATIONS

3.3 Linear models

1 u� : u� =¹₃u� + 3¹₃; u� : u� = ²₃u� + 1; ℎ : u� = −2u� + 4; u� : u� = −5u� − 2 a

2 Own answer

b The plotted points do seem to ﬁt a straight line, so yes: linear.

c Best ﬁt, in this case, is a line where an equal number of plotted points are above the line, as there are below the line.

d The line goes through the points (16, 57) and (44, 120). The formula is given by 𝐻 = 21 + 2.25𝐵.

e The diﬀerence is small.

f 𝑃 = 21 + 2.25 ⋅ 32 = 93

g Linear interpolation: ^(91+94)₂ = 93.

3 After 3 hours a decrease of 6 cm, so the gradient is −2. 2 hours after the start, the candle is 12 cm, so 2 hours earlier the length of the candle was 12 + 4 = 16 cm. The y-intercept is 16.

Formula: 𝐿 = 16 − 2u�.

a

4 𝑉(u�) = ^𝑉(0)₂₇₃ ⋅ (u� + 273) = 𝑉(0) ⋅ (₂₇₃^u� + 1)

b 𝑉 (0) is a constant coeﬃcient, therefore the formula can be seen as 𝑉 (u�) = u� ⋅ u� + u�.

The pressure must remain constant though. The domain is 𝐷 = [−273, → 〉 c Enter: Y1=1+1/273X. Window settings: −300 ≤ u� ≤ 300 and −1 ≤ 𝑌 ≤ 3.

d 𝑉 (20) = 1 +₂₇₃²⁰ = 1.073 m³

e 1.5 = 1 +₂₇₃¹ u� leads to u� = 136.5. So at 136.5°C.

a

5 u� (0) is the distance travelled at u� = 0 and u� is the speed in m/s.

b u� (u�) = 20u�. Enter: Y1=20X. Window settings: 0 ≤ u� ≤ 50 and 0 ≤ u� ≤ 1000.

c u� (u�) = 400 + 15u�, so enter Y2=400+15X.

d You may use the graphic calculator. 20u� = 400 + 15u� leads to u� = 80 a

6 In a graph you can see that it is quite reasonable to assume a linear relation: 𝑁 = 300𝐿 − 10000.

b 𝑁 = 300 ⋅ 85 − 10000 = 15500.

c You may use the graphic calculator: 300𝐿 − 10000 = 4500 leads to 𝐿 = 48.

d Yes, there is a linear relation between 𝐿 and 𝑁, so linear extrapolation is possible. The question remains whether a salmon can grow to 120 cm though.

3.4 Linear equations

a

1 𝑃 = 2, 5u� − 400

b 2, 5u� − 400 = 0 gives you u� = 160

c 2, 5u� − 400 = 1000 gives you u� = 560, so more than 560 tickets sold.

2 u� = 12 − 0, 12u� and u� = 2 + 0, 1u�

12 − 0, 12u� = 2 + 0, 1u� gives you u� = 45, 454545..., so the solution of the inequality is u� ≥ 45, 46.

a

3 𝐶 = 2, 25 + 0, 75u�

b 2, 25 + 0, 75u� = 6 gives you u� = 5, so after more than 5 minutes.

c At a speed of 60 km/h the journey would take 10 minutes, so a rail-taxi would be cheaper.

a

4 u� = 3

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b u� ≤ −4 c u� > 3, 75 d u� ≤ 5166²₃

e multiply both sides by 20 to get 24 − 8u� = 20 − 5u� and therefore u� =⁴₃ f work out the brackets to get 120 + u� = 4u� + 60 and therefore u� = 20 a

5 𝑇𝑅 = 10u� and 𝑇𝐶 = 6, 5u� + 83000

b Calculator: Y1=10X and Y2=6.5X+83000 with settings: 0 ≤ u� ≤ 30000 and 0 ≤ u� ≤ 300000 You ﬁnd: u� ≈ 23714, 3

c 10u� = 6, 5u� + 83000 gives you u� ≈ 23714, 3 d When u� > 23714

a

6 𝐶 = 1600 + 3u�

b 𝑅 = 5u�

c 5u� = 1600 + 3u� gives you u� = 800, so proﬁt is made with more than 800 vases sold.

