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Simplify the expression 7c6 3c2?
7c6
Simplify 6c3d divided by -2cd?
-3c2
How many atoms are in3C2?
Six 3x2=6 atoms, in 3 molecules of H2
What would a full size number at the start of the chemical formula reprent ex 3C2?
It represents number of moles of an element.
A bag contains 3 red balls and 2 blue balls a ball is taken at random from the bag and then put back what is the probability that both balls are red?
3c2+2c2+3c1 2c1 --------------- 7c2. I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green.
Looking for Hystar Steering Solenoid, p/n DSG-3C2-N-01, 110v a/c. Boat at sea and dead in the water. Where available?
You can find them at http://www.hystar.com.tw/hydraulic/hydraulic_2.htm there is an option to buy online even for that
How many ways can you purchase 2 CDs if there are 3 to choose from 4 cassettes if there are 4 to choose from and 2 DVDs if there are 5 to choose from?
The answer is 3C2*4C4*5C2 = [3 * 1 * (5*4)/(2*1)] = 3*1*10 = 30 ways.
Use a 3 digit number to form all two digit numbers?
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
What does c mean in math Example 5c3 and 3c2?
it means combinatorial or combination, you use the formula nCr = n!/((n-r)! x r!). example 5c3 = 5! / ((5-3)! x 3!) = 5! / (2! x 3!) = (5x4x3x2x1) / ((2x1) x (3x2x1)) = 10
What are the chances of three winning lottery tickets if the chances for one winner is 1 in 179 million and one and a half billion tickets are sold?
If the probability of winning with one lottery ticket is P(W) = 1/(179 x 106) and is independent of the number of tickets sold, and you buy 3 tickets, then you have:The probability of not winning (loosing) of one ticket is P(L) = 1 - 1/(179 x 106).The probability of all 3 tickets loosing is P(3L) = [P(L)]3 = [1 - 1/(179 x 106)]3 =0.999999983...The probability of two loose one win is P(2L1W) = 3C2 [P(L)]2 P(W)P(2L1W) = 3[1 - 1/(179 x 106)]2∙[1/(179 x 106)] = 1.675977635... x 10-8 == 1/(5.966666733... x 107), which is about 1 in 59.7 million.The probability of one loose two win is P(1L2W) = 3C2 P(L) [P(W)]2P(1L2W) = 3[1 - 1/(179 x 106)][1/(179 x 106)]2 = 9.363003599... x 10-17 == 1/(1.068033339... x 1016), which is about 1 in 10 680 333 billion, (≈ 1 in 10.7quadrillion).The probability of three win is P(3W) = [P(W)]3 = [1/(179 x 106)]3 == 1.743576099... x 10-25 = 1/(5.735339... x 1024), which is about 1 in 5.7 septillion.
What two numbers do you multiply to get 231?
1 x 231, 3 x 77, 21 x 11, 33 x 7. Add the digits and you get 6, which means it is divisible by 3. (3 x 77) 77 has two prime factors, 7 and 11. So you can multiply any combination of the factors to get 231. 3C2 (3 choose 2) in Pascal's triangle is 3, which means there are three possible combinations. Oh, yeah, and you can always just do 1 x 231.
How do you find the apothem of a hexagon with only the side length?
The apothem is the radial distance from the middle of the side to the centre of the hexagon. A hexagon is six congruent equilateral triangles joined by adjacent sides. Equilateral triangles can be divided into two equal right angled triangles. The upright of the right angled triangle is effectively the apothem of the original hexagon. Pythagoras now kicks in. The apothem (vertical) is A and half the side length (base) is B, the third (longest) side is C and is the same as the original side length. Pythagoras states A2 + B2 = C2. So by transposition, A = root (C2 - B2). As B = 1/2 C, the apothem A is given by: A = root(C2 - (C/2)2) = root(C2 - C2/4) = root(3C2/4) = C x root(3) / 2 So the apothem of a hexagon is 1/2 x root(3) x the side length.
