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Mathematics 8 Online
OpenStudy (chasebraves):

FOR A MEDAL

OpenStudy (chasebraves):

Bag A contains 3 white marbles and 2 red marbles. Bag B contains 6 white marbles and 3 red marbles. A person draws one marble from each bag. Find the probability that both marbles are white.

OpenStudy (anonymous):

I am thinking hat you would add the red to the red and the white to the white and then make a fraction out of your answers

OpenStudy (chasebraves):

ok can u help me with another question

OpenStudy (lolacole12):

2/3 * 3/5 = 6/15 = 2/5

OpenStudy (chasebraves):

A state makes auto license plates that have two letters (excluding I, 0, and Q) followed by four digits of which the first digit is not zero. How many different license plates are possible? 3,085,128 3,470,769 4,761,000

OpenStudy (lolacole12):

P(both marbles are white)=P(white from A and white from B) =P(white from A)*P(white from B)=3/5×6/9=2/5

OpenStudy (lolacole12):

26 letters - 3 letters * 26 letters - 3 letters * 10 digits - 1 digit * 10 digits * 10 digits * 10 digits = 23 * 23 * 9 * 10 * 10 * 10 = 4,761,000

OpenStudy (lolacole12):

take how many different letters the first letter could be, multiply it by the number of different letters the second could be, multiply it by the number of digits possible for the first slot, multiply it by the number of digits possible for the second slot, multiply it by the number of digits possible for the third slot, multiply it by the number of digits possible for the fourth slot, and there is your answer.

OpenStudy (anonymous):

I have answered that question for somebody i think me and the person decided that the answer is B

OpenStudy (lolacole12):

mine was c and i masterd my test idk

OpenStudy (anonymous):

Ok i tried to help

OpenStudy (chasebraves):

what about this one An astronomer wishes to measure on a photograph the distance between the image of a certain star and three other star images nearby. If eight images are nearby, how many choices of the three stars does he have?

OpenStudy (lolacole12):

mmhhmm

OpenStudy (lolacole12):

it's a combination. Using the formula for combinations: 8 C 3 = 8! / (3!(8 - 3)!) = 8! / (3! * 5!) = (8 * 7 * 6 * 5!) / (3! * 5!) = (8 * 7 * 6) / (6) = 56 56 choices. Hope this helps! :) EDIT: I'd just like to address the question a little better. Order does not matter because the order in which he chooses the three stars doesn't make a difference when measuring them. In other words, if he picks three stars in a certain order and obtains a distance, the distance will be the same if he chooses the SAME three stars in a different order. I hope that clears it up a little.

OpenStudy (anonymous):

I am going to have to think about that one

OpenStudy (anonymous):

I have no idea I can't help you with this one

OpenStudy (lolacole12):

any more

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