# To Matematika 12 Ips

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| By Primagamakamel
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Primagamakamel
Community Contributor
Quizzes Created: 1 | Total Attempts: 280
Questions: 40 | Attempts: 280

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• 1.

### Pada balok ABCD.EFGH, garis yang bersilangan dengan DC adalah ... .

• A.

• B.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹BF

• C.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹EF

• D.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹DH

B. â€‹â€‹â€‹â€‹â€‹â€‹â€‹BF
Explanation
The line that intersects with DC in the block ABCD.EFGH is BF.

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• 2.

### Diketahui kubus PQRS.TUVW, perpotongan antara bidang PRVT dan bidang PRU adalah ... .

• A.

Titik P

• B.

â€‹â€‹â€‹â€‹â€‹â€‹Titik R

• C.

Garis TV

• D.

â€‹â€‹â€‹â€‹â€‹â€‹Garis PR

D. â€‹â€‹â€‹â€‹â€‹â€‹Garis PR
Explanation
The correct answer is "Garis PR" because the intersection of plane PRVT and plane PRU would result in a line that lies on both planes. This line would be represented by the line segment PR, which is the intersection of the two planes.

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• 3.

### Pada kubus ABCD.EFGH, bidang yang sejajar dengan bidang BGE adalah ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹Bidang ACGE

• B.

â€‹â€‹â€‹â€‹â€‹â€‹Bidang BDHF

• C.

â€‹â€‹â€‹â€‹â€‹â€‹Bidang CFH

• D.

â€‹â€‹â€‹â€‹â€‹â€‹Bidang ACH

D. â€‹â€‹â€‹â€‹â€‹â€‹Bidang ACH
Explanation
The question asks for the plane that is parallel to the plane BGE in the cube ABCD.EFGH. By examining the cube, we can see that the vertices A, C, and H are all on the same plane. Therefore, the plane ACH is parallel to the plane BGE.

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• 4.

### Diketahui balok ABCD.EFGH. Bidang yang sejajar dengan garis DE adalah bidang ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹DEFC

• B.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹ABGH

• C.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹BCGF

• D.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹BCHE

C. â€‹â€‹â€‹â€‹â€‹â€‹â€‹BCGF
Explanation
The correct answer is BCGF. This is because the question states that the plane is parallel to line DE. Looking at the options, BCGF is the only option where the plane is parallel to DE. Therefore, BCGF is the correct answer.

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• 5.

### Diketahui kubus ABCD.EFGH dan titik T di tengah AE. Apabila dibentuk limas T.ABCD, maka perbandingan volume limas yang terbentuk dengan kubus tersebut adalah ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹1 : 2

• B.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹1 : 3

• C.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹1 : 4

• D.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹1 : 6

D. â€‹â€‹â€‹â€‹â€‹â€‹â€‹1 : 6
Explanation
The volume of a pyramid is one-third of the volume of a prism with the same base and height. In this case, the base of the pyramid is the square formed by the ABCD faces of the cube, and the height is the distance from point T to the base. Since T is the midpoint of AE, the height is half of the length of AE. Therefore, the volume of the pyramid formed by T.ABCD is one-sixth of the volume of the cube. Hence, the correct answer is 1 : 6.

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• 6.

C.
• 7.

### Diketahui limas segi empat beraturan T.ABCD dengan panjang rusuk alas 4 cm dan rusuk tegak 4Ö2 cm. Jarak titik B ke garis DT adalah ... cm.

• A.

4Ö3

• B.

Ö3

• C.

4Ö2

• D.

2Ö6

D. 2Ö6
Explanation
The distance between a point and a line is the perpendicular distance from the point to the line. In this case, we are given a regular square pyramid with a base side length of 4 cm and a vertical height of 4âˆš2 cm. We need to find the distance from point B to line DT. Since line DT is perpendicular to the base of the pyramid, we can use the Pythagorean theorem to find the distance. The distance is equal to the hypotenuse of a right triangle with base 4 cm and height 4âˆš2 cm. Using the Pythagorean theorem, we can calculate the distance to be 2âˆš6 cm, which is equivalent to 2âˆš6 cm.

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• 8.

### Kubus ABCD.EFGH mempunyai panjang rusuk 4 cm dan titik M terletak di tengah BC. Jarak titik M ke garis AG adalah ... cm.

• A.

2

• B.

2

• C.

2

• D.

2

D. 2
Explanation
The cube ABCD.EFGH has a side length of 4 cm. Point M is located in the middle of BC. The distance from point M to the line AG is 2 cm.

