In the 2013 National University Games, a University selected 4 students from 6 students to participate in shot put, height adjustment, long jump and sprint respectively, then the probability of student a to participate in sprint or long jump is () A. 13B. 12C. 23D. 34

Four college students are selected from six college students to take part in shot put, height adjustment, long jump and sprint. There are 360 different selection methods. Among them, there are 120 selection methods for student a to take part in sprint or long jump. Therefore, the probability of student a to take part in sprint or long jump is p = 120360 = 13

There are twelve boxes, but only one of them has cash inside. If you can choose six boxes to take away, what is the probability of getting cash?

The total situation is to take 6 cases out of 12 cases, a total of C (12,6) cases
The case of getting cash is to get the box with cash, and take 5 of the other 11 boxes, a total of C (11,5) cases
Probability of getting cash = C (11,5) / C (12,6) = 462 / 924 = 0.5

A person randomly puts four balls with the number of 1.2.3.4 into four boxes with the number of 1.2.3.4, and each box puts a ball. When the box number is the same as the ball number, it is called putting right, otherwise it is called putting wrong. Let x balls be put right, and calculate the probability of x = 0.1.2.3.4
Why use 4 * 3 * 2 * 1?

There are a total of 44 methods, namely 4 * 3 * 2 * 1 = 24
When x = 0, there is only one case, and the probability is 1 / 24
When x = 1, there are C41 cases, and the probability is 4 / 24
If x = 2, there are C42 cases, namely 4 * 3 / 2 = 6, and the probability is 6 / 24
When x = 4, there is only one case, and the probability is 1 / 24
When x = 3, the probability is 1-above, and the sum of the probabilities is 1 / 2

For rational addition and subtraction of mixed operations exercises, the more the better

1. 3/7 × 49/9 - 4/3 2. 8/9 × 15/36 + 1/27 3. 12× 5/6 – 2/9 ×3 4. 8× 5/4 + 1/4 5. 6÷ 3/8 – 3/8 ÷6 6. 4/7 × 5/9 + 3/7 × 5/9 7. 5/2 -（ 3/2 + 4/5 ） 8. 7/8 + （ 1/8 + 1/9 ） 9. 9 × 5/6 + 5/6 ...

A paradox in Lorentz's change
There is a reference frame s' with a velocity V relative to the stationary reference frame s. We have
x'=x-vt/1-(v/c)^2
x=x'+vt'/1-(v/c)^2
When t = t '= 0, we have X' = x / 1 - (V / C) ^ 2
X = x '/ 1 - (V / C) ^ 2?

No problem. X, X 'refers to the abscissa of the same point in different coordinate systems. They are different not only because of the difference in the position of the origin between the two coordinates, but also because of the difference in the relative velocity between the point and the two coordinate systems;
With different relative speeds, time and space are different, and the distance measured is of course different
When t = 0, the origin of the two coordinate systems coincides, which does not mean that the measured displacements should be equal

Help out some elementary school mental arithmetic problems

×4/5=16/5 6×3/4=18/4 10×2/5=4 18×4/9=8 3÷3/4=4 2÷1/3=6 6÷4/5=15/2 1÷5/7=7/5 3/4÷3=1/4 1/3÷2=1/6 4/5÷6=2/15 5/7÷1=5/7 2/1*2=1 3/1*3=1 3/2*3=2 3/1*6=2 4/3*8=6 5/3*20=12 7/3*14=6 8/7*40=35 4/...

Is time expansion and length shortening equivalent in special relativity
There is a problem when preparing textbooks. If an object advances at a speed close to the speed of light, then there must be a deviation from classical mechanics from a to B. if the time is advanced, is it because of time expansion or length shortening? If the time is calculated, how about using the shortened distance to get t and then converting it to t '?

The problem you have been struggling with just now is actually not so troublesome: you have mixed up the data of the two coordinate systems
Take the ground as the reference frame, because the ground does not move, so the distance and time will not change. Take the moving object as the particle, and calculate it with S = VT, that is, the original measured data. You said that the time expands and the length shortens. That is the case with the moving object as the reference frame. At this time, change the time to t / √ 1 - (V / C) ^ 2
The distance is converted into L · √ 1 - (V / C) ^ 2, and the contraction factor √ 1 - (V / C) ^ 2 is mutually reduced. In motion, the slower the time leads to the slower the speed, but at the same time, the length contraction counteracts this phenomenon. So the result is the same as the original data
Here, we use the original data to calculate. Because you have measured the data in your coordinate system, the result is certain. The event of motion is certain. The conclusion you can draw from special relativity is that the clocks on the two objects are not equal after they stop. To calculate relativity, we must understand the coordinate system

80 ways of oral arithmetic

1.(1+1/2)(1+1/3)(1+1/4).(1+1/100) 2.(1-1/2)(1-1/3)(1-1/4).(1-1/100) 3.8+2-8+2 4.25*4/25*4 5.7.26-(5.26-1.5) 6.286+198 7.314-202 8.526+301 9.223-99 10.6.25+3.85-2.125+3.875 11.9-2456*21 12.0.5/11.5-4*2...

