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Statistical Probability and Primary School Mathematics Teaching

Liu Jingli, School of Education, Beijing Normal University

A relatively large margin in the "Mathematics Curriculum Standards for Full-time Compulsory Education" (Experimental Draft) The content of "Statistics and Probability" has been added. Because in the information society, collecting, organizing, describing, displaying and interpreting data, and making decisions and predictions based on intelligence have become increasingly important skills for citizens. Therefore, it is completely necessary to add this part of primary school mathematics. The issue this article will explore is what basic concepts primary school teachers should clarify so that the teaching is both scientific and consistent with students’ cognitive characteristics; how to enable students to form and solve the real world In the process of solving problems, students develop statistical awareness, the ability to interpret data, express and communicate information using statistical methods, and the ability to collect, organize and present their thinking in a variety of ways; statistics and probability are related to other parts of primary school content How to contact.

1. Basic concepts

1. Descriptive statistics.

A large amount of data is obtained through surveys and experiments, summarized and organized using statistical methods such as grouping, tabulation, and drawing, and reflects its distribution characteristics in an intuitive and visual form, such as: in primary school mathematics Tabulations, bar charts, line charts, fan charts, etc. are all descriptive statistics. In addition, calculating the central tendency of a set of data reflected by the concentration amount, such as the arithmetic mean, median, total number, weighted arithmetic mean, etc., also falls within the scope of descriptive statistics. Its purpose is to organize, summarize, simplify and summarize a large number of scattered and disordered digital data, so that the full picture of things and their distribution characteristics can be clearly and clearly displayed.

2. Statistical definition of probability.

When people toss a coin, what kind of result will occur cannot be determined in advance, but when we toss the same uniform coin repeatedly under the same conditions, we will find that The number of "heads" or "tails" each accounts for about: about the total number of tosses. How many times does "a large number of repetitions" mean here? In history, many statisticians, such as Pearson and others, have conducted tens of thousands of coin tossing experiments. The experimental records are as follows:

Yes It can be seen that as the number of tests increases, the frequency fluctuation of the positive appearance becomes smaller and smaller. The nature of the frequency swinging around the fixed value of 0.5 is a manifestation of the inherent inevitability of the phenomenon of positive appearance. 0.5 is exactly the A numerical value that describes the likelihood of heads. 0.5 is the probability of heads when tossing a coin. This is the idea behind the definition of probability statistics. This idea also provides a method for estimating the approximate value of probability in practical problems. When the number of trials is large enough, frequency can be used as the approximate value of probability.

For example, on average, about 90 seeds germinate out of 100 seeds, then we say that the germination rate of seeds is 90%;

On average, about every 1,000 products of a certain type of product have 10 pieces of scrap, then we say that the scrap rate of this product is 1%. When using the statistical definition of probability in primary school mathematics, what is generally obtained is the approximate value of the probability. Especially when the number of times is not large enough, there is a certain error in the approximate value of the probability. For example: In the weather records of October 6th in a certain area in the past 30 years, there have been 25 times of crisp autumn weather and clear skies. What is the probability that October 6th of the next year will be sunny?

Because the first 30 times The annual frequency of sunny days is 0.83, so the probability is approximately 0.83.

3. Classical definition of probability.

For a certain type of special experiment, the probability can also be calculated from another angle. When tossing a coin, there are two experimental results: heads and tails. Since the coin is uniform, it can be seen through visual analysis that the likelihood of heads and tails is the same. Further research:

A certain test has the following properties

(1) The results of the test are limited (n)

(2) Each result appears The possibilities are the same (coins and dice are uniform, and the probability of each side appearing when tossing is the same)

If event A is composed of m of the above n results, it is called event A The probability of occurrence is m/n.

Example: Roll a uniform die and find the probability of 2 points appearing.

Since this experiment satisfies the two conditions of the classical definition of probability, and n=6, m=1, the probability of ∴ appearing 2 points is.

Also: Find the probability of an even-numbered point appearing? The occurrence of an even-numbered point contains three results, 2 points, 4 points, and 6 points. m=3

The probability of an even point appearing is, that is.

