The ratio of the 1st gear and the 4th gear is : 3:1

So the last one will rotate: 1 * 3 = 3 ( revolutions )

Choose A

Now just to find the solution! Hope you get any luck.

Hard to translate? I will translate for you.

2 , 4 , 8 FOLLOWS THE RULE . 3 , 6 , 9 FOLLOWS THE RULE . 10 , 5 , 7 FOLLOWS THE RULE . 19 , 2 , 1 FOLLOWS THE RULE. BUT 9 , 5 , 3 DON'T FOLLOWS THE RULE . WHAT IS THE RULE?

Since no box are labeled correct; box 3 must have baseballs only.So easy that box 2 contains only tennis balls.

It's depends

POTT9R

- H6RRY

=H2RRY

So easily that Y = 8 and P = 1.

 1O889R

- H6RRY

=H2RRY

4th colunm : R + R = 8; but 5th colunm have R + R = 9; so Y + Y = 1R => R = 4 and Y = 7.

1O8894

- H6447

=H2447

Hmmm... maybe there is more than 1 solution... so... anyway... done I guess?

Nope

Maybe -8

533-412=121

336+312+3=651

The answer is A.

That group will have : 12 + 8 = 20 ( students )

The probability that the 1st student is girl is: \(8:20=\dfrac{2}{5}\)

The probability that the 2nd student is girl is : \(\left(8-1\right):\left(20-1\right)=\dfrac{7}{19}\)

The probability that both student are girl is : \(\dfrac{2}{5}\cdot\dfrac{7}{19}=\dfrac{14}{95}\)

We can pour water back and forth between these containers to come up with exactly 1 gallon, because we can take water of from the container to transform to outside.

This is the way to get it:

Step 1: Pour water in to the 9 gallon container until its full.

Step 2 : Take water from the 9 gallon container to the 7 gallon container until its full.

Now, the 9 gallon container has 2 gallon of water.

Step 3 : Take all the water from the 7 gallon container on to the ground.

Step 4 : Take all the water left from the 9 gallon container to the 7 gallon container.

IF WE MAKE THIS, then each times the 9 gallon container will have 2, then increases 2 more gallon until it is full.

So in step16 , the 7 gallon container is full, so The 9 gallon container have :

         8 - 7 = 1 ( gallon )

SO WE GET 1 GALLON OF WATER.

We can easily find S = 1. Because 7 + 7 = 14 ( hundred ) => 90 + 7 + 7 = 104 ( hundred ), which that means I = 0.

But NO WORD CAN BEGIN BY 0 => Wrong question  

OMG !I don't khow what do you say!

Because the number 8 is not change, but 9 is split up, so the digit number of base₉ is the remander after divde by 9, up there is the remainder after divide by 9, etc.

So we have:

\(A\left(base_{10}\right)=\left(a\equiv x\left(mod9\right)\right)\overline{10...00}+\left(a-\left(9\overline{x0...00}\right)\equiv y\left(mod9\right)\right)\overline{10...10}+...+\left(a-\left(9\overline{x0...00}-\left(...-\left(\overline{z0...00}\right)\overline{10...00}\right)...\right)\equiv t\left(mod9\right)\right)\left(base_9\right)\)

Use this formula, we get:

\(22\left(base_{10}\right)=\left(22\equiv2\left(mod9\right)\right)10+\left(22-2\times9\right)\equiv4\left(mod9\right)\left(base_9\right)\)

So when write the number 22 in Base₉, we get 24.