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Arlecino [84]
2 years ago
14

They work-infested $18,000 into accounts, one yielding 4% interest and the other yielding 12%. If he received a total of $1,280

and interest at the end of the year, how much did he invest at 4%?​
Mathematics
1 answer:
lakkis [162]2 years ago
3 0

Answer:

<u>$11,000</u>

Step-by-step explanation:

Let amount invested for 12% yield be <em>p . </em>Amount for 4% yield is therefore (<em>18000-p)</em>

Interest for 12%:<em> 12% of p=0.12p</em>

Interest for 4%: <em>4% of </em>(<em>18000-p)=0.04(18000-p)=720-0.04p</em>

Total interest:

<em>=0.12p</em> + (<em>720-0.04p)</em>

<em>=0.12p-0.04p+720</em>

<em>=0.08p+720</em>

But we already know the total interest is 1280

Therefore: <em>0.08p+720=1280</em>

<em>0.08p=1280-720</em>

<em>0.08p=560</em>

<em>p=560÷0.08</em>

<em>p=7000</em>

Therefore at 4% the person invested:

(<em>18000-p)</em>

<em>18000-7000</em>

<em>$11,000</em>

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a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

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Here is the reduced echelon form for the augmented matrix.

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

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