d 𝑃 = 𝑅 − 𝐶 = 5u� − (1600 + 3u�) = 2u� − 1600 = 2000 when u� = 1800

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4 Exponential relations

4.1 Exponential growth

a

1 1.5

b 1.5²= 2.25, so the surface area increases with 125% in two days.

c yes, with growth rate 1.5 until the entire surface is covered.

a

2 After 1 year € 4440 and after 2 years € 4928,40.

b 1.11

c Multiply by 1.11.

d Divide by 1.11.

e ^(7279.45)_(6740.23)≈ 1.08, so the growth rate is 1.08 and the percentage of growth is 8%.

a

3 𝑁 (u�) = 5000 ⋅ 0.96^u�

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SABA PROGAM A > FORMULAS AND GRAPHS > EXPONENTIAL RELATIONS

b 𝑁 (10) ≈ 3324

c 𝑁 (17) ≈ 2498, so after 17 years.

a

4 Calculate the ratios of each two concecutive numbers. You ﬁnd approximately 1.042.

b 4.2% per year.

dc After 10 years.

e 𝐾 (5) ≈ 19254.15 and 𝐾 (10) ≈ 23425.61 f It makes no diﬀerence.

a

5 School 1 with a growth rate of 0.95, so 5% decay.

b Linearly, 45 students less per year.

c No, school 2 will have 0 students in the end and in school 1 the number of students decreases slower and slower.

d The number of students only changes incidentally during a year. There is a structural change of the number at the beginning of the school year only, depending on the applications and the exam results.

4.2 Calculations with powers

a

1 2

c

b 2¹⁰= 1024 d After 10 years.

a

2 2⁷= 128 b 2¹¹= 2028 c 2³= 8

d 2⁶²≈ 4.6117 × 10¹⁸ a

3 0.87

Formula: 𝐶 = 150 ⋅ 0.87^u�

b GR: Y1=150*0.87^X en Y2=75

using the graphic calculator you ﬁnd u� ≈ 4.98 hours.

c 0.87⁽²⁴⁾≈ 0.035, so a decay percentage of approximately 96.5%.

a

4 3³

b 3¹ c 1 = 3⁰ d ⁴₃

a

5 ₅₀₀₀⁶⁰⁰ = 0.12, so 12%.

b 0.88

dc (0.88)⁵≈ 0.528, so after 10 months € 1393.21 e (0.88)⁽¹⁰⁾≈ 0.279 and (0.88)⁽¹⁵⁾≈ 0.147

f 6 months

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SABA PROGAM A > FORMULAS AND GRAPHS > EXPONENTIAL RELATIONS

4.3 Real indices

a

1 𝐴 (10) = 25000 ⋅ 1.1¹⁰≈ 64844 b 𝐴 (10₁₂⁷) ≈ 68551

c 1.1 d 1.1⁽

1 12)

≈ 1.008 so approximately 0.8% per month.

e 𝐴 (−5) ≈ 15523 and 𝐴 (−10) ≈ 9639 f Check that 𝐴 (−5) ⋅ 1.1⁽⁻⁵⁾= 𝐴 (−10).

a

2 1-1-2001: € 7518.15 1-1-2000: € 7092.60 1-1-1999: € 6691.13 b On 1 January 1996.

c He deposited € 5000 on 1 January 1994.

a

3 u�(3 hours)= ³⁰⁰⁰₁₂₀₀ = 2.5 b u�_{(1 hour)}= (2.5)⁽

1 3)

≈ 1.357 so 35.7% per hour.

c 𝐻 (u�) = 1200 ⋅ 1.357^u�

d Around two and a quarter hours before u� = 0.

a

4 See table

period growth rate per yer growth percentage

0-1500 1.00046 0.05%

1500-1800 1.002313 0.23%

1800-1950 1.00463 0.46%

1950-1986 1.01944 1.94%

b See table

period growth rate per year growth percentage

1500-1750 1.00115 0.12%

1750-1800 1.00814 0.81%

1986-1997 1.01735 1.74%

5 The permitted content is called 𝐴, after the accident 6𝐴.

Then (¹₂)^u�⋅ 6𝐴 = 𝐴 and this gves (¹₂)^u�=¹₆.

