Examples of equality. Properties of equalities on which the solution of equations is based. way to solve the equation

“Equality” is a topic that students are taught as early as primary school. It also goes hand in hand with “Inequalities.” These two concepts are closely interrelated. In addition, they are associated with terms such as equations and identities. So what is equality?

Concept of equality

This term refers to statements that contain the “=” sign. Equalities are divided into true and false. If in the entry instead of = there is<, >, then we are talking about inequalities. By the way, the first sign of equality indicates that both parts of the expression are identical in their result or record.

In addition to the concept of equality, the school also studies the topic “Numerical Equality.” This statement refers to two numerical expressions that stand on either side of the = sign. For example, 2*5+7=17. Both parts of the record are equal to each other.

Numeric expressions of this type may use parentheses that affect the order of operations. So, there are 4 rules that should be taken into account when calculating the results of numerical expressions.

1. If there are no brackets in the entry, then the actions are performed from the highest level: III→II→I. If there are several actions of the same category, then they are performed from left to right.
2. If there are parentheses in the entry, then the action is performed in parentheses and then in steps. There may be several actions in brackets.
3. If the expression is presented as a fraction, then you need to calculate the numerator first, then the denominator, then divide the numerator by the denominator.
4. If a record has nested parentheses, then the expression in the inner parentheses is evaluated first.

So, now it is clear what equality is. In the future, the concepts of equations, identities and methods for calculating them will be considered.

Properties of numerical equalities

What is equality? Studying this concept requires knowledge of the properties of numerical identities. The text formulas below allow you to better study this topic. Of course, these properties are more suitable for studying mathematics in high school.

1. Numerical equality will not be violated if the same number is added to both parts of the existing expression.

A = B↔ A + 5 = B + 5

2. The equation will not be violated if both its parts are multiplied or divided by the same number or expression that is different from zero.

P = O↔ P ∙ 5 = O ∙ 5

P = O↔ P: 5 = O: 5

3. By adding to both sides of the identity the same function, which makes sense for any admissible values ​​of the variable, we obtain a new equality that is equivalent to the original one.

F(X) = Ψ(X) ↔ F(X) + R(X) =Ψ (X) + R(X)

4. Any term or expression can be moved to the other side of the equal sign, but the signs must be reversed.

X + 5 = Y - 20 ↔ X = Y - 20 - 5 ↔ X = Y - 25

5. By multiplying or dividing both sides of the equation by the same function, which is different from zero and has meaning for each value of X from the ODZ, we obtain a new equation that is equivalent to the original one.

F(X) = Ψ(X)↔ F(X)∙R(X) = Ψ(X)∙R(X)

F(X) = Ψ(X)↔ F(X) : G(X) = Ψ(X) : G(X)

The above rules clearly indicate the principle of equality, which exists under certain conditions.

Concept of proportion

In mathematics there is such a thing as equality of relations. In this case, the definition of proportion is implied. If you divide A by B, the result will be the ratio of the number A to the number B. A proportion is the equality of two ratios:

Sometimes the proportion is written as follows: A:B=C:D. This implies the basic property of proportion: A*D=D*C, where A and D are the extreme terms of the proportion, and B and C are the average.

Identities

An identity is an equality that will be true for all permissible values ​​of the variables included in the task. Identities can be represented as literal or numeric equalities.

Expressions that contain an unknown variable on both sides of the equality that can equate two parts of one whole are called identically equal.

If you replace one expression with another, which will be equal to it, then we are talking about an identical transformation. In this case, you can use abbreviated multiplication formulas, the laws of arithmetic and other identities.

To reduce a fraction, you need to carry out identical transformations. For example, given a fraction. To get the result, you should use abbreviated multiplication formulas, factorization, simplifying expressions, and reducing fractions.

It is worth considering that this expression will be identical when the denominator is not equal to 3.

5 ways to prove identity

To prove the identity equality, you need to transform the expressions.

Method I

It is necessary to carry out equivalent transformations on the left side. The result is the right-hand side, and we can say that the identity is proven.

II method

All expression transformation actions occur on the right side. The result of the manipulations performed is the left side. If both parts are identical, then the identity is proven.

III method

“Transformations” occur in both parts of the expression. If the result is two identical parts, the identity is proven.

IV method

The right side is subtracted from the left side. As a result of equivalent transformations, the result should be zero. Then we can talk about the identity of the expression.

V method

The left side is subtracted from the right side. All equivalent transformations are reduced to ensuring that the answer contains zero. Only in this case can we talk about the identity of equality.

Basic properties of identities

In mathematics, the properties of equalities are often used to speed up the calculation process. Thanks to basic algebraic identities, the process of calculating some expressions will take a matter of minutes instead of long hours.

