1. Applying the Wrong Measurement

2. Adaptif Hal.: 2 Aproksimasi APPROXIMATION Competence Standard Solving problem related to the concept of wrong approximation Basic Competence: Applying the wrong measurement concept. Indicators : 1. Differentiate the definition between numbering and measuring 2. Calculate the mistake (absolute and relative mistake), wrong percentage, result tolerance of a measurement

3. Adaptif Hal.: 3 Aproksimasi Scope and Limitation The definition of Approximation Rounding up Kinds of mistakes Macam-macam Kesalahan Tolerance Measurement Result Operation Chain Fraction

4. Adaptif Hal.: 4 Aproksimasi 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 The definition of Approximation What is the difference of numbers from sentences? Example: the length of a hundred thousands is 15 cm. The group member who came yesterday was 15 people.

5. Adaptif Hal.: 5 Aproksimasi Approximation State a number or size which is gained from an activity based on the approach or round up. Measuring : Approximate The result is not certain (Approximation) Numbering : The result is exact ( certain )

6. Adaptif Hal.: 6 Aproksimasi Rounding Up All measurement result state “approximation” value. The result of measuring length, mass, time, width, etc. must be gained according to carefulness. The rounding up is done by the rules: If the next number is 5 or more than 5, then the previous number is added by one. If the next number is less than 5, then that number is deleted and the previous number is still there. There are three steps to round up: a. Rounding up to the single nearest number. b. Rounding up to the amount of decimal number. c. Rounding up to the amount of significant number.

7. Adaptif Hal.: 7 Aproksimasi Rounding up to the single nearest number In case of rounding up to a single nearest number, firstly determine the smallest single number you want. Example: 165,5 cm = 166 cm, Rounded up to the nearest cm 2, 43 kg = 2 kg, Rounded up to the nearest kg 14,149 seconds = 14,15 seconds, Rounded up to the nearest hundreds seconds 14,16 seconds = 14,2 seconds, Rounded up to the nearest tenth second 14,149 seconds = 14,1 seconds, Rounded up to the nearest tenth seconds

8. Adaptif Hal.: 8 Aproksimasi Rounding Up to the amount of decimal numbers To make it easier, sometimes we need to round up a decimal numbers to the decimal place we want. 5,47035 = 5,4704 rounded up to four decimal places = 5,470 rounded up to three decimal places = 5,47 rounded up to two decimal places = 5,5 rounded up to one decimal place What is the result if 5,44735 is rounded up to one decimal place. 5,44735 = 5,4 rounded up to one decimal place

9. Adaptif Hal.: 9 Aproksimasi Pembulatan ke Banyaknya Angka-angka yang Signifikan Pembulatan dengan cara menetapkan banyaknya angka yang signifikan. Significant berarti “ bermakna “  penting 64,5 cm mempunyai 3 angka signifikan Jika diketahui suatu bilangan, berikut adalah aturan-aturan untuk menentukan angka-angka mana yang signifikan : 1). Angka yang tidak nol selalu signifikan, mis: 472,513  6 angka signifikan 2). Angka “0” signifikan jika letaknya di antara angka-angka yang signifikan, mis: 807003  6 angka signifikan 3) Angka “ 0 “ signifikan jika muncul setelah tanda tempat desimal dan angka- angka lain yang signifikan, mis: 20,080  5 angka signifikan 5) A ngka “ 0 “ signifikan jika ditandai “strip “ atau “ bar 4). Angka “ 0 “ itu tidak pernah signifikan jika mendahului angka-angka yang bukan nol meskipun muncul setelah tanda tempat desimal, mis: 043,00 m  4 angka signifikan; 0,0720 km  3 angka signifikan

10. Adaptif Hal.: 10 Aproksimasi Rounding up to the amount of significant numbers Rounding up by determining the amount of significant numbers. Significant means “ meaningful “  important 64,5 cm has 3 significant numbers If there is a number, here are the rules to determine significant numbers: 1). Except zero, all numbers are significant,ex: 472,513  6 is significant number. 2). “0” number is significant if the place is among significant numbers, ex: 807003  6 is significant number 3) “ 0 “ number is significant if it appears after decimal place symbol and the other significant numbers, ex: 20,080  5 is significant numbers 5) “ 0 “ number is significant if it is denoted by ”dash” or “ bar “ 4). “ 0 “ number is never significant if it precedes the numbers which is not zero, although it appears after decimalplace symbol, ex : 043,00 m  4 is significant number; 0,0720 km  3 is significant number.