Using the GRC you ﬁnd u� ≈ 2.58, so 2.58 periods of 8 days. That is 20.68 days. The hay should be stored 21 days.

(21)

4.4 Exponential functions

a

1 S = 10000 x 1.05^t

b Solve 10000 ⋅ 1.05^u�= 15000 that is 1.05^u�= 1.5.

u� = 8 gives 1.4774 and u� = 9 gives 1.5513, so 9 years.

c Solve 1.05^u�= 2. u� = 14 gives 1.9799 and u� = 15 gives 2.0789. So 15 years.

a

2 1 ⋅ 1.05^u�= 4000. Table: if u� = 169, then € 3810.58 and if u� = 170, then € 4001.11. So 170 years ago.

b Yes, you could choose −170 ≤ u� ≤ −160.

c No, there is a horizontal asymptote 𝑆 = 0.

a

3 u�_{(4 months)}=¹⁶³⁰₂₀₀₀ ≈ 0.815, so u�_(year)= 0.815³≈ 0.541.

1 year before 6-1-1997 the radiation was 2000 ⋅ 0.541⁽⁻¹⁾≈ 3695 Bq.

2.5 years after 6-1-1997 the readiation was 2000 ⋅ 0, 541^(2,5)≈ 431 Bq.

b 𝑆 = 2000 ⋅ 0.541^u�

c 10 years ago the radiation was 2000 ⋅ 0.541⁽⁻¹⁰⁾≈ 931231 Bq. So B = ⟨0; 931231].

d Solve 0.541^u�= 0, 5. 13 months gives 0.514 and 14 months gives 0.4883. So after 13 months and 16 days, so from22-2-1998.

a

4 u� = 2000 ⋅ 1.04^u�and u� = 1500 ⋅ 1.06^u�

b Enter: Y1=2000*1.04^X and Y2=1500*1.06^X. Window: X from 0 to 20 and Y from 0 to 1000.

c You ﬁnd u� = 15.1028... years. This is 15 years and 1.2 month after 1-1-2000, so from 1-3-2015.

5 Both curves go through (0, 10).

For curve u�: at u� = 1 the function has the value u� = 10 and at u� = 2 the value 40, so u� = 10 ⋅ 2û�. For curve u�: at u� = −1 the function has the value u� = 30 and at u� = −2 the value 90, so u� (u�) = 10⋅(¹₃)û�. 6 Intersecting 𝐻 = 650 ⋅ 1.055û�and ℎ = 650 + 50u� gives u� ≈ 12.81.

So after 13 years.

(22)

5 _Change

5.1 In graphs

1 〈 ←, −1〉: decreasingly increasing

〈 − 1, 0〉: increasingly decreasing

〈0, 1〉: decreasingly decreasing

〈1, → 〉:increasingly increasing max. u� (−1) ≈ 4 and min. u� (1) ≈ −4

In (0, 0) de rate of decrease is largest, the slope steepest.

2 〈 ←, −3〉: decreasingly decreasing

〈 − 3, 0〉: increasingly increasing

〈0, 3〉: decreasingly increasing

(23)

SABA PROGAM A > FORMULAS AND GRAPHS > CHANGE

〈3, → 〉: increasingly decreasing

max. u� (3) = 27 and min. u� (−3) = −27

In (0, 0) the increase is fastest, the slope steepest.

a

3 max. u� (0) = 8, min. u� (−2) = 0 and min. u� (2) = 0 b 〈0, 1〉

a

4 After 60 seconds, because at that point there is a sharp change in speed of descent in the graph. The speed is constant from there onwards.

b The speed is constantly increasing.

c The graph is a straight line. u� = 1000 /100 = 10 m/s.

5 There is a maximum temperature at 2:30 pm, then the curve goes from increasing to decreasing.

5.2 Per step

a 2

1 GR: Y1=0.5X^4–4X^2+8 en Y2=Y1(X)–Y1(X–1); table with step size 0,5 from u� = −3.

b 〈0, 1〉

c Three sign changes in the increments.

ba

3 Around 3:00 pm.

c No, not exactly, the measurements were done hourly only.

a

4 Make a table for 𝑉

(24)

dbc Decreasingly decreasing, the negative increments become smaller.

e The increments approach zero. The graph of 𝑁 has a horizontal asymptote.

