A savings account starts out with $0.50 and doubles every month. Write an equation that models this situation and determine how much money will be in the account in 2 years
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OpenStudy (anonymous):
its easy
OpenStudy (anonymous):
you have to multiply .50x2x24
OpenStudy (anonymous):
Because it doubles
OpenStudy (anonymous):
but like .50x2
OpenStudy (anonymous):
then x24
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OpenStudy (anonymous):
then 1.00x2
OpenStudy (sesca89):
which is one...
OpenStudy (sesca89):
2....
OpenStudy (anonymous):
yup
OpenStudy (anonymous):
and you have to do that procees 24 times
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OpenStudy (anonymous):
i messed up dont do x24
OpenStudy (sesca89):
Well, that's wrong, because it is an exponential growth problem and i need an equation...
OpenStudy (sesca89):
I mean it's right, but it's not what I need
OpenStudy (anonymous):
just do the procces 24 times to dertermine the snaswer
OpenStudy (sesca89):
Yes, but I need an equation too.
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OpenStudy (anonymous):
ok
OpenStudy (carlosjose):
hey
OpenStudy (sesca89):
Helpppp
OpenStudy (perl):
y = 0.5*2^x , where x = number of months
OpenStudy (anonymous):
I would go with Y=X exponent 2 x2
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OpenStudy (carlosjose):
imma go to english
OpenStudy (sesca89):
x is supposed to be number of years.
OpenStudy (xapproachesinfinity):
it is f(x)=(2^x)/2 like perl said
OpenStudy (perl):
y = 0.5 * 2^(12*x)
OpenStudy (perl):
because it doubles 12 times in a year
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OpenStudy (sesca89):
Yes, but wouldn't it be .5(25)^t?
OpenStudy (xapproachesinfinity):
no the amount double each month
with the initial value f(0)=0.5
OpenStudy (sesca89):
Listen, this is an exponential growth problem. SO, the equation for it would be A(1 +r)^t
A= 0.5, r = 24, and t = # of years (2)
OpenStudy (sesca89):
therefore, it should be 0.5(1+24)^2
OpenStudy (xapproachesinfinity):
this one y=0.5*2^(12x) is an exponential growth as well
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OpenStudy (sesca89):
this is confusing....
OpenStudy (xapproachesinfinity):
where did you get r=24? r repesents the rate my dear
OpenStudy (xapproachesinfinity):
in fact you can't say A(1+r)^t this interest formula you know
you need to find the interest rate r and the number of compounding
OpenStudy (xapproachesinfinity):
usually as we compound continuously we get the number e^r not 25 like you did
OpenStudy (sesca89):
it's asking for the amount after 2 years. The amount DOUBLES each MONTH 2 x 12(month) = 24
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OpenStudy (perl):
r is not 24
OpenStudy (xapproachesinfinity):
why are you making your life harder, while the problem didn't ask you anything about interest ?
OpenStudy (perl):
If something doubles that means
r = 100% ,
or in decimal
r = 100/100 = 1
OpenStudy (xapproachesinfinity):
yes @perl has given you the formula for getting that
\(y=0.5*(2)^{12x}\)
OpenStudy (sesca89):
Listen, its not asking about intrest, it's asking about what the amount will be in 2 years aka 24 months
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OpenStudy (xapproachesinfinity):
I know! you are the one who is thinking about a different approach to the problem
OpenStudy (perl):
y = A ( 1 + r)^t
here A = 0.5
r = 1
t = variable in months
y = 0.5 ( 1 + 1) ^t
and then you said t is in years , so i changed it
y = 0.5 ( 1 + 1) ^(12t)
OpenStudy (sesca89):
Because this is the approach I am being taught
OpenStudy (perl):
r is the rate it increases per time period. it increases by 100% or by a factor of 1
OpenStudy (xapproachesinfinity):
hmm, I agree with perl
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OpenStudy (sesca89):
you guys are probably right, I'm just asking if the equation, .5(25)^t is right
OpenStudy (perl):
no, it isn't
OpenStudy (perl):
for the reasons given above :)
OpenStudy (sesca89):
...
OpenStudy (sesca89):
Then what would r be?
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OpenStudy (perl):
r equals 1
OpenStudy (sesca89):
Wait, just tell me the total answer for the problem.