Hey if you could help that be great I've stuck on this for awhile

Mathematics

1 answer:

n200080 [17]3 years ago

5 0

Conversions aren't bad until you have to deal with the cubed and squared versions of them!! Remember that there are 100 cm in 1 meter. If that be the case, then there are 100cm*100cm*100cm (100cm^3) in m*m*m (1 meter cubed). We will use that in our conversion then: $250,000 \frac{cm^3}{m^2}* \frac{m*m*m}{100cm*100cm*100cm}$ . The cm^3 cancels out leaving m^3 over m^2 which is what we want. Doing the math now we get $.25 \frac{m^3}{m^2}$ , or $\frac{1}{4} \frac{m^3}{m^2}$

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How much money should be deposited today in an account that earns 6% compounded monthly so that it will accumulate to $1000000 (

aleksandr82 [10.1K]

Answer:

The principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.

Step-by-step explanation:

Given

Accrued Amount A = $1000000

Interest rate r = 6% = 0.06

Time period t = 45 years

Compounded monthly n = 12

To determine:

Principle amount P = ?

Using the formula

$A\:=\:P\left(1\:+\:\frac{r}{n}\right)^{nt}$

$P\:=\frac{A}{\left(1\:+\:\frac{r}{n}\right)^{nt}}$

substituting A = 1000000, r = 0.06, t = 45, and n = 12

$P\:=\frac{1000000}{\left(1\:+\:\frac{0.06}{12}\right)^{12\cdot 45}}\:$

$=\frac{1000000}{1.005^{540}}$

$P = 67659.17$ $

Therefore, the principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.

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