5.3 Average change

a

1 ^(Δu�)_(Δu�) =²₁ = 2 b −²₃

c 𝐷 en 𝐹 en 𝐴 en 𝐸.

d It is negative.

2 ⁴₂= 2 a

3 ⁽²⁻⁶⁾₍₂₋₀₎= −2

b 0

c The points are at the same height, same u� -value.

d P.e. on [−1, 0] with ^(Δu�)_(Δu�) =⁽⁶⁻²⁾₁ = 4 a

4 u� = 0 gives 𝑇 = 90 °C.

b ^Δ𝑇_Δu� =^{𝑇(5)−𝑇(0)}₅₋₀ ≈^49,95−90₅ ≈ −8,8 °C/min.

c Approximately 3,3°C/min.

d The difference quotient become smaller, the coffee cools slower because the temperature difference between the coffe and the room becomes smaller.

5 ^Δu�_Δu� = ^3(u�+1)₁²^−3(u�)= 6u� + 3

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6 Counting

6.1 Possibilities

a

1 -

b 4 × 4 × 4 × 4 × 4 × 4 = 4096 c 4 × 4 = 16

d 2 × 2 × 2 × 2 × 2 × 2 = 64 a

2 1000

b 24 a

3 -

b Too many (216) diﬀerent branches.

c 25 × 3 = 75

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SABA PROGAM A > PROBABILITY > COUNTING

d 5 × 3 = 15 e 1

f 16 g 215 a

4 3 × 2 × 12 = 72 b 144

c 3 × 2 × 7 = 42 a

5 20 ⋅ 20 ⋅ 20 = 8000 b 1 ⋅ 2 ⋅ 1 = 2 c 2 ⋅ 8 ⋅ 4 = 64

d 0

e There are 1 ⋅ 2 ⋅ 1 + 8 ⋅ 1 ⋅ 7 + 2 ⋅ 7 ⋅ 3 + 2 ⋅ 8 ⋅ 4 = 164 outcomes for a price (if you can play for free).

6.2 With or without repeating

1 10⁴⋅ 26²= 6760000 2 2³⁵

3 15 ⋅ 14 ⋅ 13 = 2730 a

4 8! = 40320 possibilities.

b Seat this person ﬁrst. You can do that at either end of the row. The other seven can be seated in a random order.

2 ⋅ 7! = 10080 possiblities.

c This couple can sit in 7 places. For the others there are 6 places left. But the couple can exchange places! So 2⋅ 7 ⋅6! = 10080 possibilities.

5 5

6.3 Combinations

a

1 ( (10) , (3)) = 120 b ( (10) , (9)) = 10 c 2¹⁰= 1024 a

2 The outcomes can be 0, 1, 2, 3, 4 or 5 times "head". Therefore there are 6 possibilities.

b ( (5) , (2)) = 10

c ( (50) , (20)) ≈ 4.71 ⋅ 10¹³

3 For each game two participants are selected from 24 and the order is not relevant. So there are ( (24) , (2)) = 276 games to be played.

a

4 ( (14) , (4)) = 1001

b ( (14) , (2)) ⋅ ( (12) , (2)) = 6006 a

5 8! = 40320 b 5! ⋅ 3! = 720

(27)

c 6! ⋅ 2 = 1440

d ( (8) , (3)) ⋅ 5! = 6720 a

6 6 ⋅ 6 ⋅ 6 = 216 b 19

a

7 ( (18) , (4)) = 3060

b ( (9) , (2)) ⋅ ( (9) , (2)) = 1296 c 18 ⋅ 17 ⋅ 16 ⋅ 15 = 73440

6.4 Pascal's triangle

a 1 (10

3) = 120 b 2¹⁰= 1024 c (10

8) + (10

9) + (10 10) = 56 d 1024 − [(10

9) + (10

10)] = 1013 a

2 Draw grid.

b (5 2) = 10 c 32 a 3 (14

4) = 1001 b (10

2) = 45 a

4 1 - 0, 2 - 0, 2 - 1, 3 - 1, 4 - 1, 4 - 2, 4 - 3, 5 - 3, 6 - 3, 6 - 4.

