What is the final amount of a $300 investment earning 4.5% interest compounded quarterly for 3 years?

1 answer:

wel2 years ago

7 0

The formula A=P(1+(r/n))^nt should be followed when answering
300=P
R=0.045
N=4
T=3
300(1+(0.045/4))^4x3
Answer is=343.102

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Only the a) part please!!!!

choli [55]

$(i)\\\\2A = \begin{pmatrix}3x \times 2 & 1 \times 2 & 5 \times 2\\0\times 2 & 2x \times 2 & -3 \times 2\end{pmatrix} = \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix}\\\\\\(ii)\\\\2A + B = C\\\\\implies \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix} + \begin{pmatrix}y & 0 & -4\\1 & -3y & 2\end{pmatrix} = \begin{pmatrix}31& 2 & 6\\1 & 17& -4\end{pmatrix}$

$\implies \begin{pmatrix}6x+y & 2+0 & 10-4\\0+1 & 4x-3y & -6+2\end{pmatrix} = \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\\\implies \begin{pmatrix}6x+y & 2 & 6\\1 & 4x-3y & -4\end{pmatrix} = \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\text{System of equations},\\\\6x +y =31,~~ 4x -3y = 17\\\\\\(iii)\\\\6x +y-31 =0 \\\\4x-3y -17=0\\\\\text{Solving by cross multiplication method}\\\\\\\dfrac x{1(-17) - (-31)(-3)} = \dfrac y{4(-31) - 6(-17)} = \dfrac 1{6(-3) - 4(1)}$

$\implies \dfrac{x}{-17 -93} = \dfrac{y}{-124 +102} = \dfrac 1{-18-4}\\\\\\\implies -\dfrac{x}{110} = -\dfrac{y}{22} = -\dfrac 1{22}\\\\ \implies \dfrac{x}{110} = \dfrac{y}{22} = \dfrac 1{22}\\\ \\\text{Hence,}\\\\\\x=\dfrac{110}{22}=5,~~~~ y= \dfrac{22}{22} =1$