Question #a73a3

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Question #a73a3

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Answer 1 · 2015-05-21T22:25:21

Answer: $y = 0.25 \cdot {(1.0485)}^{t}$ $y = 0.25 * (1.0485)^t$

The exponential function will be $y = a \cdot b^{t}$ $y=a*b^t$ , where $a$ $a$ is the initial value

So, $a = 0.25$ $a=0.25$ because the hourly minimum wage has the initial value of $$ 0.25$ $$0.25$ at the moment $t = 0$ $t=0$ (the year 1938).

(Verification: for $t = 0$ $t=0$ , $y = 0.25 \cdot b^{0}$ $y=0.25*b^0$ $=$ $=$ $0.25 \cdot 1 = 0.25$ $0.25*1 =0.25$ )

In 2009, after 71 years, the value of the hourly minimum wage is $7.25. Therefore, $y = 7.25$ $y=7.25$ for $t = 71$ $t=71$ .

So, ** $7.25 = 0.25 \cdot b^{71}$ $7.25=0.25*b^71$ **

In order to find our exponential function, we have to find the value of $b$ $b$ .

$b^{71} = \frac{7.25}{0.25}$ $b^71=7.25 / 0.25$ $= 29$ $= 29$ $\Rightarrow$ $=>$ $b = \sqrt[71]{29}$ $b=root71(29)$ $\approx 1.0485$ $~~1.0485$

Hence, our exponential function is $y = 0.25 \cdot {(1.0485)}^{t}$ $y=0.25*(1.0485)^t$

Based on this model, we can estimate the values of the hourly minimum wage for the other years, by substituting the values of $t$ $t$ in the general formula:

for 2015, $t = 77$ $t=77$ $\Rightarrow$ $=>$ $y = 0.25 \cdot {(1.0485)}^{77} \approx 9.5875$ $y=0.25*(1.0485)^77~~9.5875$
for 2016, $t = 78$ $t=78$ $\Rightarrow$ $=>$ $y = 0.25 \cdot {(1.0485)}^{78} \approx 10.0525$ $y=0.25 * (1.0485)^78~~10.0525$

This is the graph of the function:
enter image source here

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