Statistics

The number of combinations, denoted by 11C6 is 11!/[6!*(11-6)!]

= 11*10*9*8*7/(5*4*3*2*1) = 462

2 out of 10

2/10 is equivalent to 1/5 and 3/15.

8,754,321

0.9%

900, all numbers from 100 to 999 inclusive.

you can trust me because i am in 6 grade and i had to do that.

A chi square is square of standard normal variate, so all values are positive

0.54 or 0.0625

51/24 + 88/24 = 139/24 or 5 and 19/24

No, I don't.

8 sided

999

1/1, 1/2, 1/3, 2/1, 2/2, 3/1 = 6 out of 36 ie 16.66%

9!/(5! x 4!) = (9 x 8 x 7 x 6) / (4 x 3 x 2) =126 if all digits have to be different.

If repeated digits are allowed then it is 95 ie 59049.

Trigonometry, and mechanics for a start.

137.4 decimetres.

No.

Answer: 40000 km = 131,233,595.800 '

3

4*6+1

100

No, the digit in that position will arise from a whole range of possibilities. If N is a (integer) number taken at random then the number in the thousands place could be anything equally. In particular, if you divide 88888888 by 8 the digit in the thousands place comes from dividing 8 by 8.

64/8 is 8 so that is the digit in the thousands position. So you could have any one of numbers like xxxx8xxx where each x could be any digit, eg 11118111, 12348765, etc.