The classical definition of probability does not require a large number of experiments. As long as the results of the experiment are a limited number of equally possible situations, the probability can be found through analysis to find m and n. The advantage is that it is convenient. Calculation, but the classical definition of probability is not as broad as the statistical definition of probability. For example, when throwing a wine bottle cap, the condition that the probability of each side appearing is the same is not met. Therefore, the probability of heads appearing cannot be calculated using the classical definition of probability. To find it, use the statistical definition to approximately find its probability.

In the teaching of primary school mathematics, according to the cognitive level of primary school students, learning too many or difficult terminology should be avoided. Starting from the lower grades of primary school, probability ideas should be introduced informally instead of strict definitions and Pure calculation, therefore, the concept of "probability" is often replaced by "possibility" in elementary schools. But as a teacher, you should understand its meaning, otherwise it will be a joke. Some teachers ask students to do an experiment of tossing a coin 20 times in class, hoping that students can get the probability of heads. Because the number of tosses is small, it is difficult to get 10 heads. Probability Statistical definitions generally yield approximations of probability.

2. Develop students’ abilities in the process of learning statistics and probability

The content of statistics is to use numbers to describe and explain the world around us, which should be combined with the reality of students’ lives, such as : Can be designed as an activity that engages students proactively; asks key questions; collects and organizes data; uses charts and graphs to represent data; analyzes data; makes inferences, and communicates information in a way that is convincing to others. At the same time, you will realize that some new information will be obtained through the collection and processing of data.

For example: Organize a class meeting to enhance mutual understanding and communication among classmates. First, let the students choose their own topics and what information they want to know: "How do students come to school every day?"; "How many students have birthdays every month?"; "What kind of books do students like to read?"; "Students "What is your hobby?"; "Our favorite sport"; "Our favorite animal"... Then the students were divided into groups to investigate and collect data, summarize and organize them in tables, and make various statistical charts: such as:

From the statistical chart, we can know that most students like animal stories. Based on this statistical result, the class can organize an animal research meeting, hold an animal picture exhibition, and visit a wildlife park. The whole class can also make various charts into posters and handwritten posters to introduce their class to other students in the school.

3. The connection between statistics, probability and other content in primary schools

Example 1

The scores representing the black areas in the above figures are;;;, primary school students Even if you have not studied the concept of geometric figures, you can know through the meaning of fractions that the black area No. 2 is the easiest to hit, because according to the meaning of fractions, it accounts for the largest ratio of the total area, which is.

Example 2

Judging from the proportion of red balls, bag No. 1 is; bag No. 2 is; bag No. 3 is hit. Therefore, in comparison, bag No. 1 The bag is easiest to draw the red ball.

Example 3 The following is statistical data using a fan chart

For primary school students, the difficulty of the fan chart lies in the percentage representation and percentage representation of the parts represented by different central angles. The degree of the central angle of the circle, and for - when there is a special central angle in the above picture, you can avoid the central angle and use the meaning of fractions and percentages to get those who like English classes, science classes, and mathematics classes; join ball game interest groups 50% of the students; 18% of the students participated in the band.

As you can see from the above examples, statistics and probability can provide a background for developing and using concepts such as ratios, fractions, percentages, and decimals. Therefore, we can use a constructive approach to establish the connection between this part of the content and other primary school knowledge and construct a meaningful cognitive structure, so as to learn more deeply and more flexibly.

In short, in primary schools, the teaching of statistics and probability must be scientific and consistent with the cognitive characteristics of primary school students. At the same time, it is also a powerful tool for solving problems, and it also serves as a bridge between other content. bridge.

Sum and difference problem

Given the sum and difference of two numbers, the word problem of finding these two numbers is called the sum and difference problem.

The general relational expressions are:

(sum-difference)÷2=smaller number

(sum+difference)÷2=larger number

Example: The sum of the two numbers A and B is 24. A's number is 4 less than B's number. What are the two numbers A and B?

（24＋4）÷2

＝28÷2

＝14 →B number

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＝20÷2

＝10 →Number A

Answer: Number A is 10 and number B is 14.