• X + Y = Y + X
• X + (Y + C) = (X + Y) + C
• X + 0 = X
• X + (-X) = 0
• X ∙ (Y + C) = X∙Y + X∙C
• X ∙ (Y - C) = X ∙ Y - X ∙ C
• (X + Y) ∙ (C + E) = X∙C + X∙E + Y∙C + Y∙E
• X + (Y + C) = X + Y + C
• X + (Y - C) = X + Y - C
• X - (Y + C) = X - Y - C
• X - (Y - C) = X - Y + C
• X ∙ Y = Y ∙ X
• X ∙ (Y ∙ C) = (X ∙ Y) ∙ C
• X ∙ 1 = X
• X ∙ 1/X = 1, where X ≠ 0

Abbreviated multiplication formulas

At their core, abbreviated multiplication formulas are equalities. They help solve many problems in mathematics due to their simplicity and ease of use.

• (A + B) 2 = A 2 + 2∙A∙B + B 2 - the square of the sum of a pair of numbers;

• (A - B) 2 = A 2 - 2∙A∙B + B 2 - squared difference of a pair of numbers;

• (C + B) ∙ (C - B) = C 2 - B 2 - difference of squares;

• (A + B) 3 = A 3 + 3∙A 2 ∙B + 3∙A∙B 2 + B 3 - cube of the sum;

• (A - B) 3 = A 3 - 3∙A 2 ∙B + 3∙A∙B 2 - B 3 - cube of the difference;

• (P + B) ∙ (P 2 - P∙B + B 2) = P 3 + B 3 - sum of cubes;

• (P - B) ∙ (P 2 + P∙B + B 2) = P 3 - B 3 - difference of cubes.

Abbreviated multiplication formulas are often used if it is necessary to bring a polynomial to its usual form, simplifying it in all possible ways. The presented formulas are easy to prove: just open the brackets and add similar terms.

Equations

After studying the question of what equality is, you can proceed to the next point: An equation is understood as an equality in which unknown quantities are present. Solving an equation is finding all values ​​of a variable such that both sides of the entire expression are equal. There are also tasks in which finding solutions to an equation is impossible. In this case they say that there are no roots.

As a rule, equalities with unknowns produce integers as solutions. However, there may be cases where the root is a vector, function, or other objects.

An equation is one of the most important concepts in mathematics. Most scientific and practical problems do not allow one to measure or calculate any quantity. Therefore, it is necessary to create a ratio that will satisfy all the conditions of the task. In the process of compiling such a relationship, an equation or system of equations appears.

Typically, solving an equality with an unknown comes down to transforming a complex equation and reducing it to simple forms. It must be remembered that transformations must be carried out on both sides, otherwise the output will be incorrect.

4 ways to solve an equation

By solving an equation we mean replacing a given equality with another, which is equivalent to the first. Such a substitution is known as an identity transformation. To solve the equation, you must use one of the methods.

1. One expression is replaced by another, which will necessarily be identical to the first. Example: (3∙x+3) 2 =15∙x+10. This expression can be converted to 9∙x 2 +18∙x+9=15∙x+10.

2. Transfer of terms of equality with the unknown from one side to the other. In this case, it is necessary to change the signs correctly. The slightest mistake will ruin all the work done. Let's take the previous “sample” as an example.

9∙x 2 + 12∙x + 4 = 15∙x + 10

9∙x 2 + 12∙x + 4 - 15∙x - 10 = 0

3. Multiplying both sides of an equality by an equal number or expression that does not equal 0. However, it is worth remembering that if the new equation is not equivalent to the equality before the transformations, then the number of roots may change significantly.

4. Squaring both sides of the equation. This method is simply wonderful, especially when there are irrational expressions in the equality, that is, the expression under it. There is one nuance here: if you raise the equation to an even power, then extraneous roots may appear that will distort the essence of the task. And if you extract the root incorrectly, then the meaning of the question in the problem will be unclear. Example: │7∙х│=35 → 1) 7∙х = 35 and 2) - 7∙х = 35 → the equation will be solved correctly.

So, in this article, terms such as equations and identities are mentioned. They all come from the concept of “equality.” Thanks to various kinds of equivalent expressions, the solution of some problems is greatly facilitated.

50. Properties of equalities on which the solution of equations is based. Let's take some equation, not very complicated, for example:

7x – 24 = 15 – 3x

x/2 – (x – 3)/3 – (x – 5)/6 = 1

We see in each equation an equal sign: everything that is written to the left of the equal sign is called the left or first part of the equation (in the first equation 7x – 24 is the left or first part, and in the second x/2 – (x – 3)/ 3 – (x – 5)/6 is the first, or left, part); everything that is written to the right of the equal sign is called the right or second part of the equation (15 – 3x is the right side of the first equation, 1 is the right, or second, part of the 2nd equation).

