The document discusses permutations and combinations. It explains that permutations calculate arrangements where order matters, while combinations calculate arrangements where order does not matter. The fundamental counting principle is introduced as a way to calculate the total number of possible outcomes by multiplying the number of options for each part. An example shows there are 120 ways for 5 students to arrange themselves in 5 consecutive seats, using the fundamental counting principle.
This document discusses permutation and combination concepts in counting. It provides examples of using the multiplication principle when order matters and addition principle when order does not matter to count outcomes of events. It also defines permutation as arrangements of objects where order matters and combination as selections of objects where order does not matter. Formulas are given for counting permutations and combinations of distinct objects. Examples demonstrate calculating the number of ways to arrange or select objects in different scenarios.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Elementary Statistics Practice Test 2
Chapter 4: Probability
The document discusses permutations and combinations. It explains that permutations calculate arrangements where order matters, while combinations calculate arrangements where order does not matter. The fundamental counting principle is introduced as a way to calculate the total number of possible outcomes by multiplying the number of options for each part. An example shows there are 120 ways for 5 students to arrange themselves in 5 consecutive seats, using the fundamental counting principle.
This document discusses permutation and combination concepts in counting. It provides examples of using the multiplication principle when order matters and addition principle when order does not matter to count outcomes of events. It also defines permutation as arrangements of objects where order matters and combination as selections of objects where order does not matter. Formulas are given for counting permutations and combinations of distinct objects. Examples demonstrate calculating the number of ways to arrange or select objects in different scenarios.
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Elementary Statistics Practice Test 2
Chapter 4: Probability
El documento presenta un examen de álgebra sobre sistemas de ecuaciones lineales. El examen contiene preguntas sobre métodos para resolver sistemas de ecuaciones, identificar puntos de intersección de ecuaciones, y determinar la solución de sistemas dados gráficamente o a través de ecuaciones.
The document discusses synthetic division, providing 3 examples of dividing polynomials. The first example divides a polynomial by a monic linear divisor. The second divides a polynomial by a non-monic linear divisor. The third divides a polynomial by a monic quadratic divisor. Each example shows the division problem and solution.
The concept of trigonometric ratios is important in Maths trigonometry. Some of applications and procedures are discussed here for easy understanding the concept clarity.
1) The document discusses evaluating nth roots of numbers and expressions, including square roots, cube roots, and nth roots.
2) It provides definitions and examples of even and odd roots of positive and negative numbers. Even roots of negative numbers are undefined, while odd roots of negative numbers are negative.
3) Methods are described for evaluating nth roots of monomial expressions by splitting them into factors and taking the nth root of each factor.
The document discusses combinations and restricted combinations. Some key points:
- Combinations refer to the number of ways of selecting items without regard to order, as opposed to permutations which consider order.
- A combination formula is given to calculate the number of combinations of n items taken r at a time.
- Examples demonstrate calculating combinations in situations like selecting committee members or players for a team when certain items must or cannot be included.
- The number of combinations when p particular items must be included is written as n-pCr-p, and when p items cannot be included is n-pCr.
The document discusses expressions in mathematics. It defines expressions as calculation procedures written with numbers, variables, and operation symbols that calculate outcomes. Expressions can be combined by collecting like terms. Linear expressions take the form of ax + b, where terms can be combined by adding or subtracting the coefficients of the same variable. The example shows combining the terms of the expression 2x - 4 + 9 - 5x.
1) Διανομή σε Υποδοχές
1.1) Διανομή Ομοίων Αντικειμένων σε Υποδοχές
1.2) Διανομή Διαφορετικών Αντικειμένων Χωρίς Σειρά σε Υποδοχές
1.3) Διανομή Διαφορετικών Αντικειμένων Με Σειρά σε Υποδοχές
2) Γνωστά Προβλήματα Διατάξεων
2.1) Εξίσωση
3) Μεθοδολογία Ασκήσεων
3.1) Διανομή Ομάδων Ομοίων
3.2) Διανομή Ομοίων με Περιορισμό
3.3) Διάταξη με Εμφύτευση Υποδοχών
Ασκήσεις
O documento apresenta questões sobre números primos, perfeitos e congruências numéricas. A questão 1 discute propriedades de números primos e o Lema de Euclides. A questão 2 trata do Teorema de Euclides sobre números perfeitos. A questão 3 prova que o quinto número de Fermat não é primo usando congruências.
This document discusses 45-45-90 triangles, also known as isosceles right triangles. It provides properties of these triangles, including that the two legs are equal and each measure 45 degrees. Shortcuts are presented for calculating side lengths without using the Pythagorean theorem, such as that the hypotenuse h equals the leg length x squared, and that each leg x equals h squared over 2. Several practice problems demonstrate applying these shortcuts to find missing side lengths.