11. Adaptif Hal.: 11 Aproksimasi 23 Juni 2016 Kinds of Mistakes The length is closer to 15 cm than 14 cm or 16 cm The real length is between 14,5 cm and 15,5 cm. In this case the acceptable mistake from this measurement is 0,5 cm Or the absolute mistake is 0,5 cm. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

12. Adaptif Hal.: 12 Aproksimasi Kinds of Mistakes  Absolute Mistake = ½ x the smallest single measurement. Salah relatif x 100 %  Relative Mistake=  Persentase Kesalahan =  Upper Limit = Measurement Result + Absolute Mistake  Lower Limit = Measurement Result – Absolute Mistake

13. Adaptif Hal.: 13 Aproksimasi Tolerance Tolerance in measurement is the difference between the biggest and the smallest measurement which can be accepted.

14. Adaptif Hal.: 14 Aproksimasi Example 1 : Let the measurement result of the flag pole height is 3,5 meter, then find the smallest single measurement, absolute mistake, relative mistake, mistake percentage, and tolerance. Answer : Measurement result 3,5m The smallest single measurement : 0,1m Absolute Mistake : 0,5 x 0,1m = 0,05m Relative Mistake : 0,05 / 3,5 = 0,014 Mistake percentage : 0,014 x 100% = 1,4% Upper limit measurement: (3,5 + 0,05)m = 3,55m Lower limit measurement: (3,5 – 0,05)m = 3,45m Measurement tolerance : (3,55 – 3,45)m = 0,10m

15. Adaptif Hal.: 15 Aproksimasi Measurement Result Operation : The sum of measurement result If two measurements or more are added, so the absolute mistake is the total of the absolute mistakes from the former measurement. Maximum = b. a measurement I + b. a measurement II Minimum = b. b measurement I + b. b measurement II The difference of measurement result If two or more measurements are subtracted, then the difference of absolute mistake is the amount of absolute mistake from the former measurement. Maximum Difference = b. a Measurement I - b. b measurement II minimum Difference = b. b measurement I - b. a measurement II Multiplication result of two measurements Maximum multiplication result = b. a measurement I x b. a measurement II Minimum multiplication result = b. b measurement I x b. b measurement II

16. Adaptif Hal.: 16 Aproksimasi Example 2 : A result of measurement is stated by ( 15 ± 0,5 ) gram. Give the acceptable maximum and minimum measurement, then find the tolerance? Answer : The allowed tolerance is ( 15 ± 0,5 ) gram, means: The acceptable maximum measurement:15 + 0,5 = 15,5 gram The acceptable minimum measurement: 15 – 0,5 = 14,5 gram So the tolerance is 1 gram

17. Adaptif Hal.: 17 Aproksimasi Example Let two measurements result : 12cm and 19cm Determine : a.The total of maximum and minimum The difference between maximum and minimum The multiplication result of maximum and minimum The total of absolute mistake and the difference of the previous measurements

18. Adaptif Hal.: 18 Aproksimasi Answer : The measurement result I = 12cm The measurement result II = 19cm Absolute mistake of measurement I = 0,5cm Absolute mistake of measurement II = 0,5cm Upper limit measurement I : (12 + 0,5)cm = 12,5cm Lower limit measurement I : (12 – 0,5)cm = 11,5cm Upper limit measurement II : (19 + 0,5)cm = 19,5cm Lower limit measurement II : (19 – 0,5)cm = 18,5cm Maximum : (12,5 + 19,5)cm = 32cm Minimum : (11,5 + 18,5)cm = 30cm a. The Maximum difference : (19,5 - 11,5)cm = 8cm The minimum difference : (18,5 + 12,5)cm = 6cm b.

