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OpenStudy (anonymous):
OpenStudy (anonymous):
OpenStudy (anonymous):
i just need help how to do problem and I dont understand the ?'s
OpenStudy (anonymous):
i believe it should be doubled because of how the other ones would double if it was increased
OpenStudy (anonymous):
not 100% sure though
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OpenStudy (anonymous):
idk :(
OpenStudy (anonymous):
@King.Void. can u help us with this one please thank you
OpenStudy (king.void.):
I'm not really good at math....
OpenStudy (anonymous):
ur a human calcutator though
OpenStudy (anonymous):
@shrutipande9 can u help me brother man
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OpenStudy (king.void.):
Yea... I earned that by telling people to stop spamming in the math section...
OpenStudy (anonymous):
o ok sorry for bothering u then
OpenStudy (king.void.):
@iGreen @Data_LG2
OpenStudy (king.void.):
The 2 mathheads
OpenStudy (anonymous):
both are offline
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OpenStudy (king.void.):
They are online.. Just OS is glitchy..
OpenStudy (anonymous):
probabilities ;_;
i'm not sure how to do this, but i'll try to remember
@iGreen probably knows
OpenStudy (king.void.):
.....I thought you were a mathhead...
@iGreen you're a mathhead.. help her..
OpenStudy (anonymous):
everyone has their weakness
OpenStudy (anonymous):
@k12andstudyislandhelp
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OpenStudy (anonymous):
this is a k12 one
OpenStudy (anonymous):
ok.
OpenStudy (anonymous):
pellet what lesson is it
OpenStudy (anonymous):
i will go to the webpage and find u the answers
OpenStudy (anonymous):
data i need help as well bro
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OpenStudy (anonymous):
so what do you think?
OpenStudy (anonymous):
I dont see it anywhere.
OpenStudy (anonymous):
OpenStudy (anonymous):
OpenStudy (anonymous):
:L
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OpenStudy (anonymous):
@torifamelol do u mind if i post mine on here as well
OpenStudy (anonymous):
oh! now i have an idea.
for part A:
probability of drawing a silver coin: number of silver coin/total number of coins. now, the expectation will be probability of drawing a silver coin time the number of times drawn :P
OpenStudy (anonymous):
*times
OpenStudy (anonymous):
11/70
OpenStudy (anonymous):
and data y u block me man
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OpenStudy (anonymous):
is it 11/70
OpenStudy (anonymous):
i think so @torifamelol
OpenStudy (anonymous):
but they want a percent
OpenStudy (anonymous):
that's the number of times drawn. Do it with the actual number of coins in the bag
OpenStudy (anonymous):
oh let me see
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OpenStudy (anonymous):
0.1571428571428571 Into a percent
OpenStudy (anonymous):
10
OpenStudy (anonymous):
no, it will be: 1/10
to change this to percent: 10%
OpenStudy (anonymous):
it would be 15% because when you are changing a decimal to a percent you move 2 numbers above the secimal
OpenStudy (anonymous):
"probability of drawing a silver coin"
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OpenStudy (anonymous):
*decimal
OpenStudy (anonymous):
11/70 = 0.1571428571428571 @Data_LG2
OpenStudy (anonymous):
wait first dont I have to put it into a fraction which is 1/10. then 1/10 into a decimal?
OpenStudy (anonymous):
wait what?
OpenStudy (anonymous):
don't use the Number of times Drawn. Use the Actual number in the Bag
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OpenStudy (anonymous):
heh heh i didnt see that
OpenStudy (anonymous):
like really. I didnt.
OpenStudy (anonymous):
the actual # in bag is 10 but drawn is 70 we arent using 70 we are using 10
OpenStudy (anonymous):
so thats part A?
OpenStudy (anonymous):
:P not done with part A yet.
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OpenStudy (anonymous):
kk let me just write numbers down
OpenStudy (anonymous):
k done
OpenStudy (anonymous):
tell me what you'll put in A. I'm not sure whether you get it or not xD
OpenStudy (anonymous):
wait @Data_LG2 They said experimental probability . Which mean the probability that has been tried. Not theory. Which is what your doing by using the number in the bag.
OpenStudy (anonymous):
oh EXPERIMENTAL!!! so 11/70
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OpenStudy (anonymous):
alright for the first half of part A i would do
Since there's 10 actual numbers in a bag the fraction would be 1/10. because there's 1 silver out of ten. and 1/10 changed into a percentage would be 10%.
OpenStudy (anonymous):
nope, follow k12
OpenStudy (anonymous):
kk can we restart
OpenStudy (anonymous):
Ok :)
OpenStudy (anonymous):
XD
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OpenStudy (anonymous):
okay so we have 11/70 as a fraction
OpenStudy (anonymous):
Yes
OpenStudy (anonymous):
You need to know put that as a decimal
OpenStudy (anonymous):
what do u mean
OpenStudy (anonymous):
it says express the probability as a percent
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OpenStudy (anonymous):
o.o
OpenStudy (anonymous):
To make the percent, you first have to make it a decimal. Then you change that into a percent.
OpenStudy (anonymous):
oh ok
OpenStudy (anonymous):
0.15714285714
OpenStudy (anonymous):
but thats 15%
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OpenStudy (anonymous):
Correct
OpenStudy (anonymous):
ok let me write this down
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
done
OpenStudy (anonymous):
Ok, now on to B. If you want.
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OpenStudy (anonymous):
yes let me read the question
OpenStudy (anonymous):
not done with part A though, it also says "how would you expect it to change as the number of trials increases?"
OpenStudy (anonymous):
oh. I didnt see that.
OpenStudy (anonymous):
idk. How to do that. How bout you @Data_LG2
OpenStudy (anonymous):
@Data_LG2 can u help with this part
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OpenStudy (anonymous):
not sure with that part also, that's why i mention it xD
OpenStudy (anonymous):
:( Ugh
OpenStudy (anonymous):
let me ask my brother hes in 9th grade
OpenStudy (anonymous):
ok. But ill be here for B. Though
OpenStudy (anonymous):
My brother said its like when they pick more out of the bag how would you expect it tochange
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OpenStudy (anonymous):
@Data_LG2
OpenStudy (anonymous):
it is. But i dont know how to determine that.
OpenStudy (anonymous):
letme try finding something stay here
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Start with a shuffled deck of cards and distribute all 52 cards to 4 players, 13 cards to each.
What is the probability that each player gets an Ace? Next, assume that you are a player and
you get a single Ace. What is the probability now that each player gets an Ace?