E maths - some normal questions

bonkysleuth SG

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 SG

actually is 11+6=17.

I think.

eagle SG

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 SG

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 SG

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 SG

Hi,

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

Cheers,
Wen Shih

Forbiddensinner SG

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.