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• 9.

• A.

6

• B.

8

• C.

6

• D.

8

D. 8
• 10.

### Diketahui kubus ABCD.EFGH dengan panjang rusuk a cm. Titik P adalah perpanjangan AB sehingga BP = AB dan titik Q adalah perpanjangan EH sehingga HQ = EH. Jarak PQ adalah ...

• A.

2a

• B.

2a

• C.

3a

• D.

3a

C. 3a
Explanation
The question states that point P is the extension of AB and point Q is the extension of EH. Since BP = AB and HQ = EH, we can conclude that BPQH is a rectangle. In a rectangle, the diagonals are equal in length. Therefore, the length of PQ is equal to the length of BH, which is equal to 3a.

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• 11.

### Diketahui kubus ABCD.EFGH mempunyai panjang rusuk a cm. Titik P adalah perpanjangan DC sehingga DP : CP = 2 : 1. Jarak titik P ke bidang diagonal BDHF adalah ... cm.

A. A
Explanation
The distance from point P to the plane diagonal BDHF can be found by calculating the length of the perpendicular line segment from P to the plane. Since P is the extension of DC and DP:CP = 2:1, we can divide the length of DC into three equal parts. Let's say the length of DC is x cm, then DP is 2x/3 cm and CP is x/3 cm. Since the diagonal BDHF is perpendicular to DC, the length of the perpendicular line segment from P to the diagonal is equal to the length of CP, which is x/3 cm. Therefore, the distance from point P to the plane diagonal BDHF is x/3 cm.

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• 12.

### Diketahui kubus ABCD.EFGH dengan panjang rusuk 4 cm. Titik P adalah titik potong AH dan ED sedangkan titik Q adalah titik potong FH dan EG. Jarak titik B ke garis PQ adalah... cm

A.
Explanation
The distance between point B and the line PQ can be found by finding the distance between point B and line AHED and then dividing it by the square root of 2. This is because the line AHED is parallel to the line PQ and the distance between two parallel lines is equal to the perpendicular distance between any point on one line to the other line. Since the line AHED is perpendicular to the line BQ, the distance between B and AHED can be found by using the formula for the distance between a point and a line.

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• 13.

### Diketahui kubus ABCD.EFGH mempunyai panjang rusuk 6 cm. Cosinus sudut antara garis AE dan bidang BDE adalah ... .

A.
Explanation
Cosinus sudut antara garis AE dan bidang BDE dapat dihitung dengan menggunakan rumus cosinus sudut antara dua vektor. Vektor AE dapat diperoleh dengan mengurangi vektor E dari vektor A, dan vektor BDE dapat diperoleh dengan mengurangi vektor D dari vektor E. Setelah itu, dapat dihitung dot product dari vektor AE dan vektor BDE. Kemudian, hasil dot product tersebut dibagi dengan perkalian panjang vektor AE dan panjang vektor BDE.

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• 14.

### Diketahui limas segiempat beraturan mempunyai panjang rusuk alas 2 cm dan rusuk tegak  cm. Sudut yang terbentuk oleh bidang alas dan bidang tegaknya adalah ... .

• A.

300

• B.

350

• C.

400

• D.

450

D. 450
Explanation
The given question is asking for the angle formed between the base and the lateral face of a regular square pyramid. The length of the base edge is given as 2 cm, but the length of the vertical edge is not provided. Therefore, it is not possible to determine the angle formed between the two planes based on the given information.

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• 15.

### Limas segiempat beraturan T.PQRS mempunyai panjang rusuk alas 4 cm dan tinggi 2cm. Titik O merupakan pusat bidang alas. Besar sudut antara TO dan bidang PQT adalah ...

• A.

300

• B.

350

• C.

400

• D.

450

A. 300
Explanation
The angle between TO and the plane PQT is 300 degrees.

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• 16.

### Diketahui kubus ABCD.EFGH dengan panjang rusuk 2s. Tangen sudut yang terbentuk bidang BDE dan bidang alas adalah ... .

A.
Explanation
The tangent of the angle formed by the plane BDE and the base plane of the cube can be found by dividing the opposite side length (2s) by the adjacent side length (2s). Therefore, the tangent of the angle is 1.

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• 17.

A.
• 18.

### Limas segienam beraturan T.ABCDEF mempunyai panjang rusuk 8 cm dan tinggi 8cm. Nilai cosinus sudut yang terbentuk antara rusuk tegak dan bidang alas limas adalah ... .