Difference problem

Given the difference between two numbers and the relationship between multiples of the two numbers, solving the application problem of these two numbers is called the difference problem. The basic relationship is:

Difference between two numbers ÷ multiple difference = smaller number

Example: There are two piles of coal, and the second pile is 40 tons more than the first pile. If from the Take 5 tons of coal from the second pile and give it to the first pile. At this time, the weight of the second pile of coal is exactly three times that of the first pile. How many tons of coal are there in each of the two piles of coal?

Analysis: It turns out that the second pile of coal is 40 tons more than the first pile. After giving the first pile 5 tons, the second pile of coal is only 40-5×2 tons more than the first pile. The basic relational formula is:

(40-5×2)÷(3-1)-5

＝(40-10)÷2-5

＝30÷2-5

＝15-5

＝10 (tons) →The weight of the first pile of coal

140＝ 50 (tons) →The weight of the second pile of coal

Answer: The first pile of coal is 10 tons, and the second pile of coal is 50 tons.

Reduction problem

A problem that requires the original unknown number to be the result of certain changes in a given number is generally called a reduction problem.

Reduction problems are inverse solution word problems. Generally based on the relationship between the reciprocal operations of addition, subtraction, multiplication and division. According to the order described in the title, think in reverse order, start from the last known condition, and work backward to get the result.

Example: There is some rice in the warehouse, and the weight sold on the first day is 12 tons less than half of the total. The weight of the goods sold the next day was 12 tons less than half of what was left. As a result, there were still 19 tons left. How many tons of rice did this warehouse originally have?

Analysis: If the remaining half is sold exactly the next day, it should be 19+12 tons. After selling on the first day, the remaining tons are (19 + 12) × 2 tons. The following analogy.

Column formula: [(19＋12)×2-12]×2

＝[31×2-12]×2

＝[62-12 ]×2

＝50×2

＝100 (tons)

Answer: This warehouse originally contained 100 tons of rice.

Permutation problem

There are two unknowns in the problem. One of the unknowns is often temporarily regarded as the other unknown, and then hypothetical operations are performed based on known conditions. The results often do not meet the conditions, and appropriate adjustments are made to obtain the results.

Example: A stamp collector bought ***100 10-cent and 20-cent stamps, with a total value of 18 yuan and 80 cents. How many stamps of each type does this stamp collector buy?

Analysis: Assume that all the 100 stamps bought are 20 cents each, then the total value should be 20×100=2000 (cents), which is 2000-1880=120 more than the original total value. (point). As for the extra 120 points, a 10-cent piece is regarded as a 20-cent piece. Each extra piece is calculated as 20-10 = 10 (points). In this way, we can find out how many 10-cent pieces there are.

Column formula: (2000-1880)÷(20-10)

＝120÷10

＝12 (pieces) → 10 cents per piece Number of sheets

100-12=88 (sheets) → the number of sheets per 20-cent sheet

Or first find the number of sheets in a 20-cent sheet, and then find out the 10-cent sheet The number of sheets in one sheet is the same as above, but note that the total value is less than the original total value.

Profit and loss problem (profit and deficit problem)

There are often two allocation plans in the question. The result of each allocation plan will be more (profit) or less (loss). This type of problem is usually called a profit and loss problem (also called a profit-deficit problem).

When answering this type of question, you should first compare the two distribution plans, find out the change in the remainder caused by the change in the number of each share, and then find the total number of shares participating in the distribution, and then based on The question is to find the number of distributed items. The calculation method is:

When there is a remainder once and a deficiency the other time:

The number of each portion = (remainder + deficiency) ÷ the difference between the two times of each portion

When both times have remainders:

Total number of copies = (larger remainder - smaller number) ÷ the difference between the two times

When twice When both are insufficient:

Total number of copies = (larger deficiency - smaller deficiency) ÷ difference between the two copies

Example 1. A squad of a certain unit of the People's Liberation Army , participate in afforestation activities. If each person plants 5 saplings, there will be 14 saplings left; if each person plants 7 saplings, there will be a shortfall of 4 saplings. How many people are there in this class? How many saplings are there in one ***?

