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Mathematics 16 Online
OpenStudy (anonymous):

pls help with math @iGreen @Data_LG2

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

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

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

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

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

OpenStudy (anonymous):

so what do you think?

OpenStudy (anonymous):

I dont see it anywhere.

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

:L

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

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

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"

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

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.

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

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

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

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%

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.

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

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

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?

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