AdsPopc

Jumat, 14 Februari 2025

Cara Menghitung Peluang Kejadian pada Pelemparan Dua Dadu


Kali ini kita akan membahas cara menghitung peluang kejadian pada pelemparan dua dadu. Jika kita mempunyai dua dadu (anggap saja dadu merah dan dadu putih) dan kita akan melempar satu kali maka terdapat sebanyak 36 kemungkinan yang akan muncul. sebanyak 36 kemungkinan yang muncul tersebut dinamakan ruang sampel. Sedangkan anggota-anggota ruang sampel tersebut dinamakan titik sampel. Titik sampel pada pelemparan dua dadu sebagai berikut.

Misalkan;

(a, b) = muncul mata dadu a pada dadu merah dan mata dadu b pada dadu biru.

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Dengan demikian diperoleh n(S) adalah 36.

 

Selanjutnya kita akana menentukan peluang kejadian.

1. Jika dua dadu dilempar satu kali, tentukan peluang kejadian muncul mata dadu keduanya prima.

Jawaban:

Misalkan A = kejadian muncul mata dadu keduanya bilangan prima

A = {(2,2),(2,3),(2, 5), (3,2),(3,3),(3,5),(5,2),(5,3), (5, 5)}

n(A) = 9

Peluang kejadian A :




 

2. Jika dua dadu dilempar satu kali, tentukan peluang kejadian muncul kedua mata dadu berselisih 2.

Jawaban:

Misalkan B = kejadian muncul kedua mata dadu berselisih 3.

B = {(1, 3),(3, 1),(2, 4), (4, 2),(3, 5),(5, 3),(4, 6),(6, 4)}

n(B) = 8

 

 

Peluang kejadian B :




3. Jika dua dadu dilempar satu kali, tentukan peluang kejadian muncul kedua mata dadu berjumlah 7.

Jawaban:

Misalkan C = kejadian muncul kedua mata dadu berjumlah 7.

C = {(1, 6),(6, 1),(2, 5), (5, 2),(3, 4),(4, 3)}

n(C) = 6

Peluang kejadian C :



 

4. Jika dua dadu dilempar satu kali, tentukan peluang kejadian muncul kedua mata dadu berjumlah lebih dari 8.

Jawaban:

Misalkan D = kejadian muncul kedua mata dadu berjumlah lebih dari 8.

D = {(3, 6),(6, 3),(4, 5), (5, 4),(4, 6),(6, 4), (5, 6),(6, 5),(6, 6)}

n(D) = 9

Peluang kejadian D :



  

5. Jika dua dadu dilempar satu kali, tentukan peluang kejadian muncul salah satu mata dadu adalah mata dadu 2.

Jawaban:

Misalkan E = kejadian muncul salah satu mata dadu adalah mata dadu 2.

E = {(2, 1), (2, 2), (2, 3), (2, 4),( 2, 5),( 2, 6), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}

n(E) = 12

Peluang kejadian E :




Demikianlah skilas materi tentang peluang kejadian pada pelemparan dua dadu.

Semoga Bermanfaat.

 

 

Menyelesaikan Masalah Barisan dan Deret Aritmetika

 


Penerapan deret aritmetika dalam kehidupan keseharian sangat banyak. Selain itu dalam hal bilangan penggunaan deret aritmetika juga diperlukan. Perlu diketahui bahwa barisan aritmetika adalah barisan bilangan yang memiliki beda antarsuku selalu sama.

Jika pola barisan bilangan memiliki suku awal a dan beda (Selisih) = b, maka pola bilangan dapat dituliskan sebagai berikut.

 

a, a + b, a + 2b, a + 3b, a + 4b, .......

 

Adapun deret aritmetika adalah jumlah bilangan-bilangan yang membentuk barisan aritmetika. Deret aritmetik dituliskan sebagai berikut.

 

Sn = a + (a + b) + (a + 2b) + (a + 3b) + (a + 4b) + .......+ a + (n – 1)b

 

 

Untuk mempelajari penerapan  tentang barisan dan deret aritmetika, mari  menyelesaikan permasalahan di bawah ini.                     