How to explain the expansion of time?
Why does time stop when man's speed is close to the speed of light? Why does it take decades on earth when he travels at the speed of light for a short time

According to the theory of relativistic time expansion:
Let's set the time you feel passing as t '
At rest, time passes by T
The speed of light is C
Your speed is v
According to the formula:
T = t '/ under radical (1 - (V / C) ^ 2)
So after calculation, we know that
T = 86400 seconds
T '= 86400 * 365 seconds
C = 3 * 10 ^ 8m / S
So v = 299998874.1 meters per second
At this point, your mass approaches positive infinity due to relative expansion
And human beings can't accelerate objects as big as human beings so fast
So theoretically it is possible, but actually it is not
At present, human beings can only speed up some basic particles to 99.99% of the speed of light
And no matter how big it is, it's basically impossible
Because according to f = ma, we can get (f force, m mass, a acceleration)
When the mass is close to positive infinity, it is difficult to obtain a large acceleration
Speed can't be increased, so time travel is not allowed

Addition and subtraction within 10

5+4= 3+6= 4+2= 9+0= 2+7=
1+7= 6+4= 7+3= 2+3= 0+10=
10-2= 9-3= 8-7= 9-5= 6-2=
5-4= 10-9= 7-3= 8-4= 7-6=
3+( )=10 0+( )=4 4+( )=9 1+( )=8 6+( )=10
4+( )=8 5+( )=7 10+( )=10 2+( )=6 3+( )=5
2+( )=9 4+( )=6 3+( )=3 7+( )=8 6+( )=9
1+( )=10 8+( )=9 5+( )=10 6+( )=7 3+( )=10
( )+5=10 ( )+2=5 ( )+6=9 ( )+8=10 ( )+3=8
( )+6=6 ( )+7=8 ( )+0=5 ( )+1=7 ( )+4=6
( )+2=9 ( )+2=8 ( )+3=7 ( )+3=8 ( )+2=9
( )+3=10 ( )+2=4 ( )+6=7 ( )+7=10 ( )+0=6
9-( )=2 6-( )=1 10-( )=8 7-( )=6 5-( )=2
7-( )=7 10-( )=5 9-( )=0 10-( )=4 3-( )=2
9-( )=5 0-( )=0 8-( )=2 10-( )=3 5-( )=1
6-( )=4 7-( )=4 9-( )=7 4-( )=4 8-( )=4
( )-3=3 ( )-6=2 ( )-7=3 ( )-0=6 ( )-5=5
( )-9=1 ( )-3=0 ( )-3=5 ( )-5=1 ( )-1=8
( )-2=7 ( )-4=3 ( )-2=7 ( )-6=4 ( )-0=10
( )-3=6 ( )-6=1 ( )-5=4 ( )-7=0 ( )-1=8
( )+5=10 ( )+4=7 ( )-3=3 ( )-6=2 9-( )=2
3+( )=10 6-( )=1 ( )-7=3 ( )+2=5 0+( )=4
( )-0=6 10-( )=8 4+( )=9 7-( )=6 ( )-3=0
( )+7=8 5-( )=2 ( )-5=5 ( )+6=9 1+( )=8
7-( )=7 6+( )=10 ( )+2=8 ( )-3=4 3+( )=4
9-( )=0 ( )+6=7 4+( )=8 ( )-9=1 ( )-3=5
( )+1=4 ( )-7=3 ( )+8=10 9-( )=4 ( )-5=1
4+( )=10 ( )+5=5 ( )-2=5 10-( )=2 ( )-6=4
( )-4=2 7-( )=2 10-( )=1 ( )+3=8 5+( )=7
( )+3=9 2+( )=9 ( )-7=2 ( )-3=0 6-( )=2
3-( )=2 10-( )=4 ( )-1=8 10+( )=10 ( )+7=8
5+( )=9 ( )+5=7 10-( )=3 ( )-2=7 9-( )=5
( )+3=3 4+( )=10 ( )-9=1 ( )-3=4 8-( )=2
10-( )=5 ( )-7=2 5+( )=8 ( )+2=7 ( )-6=1
( )+1=7 4+( )=4 ( )-0=10 9-( )=1 ( )+4=6
( )+5=10 ( )-6=4 4+( )=6 10-( )=3 9-( )=8
4+( )=5 ( )+5=6 ( )-4=2 8-( )=2 ( )+6=6
( )-4=3 6+( )=7 6-( )=3 ( )+3=9 1+( )=8
( )+2=9 ( )-2=4 7-( )=7 2+( )=10 3+( )=5
0-( )=0 10-( )=3 ( )+2=7 0+( )=5 ( )-2=7

In the 2013 National University Games, a University selected 4 students from 6 students to participate in shot put, height adjustment, long jump and sprint respectively, then the probability of student a to participate in sprint or long jump is () A. 13B. 12C. 23D. 34