Analysis: It can be seen from the conditions that this question belongs to the first situation.

Formula: (14+4)÷(7-5)

＝18÷2

＝ 9 (person)

5 ×9＋14

＝45＋14

＝59 (tree)

Or: 7×9-4

＝63-4

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＝59 (trees)

Answer: There are 9 people in this class, and there are 59 saplings in one ***.

Age problem

The main feature of the age problem is that the age difference between the two people remains unchanged, but the multiple difference changes.

The commonly used calculation formula is:

The age of the youngest when doubled = the difference in age ÷ (multiple - 1)

Age a few years ago = The younger’s current age - the younger’s age when it is multiplied

The age in a few years = the younger’s age when it is doubled – the younger’s current age

Example 1. The father is 54 years old this year, and his son 12 years old this year. How many years later will the father's age be 4 times the son's?

(54-12)÷(4-1)

＝42÷3

＝14 (years old)→son’s age in a few years

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14-12=2 (years) → 2 years later

Answer: After 2 years, the father’s age will be 4 times that of the son.

Example 2. The father is 54 years old this year and the son is 12 years old this year. How many years ago was the father's age 7 times his son's?

(54-12)÷(7-1)

＝42÷6

＝7 (years old)→son’s age a few years ago

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12-7=5 (years) → 5 years ago

Answer: 5 years ago, the father’s age was 7 times that of his son.

Example 3. The sum of the ages of Wang Gang's parents this year is 148 years old. The difference between his father's age and his mother's age is 3 times greater than the sum of their ages by 4 years. How old are Wang Gang's parents this year?

(148×2+4)÷(3+1)

＝300÷4

＝75 (years)→father’s age

148-75=73 (years) → mother’s age

Answer: Wang Gang’s father is 75 years old and his mother is 73 years old.

Or: (148+2)÷2

＝150÷2

＝75 (years old)

75-2＝73 ( years)

Chicken and Rabbit Problem

Given the total number of chickens and rabbits and the total number of rabbits, it is a type of word problem to find the number of chickens and rabbits. It is called the Chicken and Rabbit Problem. Also called "turtle-crane problem" and "replacement problem".

Generally, assume that they are all chickens (or rabbits), and then replace the chickens (or rabbits) with rabbits (or chickens). The commonly used basic formulas are:

(Total number of chicken feet - number of chicken feet × total number of chickens) ÷The difference between the number of chicken and rabbit feet = number of rabbits

(Number of rabbit feet × The difference between the total number of chickens and rabbits (total number of chickens and rabbits) ÷ the number of chickens and rabbits = number of chickens

Example: There are 24 chickens and rabbits in the same cage.

There are 64 legs. How many chickens and rabbits are there in the cage?

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(64-2×24)÷(4-2)

＝(64-48)÷(4-2 )

＝16 ÷2

＝8 (only) → the number of rabbits

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Answer: There are 8 rabbits and 16 chickens in the cage

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<. p>Cattle grazing problem (ship leakage problem)

A number of cattle graze on a limited range of grass. When the cattle graze, the grass grows on the other side. When the quantity is , how long will it take for the grass on this grassland to be eaten up?

Example 1. A piece of grassland can be eaten by 15 cows for 10 days, but it can be eaten by 25 cows? Eat for 5 days. If the grass grows at the same rate every day, how many days can this grassland be eaten by 10 cows?

Analysis: Generally, the amount of grass eaten by one cow per day is regarded as one serving. Count, then 15 cows eat for 10 days, including the original grass on the meadow, plus the grass grows in this meadow for 10 days, and so on... It can be found that the amount of grass eaten by 25 cows in 5 days is greater than that of 15 cows in 10 days. The amount of grass eaten in a day is less. The reason is that firstly, it takes less time; secondly, the corresponding grass grows less. This difference is the amount of grass grown in this meadow every day. 5 cows eat one day. So when 10 cows are fed, 5 cows are dedicated to eating the grass that grows every day, and the remaining cows eat the original grass on the grassland.