b 210 c 40 5 11 routes.

a

6 2⁵= 32 b 30

c There are exactly (5

2) = 10 possibilities:

1: - - — — — 2: - — - — — 3: - — — - — 4: - — — — - 5: — - - — — 6: — - — - — 7: — - — — - 8: — — - - — 9: — — - — - 0: — — — - -

(28)

7 Probability

7.1 Experiments

a

1 Yes, with fair dice.

b Not possible, this die is not fair.

c Is possible with a fair die.

2 ¹₄

Simulate using the random numbers 1 through 4. There are diﬀerent ways to do this:

- there are 16 possible pairs, so you can simulate with random numbers 1 through 16;

- two ’separate’ dice, ﬁrst throws (20 or so) for the ﬁrst list, second throws (20 as well) for the second list.

(29)

SABA PROGAM A > PROBABILITY > PROBABILITY

a

3 There are 9 possible pairs, that are all randomly chosen (assuming they are playing without any strat- egy).Number the possibilities, 1 t/m 9. For example: the numbers 2, 4, 6, 8 mean A wins; the other numbers mean that B wins.

b Zie ﬁguur.

c No, B has a greater chance of winning.

a

4 300

c

b ₃₀₀³² ≈ 11%

d ₃₀₀⁷⁰ ≈ 23%

e That is a life of less than 1350 and more than 1650 hours. So approximately ₃₀₀⁶⁸ +₃₀₀⁷⁰ ≈ 46%.

a 5 ₁₈⁷

b ₁₀₀²⁹

7.2 Reasoning

a 1 ₁₀³

b ¹₅ c ₁₀¹ d ²₅

a

2 0.001

b 0.001 if she has only one ticket.

c 0.5

d At b only your friend’s number is right, at c is every even number is right.

e ₉₉₉¹ f ⁵⁰⁰₉₉₉ a

3 ₃₆⁶ =¹₆ b ²¹₃₆=₁₂⁷ c ₃₆¹

(30)

SABA PROGAM A > PROBABILITY > PROBABILITY

d ¹₂ a 4 ¹₂

b ₁₆¹ c ₁₆⁴ 5 ³₈

7.3 Tree diagrams

a 1 ³₈

b ³₈ c ₁₄⁵ d ¹₈

e ₆₄³

f Yes, it is now ²⁵₆₄ a

2 Own answer.

b ₁₈⁵ a

3 0.04

b 0.08 c ₉₀⁴ d ₉₀⁸

e Own solution.

f u�u�u�ℎu�u�u�u�u�u�u�u�u�u�u� : ²¹₅₀; u�u�u�ℎu�u�u�u�u�u�u�u�u�u�u�u�u�u� : ²¹₄₅

g If you draw a red (green) marble, the probability of an green (red) marble increases slightly.

a 4 ₅₄¹

b ₃₆¹ c ₂₇²

d 5 red (not six) with 1 white marble; three draws u�u�u�ℎu�u�u�u�u�u�u�u�u�u�u�

7.4 Probability distribution

ba

1 Expected value is 2 c 300

a

3 ba

2 0, 1, 2, 3, 4 c

b Approximately 1.3 ba

4 The expected value is 12,25 12,00¹³₃₆ a

5 𝑃 (𝑆 = 3) = 0.25; 𝑃 (𝑆 = 4) = 0.375 and 𝑃 (𝑆 = 5) = 0.375. The expected value is 4.125 sets.

b 412 or 413 sets

c 𝑃 (𝑆 = 3) ≈ 0.49; 𝑃 (𝑆 = 4) ≈ 0.24 and 𝑃 (𝑆 = 5) ≈ 0.27. The expected value is approximately 3.8 sets.

