The students who participate in the Mathematics Olympiad are based on the school, and only the top 5

10 9 8 (3 2 1) 120, the final result may be 120.

Obtain. The number of first and second prizes accounts for 2 5 of the total number of winners.

The number of second and third prizes accounts for 9 10 of the total number of winners

2 5 + 9 10 = 13 10 Here the second prize is added twice.

13 10-1 = 3 10 is the share of the second prize.

12 people won the second prize, and there were 12*10 3=40 people.

Prizes are awarded according to 1 5 of the number of entries, with 40 * 5 = 200 people.

The proportion of people who have won the third prize is 1 - the proportion of the first and second prizes added up (2 5) = 3 5, or the proportion of the 2nd prize = 2, the proportion of the 3rd prize - the proportion of the 3rd prize = 3 10, the number of people is 12, then the total number = 12 divided by 3 10 = 40 people.

9 10 + 2 5 -1 = 3 10 (the proportion of the number of third prizes to the number of winners).

9 10 -3 10 = 6 10 (the proportion of the number of second prize winners) 12 divided by 6 10 = 20 (total number of winners) 20 divided by 1 5 = 100 (class size).

Obtain. The number of first and third prizes accounts for 2 5 of the total number of winners, and the number of second prize winners accounts for 3 5 of the total number of winners, and the equation is: 12 3 5 = 12 5 3 = 20 (people) (total number of winners) 20 5 1 = 20 5 = 100 (people).

Equation: Let the number of winners be y

2/5y+12=y

2 5y—y=-12 20 5=100 (person) 3 5y=-12

y=20 A: 100 people will participate (for example, the number of first prize winners is 2, and the third prize is 6 people.) ）

The title is wrong. Because it should be 2 5 for the grand prize.

If so.

1st prize: 1-9 10=1 10

1st and 2nd prizes: 2 5 = 4 10

Second prize: 4 10-1 10 = 3 10 12 divided by 3 10 = 40 Total number of people: 40 divided by 1 5 = 200 (people).

Solution: The number of students in the second school is 3 4x7 (7+8)=7 20, and the number of students in the second school is 3 4x8 (7+8)=2 5, and the total number of students is: 63 (2 5-1 4)=63x20 3=420 (people).

If the number of two or three schools is 7t8t, then the first school is 5t, and the total number of students is 20t, and 8t-5t=63 t=21. The total is 420

Four students signed up for the competition of mathematics, physics and chemistry organized by the school, and each person was limited to one of them, and the number of different reports was 34, so it was incorrect

4 students are divided into 3 football tickets, each person can score up to l, and they must be divided, so the different division is to choose three people from the four people The number is C4

Therefore, correctly 3 pieces of 100 products containing 98 pieces of ** and 2 pieces of defective products are randomly selected, and the probability of at least l defective products in these 3 products extracted is c

cc+ccc100, so incorrect, according to the nature of the binomial coefficient, is known at (1-x)2n+1

n n*), there are 2n+2 terms.

The term with the largest coefficient is the n+2 term, and the term with the smallest coefficient is the n+1 term, so it is incorrect, and it can be seen that only is correct, so the answer is: