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saveliy_v [14]
3 years ago
15

Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals

are made, how much money, to the nearest dollar, would be in the account after 6 years?
Physics
2 answers:
Stels [109]3 years ago
5 0

Answer:

15448

Explanation:

A=11000(1.01425)^{24}

A=11000(1.01425)  

24

daser333 [38]3 years ago
3 0

Answer:

15448

Explanation:

Compounded Quarterly:

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

A=P(1+

n

r

​

)

nt

Compound interest formula

P=11000\hspace{35px}r=0.057\hspace{35px}t=6\hspace{35px}n=4

P=11000r=0.057t=6n=4

Given values

A=11000\left(1+\frac{0.057}{4}\right)^{4(6)}

A=11000(1+

4

0.057

​

)

4(6)

Plug in values

A=11000(1.01425)^{24}

A=11000(1.01425)

24

Simplify

A=15448.0290759

A=15448.0290759

Use calculator

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Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

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\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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