5.1 Introduction 5.2 Ratio And Proportionality 5.3 Similar Polygons 5.4 Basic Proportionality Theorem 5.5 Angle Bisector Theorem 5.6 Similar Triangles 5.7 Properties Of Similar Triangles
This document discusses how to solve absolute value inequalities by:
1) Determining whether the absolute value is greater than or less than the variable, which indicates a disjunction or conjunction graph.
2) Solving the inequality for both possibilities of the expression inside the absolute value being positive or negative.
3) Combining the solutions from both possibilities using the appropriate inequality symbol (>, <, etc.) to obtain the final solution set.
Solving systems of linear equations by graphing lectureKaiya Duppins
This document discusses solving systems of linear equations by graphing. It defines a system of linear equations as two or more equations containing at least one variable. It explains that to solve a system by graphing, each equation should be written in slope-intercept form and then graphed on the same set of axes. The type of solution can be determined by examining the graphs - one solution if the lines intersect, no solution if the lines are parallel, or infinitely many solutions if the lines coincide. It provides examples to illustrate each type of solution and explains that the point of intersection or coinciding point(s) will be the solution(s) to plug back into the original system to check.
G10 Math Q2- Week 8- Equation of a Circle.pptxRodolfoHermo3
This document discusses the standard and general forms of equations for circles. It provides examples of writing the standard and general form equations given the center and radius of various circles. Key points covered include:
- The standard form of a circle equation is (x-h)2 + (y-k)2 = r2, where (h,k) is the center and r is the radius.
- The general form is Ax2 + By2 + Cx + Dy + E = 0, where the coefficients can be used to find the center and radius.
- Examples are worked through of writing circle equations in both standard and general form given attributes of the circle like the center and radius.
This document provides instructions on how to solve radical equations. It begins by defining a radical equation and radicand. It then explains the main steps to solve radical equations: isolate the radical to one side, raise both sides to the same power, and simplify. Several examples are worked through to demonstrate this process. The document emphasizes checking solutions, as raising both sides to an even power can result in extraneous solutions. TI instructions and additional practice problems are also included.
This document contains examples of expanding double brackets using the grid method. It provides step-by-step workings for expressions like (x + 5)(x + 6), (x + 5)(3x - 2), and (2x + 1)(x - 3). The document includes a game where students roll dice to select a bracket expression from a grid to expand. The worked solutions show expanding the brackets, combining like terms, and simplifying the expression.
Here is the system of inequalities for the cross-country team problem:
Let x = number of water bottles sold to students
Let y = number of water bottles sold to others
x + y ≤ 100 (they have 100 bottles total)
3x + 5y ≥ 400 (they need at least $400)
Graph the regions defined by these inequalities and the overlapping region is the solution.
To solve quadratic inequalities:
1. Express the inequality in standard form ax^2 + bx + c > 0.
2. Determine the boundary points by setting the equation equal to 0 and solving for x.
3. Set intervals to the left and right of the boundary points.
4. Pick a test point in each interval and check if it makes the inequality true.
5. Write the solution intervals.
The document discusses permutations and combinations. It defines permutations as arrangements where order is important, and provides an example of finding the number of permutations of selecting 2 letters from a set of 3 letters. Combinations are arrangements where order is not important, and gives an example of finding the number of combinations of selecting 2 letters from a set of 3. It also presents the factorial formulas for permutations and combinations and provides examples of using them to calculate permutation and combination problems. Finally, it highlights the difference between permutations and combinations and provides 4 example problems to practice calculating permutations and combinations.
El documento presenta un examen de álgebra sobre sistemas de ecuaciones lineales. El examen contiene preguntas sobre métodos para resolver sistemas de ecuaciones, identificar puntos de intersección de ecuaciones, y determinar la solución de sistemas dados gráficamente o a través de ecuaciones.
The document discusses synthetic division, providing 3 examples of dividing polynomials. The first example divides a polynomial by a monic linear divisor. The second divides a polynomial by a non-monic linear divisor. The third divides a polynomial by a monic quadratic divisor. Each example shows the division problem and solution.
The concept of trigonometric ratios is important in Maths trigonometry. Some of applications and procedures are discussed here for easy understanding the concept clarity.
1) The document discusses evaluating nth roots of numbers and expressions, including square roots, cube roots, and nth roots.
2) It provides definitions and examples of even and odd roots of positive and negative numbers. Even roots of negative numbers are undefined, while odd roots of negative numbers are negative.