d 𝑃 (𝑆 = 3) = 0.49; 𝑃 (𝑆 = 4) = 0.273 and 𝑃 (𝑆 = 5) = 0.237 e Approximately 3.75

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ba

6 3 euros c 0.046875 a

7 𝑃 (𝑀 = 0) ≈ 12.17% and 𝑃 (𝑀 = 1) ≈ 37.46%

b 𝑃 (𝑀 = 2) ≈ 36.12%; 𝑃 (𝑀 = 3) ≈ 12.84% en 𝑃 (𝑀 = 4) ≈ 1.40%

c 𝑃 (𝑉 = 0) = 𝑃 (𝑀 = 4) etcetera.

d 1.54 and 2.46

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8 Normal distribution

8.1 Normal curve

a

1 You don’t get a perfectly symmetric bell-shaped curve, but don’t worry about that...

b 𝜇 ≈ 1005 and 𝜎 ≈ 2.4 grams.

c According to the histogram: 6%.

d The mean minus 2 times the standard deviation gives you a value of 1000 grams. According to the rules of thumb, 2.5% of the observations should be lower than this value.

e About 16%.

f About 68%.

a

2 50%

b 85%

c 𝜎𝐵= 50 and 𝜇𝐵= 1150(hours).

(33)

SABA PROGAM A > PROBABILITY > NORMAL DISTRIBUTION

d Because the distribution of type B is more spread out, but its total area still has to be 100%, then the height of the peak has to be lower.

e 2.5%

a

3 Own answer b 68%

c 2.5%

d 97.5%

e 2.5%

a

4 𝜇 = 3.0 and 𝜎 = 0.2 𝜇 = 82 and 𝜎 = 6.

b Top curve: the area corresponds to 16%.

Bottom curve: area corresponds to 84%.

ba

5 32%

c 84%

d No, because you cannot use the rules of thumb for this.

e About 16% + 20% = 36%. Yes, their complaint was probably justiﬁed.

a

6 68%

b 16%

c 84%

d Less than 85.

a

7 Between 32 and 64.

b Between 32 and 80.

c 16%

8.2 Normal probabilities

a

1 13.33%

b 86.67%

c 71.07%

d 13.33%

a

2 20000 ⋅ 0.35806... ≈ 7161 b 20000 ⋅ 0.2927... ≈ 5854 c 20000 ⋅ 0.1456... ≈ 219 a

3 9.12%

b 25.25%

c 0.38 d 11.09%

(34)

SABA PROGAM A > PROBABILITY > NORMAL DISTRIBUTION

e 0.9953 litres f 1.0392 litres a

4 10.9%

b 51.2%

c 5.8%

d No taller than 153.7 cm, therefore no taller than 153 cm.

e At least 170.3, therefore at least 171 cm.

a

5 𝜇 = 43.6 cm and 𝜎 = 2.7 cm.

b Own answer.

c P (𝐾 < u�) = 0.05 gives you u� ≈ 39.2 cm.

d P (𝐾 < u�) = 0.80 gives you u� ≈ 45.9 cm. This means the longest 20% have a knee height of 45.9 cm or more.

8.3 Standardising

a

1 1006.6 gram.

b 𝜇 = 1008.215 gram that is approximately 1008.2 gram.

c 𝜎 = 1.4607 gram or approximately 1.5 gram.

a

2 0.1056 ⋅ 1200 ≈ 127 cars.

b 48.4206 or about 48.4 seconds.

c 2.1493 or about 2.1 seconds.

a

3 66.87% (about 67%) b 84.13% (about 84%)

c There are several possibilities; for example 𝜇 = 10.01 and 𝜎 = 0.008 will result in 99.37% of screws being accepted.

a

4 P (u� < 60 | 𝜇 = u� and 𝜎 = u�) = 0.875 gives you^(60−u�)_u� ≈ 1.15.

P (u� < 30 | 𝜇 = u� and 𝜎 = u�) = 0.39 gives you ^(30−u�)_u� ≈ −0.28.

Therefore: 60 − u� = 1.15u� and 30 − u� = −0.28u�. Solve the equations to get u� ≈ 39.5 and u� ≈ 21.0.

