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prisoha [69]
3 years ago
14

A person invested $3,500 in an account growing at a rate allowing the money to double every 15 years. How long, to the nearest t

enth of a year would it take for the value of the account to reach$7,700?
Mathematics
2 answers:
nignag [31]3 years ago
8 0

Answer:

30 years and a half

Step-by-step explanation:

well let's just say 3.5k x hmm 2? at an estimate and that will get us to 7000.. so in order to get to 7000 we have to wait 15 years which is 30 years and a half. From Xthermyx. A brainly user.

Maslowich3 years ago
7 0

Answer:

t = 17.1

Step-by-step explanation:

y = a(2) ^{ \frac{t}{d} }

y = 7700 \:  \: a = 3500 \:  \: d = 15

\frac{7700}{3500}  =  \frac{3500(2) ^{ \frac{t}{15} } }{3500}

2.2 =  { 2}^{ \frac{t}{15} }

\frac{ log(2.2) }{ log(2) }  =  \frac{t}{15}

1.1375352 =  \frac{t}{15}

15(1.13750352) = t

17.062552856 = t

t = 17.1

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Simple interest is calculated using the formula:

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