 

Permasalahan 1

Diketahui jumlah 3 bilangan ganjil berurutan adalah 5.001. Tentukan bilangan-bilangan itu.

Penyelesaian

Ingat bahwa untuk n = 1, 2, 3, 4,.... Bilangan ganjil dapat disimbolkan atau dimisalkan dengan 2n + 1.

Jika terdapat tiga bilangan ganjil berurutan maka dapat dituliskan: 2n + 1, 2n + 3, 2n + 5.

 

Jumlah 3 bilangan ganjil berurutan adalah 5.001.

(2n + 1) + (2n + 3) + (2n + 5) = 5.001

                                 6n + 9 = 5.001

                                       6n = 5.001 – 9

                                       6n = 4.992

                                        n = 4.992 : 6

                                        n = 832

Sehingga diperoleh bilangan-bilangan itu sebagai berikut.

2n + 1 = 2(832) + 1 = 1.664 + 1 = 1.665

2n + 3 = 2(832) + 3 = 1.664 + 3 = 1.667

2n + 5 = 2(832) + 5 = 1.664 + 5 = 1.669

Jadi, ketiga bilangan itu adalah 1.665, 1.667, dan 1.669.

 

Permasalahan 2

Diketahui jumlah 3 bilangan genap berurutan adalah 12.000. Tentukan bilangan-bilangan itu.

Penyelesaian

Ingat bahwa untuk n = 1, 2, 3, 4,.... Bilangan genap dapat disimbolkan atau dimisalkan dengan 2n, atau 2n + 2.

Jika terdapat tiga bilangan genap berurutan maka dapat dituliskan: 2n, 2n + 2, 2n + 4.

 

Jumlah 3 bilangan genap berurutan adalah 12.000.

2n + (2n + 2) + (2n + 4) = 12.000

                         6n + 6 = 12.000

                               6n = 12.000 – 6

                               6n = 11.994

                                 n = 11.994 : 6

                                 n = 1.999

Sehingga diperoleh bilangan-bilangan itu sebagai berikut.

2n = 2(1.999) = 3.998

2n + 2 = 2(1.999) + 2 = 3.998 + 2 = 4.000

2n + 4 = 2(1.999) + 4 = 3.998 + 4 = 4.002

Jadi, ketiga bilangan itu adalah 3.998, 4.000, dan 4.002.

 

 

Permasalahan 3

Diketahui jumlah 5 bilangan ganjil berurutan adalah 9.005. Tentukan bilangan-bilangan itu.

Penyelesaian

Ingat bahwa untuk n = 1, 2, 3, 4,.... Bilangan ganjil dapat disimbolkan atau dimisalkan dengan 2n + 1.

Jika terdapat lima bilangan ganjil berurutan maka dapat dituliskan: 2n + 1, 2n + 3, 2n + 5, 2n + 7, 2n + 9.

 

Jumlah 5 bilangan ganjil berurutan adalah 9.005.

(2n + 1) + (2n + 3) + (2n + 5) + (2n + 7) + (2n + 9) = 9.005

                   10n + 15 = 9.005

                        10n = 9.005 – 15

                        10n = 8.990

                          n = 8.990 : 10

                          n = 899

Sehingga diperoleh bilangan-bilangan itu sebagai berikut.

2n + 1 = 2(899) + 1 = 1.798 + 1 = 1.799

2n + 3 = 2(899) + 3 = 1.798 + 3 = 1.801

2n + 5 = 2(899) + 5 = 1.798 + 5 = 1.803

2n + 7 = 2(899) + 7 = 1.798 + 7 = 1.805

2n + 9 = 2(899) + 9 = 1.798 + 9 = 1.807

 

Jadi, kelima bilangan itu adalah 1.799, 1.801, 1.803, 1.805, dan 1.807.

 

 

Permasalahan 4

Diketahui jumlah 5 bilangan genap berurutan adalah 100.000. Tentukan bilangan-bilangan itu.