The values are therefore 𝜇 ≈ 39.5 and 𝜎 ≈ 21.0.

b P (u� < u� | 𝜇 = 35.9 and 𝜎 = 21.0) = 0.30 gives you u� ≈ 24.9. Therefore plants up to a length of about 25 cm are being destroyed.

a

5 𝜎 ≈ 60.8 gram b 4,8%

c 1006.9 gram (about 1007 gram)

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8.4 Normal or not

a

1 Men: 𝜇 ≈ 128.5 and 𝜎 ≈ 12.6.

Women: 𝜇 ≈ 131.7 and 𝜎 ≈ 13.7.

b Class width of 5 and the boundaries of the ﬁrst class are 102.5 ≺ 107.5.

c The points don’t lie on a straight line.

d You can always draw a straight line through some points, but the diﬀerences from the calculated values will be quite large.

e No, this distribution is also skewed.

a

2 The average height is 162 cm and the standard deviation is 6.5 cm.

b Own answer.

c Yes, the heights of these 5001 women are fairly normally distributed.

d The boundaries are approximately 149 and 175 cm. Thus u� ≈ 13 cm.

e Taller than about 169 cm.

(36)

9 Probability models

9.1 Yes/No probabilities

a 1 (10

4) ⋅ (0.5)⁰⋅ (0.5)¹⁰≈ 0.2051 b See table.

c 0.0547

d 1 − 0.0108 = 0.9892 a

2 It is a binomial experiment because you have 15 repeats of a chance experiment (rolling the die) wherein you can have one of two outcomes, 1 or ’not 1’.

b 0.2363 c 0.5322

(37)

SABA PROGAM A > PROBABILITY > PROBABILITY MODELS

d Because you now don’t have a choice out of two outcomes with every roll of the die, but you count the total number of pips over 15 rolls.

e To get a total of 17 there are two possibilities: you throw a 2 twice and a 1 the rest of the time, or you throw a 3 once and a 1 the rest of the time.

The probability is: (15

2) ⋅ (¹₆)¹⁵+ (15

1) ⋅ (¹₆)¹⁵≈ 0.000000000255.

2 2 3 0.1052

9.2 Binomial distribution

a

1 0.4166 b 0.9761 c 0.8270 d 0.2503 e 0.8382

2 P (𝑋 ≤ 1 | u� = 50 and u� =₃₇¹ ) ≈ 0.061.

a

3 P (𝑋 ≤ 3 | u� = 15 and u� = 0.25) ≈ 0.46128

b P (𝑋 > 10) = 1 − P (𝑋 ≤ 10) ≈ 1 − 0.99988 = 0.00012 c 15 ⋅ 0.25 = 3.75

4 P(diﬀerent color) = P(r,w) + P(w,r) = ₁₀⁸ ⋅⁶₉+₁₀² ⋅³₉ =⁵⁴₉₀=³₅.

P (𝑋 > 5 | u� = 10 and u� =³₅) = 1 − P (𝑋 ≤ 5) ≈ 1 − 0.36689 = 0.63311 ≈ 0.633.

a

5 P (𝑋 ≥ 1 | u� = 100 and u� = 0.01) = 1 − P (𝑋 = 0) ≈ 1 − 0.36603 = 0.63397.

b P (𝑋 = 3 | u� = 100 and u� = 0.01) ≈ 0.06099.

a

6 After 6 trials.

b 30 ⋅¹₆ = 5.

c P (𝑋 ≤ 4 | u� = 30 and u� =¹₆) ≈ 0.42433

9.3 Non-binomial distributions

a

1 ₁₂⁵ ⋅₁₁⁴ ⋅₁₀⁷ ⋅⁶₉⋅ 6 ≈ 0.4242

b 1 − (₁₂⁸ ⋅₁₁⁷ ⋅₁₀⁶ ⋅⁵₉ +₁₂⁴ ⋅₁₁⁸ ⋅₁₀⁷ ⋅⁶₉⋅ 4) ≈ 0.4061 c ₁₂⁹ ⋅₁₁⁸ ⋅₁₀⁷ ⋅⁶₉ ≈ 0.2545

d 4 ⋅₁₂⁵ = 1²₃ a

2 A can will not appear twice in the same sample. It therefore is sampling without replacement.

b ₁₀₀₀¹⁰⁰ ⋅₉₉₉⁹⁹ ⋅₉₉₈⁹⁸ ⋅⁹⁰⁰₉₉₇⋅⁸⁹⁹₉₉₆⋅⁸⁹⁸₉₉₅⋅⁸⁹⁷₉₉₄⋅⁸⁹⁶₉₉₃⋅ (8

3) ≈ 0.0326

c P (𝐵 = 3 | u� = 8 and u� = 0.10) ≈ 0.0331. The diﬀerence is is 0.0005, or 0.05%.