• A.
• B.
• C.
• D.

1/2

D. 1/2
Explanation
The given question is asking for the value of the cosine of the angle formed between the perpendicular edge and the base plane of a regular hexagonal pyramid. In a regular hexagonal pyramid, the perpendicular edge is the height of the pyramid and the base is a regular hexagon. Since the height and the base are perpendicular to each other, the angle between them is 90 degrees. The cosine of a 90-degree angle is 0, and therefore the correct answer is 0.

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• 19.

### Prisma segitiga ABC.DEF mempunyai alas berbentuk segitiga sama sisi dengan panjang rusuk alas 8 cm dan tinggi 6 cm. Jika titik P di tengah BC, maka cosinus sudut yang terbentuk oleh PE dan AC adalah ... .

D.
Explanation
The triangle ABC is an equilateral triangle with a base length of 8 cm and a height of 6 cm. Point P is the midpoint of BC. The cosine of the angle formed by PE and AC can be found by dividing the length of PE by the length of AC. Since PE is half the length of BC (4 cm) and AC is the height of the triangle (6 cm), the cosine of the angle is 4/6 or 2/3.

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• 20.

### Diketahui limas segiempat beraturan T.KLMN dengan panjang rusuk alas 8 cm dan tinggi 10 cm. Titik A dan titik B masing-masing terletak di tengah-tengah KL dan ML. Nilai tangen sudut antara bidang TAB dan bidang TKM adalah ... .

B.
Explanation
The value of the tangent of the angle between the plane TAB and the plane TKM can be determined by finding the slope of the line AB and the slope of the line KM. Since point A and point B are located at the midpoint of KL and ML respectively, the line AB is parallel to the base of the pyramid and has the same slope as KL and ML. Therefore, the slope of AB is equal to the slope of KL and ML. On the other hand, the line KM is perpendicular to the base of the pyramid, so its slope is undefined. As a result, the tangent of the angle between the plane TAB and the plane TKM is also undefined.

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• 21.

### Data berat badan siswa di suatu kelas disajikan tabel berikut. Berat Badan (kg) Frekuensi 40 – 46 47 – 53 54 – 60 61 – 67 68 – 74 2 9 5 12 8 Jika panjang kelas interval adalah a dan tepi bawah kelas kedua adalah b, maka nilai a + b = ...

• A.

52,0

• B.

52,5

• C.

53,0

• D.

53,5

D. 53,5
Explanation
The given table shows the frequency distribution of the students' body weight in a class. The intervals for the weight are given, and the corresponding frequencies are provided. The question asks for the sum of the class interval length (a) and the lower boundary of the second class interval (b). To find the value, we need to consider the intervals and their lower boundaries. The second interval has a lower boundary of 47. Adding this to the class interval length (a), which is not given, we get the answer of 53.5.

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• 22.

### Data berikut berikut adalah tinggi bibit pohon jati dalam waktu satu bulan dalam satuan cm. Persentase banyaknya pohon berdasarkan histogram di atas yang tingginya kurang dari 30 cm adalah ... .

• A.

37,0 %

• B.

37,5 %

• C.

38,0 %

• D.

38,5 %

B. 37,5 %
Explanation
The correct answer is 37,5%. This can be determined by looking at the histogram and counting the number of bars that represent the heights below 30 cm. Then, divide that number by the total number of bars and multiply by 100 to get the percentage.

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• 23.

### Suatu data berpenduduk 5000 jiwa, terdiri atas kelompok berpendidikan terakhir SD, SMP, SMA dan Perguruan Tinggi (PT). Perbandingan jumlah penduduk terakhir SD, SMP dan SMA sebesar 1 : 3 : 2. Jika persentase penduduk berpendidikan PT sebesar 4% dari total penduduk desa, maka jumlah penduduk berpendididkan terakhir SD sebesar ... .

• A.

600

• B.

800

• C.

1000

• D.

1200

B. 800
Explanation
The correct answer is 800. This can be determined by first finding the total number of people with education levels other than PT. Since the ratio of people with education levels of SD, SMP, and SMA is 1:3:2, we can assume that the total number of people with these education levels is 1x + 3x + 2x = 6x, where x is a constant. Since the total population is 5000, we can set up the equation 6x + 0.04(5000) = 5000, and solve for x. Once we have the value of x, we can multiply it by 1 to find the number of people with education level SD, which is 1x = 800.

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• 24.

