Angelina deposits $3000 in a savings account on 1 January 2019, earning compound interest of 1.5% per year.

Mathematics

1 answer:

jekas [21]2 years ago

4 0

The interest earned would be $481.62.

The total amount of money in her account would be $16,377.12.

<h3>Determining interest after 10 years </h3>

Interest = future value - amount deposited

$3000 (1.015)^10 = $3,481.62

Interest = $3,481.62 - $3000 = $481.62

<h3>Determining the future value of the account.</h3>

Future value of the lump sum = $3000 x (1.015)^11 = $3,533.85

Future value of the annuity = amount deposited x annuity factor

Annuity factor = {[(1+r)^n] - 1} / r = [(1.015)^10 - 1] / 0.015 = 10.702722

Future value = 10.702722 x $1,200 = $12,843.27

Total future value = $12,843.27 + $3,533.85 = $16,377.12

To learn more about future value, please check: brainly.com/question/18760477

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The formula to compute the combined standard deviations of two different data sets is:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}$

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$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}$

It is provided that the sample size is same for both the data sets, i.e. $n_{X} = n_{Y}=n$

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$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}\\=\frac{(n\times10)+(n\times10)}{n+n}}\\=\frac{20n}{2n}\\ =10$

Compute the combined standard deviation as follows:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}}\\=\sqrt{\frac{(n\times1^{2})+(n\times2^{2})+(n(10-10))+(n(10-10))}{n+n}}\\=\sqrt{\frac{n+4n}{2n} } \\=\sqrt{\frac{5n}{2n} } \\=\sqrt{\frac{5}{2}} \\=1.58114$