(15×10-25. ×5)÷(10-5)

＝(150-125)÷(10-5)

＝25÷5

＝5 (head )→can feed 5 cows for one day

150-10×5

=150-50

=100 (head)→original on the grassland. Grass can feed 100 cows for one day

100÷(10-5)

＝100÷5

＝20 (day)

< p>Answer: If 10 cows are fed, they can eat it for 20 days.

Example 2: A well pours water at a constant speed and can be drained by 4 pumps in 100 minutes; if 6 identical pumps are used. It takes 50 minutes to drain the well. How many minutes will it take to drain the well using 7 identical pumps?

(100×4-50×6)÷(100-50) < /p>

＝（400－300）÷（100－50）

＝100÷50

＝2

400－100×2

＝400-200

＝200

200÷(7-2)

＝200÷5

< p>=40 (points)

Answer: Using 7 identical pumps, the water in this well can be drained in 40 minutes.

Common divisors and common multiples problems

Using the greatest common divisor or least common multiple to solve word problems is called common divisors and common multiples problems.

Example 1: A rectangular piece of wood is 2.5 meters long, 1.75 meters wide, and 0.75 meters thick. If this piece of wood is sawn into cubes of the same size, with no leftovers allowed, and the volume of each piece is as large as possible, what is the edge length of the cube? ***How ??many pieces were sawed?

Analysis: 2.5=250 cm

1.75=175 cm

0.75=75 cm

Of which 250 The greatest common divisor of , 175, and 75 is 25, so the edge length of the cube is 25 cm.

(250÷25)×(175÷25)×(75÷25)

＝10×7×3

＝210 (block) < /p>

Answer: The edge length of the cube is 25 cm, and *** saw 210 pieces.

Example 2. Two meshing gears, one with 24 teeth and the other with 40 teeth. Find the minimum number of rotations of each gear from the first contact to the second contact of a pair of teeth. week?

Analysis: Because the least common multiple of 24 and 40 is 120, that is, when both gears rotate 120 teeth, the pair of teeth that come into contact for the first time will exactly contact for the second time.

120÷24＝5 (weeks)

120÷40＝3 (weeks)

Answer: Each gear rotates 5 and 3 times respectively .

Fraction word problems

Refers to word problems that are solved by using fraction calculations, called fraction word problems, also called fraction problems.

Fraction problems are generally divided into three categories:

1. Find what fraction of one number is another number.

2. Find what fraction of a number is.

3. If you know what fraction of a number is, find the number.

Each category is divided into two types, one: general fraction word problems; the other: more complex fraction word problems.

Example 1: Yucai Primary School has 1,000 students, including 250 good students. What percentage of students in the school are students with good grades?

Answer: Three good students account for 10% of all students in the school.

Example 2: A pile of coal weighing 180 tons was transported away. How many tons were gone?

180×=80 (tons)

Answer: 80 tons were transported.

Example 3: An agricultural machinery factory produced 1,800 agricultural machinery last year, and plans to increase the number this year compared to last year. How many units are planned to be produced this year?

1800× (1+)

＝1800×

＝2400 (unit)

Answer: We plan to produce 2,400 units this year.

Example 4: Build a 2,400-meter-long highway. Build the entire length on the first day and the rest on the second day. How many meters are left?

2400×(1-)×(1-)

＝2400××

＝1200 (meter)

Answer: There are still 1200 meters left.

Example 5: There are 168 three-good students in a school, accounting for the total number of students in the school. How many students are there in the school?

168÷＝840 (person)

Answer: There are 840 students in the school.

Example 6: A has 120 tons of grain in stock, which is less than B’s grain in stock. How many tons of grain does B have in stock?

120÷=120×=180 (tons)

Answer: B has 180 tons of grain in stock.

Example 7: A pile of coal, all of it was transported for the first time, and all of it was transported for the second time. The second time was 8 tons less than the first time. How many tons did this pile of coal originally weigh?

8÷(-)

＝ 8÷

＝48 (tons)

Answer: This pile of coal originally contained 48 tons .

Engineering problem

It is a special case of fraction word problem. It is a problem of finding the third quantity from two of the three quantities given the workload, working time and work efficiency.