Each part of any equation represents a number. The numbers expressed by the left and right sides of the equation must be equal to each other. It is clear to us: if we add the same number to each of these numbers, or subtract the same number from them, or multiply each of them by the same number, or, finally, divide by the same number, then the results of these actions should also be equal to each other. In other words: if a = b, then a + c = b + c, a – c = b – c, ac = bc and a/c = b/c. Regarding division, it should be borne in mind, however, that in arithmetic there is no division by zero - we cannot, for example, divide the number 5 by zero. Therefore, in the equality a/c = b/c, the number c cannot be equal to zero.

1. The same number can be added to or subtracted from both sides of the equation.
2. Both sides of an equation can be multiplied or divided by the same number, unless the number is zero.

Using these properties of the equation, we can find a convenient way to solve the equations. Let's clarify this case with examples.

Example 1. Suppose we need to solve the equation

5x – 7 = 4x + 15.

We see that the first part of the equation contains two terms; one of them 5x, containing the unknown factor x, can be called the unknown term, and the other -7 - known. The second part of the equation also has 2 terms: unknown 4x and known +15. Let's make sure that on the left side of the equation there are only unknown terms (and the known term –7 would be destroyed), and on the right side there would be only known terms (and the unknown term +4x would be destroyed). For this purpose, we add the same numbers to both sides of the equation: 1) add +7 each (so that the –7 term is destroyed) and 2) add –4x each (so that the +4x term is destroyed). Then we get:

5x – 7 + 7 – 4x = 4x + 15 + 7 – 4x

Having reduced similar terms in each part of the equation, we get

This equality is the solution to the equation, since it indicates that for x we ​​must take the number 22.

Example 2. Solve the equation:

8x + 11 = 7 – 4x

Again we add –11 and +4x to both sides of the equation, we get:

8x + 11 – 11 + 4x = 7 – 4x – 11 + 4x

Reducing similar terms, we get:

Now divide both sides of the equation by +12, we get:

x = –4/12 or x = –1/3

(the first part of the equation 12x divided by 12 - we get 12x/12 or just x; the second part of the equation -4 divided by +12 - we get -4/12 or -1/3).

The last equality is the solution to the equation, since it indicates that for x we ​​need to take the number –1/3.

Example 3. Solve with equation

x – 23 = 3 (2x – 3)

Let's first open the brackets and get:
x – 23 = 6x – 9

Add +23 and –6x to both sides of the equation, we get:

x – 23 + 23 – 6x = 6x – 9 + 23 – 6x.

Now, in order to subsequently speed up the process of solving the equation, we will not immediately perform the reduction of all similar terms, but only note that the terms –23 and +23 on the left side of the equation cancel each other out, and the terms +6x and –6x in the first part cancel each other out. are destroyed - we get:

x – 6x = –9 + 23.

Let's compare this equation with the initial one: in the beginning there was an equation:

x – 23 = 6x – 9

Now we have the equation:

x – 6x = –9 + 23.

We see that in the end it turned out that the term –23, which was initially on the left side of the equation, now seemed to move to the right side of the equation, and its sign changed (there was a term –23 on the left side of the initial equation, but now it is not there , but on the right side of the equation there is a term + 23, which was not there before). Similarly, on the right side of the equation there was a term +6x, now it is not there, but on the left side of the equation a term –6x has appeared, which was not there before. Considering examples 1 and 2 from this point of view, we come to a general conclusion:

You can transfer any term of the equation from one part to another by changing the sign of this term(we will use this in further examples).

So, returning to our example, we have the equation

x – 6x = –9 + 23

Divide both sides of the equation by –5. Then we get:

[–5x: (–5) we get x] – this is the solution to our equation.

Example 4. Solve the equation:

Let's make sure there are no fractions in the equation. For this purpose, we will find a common denominator for our fractions - the common denominator is the number 24 - and multiply both sides of our equation by it (it is possible, after all, to ensure equality is not violated, we can multiply only both sides of the equation by the same number). The first part has 3 terms, and each term is a fraction - it is necessary, therefore, to multiply each fraction by 24: the second part of the equation is 0, and multiply zero by 24 - we get zero. So,

We see that each of our three fractions, due to the fact that it is multiplied by the common least multiple of the denominators of these fractions, will be reduced and become a whole expression, namely we get:

(3x – 8) 4 – (2x – 1) 6 + (x – 7) 3 = 0

Of course, it is advisable to do all this in our minds: we need to imagine that, for example, the numerator of the first fraction is placed in parentheses and multiplied by 24, after which our imagination will help us see the reduction of this fraction (by 6) and the final result, i.e. (3x – 8) · 4. The same holds for other fractions. Let us now open the brackets in the resulting equation (on its left side):

12x – 32 – 12x + 6 + 3x – 21 = 0

(please note that here it was necessary to multiply the binomial 2x – 1 by 6 and subtract the resulting product 12x – 6 from the previous one, due to which the signs of the terms of this product should change - above it is written –12x + 6). Let's move the known terms (i.e. –32, +6 and –21) from the left side of the equation to its right side, and (as we already know) the signs of these terms should change - we get:

12x – 12x + 3x = 32 – 6 + 21.