3) Methods are described for evaluating nth roots of monomial expressions by splitting them into factors and taking the nth root of each factor.
The document discusses combinations and restricted combinations. Some key points:
- Combinations refer to the number of ways of selecting items without regard to order, as opposed to permutations which consider order.
- A combination formula is given to calculate the number of combinations of n items taken r at a time.
- Examples demonstrate calculating combinations in situations like selecting committee members or players for a team when certain items must or cannot be included.
- The number of combinations when p particular items must be included is written as n-pCr-p, and when p items cannot be included is n-pCr.
The document discusses expressions in mathematics. It defines expressions as calculation procedures written with numbers, variables, and operation symbols that calculate outcomes. Expressions can be combined by collecting like terms. Linear expressions take the form of ax + b, where terms can be combined by adding or subtracting the coefficients of the same variable. The example shows combining the terms of the expression 2x - 4 + 9 - 5x.
1) Διανομή σε Υποδοχές
1.1) Διανομή Ομοίων Αντικειμένων σε Υποδοχές
1.2) Διανομή Διαφορετικών Αντικειμένων Χωρίς Σειρά σε Υποδοχές
1.3) Διανομή Διαφορετικών Αντικειμένων Με Σειρά σε Υποδοχές
2) Γνωστά Προβλήματα Διατάξεων
2.1) Εξίσωση
3) Μεθοδολογία Ασκήσεων
3.1) Διανομή Ομάδων Ομοίων
3.2) Διανομή Ομοίων με Περιορισμό
3.3) Διάταξη με Εμφύτευση Υποδοχών
Ασκήσεις
O documento apresenta questões sobre números primos, perfeitos e congruências numéricas. A questão 1 discute propriedades de números primos e o Lema de Euclides. A questão 2 trata do Teorema de Euclides sobre números perfeitos. A questão 3 prova que o quinto número de Fermat não é primo usando congruências.
This document discusses 45-45-90 triangles, also known as isosceles right triangles. It provides properties of these triangles, including that the two legs are equal and each measure 45 degrees. Shortcuts are presented for calculating side lengths without using the Pythagorean theorem, such as that the hypotenuse h equals the leg length x squared, and that each leg x equals h squared over 2. Several practice problems demonstrate applying these shortcuts to find missing side lengths.
5.1 Introduction 5.2 Ratio And Proportionality 5.3 Similar Polygons 5.4 Basic Proportionality Theorem 5.5 Angle Bisector Theorem 5.6 Similar Triangles 5.7 Properties Of Similar Triangles
This document discusses how to solve absolute value inequalities by:
1) Determining whether the absolute value is greater than or less than the variable, which indicates a disjunction or conjunction graph.
2) Solving the inequality for both possibilities of the expression inside the absolute value being positive or negative.
3) Combining the solutions from both possibilities using the appropriate inequality symbol (>, <, etc.) to obtain the final solution set.
Solving systems of linear equations by graphing lectureKaiya Duppins
This document discusses solving systems of linear equations by graphing. It defines a system of linear equations as two or more equations containing at least one variable. It explains that to solve a system by graphing, each equation should be written in slope-intercept form and then graphed on the same set of axes. The type of solution can be determined by examining the graphs - one solution if the lines intersect, no solution if the lines are parallel, or infinitely many solutions if the lines coincide. It provides examples to illustrate each type of solution and explains that the point of intersection or coinciding point(s) will be the solution(s) to plug back into the original system to check.
G10 Math Q2- Week 8- Equation of a Circle.pptxRodolfoHermo3
This document discusses the standard and general forms of equations for circles. It provides examples of writing the standard and general form equations given the center and radius of various circles. Key points covered include:
- The standard form of a circle equation is (x-h)2 + (y-k)2 = r2, where (h,k) is the center and r is the radius.
- The general form is Ax2 + By2 + Cx + Dy + E = 0, where the coefficients can be used to find the center and radius.
- Examples are worked through of writing circle equations in both standard and general form given attributes of the circle like the center and radius.
This document provides instructions on how to solve radical equations. It begins by defining a radical equation and radicand. It then explains the main steps to solve radical equations: isolate the radical to one side, raise both sides to the same power, and simplify. Several examples are worked through to demonstrate this process. The document emphasizes checking solutions, as raising both sides to an even power can result in extraneous solutions. TI instructions and additional practice problems are also included.
This document contains examples of expanding double brackets using the grid method. It provides step-by-step workings for expressions like (x + 5)(x + 6), (x + 5)(3x - 2), and (2x + 1)(x - 3). The document includes a game where students roll dice to select a bracket expression from a grid to expand. The worked solutions show expanding the brackets, combining like terms, and simplifying the expression.
