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About 73.
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How many multiples of 3 are there between 10 and 226?
1st do 226-10.This is 216.Then do 216 /3 which is 72.So there are 72 multiples of 3 between 10 and 226.
When was Not for Threes created?
Not for Threes was created on 1997-10-27.
How many threes are in 10?
Three and three-tenths.
How many feet are there in 226 inches?
There are 12 inches in one foot. Therefore, 226 inches is equal to 226/12 = 18.83 recurring (that is, 18.83333...) feet or 18 feet 10 inches.
How many yards did terrell Owens get on 10-3-2010?
226 yards
How do you use symbols to show how five threes equal 10?
Five threes do not equal ten, so you can't use symbols to show that.
What are the release dates for Jeopardy - 1984 10-226?
Jeopardy - 1984 10-226 was released on: USA: 18 July 1994
How do you get the answer ten using four threes?
(3x3) + (3/3) = 10
How many decimetres are there in 22.6 meters?
There are 10 decimetres in one metre. Therefore, 22.6 metres is equal to 22.6 x 10 = 226 decimetres.
How many threes are there in the first one million digits?
We only use 10 digits (0 through 9), so I would say, there is one of them.
What is ten divided by three into a decimal?
10/3 is 3.33333333333333333 (the threes go on forever).
What is the probability that 13 rolls of a fair die will show three threes?
Not an easy question. Here's a long but logical answer. The chance of not getting a 3 on any throw is 5/6, so the chance of not getting any threes in the first 10 throws is (5/6)^10. Now, having reached that point, the chance of throwing three threes in the last three throws is 1/6 x 1/6 x1/6 or 1/216. So the probability of throwing three threes in any chosen three throws is (5/6)^10 x 1/216. This should seem logical because the probability of throwing three threes in the first three throws followed by ten failures to throw threes, should be the same as just getting threes in the last three throws. The probability is the same for getting threes in only the second, fourth and sixth throw, or any other combination of three throws. Now, there are a set number of combinations of 13 things taken in groups of three. This is given by the formula for C(13,3) = (13x12x11x10x9x8x7x6x5x4)/((10x9x8x7x6x5x4x3x2x1)x(3x2x1)). That cancels down to (13x12x11)/(3x2x1), and equals 286. So, there are this many ways you can pick out three throws which you want to be the threes, and the probability of them actually being threes is the first calculation we did. The answer to the question is 286 x (5/6)^10 x 1/216 which is 0.2138453, about.
