Mathematics detailed problem solving process .

From the five numbers of 1, 2, 3, 4, and 5, first extract one arbitrarily, and then extract one arbitrarily from the remaining four numbers, and find the probability that the numbers drawn twice are odd and the probability that the sum of the numbers drawn twice is even

1, 3, 5 are odd numbers, 2, 4 are even numbers.

Odd number drawn for the first time = 3 5

The probability of drawing an odd number again the second time = 2 4

So the probability that the numbers drawn twice are odd = (3 5) * (2 4) = 3 10 = 30%.

Probability of drawing an even number for the first time = 2 5

The probability of drawing an even number again the second time = 1 4

So the probability that the numbers drawn twice are even = (2 5) * (1 4) = 1 20 = 5%.

The probability that the number drawn twice is an odd number = 3 5 * 1 2 = 3 10

The probability that the sum of the numbers drawn twice is an even number = (3 5 * 1 2) + (2 5 * 1 4) = 2 5

Probability that both draws are odd:

There are 3 possibilities for the first draw, 2 possibilities for the second draw, and 3*2 6 possibilities for drawing odd numbers, so the probability that the numbers drawn twice are odd is 6 (5*4) 3 10.

Probability that the sum of the numbers drawn twice is an even number:

1. If the two draws are odd numbers, there is a possibility of 3 10.

2. If the two draws are even, there is a possibility of (2*1) (5*4) 1 10.

So, the probability that the sum of the numbers drawn twice is an even number is 4 10.

The number of basic events is 5 x 4 = 20

The number of the two draws is an odd number, and the number of events is 3*2=6, so the probability that the numbers drawn twice are odd is 6, 20=3, 10, and the sum of the numbers drawn twice is an even number, and the number of basic events is.

The probability that the sum of the numbers drawn twice is an even number is 7 20

Do you understand? I don't know how to ask again.

Good luck with your studies!

This is a question that examines the concept, and binary means that there are two unknowns, and the number of each unknown is 1So.

a+5b-5)=1 (3a-6b-3)=1 solves the system of equations to obtain a=8 3 b=2 3 a+2b=8 3+2 2 3=12 3=4

Solution: From the meaning of the question: a 5b 5 1

3a－6b－3＝1

Solution: a 8 3

b＝2/3a＋2b＝8/3＋2×2/3＝4.

Because it is a binary equation, x and y are powers of the first degree, i.e., a+5b-5=1, 3a-6b-3=1

Since 3x (a+5b-5)-2y (3a-6b-3)=5 is a binary linear equation with respect to x,y, we know that:

a+5b-5＝1 (1)

3a-6b-3＝1 (2) ＝1)×3 => 3a+15b-15＝3 (4) => （4）－（2） =>

21b-12=2 so we get b=2 3 substituting the value of b into (1) gives a=8 3

So a+2b 8 3+2 (2 3)=4

Because E makes the parallel line of AB, and AB CD, so according to the transmissibility of parallelism, you can get it!

It is equivalent to what we encountered when we were in elementary school, the weight of two pears is equal to the weight of one apple, and the weight of one apple is equal to the weight of three oranges, so we get the same meaning as the weight of two pears is equal to the weight of three oranges!

Cross the point e to make the parallel line EF of AB

Because ab cd ab ef

So EF cd (transitivity of parallel lines).

If both lines are parallel to the third line, then the two lines are also parallel to each other.

A square with a side length of 2 and an area of 4. Two squares with the same side length of 2, the area of a large square is 4x2=8, the side length = 8=2 2, keep two yes, keep three yes.

Note plagiarism: In the process of writing, you must not use the assignment method to solve the problem, this method is only suitable for choosing to fill in the blanks!

1) solution, because (a-1)x+(a+2)y+5-2a=0, so ax-x+ay+2y-2a=-5

So 2y-x+(x+y-2)a=-5

And because the value of the solution has nothing to do with a.

So x+y-2=0

So [2y-x=-5

x+y-2=0

The solution is [x=3

y=-12), solution, put [x=3

y=-1 is substituted into (a-1)x+(a+2)y+5-2a=0 to obtain the original formula = 3a-3-a-2-2a+5

0 is really hard to beat! Satisfied.

E-e, the choreography didn't pay attention, o( o

^(a-1)x+(a+2)y+5-2a=0ax-x+ay+2y+5-2a=0

a(x+y-2)=x-2y-5

When x+y-2=0 and x-2y-5=0, the compound is the same x+y-2=0 regardless of the value of a

x-2y-5=0

Solve the system of equations bai:

dux=3y=-1

That is, the straight line is constantly passing the point zhi(3,-1). So dao:m=3,n=-1mn 4=3*(-1) 4=3

(1) Take any two A values and bring them into the calculation

zhia=0 then -x+2y+5=0, a=1 then 3y+3=0, and bison can be solved to obtain daox=3, y=-1

2)(a-1)x+(a+2)y+5-2a=0 can be reduced to a(x+y-2)-x+2y+5=0, and it can be seen from the problem that the solution is irrelevant to tolerance a when x+y-2=0, so the simultaneous equation: x+y-2=0, -x+2y+5=0, the solution is x=3, y=-1

Solution: (x-3)(x+8)=x +5x-24 and because (x-3)(x+8)=x +mx+n, that is, x +5x-24=x +mx+n

Contrast coefficient.

m=5n=-24

Happy learning.

o(∩_o~

Original = 1-1 2+1 2-1 3+1 3-1 4+...1/2016-1/2017

1-1 2017 All in the middle is added and subtracted.

The absolute value is a positive number, and the absolute value symbol can be eliminated directly.

One subtract one plus and it disappears.