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Bess [88]
3 years ago
7

if $11,100 is invested at 3.1% interest compounded monthly, how much will the investment be worth in 19 years?

Mathematics
2 answers:
quester [9]3 years ago
7 0

Answer:

<em>In 19 years, the investment will be worth $19973.34</em>

Step-by-step explanation:

The compound interest formula is :

A=P(1+\frac{r}{n})^{nt}

p = 11100

r = 3.1% or 0.031

t = 19

n = 12 (compounded monthly)

Now, putting these values in the formula we get;

A=11100(1+\frac{0.031}{12})^{12*19}

=> A=11100(1.00258)^{228}

<em>A = $19973.34</em>

Advocard [28]3 years ago
3 0
A=p(1+i/m)^mn
A=11,100×(1+0.031÷12)^(12×19)
A=19,989.07
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P(A|B) =\frac{P(A and B)}{P(B)}

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And if we replace we got:

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b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

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B= "student was given $1 Bill"

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For this case we want this conditional probability:

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P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

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For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

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