d The diﬀerence is very small because you are taking a very small sample (8 cans) out of a large population (1000 cans).

e P (𝐵 ≤ 3 | u� = 8 and u� = 0.10) ≈ 0.9950

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a

3 4 ⋅ 0.4 = 1.6

b ₂₀⁸ ⋅₁₉⁷ ⋅₁₈⁶ ⋅¹²₁₇⋅ 4 ≈ 0.1387

c P (𝑋 = 3 | u� = 4 and u� = 0.40) ≈ 0.1536. This probability diﬀers from the actual chance by 0.0149, which is quite a large diﬀerence.

d ₁₀₀⁴⁰ ⋅³⁹₉₉⋅³⁸₉₈⋅⁶⁰₉₇⋅ 4 ≈ 0.1512

e P (𝑋 = 3 | u� = 4 and u� = 0.40) ≈ 0.1536. The diﬀerence is now much smaller. The larger the sampling population, the smaller the diﬀerence.

a

4 ₃₀⁵ ⋅²⁵₂₉⋅²⁴₂₈⋅²³₂₇⋅ 4 ≈ 0.4196

b 1 − (²⁵₃₀⋅²⁴₂₉⋅²³₂₈⋅²²₂₇+₃₀⁵ ⋅²⁵₂₉⋅²⁴₂₈⋅²³₂₇⋅ 4) ≈ 0.3872 c ₃₀⁵ ⋅₂₉⁴ ⋅₂₈³ ⋅²⁵₂₇⋅ 4 ≈ 0.0091

5 P (𝑋 ≤ 15 | u� = 20 and u� = 0.90) ≈ 0.0432

9.4 Probability models

ba

1 The expected value is ⁸⁰₁₆= 5.

a

2 If the side facing up is white, and Bert bets on the other side being red, then only one of the three cards would make him win. His chance is therefore ¹₃.

b P(3 bets) = (¹₃)³+ (²₃)³=₂₇⁹ =¹₃.

c This cannot happen (after three bets a new series starts and they get new cookies), so the probability is 0.

a

3 P (𝑇 < 8.5 | 𝜇 = 9.2 and 𝜎 = 0.6) = 0.12167

You would expect 0.12167 ⋅ 90 = 10.95, or 11 years.

b 𝑃 (𝑇 > 10.3 | 𝜇 = 9.2 and 𝜎 = 0.6) = 0.033376 for a century, which is about 3 years per century.

4 ₅₂⁴ ⋅³⁶₅₁⋅³⁵₅₀⋅³⁴₄₉⋅ 4 + ₅₂⁴ ⋅₅₁⁴ ⋅³⁶₅₀⋅³⁵₄₉⋅ 12 +₅₂⁴ ⋅₅₂³ ⋅³⁶₅₀⋅³⁵₄₉⋅ 6 +₅₂⁴ ⋅₅₁⁴ ⋅₅₀³ ⋅³⁶₄₉⋅ 12 ≈ 0.1570 a

5 1 −₁₀⁵ ⋅⁴₉⋅³₈⋅²₇⋅¹₆ ≈ 0.9960

b 1 − P (𝑉 = 5 | u� = 5 and u� = 0.50) = 0.96875 a

6 P (𝑉 < 0.25 | 𝜇 = 0.27 and 𝜎 = 0.01) ≈ 0.0228 b P (𝐴 ≤ 2 | u� = 24 and u� = 0.0228) ≈ 0.9832

c P (𝑉 < 0.25 | 𝜇 = u� and 𝜎 = 0.01) ≈ 0.01 gives you ^{(0.25−u�)}_0.01 ≈ −2.326 and therefore u� = 𝜇 ≈ 0.273.

On average, the machine needs to dispense 0.273 litres into each bottle.

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