Here is the system of inequalities for the cross-country team problem:
Let x = number of water bottles sold to students
Let y = number of water bottles sold to others
x + y ≤ 100 (they have 100 bottles total)
3x + 5y ≥ 400 (they need at least $400)
Graph the regions defined by these inequalities and the overlapping region is the solution.
To solve quadratic inequalities:
1. Express the inequality in standard form ax^2 + bx + c > 0.
2. Determine the boundary points by setting the equation equal to 0 and solving for x.
3. Set intervals to the left and right of the boundary points.
4. Pick a test point in each interval and check if it makes the inequality true.
5. Write the solution intervals.
The document discusses permutations and combinations. It defines permutations as arrangements where order is important, and provides an example of finding the number of permutations of selecting 2 letters from a set of 3 letters. Combinations are arrangements where order is not important, and gives an example of finding the number of combinations of selecting 2 letters from a set of 3. It also presents the factorial formulas for permutations and combinations and provides examples of using them to calculate permutation and combination problems. Finally, it highlights the difference between permutations and combinations and provides 4 example problems to practice calculating permutations and combinations.
7. مفهوم أساسي
الحل بالتعويض:
الخطوة1:
حل إحدى المعادلتين على القل باستعمال أحد المتغيرين إذا كان ذلك ضروريا.
ً
الخطوة2:
عوض المقدار الناتج من الخطوة )1( في المعادلة الثانية ، ثم حلها.
الخطوة3:
عوض القيمة الناتجة من الخطوة )2( في أي من المعادلتين
وحلها ليجاد قيمة المتغير الثاني ، واكتب الحل كزوج مرتب.
8. حل نظام من معادلتين بالتعويض
مثال1
استعمل التعويض لحل النظام التي:
ص = 2س + 1
3 س + ص = -9
9. ص = 2س + 1
3س + ص = -9
الخطوة1: إحدى المعادلتين مكتوبة أساسا بدللة ص
ً
الخطوة2: عوض 2س + 1 بدل من ص في المعادلة الثانية
ً
3س + ص = -9
المعادلة الثانية
3س + 2س + 1= -9 عوض عن ص بـ 2س + 1
5س + 1= -9 اجمع الحدود المتشابهة
5س = -01
س = -2
اقسم كل طرف على 5
اطرح )1( من كل طرف
الخطوة3:
ص = 2س + 1 المعادلة الأولى
= 2 )-2( + 1 عوض عن س بـ )-2(
بسط
طّ
= -3
عوض -2 بدل من س في أي من المعادلتين ليجاد قيمة ص.
ً
إذن ، الحل هو: )-2 ،-3(.
12. وإذا لم يكن أحد المتغيرين مكتوبا وحده في طرف إحدى
ً
المعادلتين في النظام، فحل إحدى المعادلتين أوال بالنسبة
ً
لهذا المتغير، ثم عوض لحل النظام.
13. إرشادات للدراسة
صيغة الميل والمقطع
إذا كتبت كل من المعادلتين بصيغة الميل والمقطع
)ص = م س + ب(، فيمكن مساواتهما معا، ثم إيجاد
ً
قيمة س، وتعويضها ليجاد قيمة ص.
15. س + 2ص = 6
3 س – 4 ص = 82
الخطوة1: حل المعادلة اللولى بالنسبة للمتغير س ل ن معامل س = 1.
المعادلة الولى
س + 2ص = 6
س + 2ص - 2ص
= 6 – 2ص
اطرح 2ص من كل طرف
س = 6 - 2ص
بسط
طّ
الخطوة2: عوض عن س بـ )6- 2ص( في المعادلة الثانية ليجاد قيمة ص.
3)2-6ص(-4ص=82 عوض عن س بـ )6- 2ص(
خاصية التوزيع
6-81ص - 4ص=82
اجمع الحدود المتشابهة
81 – 01ص =82
81–01ص -81 =82-81 اطرح 81 من كل طرف
16. -01 ص = 01
ص = -1
بسط
طّ
اقسم كل طرف على -01
الخطوة3: أوجد قيمة س بالتعويض في المعادلة الولى.
س+2ص=6
المعادلة الولى
أوجد قيمة س بالتعويض في المعادلة الولى.