Let's cast similar terms:

(with skill, you should immediately transfer the necessary terms from one part of the equation to another and bring similar terms), finally, divide both sides of the equation by 3 - we get:

x = 15(2/3) - this is the solution to the equation.

Example 5. Solve the equation:

5 – (3x + 1)/7 = x + (2x – 3)/5

There are two fractions here, and their common denominator is 35. To free the equation from fractions, we multiply both sides of the equation by the common denominator 35. Each part of our equation has 2 terms. When multiplying each part by 35, each term must be multiplied by 35 - we get:

The fractions are reduced and we get:

175 – (3x + 1) 5 = 35x + (2x – 3) 7

(of course, if you had the skill, you could write this equation right away).

Let's do all the steps:

175 – 15x – 5 = 35x + 14x – 21.

Let's move all unknown terms from the right side (i.e., terms +35x and +14x) to the left, and all known terms from the left side (i.e., terms +175 and –5) to the right - we should not forget transferred members change sign:

–15x – 35x – 14x = –21 – 175 + 5

(the term –15x, as it used to be on the left side, remains in it now - therefore, it should not change its sign at all; a similar thing occurs for the term –21). Having reduced similar terms, we get:

–64x = –191.

[It is possible to make sure that there is no minus sign on both sides of the equation; To do this, we multiply both sides of the equation by (–1), we get 64x = 191, but we don’t have to do this.]
We then divide both sides of the equation by (–64), and obtain a solution to our equation

[If we multiplied both sides of the equation by (–1) and got the equation 64x = 191, then now we need to divide both sides of the equation by 64.]

Based on what we had to do in examples 4 and 5, we can establish: it is possible to free the equation from fractions - to do this, we need to find the common denominator for all fractions included in the equation (or the least common multiple of the denominators of all fractions) and multiply both by it parts of the equation - then the fractions should disappear.

Example 6. Solve the equation:

Moving the 4x term from the right side of the equation to the left, we get:

5x – 4x = 0 or x = 0.

So, the solution has been found: for x we ​​need to take the number zero. If we replace x in this equation with zero, we get 5 0 = 4 0 or 0 = 0, which indicates that the requirement expressed by this equation is met: find a number for x such that the monomial 5x is equal to the same number as monomial 4x.

If one notices from the very beginning that both sides of the equation 5x = 4x can be divided by x and performs this division, the result is a clear inconsistency: 5 = 4! The reason for this is that dividing 5x/x cannot be done in this case, since, as we saw above, the question expressed by our equation requires that x = 0, and division by zero is not possible.

Let us also note that multiplying by zero requires some care: multiplying by zero and two unequal numbers, as a result of these multiplications we will obtain equal products, namely zeros.

If, for example, we have the equation

x – 3 = 7 – x (his solution: x = 5)

and if someone wants to apply to it the property “both sides of the equation can be multiplied by the same number” and multiply both sides by x, they will get:

x 2 – 3x = 7x – x 2.

After this, you may notice that all terms of the equation contain a factor x, from which we can conclude that to solve this equation we can take the number zero, that is, put x = 0. And indeed, then we get:
0 2 – 3 0 = 7 0 – 0 2 or 0 = 0.

However, this solution x = 0 is obviously not suitable for the given equation x – 3 = 7 – x; replacing x with zero, we get an obvious inconsistency: 3 = 7!

Let the event IN is that the second ball drawn will be white. Probability of event IN can be determined using the total probability formula, and the conditional probabilities R(H1 /A) And R(H2 /A) become a priori for the event IN, That's why

P(B) = P(H1 /A)∙P(B/H1 ) + P(H2 /A)∙P(B/H2 ) = 1/4∙4/5 + 3/4∙2/5 = 1/2.

2.6. Tasks

1. When is the equality AB = A possible?

Answer: event A– a special case of an event IN.

2..gif" width="15" height="21 src=">+ C).

Answer: A = BC.

3. Prove that = A + B and .

4. When are equalities possible: a) A + B = , b) AB = , c) A + B = AB?

Answer: a) A is impossible, and B is certain;

b) A is reliable, and B is impossible;

5. Find a random event X from equality: https://pandia.ru/text/80/003/images/image050_0.gif" width="12" height="23 src=">.gif" width="120 height=23" height=" 23"> so what A, https://pandia.ru/text/80/003/images/image128_0.gif" width="16" height="16 src="> and through A, Bk And WITHJ.