When solving engineering problems, generally all projects should be regarded as "1", and then the answer should be based on the following quantitative relationship:

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Workload ÷ working time = work efficiency

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Workload ÷ work efficiency = working time

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Example 1: A project, It takes 18 days for Team A to do it alone, and 24 days for Team B to do it alone. If the two teams cooperate for 8 days and Team A does the remaining work alone, how many days will it take to complete the project?

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Example 2: A pool is equipped with two inlet pipes A and B and one outlet pipe. It can be filled in 2 hours by opening pipe A alone; it can be filled in 3 hours by opening pipe B alone; it can be filled by opening pipe B alone in 6 hours. It can be filled in hours. Now that the three tubes are opened when the pool is empty, how many hours can it take to fill the pool?

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Percent word problems

The answers to this type of word problems are roughly the same as those of fractions. When only asking for "rate", the expressions are different and the meanings are different.

Example 1. An agricultural science institute conducted a germination test and planted 250 seeds. There are 230 seeds that sprouted. Find the germination rate.

Answer: The germination rate is 92%.

1. Perimeter of rectangle = (length + width) × 2 C = (a + b) × 2

2. Perimeter of square = side length × 4 C= 4a

3. The area of ??the rectangle = length × width S = ab

4. The area of ??the square = side length × side length S=a.a= a

5. The area of ??the triangle = base × height ÷ 2 S = ah ÷ 2

6. The area of ??the parallelogram = base × height S = ah

7. The area of ??the trapezoid = ( Upper bottom + lower bottom) × height ÷ 2 S = (a + b) h ÷ 2

8. Diameter = radius × 2 d=2r Radius = diameter ÷ 2 r= d ÷ 2

< p>9. Circumference of a circle = pi × diameter = pi × radius × 2 c=πd =2πr

10. Area of ??a circle = pi × radius × radius? = πr

11. Surface area of ??cuboid = (length × width + length × height + width × height) × 2

12. Volume of cuboid = length × width × height V =abh

13. Surface area of ??cube = edge length × edge length × 6 S = 6a

14. Volume of cube = edge length × edge length × edge length V = a.a.a= a

15 , Side area of ??cylinder = Circumference of base circle × Height S = ch

16. Surface area of ??cylinder = Area of ??upper and lower bases + Side area

S=2πr +2πrh=2π( d÷2) +2π(d÷2)h=2π(C÷2÷π) +Ch

17. Volume of cylinder = base area × height V = Sh

V=πr h=π(d÷2) h=π(C÷2÷π) h

18. The volume of the cone = base area × height ÷3

V= Sh÷3=πr h÷3=π(d÷2) h÷3=π(C÷2÷π) h÷3

19. Cuboid (cube, cylinder) body

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1. The number of copies per copy 1 multiple = multiple multiples ÷ multiples = 1 multiple

3. Speed ??× time = distance ÷ speed = time distance ÷ time = speed

4. Unit price × quantity = total price Total price ÷ unit price = quantity Total price ÷ quantity = unit price

5. Work efficiency × working time = total amount of work total amount of work ÷ work efficiency = total working time ÷ working time = work efficiency

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6. Addend + Addend = sum and sum - one addend = another addend

7. Minuend - Minuend = Difference Minuend - Difference = Difference of the minuends ＋Minuum = Minuend

8. Factor × factor = product ÷ one factor = another factor

9. Divisor ÷ divisor = quotient Divisor ÷ quotient = divisor quotient × Divisor = dividend

Primary school mathematics graph calculation formula

1. Square C perimeter S area a side length perimeter = side length × 4 C = 4a area = side length × side length S=a×a

2. Cube V: Volume a: Edge length Surface area = Edge length × Edge length × 6 S table = a × a × 6 Volume = Edge length × Edge length × Edge length V =a×a×a

3, rectangle

C perimeter S area a side length

Perimeter = (length + width) × 2

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C=2(a+b)

Area = length × width

S=ab

4, cuboid

< p>V: Volume s: Area a: Length b: Width h: Height

(1) Surface area (length × width + length × height + width × height) × 2

S=2(ab+ah+bh)

