# Model the data using an exponential function f(x)=Ab^x begin{array}{|l|l|l|}hline x&0&1&2hline f(x)&125&25&5hlineend{array}

Question
Exponential models
Model the data using an exponential function $$f(x)=Ab^x$$
$$\begin{array}{|l|l|l|}\hline x&0&1&2\\\hline f(x)&125&25&5\\\hline\end{array}$$

2021-03-08

Given- Consider the data,
$$\begin{array}{|l|l|l|}\hline x&0&1&2\\\hline f(x)&125&25&5\\\hline\end{array}$$
To model- The above data using an exponential function $$f(x)=Ab^x$$
Explanation- As per the question, we have to model the above data using the given exponential expression as,
$$f(x)=Ab^x$$ (1)
Now, using first value $$x=0$$ and $$f(x)=125$$, we get from the above expression,
$$f(0)=Ab^0$$
$$125=A\cdot1$$
$$125=A$$
So, equation (1) becomes,
$$f(x)=125b^x$$
Now, using second value $$x=1$$ and $$f(x)=25$$, we get from the above expression,
$$f(1)=125b^1$$ (2)
$$25=125\cdot b$$
The above expression can be written as,
$$b=\frac{25}{125}$$
$$b=\frac{1}{5}$$
So, equation (2), becomes,
$$f(x)=125(\frac{1}{5})^x$$ (3)
Now, to check the above expression, we can use the third value as $$x=2$$ and $$f(x)=5$$, we get,
$$5=125\cdot(\frac{1}{5})^2$$
$$5=125\cdot\frac{1}{25}$$
$$5=5$$
So, the expression $$f(x)=125(\frac{1}{5})^x$$ is correct expression.
Answer- Hence, the exponential expression which models the given data is
$$f(x)=125(\frac{1}{5})^x$$

### Relevant Questions

Model the data using an exponential function $$f(x)=Ab^x$$
$$\begin{array}{|l|l|l|}\hline x&0&1&2\\\hline f(x)&125&25&5\\\hline\end{array}$$

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$$\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
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$$\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
​(Round to one decimal place as​ needed.)
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$$B. 20602060xf(x)$$
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$$C. 20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
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$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ and for 35
$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$
The mileage for the tire pressure $$\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ is
The mileage for the tire pressure $$\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}$$ in. is
(Round to two decimal places as​ needed.)
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