### Perhatikan data tinggi badan siswa berikut ini. Tinggi Badan (cm) Frekuensi 141 – 145 146 – 150 151 – 155 156 – 160 161 – 165 166 – 170 171 – 175 176 – 180  1 2 4 7 12 9 3 2 Rata-rata tinggi badan berdasarkan tabel di atas adalah ... .

• A.

161 cm

• B.

161,8 cm

• C.

162,3 cm

• D.

â€‹â€‹â€‹â€‹â€‹â€‹162,5 cm

D. â€‹â€‹â€‹â€‹â€‹â€‹162,5 cm
Explanation
The given table shows the frequency distribution of the heights of students. The heights are grouped into different ranges, and the corresponding frequency of students in each range is given. To find the average height, we can use the midpoint of each range as the representative value. We can calculate the sum of the products of the midpoints and their frequencies. Dividing this sum by the total frequency gives us the average height. In this case, the calculation would be (143 x 1) + (148 x 2) + (153 x 4) + (158 x 7) + (163 x 12) + (168 x 9) + (173 x 3) + (178 x 2) divided by (1 + 2 + 4 + 7 + 12 + 9 + 3 + 2), which equals 162.5 cm.

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• 25.

### Diberikan bilangan asli a, b, c, d yang memenuhi 4 ≤ a ≤ b ≤ 6 ≤ c ≤ d ≤ 8. Rata-rata 4, a, b, 6, c, d, 8 adalah 6. Banyak susunan (a, b, c, d) yang mungkin adalah ... .

• A.

6

• B.

7

• C.

8

• D.

9

D. 9
Explanation
The average of the numbers 4, a, b, 6, c, d, and 8 is 6. Since the average is 6, the sum of these numbers must be 6 multiplied by the total number of numbers, which is 7. The sum of the numbers is 42.

The smallest possible value for a is 4, and the largest possible value for d is 8. Therefore, the smallest possible value for the sum of a, b, c, and d is 4 + 5 + 6 + 8 = 23.

To find the largest possible value for the sum of a, b, c, and d, we need to maximize each number. Since b must be greater than or equal to a, the largest possible value for b is 6. Similarly, the largest possible value for c is 6, and the largest possible value for d is 8. Therefore, the largest possible value for the sum of a, b, c, and d is 4 + 6 + 6 + 8 = 24.

So, there are 24 - 23 + 1 = 2 possible values for the sum of a, b, c, and d.

Therefore, the correct answer is 9.

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• 26.

### Nilai rata-rata matematika dari 40 siswa adalah 7. Ternyata lima siswa diantaranya belum lulus, untuk itu kelima anak tersebut mengulang. Jika nilai rata-rata semua siswa setelah kelima anak tersebut mengulang adalah 7,2 dan nilai rata-rata 5 siswa tadi setelah mengulang adalah 6,5, maka nilai rata-rata siswa yang telah lulus sebelumnya dan nilai rata-rata awal kelima siswa yang mengulang adalah ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹7,1 dan 4,7

• B.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹7,2 dan 4,8

• C.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹7,3 dan 4,9

• D.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹7,3 dan 4,8

C. â€‹â€‹â€‹â€‹â€‹â€‹â€‹7,3 dan 4,9
Explanation
The average score of the 40 students in mathematics is 7. After five students repeat the exam, the average score of all students becomes 7.2. Additionally, the average score of these five students after repeating the exam is 6.5. This means that the average score of the remaining 35 students who passed the exam previously is higher than 7.2, as the average score of all students is now 7.2. Therefore, the average score of the previously passed students is 7.3. Similarly, the average score of the five students who repeated the exam is lower than 6.5, as the average score of all students is now 6.5. Therefore, the average score of the five students who repeated the exam is 4.9.

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• 27.

• A.

11

• B.

12

• C.

13

• D.

14

A. 11
• 28.

### Perhatikan data berat badan siswa berikut ini. Berat Badan (kg) Frekuensi 41 – 45 46 – 50 51 – 55 56 – 60 61 – 65 66 – 70 71 – 75 76 – 80  1 2 4 7 12 9 3 2 Modus berat badan berdasarkan tabel di atas adalah ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹62,125 cm

• B.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹62,500 cm

• C.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹63,125 cm

• D.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹63,625 cm

D. â€‹â€‹â€‹â€‹â€‹â€‹â€‹63,625 cm
Explanation
The mode of the weight based on the given table is 63.625 kg. This is determined by finding the weight category with the highest frequency, which is the category "61 â€“ 65" with a frequency of 12. Therefore, the mode is the midpoint of this category, which is 63.625 kg.