Answer: D= A(B1 + B2 + B3 + B4) (C1 + C2),

8. The student knows 20 out of 25 questions in the program. The test is considered passed if the student answers 3 out of 4 questions posed. What is the probability that the student will pass the test?

Answer: R = 2109/2530 ≈ 0,834.

9. Two shooters, for whom the probabilities of hitting the target are 0.7 and 0.8, respectively, fire one shot each. Determine the probability of at least one hit on the target.

Answer: R = 0,94.

10. The probability of hitting the first target for the shooter is 2/3. If a hit is recorded on the first shot, then the shooter is entitled to a second shot at another target. The probability of hitting both targets with two shots is 0.5. Determine the probability of hitting the second target.

Answer: R = 0,75.

11. A student looks for the formula he needs in three reference books. The probabilities that the formula is contained in the first, second, third directory, respectively, are 0.6; 0.7; 0.8. Find the probability that the formula is contained: a) in only one reference book; b) only in two directories; c) in all three directories.

Answer: a) R= 0.188; b) R= 0.452; V) R = 0,336.

12. Students perform a test in the class of controlling machines. The work consists of three tasks. To receive credit, it is enough to solve two problems. For each problem, five different answers are encrypted, of which only one is correct. Student Petrov does not know the material well and therefore chooses answers for each problem at random. What is the probability that he will receive credit?

Answer: R = 0,104.

Problems 13–17 show connection diagrams for elements that form a circuit with one input and one output. It is assumed that element failures are collectively independent events. Reliability is considered known pk k th element and, accordingly, qk = (1 - pk) is the probability of its failure. The failure of any element leads to an interruption of the signal in the branch of the circuit where this element is located. Calculate reliability p each of the schemes.

13.

Answer: R = 1 – q1 q2 q 3.

Answer: R= 1 – (1 – р1р2р 3) (1 – p4p5p 6).

15.

Answer: R = р1р4(1 – q2 q3).

16.

Answer: R = (1 – q1 q2 ) (1 – q3 q4 ).

17.

Answer: R= p5(1 – q1 q2 ) (1 – q3 q4 ) + q5 (р1р3 + р2р4 – р1р2р3р4).

18. Over a certain period of time, a bacterium can die with a probability of 1/4, survive with a probability of 1/4, and split into two with a probability of 1/2. In the next similar period of time, the same thing happens to each bacterium, regardless of its origin. How many bacteria and with what probabilities can exist by the end of the second period of time?

Answer: 0, 1, 2, 3, 4 bacteria can exist, respectively, with probabilities of 11/32, 4/32, 5/32, 4/32 and 4/32.

19. Ivan and Peter take turns each m Two dice are thrown once. The winner is the one who first gets the sum of points on both dice equal to 8. Ivan throws first. Find probabilities p1 And p2 winnings for each player and determine how many times Ivan’s chances of winning are higher than Peter’s if: a) the number of throws is not limited and m=1; b) the number of throws is not limited, but m = 2.

Answer: a) p1 = 36/67; p2 = 31/67; р1/р2 = 36/31 ≈ 1,16;

b) p1 =362/(362 + 312) ≈ 0,574; p2 = 312/(362 + 312) ≈ 0,426; р1/р2 = 62/312 ≈ ≈ 1,35.

20. One aerial bomb is enough to destroy a bridge. Find the probability that the bridge will be destroyed if 4 bombs are dropped on it, the hit probabilities of which are respectively equal to 0.3; 0.4; 0.5 and 0.6.

Answer: R= 0,916.

21. The probability of at least one hit on the target with four shots is 0.9919. Find the probability of hitting the target with one shot.

Answer: R = 0,7.

22. Televisions from three factories go on sale. The products of the first plant contain 20% of TVs with hidden defects, the second - 10%, the third - 5%. What is the probability of purchasing a working TV if the store received 30% of TVs from the first factory, 20% from the second and 50% from the third?

Answer: R = 0,895.

23. Three single shots are fired at the aircraft. The probability of a hit on the first shot is 0.4, on the second – 0.5, on the third – 0.7. Three hits are obviously enough for an aircraft to fail; with one hit, the aircraft fails with a probability of 0.2, and with two hits, with a probability of 0.6. Find the probability that the plane will be disabled as a result of three shots.

Answer: R = 0,458.

24. The first urn contains 10 balls, 8 of which are white; The second urn contains 20 balls, 4 of which are white. One ball is drawn at random from each urn, and then one ball is drawn at random from these two balls. Find the probability that a non-white ball will be taken.

Answer: R = 0,5.

25. The first urn contains 6 white and 4 black balls, the second urn contains 3 white and 2 black balls, 3 balls are drawn at random from the first urn, and the balls of the color that are in the majority are dropped into the second urn and mixed thoroughly. After this, 1 ball is drawn at random from the second urn. What is the probability that this ball will be white?