س + 2 )-1( = 6
س-2=6
س=8
الحل هو )8، -1(
عوض عن ص بـ )-1(
بسط
طّ
أضف 2 إلى كل طرف
19. وبصورة عامة، إذا كانت نتيجة حل نظام
من معادلتين جملة خطأ مثل 3 = -2، فل
يوجد حل للنظام في هذه الحالة، أما إذا
كانت النتيجة متطابقة مثل 3 = 3 فهناك
عدد النهائي من الحلول.
20. إرشادات للدراسة
النظام غير المستقل
هناك عدد النهائي من الحلول للنظام
في المثال3؛ لنه عند كتابة المعادلتين
بصيغة الميل والمقطع تكونان
متكافئتين ولهما التمثيل البياني نفسه.
21. عدم وجود حل للنظام، أو وجود عدد ال
نهائي من الحلول:
مثال3
حل النظام التي مستعمل التعويض:
ً
ص = 2س – 4
-6س + 3ص = -21
22. ص = 2س – 4
6س + 3ص = -21-6س + 3ص = -21
عوض عن ص بـ )2س – 4( في
المعادلة الثانية.
المعادلة الثانية
-6س + 3)2س-4( = -21 عوض عن ص بـ )2س – 4(
خاصية التوزيع
6س + 6س – 21 = -21 21 = - 21اجمع الحدود المتشابهة
بما أن الجملة الناتجة تشكل متطابقة، لذا يوجد عدد ال
نهائي من الحلول.
25. ا ً
حل كال من النظامين التيين مستعمال
ا ً
التعويض.
حل مسائل من واقع الحياة:
يمكنك استعمال التعويض لحل
مسألة من واقع الحياة تتضمن
نظاما من معادلتين.
ا ً
26. كتابة نظام من معادلتين وحله
مثال 4
من واقع الحياة
أجهزة: باع متجر 521 جهاز تسجيل وسماعة، بسعر 59.401 ريال ت
اّ
لجهاز التسجيل الواحد، و59.81 ريال للسماعة الواحدة، فإذا كان ثمن
ا ً
مبيعاته من الجهزة 57.6296 ريال. فكم جهازا باع من كل نوع؟
ا ً
ا ً
لتكن جـ = عدد أجهزة التسجيل، ت = عدد السماعات.
عدد الوحدات
المبيعة
السعر
جـ
ت
521
59.401جـ
59.81ت
57.6296
27. أجهزة: باع متجر 521 جهاز تسجيل وسماعة، بسعر 59.401 ريال ت
اّ
لجهاز التسجيل الواحد، و59.81 ريال للسماعة الواحدة، فإذا كان ثمن
ا ً
مبيعاته من الجهزة 57.6296 ريال. فكم جهازا باع من كل نوع؟
ا ً
ا ً
عدد الوحدات
المبيعة
السعر
جـ
ت
521
59.401جـ
59.81ت
57.6296
فتكون المعادلتان هما: جـ + ت = 521، 59.401جـ
+ 59.81 ت = 57.6296
28. أجهزة: باع متجر 521 جهاز تسجيل وسماعة، بسعر 59.401 ريال ت
اّ
لجهاز التسجيل الواحد، و59.81 ريال للسماعة الواحدة، فإذا كان ثمن
ا ً
مبيعاته من الجهزة 57.6296 ريال. فكم جهازا باع من كل نوع؟
ا ً
ا ً
عدد الوحدات
المبيعة
السعر
جـ
ت
521
59.401جـ
59.81ت
57.6296
29. الخطوة1:
حل المعادلة الولى بالنسبة للمتغير جـ
جـ + ث = 521
جـ + ت- ت =521- ت
جـ =521- ت
المعادلة الولى
اطرح ت من كل طرف
بس ط
اّ
الخطوة2: عوض عن جـ بـ )521- ت( في المعادلة الثانية.
59.401جـ + 59.81 ت = 57.6296 المعادلة التانية
59.401)521- ت( + 59.81 ت = 57.6296
عوض عن جـ بـ )521- ت(
59.401-57.81131 ت + 59.81 ت = 57.6296
خاصية التوزيع
31. الخطوة3:
عوض عن ت بـ )27( في إحدى المعادلتين
ليجاد قيمة جـ.
جـ + ت = 521
المعادلة الولى
جـ + 27 = 521
عوض عن ت بـ )27(
جـ = 35
اطرح 27 من كل طرف
إذن، باع المتجر 35 جهاز تسجيل، 27 سماعة
اّ
32. تحقق من فهمك
4( رياضة: مجموع النقاط التي سجلها فريقان في إحدى
مباريا ت كرة اليد 13 نقطة. فإذا كان عدد نقاط الفريق
ُ
الول يساوي 2.5 مرة عدد نقاط الفريق الثاني. فما عدد
نقاط كل فريق؟