SOLUTION: there are 45 coins in a jar. the probability of drawing a coin randomly that is not a dime is 11/15. how many nickels should be added to have the probability of drawing a dime to b

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Question 980019: there are 45 coins in a jar. the probability of drawing a coin randomly that is not a dime is 11/15. how many nickels should be added to have the probability of drawing a dime to be 2/9?
the probability of drawing a yellow marble from a bag with 42 marbles is 2/3. how many orange marbles should be taken away to have the probability of drawing a yellow marble from the bag to be 14/15?
there is a stack of folders. the probability of drawing out a yellow folder from the stack is 7/12. after 18 yellow folders are taken out, the probability of drawing out a yellow folder is 4/9. how many yellow folders are there initially?
there are 9 girls in a team. if 6 boys joined the team then the probability of choosing a boy from the team is 4/7. what is the probability of choosing a boy from the team before the six boys join the team?


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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
If 11/15 chance a coin isn't a dime, there is a 4/15 chance it is, and with 45 coins, there are 12 dimes.
Let x=number of nickels
12/(45+x) =2/9
Cross-multiply
90+2x=108
2x=18
x=9
9 nickels
54 coins in bag, 12 are dimes. Therefore, probability of drawing a dime is 12/54=2/9
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There are 28 yellow marbles out of 42
28/(42-x) = 14/15; x = orange marble number;
cross-multiply
588-14x=420
-14x=-168
x=12 orange marbles must be removed
bag now has 30 marbles with 28 of them yellow. That is 14/15
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x=multiple of numerator and denominator.
(7x/12x)= fraction of number of folders that are yellow.
(7x-18)/(12x-18) = 4/9
cross multiply
48x-72=63x-162; collect like terms on same side, will both be positive or both be negative
-15x=-90
x=6
12x=12*6=72 folders
(7/12) or 42 folders are yellow
remove 18 yellow folders and have 24 yellow folders/54 (72-18) total.
24/54=4/9, if you divide numerator and denominator by 6.
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team originally had 9 girls +x boys.
now the team has 9 girls and (6+x) boys, or 15+x total members
probability of choosing a boy is (6+x)/(15+x) =(4/7)
cross-multiply
60+4x=42+7x
18=3x
x=6
The team originally had 15 members, 6 boys, and the probability of choosing a boy was 6/15=2/5. This is the answer.
Now the team has 21 members, 12 boys, and the probability of choosing a boy is 12/21=(4/7)
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