Answer: R = 349/560 ≈ 0,623.

26. To search for an oil field in a given territory, a n geological parties, each of which, regardless of the others, discovers a deposit with a probability R. After processing and analysis of seismographic records, the entire territory was divided into two areas. In the first area, oil may occur with a probability p1, and in the second – with probability 1 - p1. How should it be distributed? n geological parties in two areas so that the probability of discovering oil is maximized?

Answer: you should send to the first district k0 geological parties, where k0 – the closest integer to the number [ n/2 + (ln((1 – p1)/p1))/2ln(1 – R)]. Let the event A– oil has been discovered in a given area. Then

P(A) = 1 – p1(1 – R)k – (1 – p1) (1 –R)n- k, Where k– the number of geological parties sent to the first region. Next, consider the function

f(x) = 1 – p1(1 – R)X – (1 – p1) (1 - R)n-X and find its maximum at XÎ.

27. There are 10 rifles in the pyramid, 4 of which are equipped with an optical sight. The probability that a shooter will hit the target when firing a rifle with a telescopic sight is 0.95; for a rifle without an optical sight, this probability is 0.8. The shooter hit the target with a rifle taken at random. What is more likely: the shooter shot from a rifle with or without an optical sight?

Answer: it is more likely that the rifle was without an optical sight (the probability that the rifle was without an optical sight is 24/43, and with an optical sight is 19/43).

28. Three shooters fire one shot at the same target. The probabilities of hitting the target with one shot for each of the shooters are respectively equal p1, p2, p3. What is the probability that the second shooter missed if after shooting there were two holes in the target?

Answer: R = [(1 – p2) p1 p3] / [(1 – p1)p2 p3 + (1 – p2) p1 p3 + (1 – p3) p1 p2].

29. In a group of 25 people who came to take exams in probability theory, there are 10 excellent students, 7 well prepared, 5 satisfactorily prepared and 3 poorly prepared. Excellent students know all 25 questions of the program, well prepared ones - 20, satisfactorily prepared ones - 15, poorly prepared ones know only 10 questions. A student called at random answered 2 questions asked. Find the probabilities of the following events: S1 = (student is excellent or well prepared), S2 = (student is prepared satisfactorily), S3 = (student is poorly prepared).

Answer: R(S1) ≈ 0,8677, R(S2) ≈ 0,1052, R(S3) ≈ 0,0271.

30. Out of 18 shooters, 5 hit the target with a probability of 0.8; 7 – with probability 0.7; 4 – with probability 0.6; 2 – with probability 0.5. A randomly selected shooter fired a shot, but missed the target. Which group did this shooter most likely belong to?

Answer: shooter from the second group.

§ 3. SEQUENCE OF INDEPENDENT TESTS

3.1. Repeating experiments. Bernoulli's formula

In the practical application of probability theory, one often encounters problems in which the same experiment or similar experiments are repeated repeatedly.

As a result of each experience, an event may or may not appear A, and we will be interested not in the result of each experiment, but in the overall result, that is, the number of occurrences of the event A in this series of experiments.

For example, if several shots are fired at a target, then we will be interested not in the result of each shot, but in the total number of hits. In such problems, you need to be able to find the probability of any number of occurrences of an event A. These problems can be solved quite simply if the experiments are independent. Experiments are independent if the outcome of each experiment does not depend on the outcome of the others. For example, several successive coin tosses constitute independent experiments. If the probability of an event occurring A in each experiment is unchanged, i.e., the conditions of the experiments are the same, then a particular theorem on the repetition of experiments applies to this case. If the probability of an event occurring A changes from experiment to experiment, i.e. the experimental conditions are different, then the general theorem applies to this case. Experiments (tests) in which the probability of an event occurring A remains unchanged, called Bernoulli tests. In each Bernoulli test, two and only two outcomes are possible - the occurrence of the event A(“success”) and non-occurrence of the event A("failure"). The probabilities of “success” and “failure” are indicated by letters respectively p And q. It's obvious that p + q = 1.

Let it be produced n independent experiments, in each of which an event may appear A with a probability equal to R and, therefore, with probability equal to q = 1 – R, event A may not appear. Let's determine the probability Rn(m) what's in these n testing event A will appear exactly m once. Consider the event Bm, consisting in the fact that in n testing event A will appear exactly m times and therefore n – m times event A will not appear.

Let us denote by Ai occurrence of an event A V i experience, and through https://pandia.ru/text/80/003/images/image138_0.gif" width="314" height="29">

and in each work there is an event A must be included m times, but it should be included n – m once. The number of such terms is equal, that is, the number

lu ways in which you can n experiments to choose m, in which the event occurred A. According to the theorems of multiplication and addition of probabilities we have:

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Thus, we have the following theorem: if producedn independent experiments, in each of which an event A will appear with probability equal to R, then the probability that the event A will appear exactlym times, expressed by Bernoulli's formula

, (3.1)

Where q = 1 – p,

.

