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OpenStudy (ashontae19):
Can someone please help me
OpenStudy (ashontae19):
@mastermindkakashi
OpenStudy (ashontae19):
@misssunshinexxoxo
OpenStudy (anonymous):
i think its a
OpenStudy (ashontae19):
thank you can you answer one more
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OpenStudy (anonymous):
lol ok
OpenStudy (anonymous):
also that soda pop question: you'd take the chance of each outcome and multiply
OpenStudy (anonymous):
so that would be 0.02*0.98 = ?
(the 0.98 comes from the chance that the soda is filled, so .100-.02)
OpenStudy (ashontae19):
you want me to answer?
OpenStudy (ashontae19):
0.0196
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OpenStudy (anonymous):
yep! you were right the first time
OpenStudy (anonymous):
and the second one you sent, you basically just combine all the sets
OpenStudy (anonymous):
just as long as it includes all the numbers
OpenStudy (ashontae19):
so it not a???? and can you help me because im in a rush and i dont want to miss anything out
OpenStudy (anonymous):
which question are you speaking of?
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OpenStudy (ashontae19):
the sets
OpenStudy (anonymous):
the set one should be C, but thats because it includes the 6
OpenStudy (ashontae19):
so how i did you do that
OpenStudy (anonymous):
you just look at the numbers in all the sets and combine them together since it asks for A U B U C. that means the union of all those, so basically all the numbers included in those sets
OpenStudy (ashontae19):
so add all the sets and what
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OpenStudy (anonymous):
well not add them as in add the values, just combine the sets, excluding duplicates
OpenStudy (anonymous):
in all those sets, we have a 1, 2, 3, 4, 5, 6, and a 7. so the answer should include all of those
OpenStudy (anonymous):
for the first one, you multiply the probability of him making the goal by the probability of him making the goal each time he tries
OpenStudy (anonymous):
sorry if that doesnt make sense
OpenStudy (anonymous):
basically since he has the same probability of making the goal each time, it would be 0.75*0.75*0.75
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OpenStudy (ashontae19):
im really in rush to be honest i did all the work i just want to know if i am right
OpenStudy (anonymous):
what did you get for that one?
OpenStudy (ashontae19):
3/8
OpenStudy (anonymous):
I got 27/64th
OpenStudy (ashontae19):
for the sets a and c i got c is that correct?
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OpenStudy (ashontae19):
it was the second question i posted
OpenStudy (anonymous):
ah for the set one, since it's not union this time, its the intersection, so between the two sets A and C you just want to find what numbers the two sets have in common
OpenStudy (ashontae19):
so the answet would be???
OpenStudy (anonymous):
so just 2 and 3
OpenStudy (ashontae19):
ok thank you i was just about to type that
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OpenStudy (ashontae19):
for the third one a and b i got d
OpenStudy (anonymous):
and for the third set one, it would be the combination
OpenStudy (anonymous):
thats exactly right
OpenStudy (ashontae19):
thank you for the deck of cards the last one i got 1/8
OpenStudy (anonymous):
now for the last one
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OpenStudy (anonymous):
since there's 8 outcomes that fit for that and 52 total cards, you'd take 8 over 52 and just reduce it
OpenStudy (ashontae19):
2/3?
OpenStudy (anonymous):
2/13
OpenStudy (ashontae19):
2/13
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
is that all?
OpenStudy (anonymous):
it would be B on that one. as long as x is a vowel, which is what that set contains