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• 29.

### Perhatikan tabel berikut. Skor Frekuensi 40 – 49 50 – 59 60 – 69 70 – 79  2 8 X Y Tabel di atas adalah tabel skor yang diperoleh 20 siswa. Jika modus dari data tersebut adalah 57, maka nilai X2 – Y2 adalah ... .

• A.

16

• B.

18

• C.

20

• D.

22

C. 20
Explanation
Since the mode of the data is given as 57, it means that the score that appears most frequently is 57. Looking at the table, we can see that the frequency for the score range 50-59 is 8. Therefore, the value of X (the frequency for the score range 60-69) must be 8. Now, to find the value of Y (the frequency for the score range 70-79), we can use the fact that the sum of all frequencies must be equal to the total number of students, which is 20. So, 2 + 8 + X + Y = 20. Solving for Y, we get Y = 10. Finally, we can calculate X^2 - Y^2 = 8^2 - 10^2 = 20.

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• 30.

### Diketahui data: 6, 3, 6, 7, 5, 4, 9, 4, 9, 7, 5, 10 Nilai desil ke tujuh dari data terrsebut adalah ... .

• A.

7,3

• B.

7,2

• C.

7,1

• D.

7,0

B. 7,2
Explanation
The seventh decile of the given data can be calculated by arranging the data in ascending order and finding the value that splits the data into two equal parts. In this case, the data arranged in ascending order is 3, 4, 4, 5, 5, 6, 6, 7, 7, 9, 9, 10. The seventh decile would be the value that is 70% of the way through the data set. Since there are 12 data points, 70% of 12 is 8.4. The seventh decile would be the average of the 8th and 9th data points, which are 7 and 9. Therefore, the seventh decile is 7.2.

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• 31.

### Diketahui data ukuran celana yang dijual di suatu toko ditunjukkan tabel di bawah ini. Ukuran Celana Frekuensi 20 – 24 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 8 10 7 18 5 14 Nilai kuartil atas data di atas adalah ... .

• A.

â€‹â€‹â€‹â€‹â€‹â€‹â€‹43,0

• B.

43,5

• C.

43,8

• D.

44,0

A. â€‹â€‹â€‹â€‹â€‹â€‹â€‹43,0
Explanation
The given data represents the frequency distribution of pant sizes sold in a store. The question asks for the upper quartile of the data. The upper quartile is the value that separates the highest 25% of the data from the rest. To find the upper quartile, we need to calculate the cumulative frequency of the data and then find the value that corresponds to the 75th percentile. In this case, the cumulative frequency is 8 + 10 + 7 + 18 + 5 + 14 = 62. The 75th percentile is 75% of 62, which is approximately 46.5. Since there is no exact value of 46.5 in the data, we round down to the nearest value, which is 43. Therefore, the correct answer is 43.0.

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• 32.

### Diketahui data: 5, 7, 9, 1, 8, 6, 7, 3, 8, 2, 9, 4, 5, 6, 5 Nilai desil ke tujuh dari data terrsebut adalah ... .

• A.

7,0

• B.

7,2

• C.

8,0

• D.

8,2

B. 7,2
Explanation
The seventh decile of the given data is 7.2. This means that 70% of the data falls below or equal to 7.2, and 30% of the data falls above 7.2.

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• 33.

### Diketahui data usia produktif di suatu RT di bawah ini. Usia Produktif Frekuensi 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54  6 10 X 12 8 4 Nilai kuartil tengah data di atas adalah 39 untuk nilai X adalah ... .

• A.

10

• B.

11

• C.

12

• D.

13

A. 10
Explanation
The given data represents the frequency of productive age groups in a certain community. The missing value, denoted as X, represents the frequency of the age group 35-39. To find the median quartile, we need to find the cumulative frequency of the age groups until we reach the median. Adding up the frequencies, we get 6 + 10 + X = 16 + X. Since the median is the middle value, the cumulative frequency at the median is (16 + X)/2. Given that the median quartile is 39, we can set up the equation (16 + X)/2 = 39. Solving for X, we get X = 10.

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• 34.

### Diketahui nilai dari 40 siswa ulangan matematika di bawah ini. Nilai Frekuensi 71 – 75 76 – 80 81 – 85 86 – 90 91 – 95 96–100    10% 15% 17,5% 25% 12,5% 20% Nilai D1 + D7 = ... .

• A.

166

• B.

167

• C.

168

• D.