(2) Volume = length × width × height

V=abh

5 Triangle

s area a base h height

Area=base×height÷2

s=ah÷2

Height of triangle = area×2÷base

Triangle base = area × 2÷height

6 Parallelogram

s area a base h height

Area = base × height

s=ah

7 Trapezoid

s area a upper base b lower base h height

Area = (upper base + lower Bottom) × height ÷ 2

s=(a+b)× h÷2

8 Circle

S area C perimeter ∏ d=diameter r=radius

(1)Perimeter=diameter×∏=2×∏×radius

C=∏d=2∏r

(2) Area = radius × radius × ∏

9 cylinder

v: volume h: height s; base area r: base radius c: base perimeter

(1) Side area = bottom perimeter × height

(2) Surface area = side area + bottom area × 2

(3) Volume = bottom area × height

(4) Volume = side area ÷ 2 × radius

10 cone

v: volume h: height s; base area r: base radius

< p>Volume = base area × height ÷ 3

Total number ÷ total number of copies = average

Sum and difference problem

(Sum + difference) ÷ 2 = Large numbers

(Sum-Difference) ÷2 = Decimal

Sum and multiples problem

Sum ÷(Multiple-1) = Decimal

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Decimal × multiple = large number

(or decimal + difference = large number)

Tree planting problem

1 On non-closed lines The tree planting problem can be mainly divided into the following three situations:

⑴ If trees are to be planted at both ends of the non-closed line, then:

Number of trees = number of sections + 1 = full length ÷ spacing between trees - 1

Total length = plant spacing × (number of plants - 1)

Plant spacing = full length ÷ (number of plants - 1)

⑵ If it is at one end of a non-closed line If you want to plant a tree but do not plant a tree at the other end, then:

Number of plants = number of segments = total length ÷ spacing between plants

Full length = spacing between plants × number of plants

Spacing between plants = total length ÷ Number of trees

⑶ If no trees are planted at both ends of the non-closed line, then:

Number of trees = number of sections - 1 = total length ÷ spacing between trees - 1

Full Length=plant spacing = Number of segments = total length ÷ plant spacing

Full length = plant spacing × number of plants

Plant spacing = total length ÷ number of plants

Profit and loss issues

( Profit + loss) ÷ The difference between the two distribution amounts = The number of shares participating in the distribution

(Big profit - Small profit) ÷ The difference between the two distribution amounts = The number of shares participating in the distribution

(Big loss - Small loss) ÷The difference between the two distribution amounts = The number of shares participating in the distribution

Meeting problem

Meeting distance = speed and × meeting time

< p>Meeting time=Meeting distance ÷Speed ??sum

Speed ??sum=Meeting distance ÷Meeting time

Catching problem

Catching distance=Speed ??difference×Catching time< /p>

Catching time = catching distance ÷ speed difference

Speed ??difference = catching distance ÷ catching time

Flowing water problem

Downstream speed = still water Speed ??+ water flow speed

Counter flow speed = static water speed - water flow speed

Static water speed = (downstream speed + countercurrent speed)÷2

Water flow speed = ( Downstream velocity - countercurrent velocity) ÷ 2

Concentration problem

Weight of solute + weight of solvent = weight of solution

Weight of solute ÷ weight of solution ×100%=concentration

Weight of solution × concentration = weight of solute

Solute

Weight ÷ concentration = weight of solution

Profit and discount issues

Profit = selling price - cost

Profit rate = profit ÷ cost × 100% = ( Selling price ÷ cost - 1) × 100%

Increase or decrease amount = principal × increase or decrease percentage

Discount = actual selling price ÷ original selling price × 100% (discount < 1)

Interest = principal × interest rate × time

After-tax interest = principal × interest rate × time × (1-20%)

Time unit Conversion

1 century = 100 years 1 year = 12 months

The big month (31 days) has: 1\3\5\7\8\10\12 months

Small months (30 days) include: April\6\9\November

February has 28 days in ordinary years and February has 29 days in leap years

365 days in ordinary years days, there are 366 days in a leap year

1 day = 24 hours 1 hour = 60 minutes

1 minute = 60 seconds 1 hour = 3600 seconds Product = base area × height V = Sh