Due to the fact that the probabilities determined by formula (3.1) are terms of the binomial expansion ( q + p)n, then distribution (3.1) is called binomial distribution.

This article brings together information that shapes the idea of ​​equality in the context of mathematics. Here we will find out what equality is from a mathematical point of view, and what they are. Let's also talk about writing equalities and the equal sign. Finally, we list the main properties of equalities and give examples for clarity.

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What is equality?

The concept of equality is inextricably linked with comparison - the comparison of properties and characteristics in order to identify similar features. And comparison, in turn, presupposes the presence of two objects or objects, one of which is compared with the other. Unless, of course, you compare an object with itself, and then this can be considered as a special case of comparing two objects: the object itself and its “exact copy”.

From the above reasoning it is clear that equality cannot exist without the presence of at least two objects, otherwise we simply will have nothing to compare. It is clear that you can take three, four or more objects for comparison. But it naturally comes down to comparing all possible pairs made up of these objects. In other words, it comes down to comparing two objects. So equality requires two objects.

The essence of the concept of equality in the most general sense is most clearly conveyed by the word “identical.” If we take two identical objects, then we can say about them that they equal. As an example, we give two equal squares and . The different objects, in turn, are called unequal.

The concept of equality can apply both to objects as a whole and to their individual properties and characteristics. Objects are equal overall when they are equal in all respects inherent to them. In the previous example, we talked about the equality of objects in general - both objects are squares, they are the same size, the same color, and in general they are completely the same. On the other hand, objects may be unequal overall, but may have some equal characteristics. As an example, consider such objects and . Obviously they are equal in shape - they are both circles. And in color and size they are unequal, one of them is blue and the other is red, one is small and the other is large.

From the previous example, we note for ourselves that we need to know in advance what exactly we are talking about equality.

All of the above arguments apply to equalities in mathematics, only here equality refers to mathematical objects. That is, when studying mathematics, we will talk about the equality of numbers, the equality of expression values, the equality of any quantities, for example, lengths, areas, temperatures, labor productivity, etc.

Writing equalities, =

It's time to look at the rules for writing equalities. For this purpose it is used =(it is also called the equal sign), which has the form =, that is, it represents two identical lines located horizontally one above the other. The equal sign = is considered generally accepted.

When writing equalities, write equal objects and put an equal sign between them. For example, writing the equal numbers 4 and 4 would look like 4=4 and could be read as “four equals four.” Another example: the equality of the area S ABC of triangle ABC to seven square meters will be written as S ABC = 7 m 2. By analogy, we can give other examples of writing equalities.

It is worth noting that in mathematics, the considered notations of equalities are often used as a definition of equality.

Definition.

Records that use the equal sign to separate two mathematical objects (two numbers, expressions, etc.) are called equalities.

If you need to indicate in writing the inequality of two objects, then use not equal sign≠. We see that it represents a crossed out equal sign. As an example, let's take the entry 1+2≠7. It can be read like this: “The sum of one and two is not equal to seven.” Another example is |AB|≠5 cm – the length of segment AB is not equal to five centimeters.

True and false equalities

The written equalities may correspond to the meaning of the concept of equality, or they may contradict it. Depending on this, equalities are divided into true equalities And false equalities. Let's understand this with examples.

Let's write the equality 5=5. The numbers 5 and 5 are undoubtedly equal, so 5=5 is a true equality. But the equality 5=2 is incorrect, since the numbers 5 and 2 are not equal.

Properties of equalities

From the way the concept of equality is introduced, its characteristic results—the properties of equalities—follow naturally. There are three main ones properties of equalities:

• The property of reflexivity, which states that an object is equal to itself.
• The property of symmetry, which states that if the first object is equal to the second, then the second is equal to the first.
• And finally, the property of transitivity, which states that if the first object is equal to the second, and the second is equal to the third, then the first is equal to the third.

Let's write down the voiced properties in the language of mathematics using letters:

• a=a ;
• if a=b then b=a ;
• if a=b and b=c then a=c .

Separately, it is worth noting the merit of the second and third properties of equalities - the properties of symmetry and transitivity - in the fact that they allow us to talk about the equality of three or more objects through their pairwise equality.

Double, triple equalities, etc.

Along with the usual notations for equalities, examples of which we gave in the previous paragraphs, the so-called double equalities, triple equalities and so on, representing, as it were, chains of equalities. For example, the notation 1+1+1=2+1=3 is a double equality, and |AB|=|BC|=|CD|=|DE|=|EF| - an example of a quadruple equality.