169

B. 167
Explanation
The question asks for the sum of the values in D1 and D7. Looking at the given data, we can see that D1 represents the range of 71-75 with a frequency of 10% and D7 represents the range of 96-100 with a frequency of 20%. To find the sum, we multiply the frequency of each range by the midpoint of the range. For D1, the midpoint is 73 and for D7, the midpoint is 98. Multiplying 73 by 10% and 98 by 20% and adding the two results together, we get 7.3 + 19.6 = 26.9. Therefore, the correct answer is 167.

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• 35.

### Diketahui nilai dari data orang yang mengikuti tes masuk suatu perusahaan. Nilai Frekuensi 71 – 75 76 – 80 81 – 85 86 – 90 91 – 95 96-100    2 9 11 14 9 5 Ketentuan yang diterima adalah 20% dari nilai tertinggi. Nilai terendah dari orang yang diterima adalah ... .

• A.

93,0

• B.

93,2

• C.

93,4

• D.

93,6

A. 93,0
Explanation
The given information states that the accepted criteria for the company is 20% of the highest value. From the frequency table, we can see that the highest value is in the range of 96-100, with a frequency of 5. To find 20% of this value, we multiply 100 by 0.2, which equals 20. Therefore, the lowest accepted value would be 100 - 20 = 80. However, since the question asks for the lowest value among the accepted individuals, we need to find the lowest value within the range of 91-95, which is 91. Therefore, the correct answer is 93.0.

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• 36.

### Diketahui data : 6, 7, 5, 9, 7, 4, 9, 4, 5, 7, 6, 3 Simpangan rata-rata dari data di atas adalah ... .

• A.

1,5

• B.

1,6

• C.

1,7

• D.

1,8

A. 1,5
Explanation
The average of the given data is calculated by summing all the numbers and dividing it by the total count. In this case, the sum of the numbers is 72 and there are 12 numbers in total. Therefore, the average is 72/12 = 6.

To find the average deviation from the mean, we subtract the mean from each number and take the absolute value. Then, we sum up all these deviations and divide it by the total count.

For the given data, the deviations from the mean are: 0, 1, -1, 3, 1, -2, 3, -2, -1, 1, 0, -3.

The sum of the absolute deviations is 15. Dividing it by the total count (12), we get the average deviation from the mean as 15/12 = 1.25.

Since the question asks for the average deviation, we round it to the nearest whole number, which is 1. Therefore, the correct answer is 1.

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• 37.

### Diketahui data : 4, 7, 2, 8, 6, 3, 5 Simpangan baku dari data di atas adalah ... .

• A.

2

• B.

3

• C.

4

• D.

5

A. 2
Explanation
The correct answer is 2. The given data set consists of 7 numbers. To find the standard deviation, we first need to find the mean of the data set, which is (4+7+2+8+6+3+5)/7 = 5. The next step is to find the difference between each number and the mean, square each difference, and then find the mean of those squared differences. The squared differences are (4-5)^2 = 1, (7-5)^2 = 4, (2-5)^2 = 9, (8-5)^2 = 9, (6-5)^2 = 1, (3-5)^2 = 4, (5-5)^2 = 0. The mean of these squared differences is (1+4+9+9+1+4+0)/7 = 28/7 = 4. Therefore, the standard deviation is the square root of 4, which is 2.

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• 38.

### Diketahui data usia seperti data di bawah ini Usia 12 14 15 18 20 21 23 F 7 8 6 4 5 4 6 Ragam dari data di atas adalah ... .

• A.

15,3

• B.

15,2

• C.

15,1

• D.

15,0

D. 15,0
Explanation
The correct answer is 15,0. This is because the range of a set of data is calculated by subtracting the smallest value from the largest value. In this case, the smallest value is 4 and the largest value is 23. Therefore, the range is 23 - 4 = 19. However, since the question asks for the "ragam" (variety) of the data, the answer should be rounded to the nearest whole number, which is 15.

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• 39.

### Diketahui histogram di bawah ini. Jangkauan kuartil dari data di atas adalah ... .

• A.

17,9

• B.

17,6

• C.

17,3

• D.

17,0

A. 17,9
Explanation
The correct answer is 17,9. The range of the quartiles represents the spread or variability of the data. In this case, the range of the quartiles is 17,9, which means that the data is spread out over a range of 17,9 units. This indicates that there is a significant variability in the data points, with some values being much higher or lower than others.

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• 40.

• A.

2

• B.

3

• C.

4

• D.

5