Penyelesaian

Ingat bahwa untuk n = 1, 2, 3, 4,.... Bilangan genap dapat disimbolkan atau dimisalkan dengan 2n atau 2n + 2.

Jika terdapat lima bilangan genap berurutan maka dapat dituliskan: 2n, 2n + 2, 2n + 4, 2n + 6, 2n + 8.

 

Jumlah 5 bilangan genap berurutan adalah 100.000.

(2n) + (2n + 2) + (2n + 4) + (2n + 6) + (2n + 8) = 100.000

                   10n + 20 = 100.000

                        10n = 100.000 – 20

                        10n = 99.980

                          n = 99.980 : 10

                          n = 9.998

Sehingga diperoleh bilangan-bilangan itu sebagai berikut.

2n = 2(9.998) = 19.996

2n + 2 = 2(9.998) + 2 = 19.996 + 2 = 19.998

2n + 4 = 2(9.998) + 4 = 19.996 + 4 = 20.000

2n + 6 = 2(9.998) + 6 = 19.996 + 6 = 20.002

2n + 8 = 2(9.998) + 8 = 19.996 + 8 = 20.004

 

Jadi, kelima bilangan itu adalah 19.996, 19.998, 20.000, 20.002, dan 20.004.

 

 

Permasalahan 5

Diketahui panjang tali mula-mula adalah 950 cm. Tali itu akan dipotong menjadi 5 tali dan panjang tali membentuk barisan aritmetika. Tentukan panjang setiap tali jika selisih antartali adalah 5 cm.

Penyelesaian

Permasalahan tentang deret aritmetika dengan jumlah 5 bilangan.

Diketahui Jumlah lima tali (Sn) = 950 dan beda (b) = 5.

Sehingga dapat ditulis:

a + (a + b) + (a + 2b) + (a + 3b) + (a + 4b) = S5

a + (a + 5) + (a + 2(5)) + (a + 3(5)) + (a + 4(5)) = 950

 

a + (a + 5) + (a + 10) + (a + 15) + (a + 20) = 950

      5a + 50 = 950

             5a = 950 – 50

             5a = 900

               a = 900 : 5

               a = 180

Sehingga diperoleh bilangan-bilangan itu sebagai berikut.

a ; (a + 5) ; (a + 10) ; (a + 15)  dan  (a + 20)

180 ; (180 + 5) ; (180 + 10) ; (180 + 15) dan  (180 + 20)

180, 185, 190, 195, dan 200.

 

Jadi, kelima bilangan itu adalah 180, 185, 190, 195, dan 200.

 

Demikianlah sekilas materi tentang penerapan barisan dan deret aritmetika dalam menyelesaikan masalah.

Semoga bermanfaat.



Kamis, 13 Februari 2025

HOW TO SOLVING VOLUME OF CYLINDER

 

Volume of Cylinder

The volume of a cylinder is the capacity of the cylinder which calculates the amount of material quantity it can hold. In geometry, there is a specific formula to calculate the volume of a cylinder that is used to measure how much amount of any quantity whether liquid or solid can be immersed in it uniformly. A cylinder is a three-dimensional shape with two congruent and parallel identical bases. There are different types of cylinders. They are:

 

Right circular cylinder: A cylinder whose bases are circles and each line segment that is a part of the lateral curved surface is perpendicular to the bases.

Oblique Cylinder: A cylinder whose sides lean over the base at an angle that is not equal to a right angle.

Elliptic Cylinder: A cylinder whose bases are ellipses.

Right circular hollow cylinder: A cylinder that consists of two right circular cylinders bounded one inside the other.

The formula to find the volume of a cylinder is V = πr2h. Let us learn more about this formula in the upcoming sections.

 

What is the Volume of a Cylinder?

The volume of a cylinder is the number of unit cubes (cubes of unit length) that can be fit into it. It is the space occupied by the cylinder as the volume of any three-dimensional shape is the space occupied by it. The volume of a cylinder is measured in cubic units such as cm3, m3, in3, etc. Let us see the formula used to calculate the volume of a cylinder.