Using doubles, triples, etc. For equalities, it is convenient to write the equality of three, four, etc. objects accordingly. These records inherently denote the equality of any two objects that make up the original chain of equalities. For example, the above double equality 1+1+1=2+1=3 essentially means the equality 1+1+1=2+1, and 2+1=3, and 1+1+1=3, and in due to the property of symmetry of the equalities and 2+1=1+1+1, and 3=2+1, and 3=1+1+1.

In the form of such chains of equalities, it is convenient to formulate a step-by-step solution to examples and problems, while the solution looks brief and the intermediate stages of transforming the original expression are visible.

Bibliography.

• Moro M.I.. Mathematics. Textbook for 1 class. beginning school In 2 hours. Part 1. (First half of the year) / M. I. Moro, S. I. Volkova, S. V. Stepanova. - 6th ed. - M.: Education, 2006. - 112 p.: ill.+Add. (2 separate l. ill.). - ISBN 5-09-014951-8.
• Mathematics: textbook for 5th grade. general education institutions / N. Ya. Vilenkin, V. I. Zhokhov, A. S. Chesnokov, S. I. Shvartsburd. - 21st ed., erased. - M.: Mnemosyne, 2007. - 280 pp.: ill. ISBN 5-346-00699-0.

Two numerical mathematical expressions connected by the sign “=” are called equality.

For example: 3 + 7 = 10 - equality.

Equality can be true or false.

The point of solving any example is to find a value of the expression that turns it into a true equality.

To form ideas about true and false equalities, examples with a window are used in the 1st grade textbook.

For example:

Using the selection method, the child finds suitable numbers and checks the accuracy of the equality by calculation.

The process of comparing numbers and indicating the relationships between them using comparison signs leads to inequalities.

For example: 5< 7; б >4 - numerical inequalities

Inequalities can also be true or false.

For example:

Using the selection method, the child finds suitable numbers and checks the accuracy of the inequality.

Numerical inequalities are obtained by comparing numerical expressions and numbers.

For example:

When choosing a comparison sign, the child calculates the value of the expression and compares it with a given number, which is reflected in the choice of the corresponding sign:

10-2>7 5+K7 7 + 3>9 6-3 = 3

Another way to select a comparison sign is possible - without reference to calculating the value of the expression.

Nappimep:

The sum of the numbers 7 and 2 will obviously be greater than the number 7, which means 7 + 2 > 7.

The difference between the numbers 10 and 3 will obviously be less than the number 10, which means 10 - 3< 10.

Numerical inequalities are obtained by comparing two numerical expressions.

To compare two expressions means to compare their meanings. For example:

When choosing a comparison sign, the child calculates the meanings of expressions and compares them, which is reflected in the choice of the corresponding sign:

Another way to select a comparison sign is possible - without reference to calculating the value of the expression. For example:

To set comparison signs, you can carry out the following reasoning:

The sum of the numbers 6 and 4 is greater than the sum of the numbers 6 and 3, since 4 > 3, which means 6 + 4 > 6 + 3.

The difference between the numbers 7 and 5 is less than the difference between the numbers 7 and 3, since 5 > 3, which means 7 - 5< 7 - 3.

The quotient of 90 and 5 is greater than the quotient of 90 and 10, because when dividing the same number by a larger number, the quotient is smaller, which means 90: 5 > 90:10.

To form ideas about true and false equalities and inequalities, the new edition of the textbook (2001) uses tasks of the form:

To check, the method of calculating the meaning of expressions and comparing the resulting numbers is used.

Inequalities with a variable are practically not used in the latest editions of the stable mathematics textbook, although they were present in earlier editions. Inequalities with variables are actively used in alternative mathematics textbooks. These are inequalities of the form:

 + 7 < 10; 5 -  >2;  > 0;  > O

After introducing a letter to denote an unknown number, such inequalities take on the familiar form of inequalities with a variable:

a + 7>10; 12-d<7.

The values ​​of the unknown numbers in such inequalities are found by selection, and then each selected number is checked by substitution. The peculiarity of these inequalities is that several numbers can be selected that fit them (giving the correct inequality).

For example: a + 7 > 10; a = 4, a = 5, a = 6, etc. - the number of values ​​for the letter a is infinite, any number a > 3 is suitable for this inequality; 12 - d< 7; d = 6, d = 7, d = 8, d = 9, d = 10, d = 11, d = 12 - количество значений для буквы d конечно, все значения могут быть перечислены. Ребенок подставляет каждое найденное значение переменной в выражение, вычисляет значение выражения и сравнивает его с заданным числом. Выбираются те значения переменной, при которых неравенство является верным.

In the case of an infinite number of solutions or a large number of solutions to an inequality, the child is limited to selecting several values ​​of the variable for which the inequality is true.