19. Adaptif Hal.: 19 Aproksimasi The measurement result I = 12cm The measurement result II = 19cm Salah mutlak pengukuran I = 0,5cm Salah mutlak pengukuran II = 0,5cm Batas atas pengukuran I : (12 + 0,5)cm = 12,5cm Batas bawah pengukuran I : (12 – 0,5)cm = 11,5cm Batas atas pengukuran II : (19 + 0,5)cm = 19,5cm Batas bawah pengukuran II : (19 – 0,5)cm = 18,5cm Hasil kali Maksimum : (12,5 x 19,5)cm = 243,75cm 2 Hasil kali Minimum : (11,5 x 18,5)cm = 212,75cm 2 c. Jumlah salah mutlak : 0,5cm + 0,5cm = 1 cm Selisih salah mutlak : 0,5cm + 0,5cm = 1 cm d. Answer :

20. Adaptif Hal.: 20 Aproksimasi APPROXIMATION BBase Competence Applying operation concept of measurement result IIndicators 1. Calculating the sum and the measurement difference result 2. Calculating the maximum multiplication result and the minimum of measurement result

21. Adaptif Hal.: 21 Aproksimasi The Fraction Aproximation The fraction can be approached with the other fraction by the chain fraction way Ex : can be written by the chain fraction: To close to-n by p n =a n p n-1 +p n-2 and q n =a n q n-1 +q n-2

22. Adaptif Hal.: 22 Aproksimasi To determine the approach by the table The division result of chain fraction a1a1 a2a2 a3a3 …...a n-1 anan 0 1 1 0 a1a1 a 2. a 1 +1 pnpn qnqn

23. Adaptif Hal.: 23 Aproksimasi Example: 1 Determine fraction which closes to:

24. Adaptif Hal.: 24 Aproksimasi Make level division 99 / 224 \ 2 = a 1 198 26 / 99 \ 3 = a 2 78 21 / 26 \ 1 = a 3 21 5 / 21 \ 4 = a 4 20 1 / 5 \ 5 = a 5 5 0 Can be written: So the approachable fraction is: 2

25. Adaptif Hal.: 25 Aproksimasi To determine the approach by the table Division result of chain fraction 23145 0 1 1 0 27943224 1341999 x 2 + 2 x 0 + 1 So the approachable fraction is: 2

26. Adaptif Hal.: 26 Aproksimasi Example: 2 Determine the approachable fraction : 79 / 213 \ 2 158 55 / 79 \ 1 55 24 / 55 \ 2 48 7 / 24 \ 3 21 3 / 7 \ 2 6 1 / 3\ 3 3 0

27. Adaptif Hal.: 27 Aproksimasi By table Division result of chain fraction 212323 1 0 0 1 1131023 2382762 x 0 + 1 x 2 + 2 79 213 So the approachable fraction is :

28. Adaptif Hal.: 28 Aproksimasi Example: Pada mesin bubut yang mempunyai kisar transportir 5 mm akan dibuat ulir dengan kisar 2,06 mm. Persediaan roda gigi pengganti mempunyai gigi 20-120 dan merupakan kelipatan dari 5. Tentukan perbandingan roda gigi penggantinya sehingga menghasilkan ulir yang paling mendekati ukuran sebenarnya dan berapa persentase kesalahannya!

29. Adaptif Hal.: 29 Aproksimasi Solution: Jika roda gigi yang menggerakkan adalah DR dan roda gigi yang digerakkan DN, maka: To find the approachable price from the previous price, we can use the chain fraction, as follow: 103 / 250 \ 2 206 44 / 103 \ 2 88 15 / 44 \ 2 30 14 / 15 \ 1 14 1 / 14 \ 14 14 0

30. Adaptif Hal.: 30 Aproksimasi Determining the approach by table Division result of chain fraction 222114 1 0 0 1 1257103 251217250 x 0 + 1 x 1 + 0 So the approachable fraction is:

31. Adaptif Hal.: 31 Aproksimasi