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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

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

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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?

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

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It's depends

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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?

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Nope
Maybe 8

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533412=121
336+312+3=651

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The answer is A.

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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 : \(\left(81\right):\left(201\right)=\dfrac{7}{19}\)
The probability that both student are girl is : \(\dfrac{2}{5}\cdot\dfrac{7}{19}=\dfrac{14}{95}\)

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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.

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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

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OMG !I don't khow what do you say!

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Because the number 8 is not change, but 9 is split up, so the digit number of base_{9} 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:
\(22\left(base_{10}\right)=\left(22\equiv2\left(mod9\right)\right)10+\left(222\times9\right)\equiv4\left(mod9\right)\left(base_9\right)\)
So when write the number 22 in Base_{9}, we get 24.
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