E maths - some normal questions

• bonkysleuth

  17 Oct 09, 19:16

  A spherical particle from a specimen has a volume of 4.7 cubic nanometres. Find the volume of 1.5 X 10^11 million of such particles in cubic metres, giving your answer in standard form.

  I found that 1 volume of the spherical particle is 4.7 X 10^-27 m3

  So to find the total volume, i used 4.7 X 10^-27 m3 X (1.5 X 10^11) = 7.05 X 10^-16.. answer given is 7.05X 10^-17 m3

  By the way the question said 1.5 X 10^11 million, so should i used 4.7 X 10^-27 m3 X (1.5 X 10^11 X 10 ^6) instead? If not, why?

  Thanks

• yiha093

  18 Oct 09, 01:30

  actually is 11+6=17.

  I think.

• eagle

  18 Oct 09, 01:53

  bonkysleuth, I see nothing wrong with your solution (2nd part).

  Yes, it should be 4.7 X 10^-27 m3 X (1.5 X 10^11 X 10 ^6) instead.

• bonkysleuth

  19 Oct 09, 11:28

  I have a probability question to ask.

  There are 5 groups of balls in a bag. 2 groups are green, 2 groups are red, and one group yellow. Each group has 3 balls labeled A,B and C

  2 balls are chosen at random, without replacement, from the bag

  Find the probability that the balls chosen are

  both labeled A

  My solution : p(both labeled A) = 5/15 X 4/14

                                                       = 2/21

  Answer given is 2/35. Did I make any mistake somewhere?

• yiha093

  19 Oct 09, 12:55

  Originally posted by bonkysleuth:

  I have a probability question to ask.

  There are 5 groups of balls in a bag. 2 groups are green, 2 groups are red, and one group yellow. Each group has 3 balls labeled A,B and C

  2 balls are chosen at random, without replacement, from the bag

  Find the probability that the balls chosen are

  both labeled A

  My solution : p(both labeled A) = 5/15 X 4/14

                                                       = 2/21

  Answer given is 2/35. Did I make any mistake somewhere?

  2 group green = 2 A .  2 group red = 2A one yellow = 1 A

  5/15 x 4/14 .

  i also will do likethat leii o.O

  but my maths not v.good.

   

• wee_ws

  19 Oct 09, 15:05

  Hi,

    The working looks good. I can't see why it should lead to 2/35. Thanks!

  Cheers,
  Wen Shih

• Forbiddensinner

  19 Oct 09, 15:10

  Originally posted by bonkysleuth:

  I have a probability question to ask.

  There are 5 groups of balls in a bag. 2 groups are green, 2 groups are red, and one group yellow. Each group has 3 balls labeled A,B and C

  2 balls are chosen at random, without replacement, from the bag

  Find the probability that the balls chosen are

  both labeled A

  My solution : p(both labeled A) = 5/15 X 4/14

                                                       = 2/21

  Answer given is 2/35. Did I make any mistake somewhere?

  Your answer is correct.