Application question three who can help me answer, give 30 points 1. A walks 50 meters per minute, 70 meters per minute, and C walks 80 meters per minute. A and B start from city a, and C start from city B at the same time. Two minutes after C meets B, C meets a again, seeking the distance between city a and B 2. A starts from a certain place at the speed of 60 meters per minute. Five minutes later, B also starts from this place at the speed of 100 meters per minute to pursue A. when they meet, how far are they from the starting place? 3. Xiao Ming goes to school by bike in the morning. If he rides 200 meters per minute, he will be 10 minutes late. If he rides 250 meters per minute, he can get to school five minutes in advance. How many meters is Xiao Ming's home from school? Don't be too complicated. How to solve the equation? I haven't learned it

Application question three who can help me answer, give 30 points 1. A walks 50 meters per minute, 70 meters per minute, and C walks 80 meters per minute. A and B start from city a, and C start from city B at the same time. Two minutes after C meets B, C meets a again, seeking the distance between city a and B 2. A starts from a certain place at the speed of 60 meters per minute. Five minutes later, B also starts from this place at the speed of 100 meters per minute to pursue A. when they meet, how far are they from the starting place? 3. Xiao Ming goes to school by bike in the morning. If he rides 200 meters per minute, he will be 10 minutes late. If he rides 250 meters per minute, he can get to school five minutes in advance. How many meters is Xiao Ming's home from school? Don't be too complicated. How to solve the equation? I haven't learned it

1. Suppose that the distance between the two cities is s, and the time taken by a and C to get together is x, then 50x + 80x = s, 70 (X-2) + 80 (X-2) = s, and the solution is s = 1950 meters, x = 15 minutes 2. Suppose that the time taken by a and C to get together is x, then 60x = 100 (X-5), and the solution is x = 12.5 minutes

2X + 4 (15-x) = 100 compile an application problem of solving equation

The topics are as follows:
There are two numbers a and B, and their sum is 15. The sum of two times a and four times B is 100
What are the values of a and B?
Let a be x, then B be 15-x
Formula 2x + 4 (15-x) = 100
The solution is x = - 20
So a = - 20, B = 35
Finally, I wish LZ academic progress!

Write a practical problem according to 2x + 4 (15-x) = 100
How about the leg?

Chickens and rabbits are in the same cage, with 15 heads and 100 legs. How about chickens and rabbits?
But the title is wrong
If it's all rabbits, it's only 60 feet

From the five numbers 1, 2, 3, 4 and 5, any three different numbers are arranged into a three digit number to find: (1) the probability that the three digits obtained are even numbers（

Choose any three, a (5,3) = 60
Choose a place first, 2 kinds
Then choose ten and hundred, a (4,2) = 12
There are even numbers: 2 × 12 = 24
The probability is: 24 △ 60 = 0.4

If you take any three of the ten natural numbers 1, 2,..., 10, the probability that the largest of the three numbers is 3 is zero
The answer is 1 / 120
Please give a detailed solution process
thank you!

Among 123 with the largest number of 3, C33 = 1 species
Take any three numbers c10,3 = 120
So the probability is C 33 / C 10 3 = 1 / 120

The probability that any one of the 10 natural numbers 0-9 is a multiple of 2 or a multiple of 3 is ()
A. 12B. 25C. 910D. 710

P (is a multiple of 2 or a multiple of 3) = 710

What is the probability that any one of the nine natural numbers from 1 to 9 is a multiple of 2 or a multiple of 3
6 is a multiple of 2 and 3. Is it twice or once

If 2, 3, 4, 6, 8 and 9 are integral multiples of 2 or 3, the probability is 2 / 3

Take any three of the ten natural numbers 0, 1, 2... 9. Find the probability that at least two of the three numbers are adjacent numbers
(for example: take 1,2,3, then there are two adjacent numbers)
Who said that the next process should not just say the answer
The total number is 84, that is, the denominator is 84)
Scores are not stingy
If they are not adjacent to each other, what is the probability?

At least two numbers are adjacent. There are two cases
1: 2 adjacent numbers
There are also two cases: (1) in one end and one end, that is, the two adjacent numbers are 0, 1 and 8, 9. At this time, the third number has seven choices. Therefore, there are a total of 2 * 7 = 14 choices in this case
(2) In the middle, there are seven choices for two adjacent numbers (i.e. 1 and 2, 2 and 3, 3 and 4.4 and 5.5 and 6.6 and 7, 7 and 8). At this time, there are two adjacent numbers that cannot be selected. There are only six choices for the third number. Therefore, there are a total of 42 choices for 7 * 6
2: Three adjacent numbers
That is 01212334,. 789
Take 3 out of 10 numbers, total selection = C3 / 10 = 120
So the final result is (14 + 42 + 8) / 120 = 8 / 15

At 1, 2, 3 Let X be the number of two adjacent groups of these three numbers (for example, if the number 1, 2, 3 is taken out, then there are two adjacent groups of numbers 1, 2 and 2, 3, and the value of X is 2). Find the distribution column of random variable x
But I don't understand why,
Please don't copy and paste.

There are seven ways to take three numbers with just two continuous natural numbers, 2C (6,1) + 6C (5,1) = 42, and there are 84-7-42 = 35, so p (x = 2) = 7 / 8

Can the natural number 123456789 be filled in the box so that the sum of the three numbers in each row is even?

“zhenghui730402”：
This is impossible, because the sum of nine natural numbers from 1 to 9 is 45, 45 △ 3 = 15, and 15 is odd, so it is impossible
Every three numbers can be even, only three numbers can be odd, or two even numbers, one odd number
Are you right? Good luck. Goodbye