 

Definition of a Cylinder

A cylinder is a three-dimensional solid shape that consists of two parallel bases linked by a curved surface. These bases are like a circular disk in a shape. The line passing from the center or joining the centers of two circular bases is called the axis of the cylinder.

 

Volume of Cylinder Formula

We know that a cylinder resembles a prism (but note that a cylinder is not a prism as it has a curved side face), we use the same formula of volume of a prism to calculate the volume of a cylinder as well. We know that the volume of a prism is calculated using the formula,

 

V = A × h

 

where

A = area of the base

h = height

Using this formula, the formulas of volume of cylinder are:

The formula for volume of a right circular cylinder is,

V = πr2h (r = radius, h = height)

The formula for volume of an oblique cylinder is,

V = πr2h (r = radius, h = height)

The formula for volume of an elliptic cylinder is,

V = πabh (a and b = radii, h = height)

The formula for volume of a right circular hollow cylinder is,

V = π(R2 - r2)h (R = outer radius, r = inner radius, h = height)

 

Now we will apply the formula V = A × h to calculate the volume of different types of cylinders.

 

Volume of a Right Circular Cylinder Formula

We know that the base of a right circular cylinder is a circle and the area of a circle of radius 'r' is πr2. Thus, the volume (V) of a right circular cylinder, using the above formula (V = A × h), is,

V = πr2h

Here,

'r' is the radius of the base (circle) of the cylinder

'h' is the height of the cylinder

π is a constant whose value is either 22/7 (or) 3.142.

Thus, the volume of cylinder directly varies with its height and directly varies with the square of its radius. i.e., if the radius of the cylinder becomes double, then its volume becomes four times.

 

Formula to Find Volume of an Oblique Cylinder

The formula to calculate the volume of cylinder (oblique) is the same as that of a right circular cylinder. Thus, the volume (V) of an oblique cylinder whose base radius is 'r' and whose height is 'h' is,

 

V = πr2h

 

Formula to Calculate Volume of an Elliptic Cylinder

We know that an ellipse has two radii. Also, we know that the area of an ellipse whose radii are 'a' and 'b' is πab. Thus, the volume of an elliptic cylinder is,

 

V = πabh

 

Here,

'a' and 'b' are the radii of the base (ellipse) of the cylinder.

'h' is the height of the cylinder.

π is a constant whose value is either 22/7 (or) 3.142.

 

Volume of a Right Circular Hollow Cylinder Formula

As a right circular hollow cylinder is a cylinder that consists of two right circular cylinders bounded one inside the other, its volume is obtained by subtracting the volume of the inside cylinder from that of the outside cylinder. Thus, the volume (V) of a right circular hollow cylinder is,

 

V = π(R2 - r2)h

 

Here,

'R' is the base radius of the outside cylinder.

'r' is the base radius of the inside cylinder.

'h' is the height of the cylinder.

π is a constant whose value is 22/7 (or) 3.142.

 

How To Find the Volume of Cylinder?

Here are the steps to find the volume of cylinder:

Identify the radius to be 'r' and height to be 'h' and make sure that they both are of the same units.

Substitute the values in the volume formula V = πr2h.

Write the units as cubic units.

 

Example: Find the volume of a right circular cylinder of radius 50 cm and height 1 meter. Use π = 3.142.

 

Solution:

The radius of the cylinder is, r = 50 cm.

Its height is, h = 1 meter = 100 cm.

Its volume is, V = πr2h = (3.142)(50)2(100) = 785,500 cm3.

 

Note: We need to use the formula to find the volume of a cylinder depending on its type as we discussed in the previous section. Also, assume that a cylinder is a right circular cylinder if there is no type given and apply the volume of a cylinder formula to be V = πr2h.

 

Important Notes on Volume of Cylinder:

The volume of a cylinder is calculated using the formula, V = πr2h, where r is the radius of its circular base and 'h' is the perpendicular distance (height) between the centres of the bases.

If diameter (d) is given, then find the radius (r) using r = d/2 and then substitute in the above formula to find the volume of cylinder.

 





Understanding Algebraic Forms

 

What is Algebra?

Algebra is a branch of mathematics that uses symbols, letters, and numbers to express mathematical relationships and solve problems. Instead of only using numbers, algebra allows us to represent unknown values with letters like x, y, or a.

What is an Algebraic Form?

An algebraic form is a mathematical expression made up of numbers, variables, and operations such as addition, subtraction, multiplication, and division.

Example of an algebraic form:

2x + 7

In this example, 2x + 7 is an algebraic form. It consists of several important parts that we need to understand.

 

Understanding Terms in an Algebraic Form

When working with algebra, we often hear words like terms, variables, coefficients, and constants. Let’s break these down:

1. Terms:

A term is a part of an algebraic form separated by + or - signs. Each term may contain numbers, variables, or both.

Example: In 2x + 7, there are two terms:

  • 2x,   and
  • 7

2. Variable:

A variable is a letter that represents an unknown value. It can take different values.

Example: In 2x + 7, the variable is x.

 

3. Coefficient:

A coefficient is the number that multiplies a variable.

Example: In 2x, the coefficient is 2 because it is multiplying x.

 

4. Constant:

A constant is a number that does not change. It has no variable attached to it.

Example: In 2x + 7, the constant is 7.

 

Putting it Together:

Let’s look at another example:

2y - 4 + 7x

  • Terms: 2y, -4, and 7x
  • Variables: y and x
  • Coefficients: 2 (with y), 7 (with x)
  • Constant: -4

 

5a + 7b - 4c + 9

  • Terms: 5a, 7b, -4c, 9
  • Variables: a, b, and c
  • Coefficients: 5 (with a), 7 (with b), -4 (with c)
  • Constant: 9

 

What are Similar Terms?

Similar terms (like terms) are terms that have the same variable raised to the same power. Only similar terms can be combined through addition or subtraction.

Examples of similar terms:

  • 2x and 5x are similar terms because they both have x.
  • 3y and -7y are similar terms because they both have y.

Examples of terms that are NOT similar:

  • 3x and 4y are not similar because one has x and the other has y.
  • 2x and x2 are not similar because x and x2 are different powers of x.

 

Combining Similar Terms:

When we combine similar terms, we add or subtract the coefficients and keep the variable the same.

Example:

3x + 5x = (3 + 5)x = 8x

Example with subtraction:

7y - 2y = (7 - 2)y = 5y

Example with a mix of terms:

3x + 4y - 2x + y

Combine similar terms:

  • 3x - 2x = x
  • 4y + y = 5y

Final result: x + 5y

 

Conclusion

Understanding algebraic forms is important in learning mathematics. Remember these key points:

  • Terms: Parts of an expression.
  • Variables: Represent unknown values.
  • Coefficients: Numbers multiplying variables.
  • Constants: Fixed numbers.
  • Similar Terms: Terms with the same variables and powers.

Knowing how to identify and combine similar terms will help you simplify algebraic expressions and solve equations more easily!




Rabu, 12 Februari 2025

Cara Membuktikan Tan 5x - Tan 3x - Tan 2x = Tan 5x Tan 3x Tan 2x

 

Dalam kesempatan  ini kita membuktikan fungsi trigonometeri dengan bentuk

 Tan 5x - tan 3x - tan 2x menjadi tan 5x tan 3x tan 2x.

 

Bagaimana cara membuktikan bentuk trigonometri di atas?

Begini caranya:

 

Ingat rumus dasar berikut.



           tan 3x + tan 2x = tan 5x (1 – tan 3x tan 2x)

           tan 3x + tan 2x = tan 5x  –  tan 5x tan 3x tan 2x

    tan 5x tan 3x tan 2x = tan 5x - (tan 3x + tan 2x)

    tan 5x tan 3x tan 2x = tan 5x - tan 3x - tan 2x

tan 5x - tan 3x - tan 2x = tan 5x tan 3x tan 2x

 

Jadi, terbukti bahwa tan 5x - tan 3x - tan 2x = tan 5x tan 3x tan 2x.