Problem

Data points $(x, y)$ are given. (a) Draw a scatt…

Problem 62 Easy Difficulty

Data points $(x, y)$ are given.
(a) Draw a scatter plot of the data points.
(b) Make semilog and log-log plots of the data.
(c) Is a linear, power, or exponential function appropriate for modeling these data?
(d) Find an appropriate model for the data and then graph the model together with a scatter plot of the data.
$$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} & {12} \\ \hline y & {0.08} & {0.12} & {0.18} & {0.26} & {0.35} & {0.53} \\ \hline\end{array}$$

$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {10} & {20} & {30} & {40} & {50} & {60} \\ \hline y & {29} & {82} & {150} & {236} & {330} & {430} \\ \hline\end{array}$

Answer

(a) See explanation for graph
(b) See explanation for graph
(c) Power
(d) $y=0.894488 \cdot x^{1.50922981}$

Related Courses

Calculus 1 / AB

Biocalculus Calculus for the Life Sciences

Chapter 1

Functions and Sequences

Section 5

Logarithms; Semilog and Log-Log Plots

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Video Transcript

this question shows us some data about the number of single dads in the US with Children under 18. It shows us this data over a course of several years and were asked first to plot it and then to describe the relationship between the year and the number of single fathers. Let's start by Drawing some axes is so we can plot our data. My Y axis will be the number of single dads and millions. My X axis will be the year on the left. I know that I need to go all the way up to 2.5. No, partition this off into fifth, so we have to 1.5 one and 0.5. And that is the number of single fathers in millions. And then on my ex excess, I will go from the year 1980. So too little squiggle here to show I'm not showing it starting at 0 1980 all the way up to go to 2010 and we will stop at 1990 and 2000. So let's start by plotting. In 1980 there were 19800.6 million single fathers with Children under 18 in the U. S. So in 1980 were at 19800.6, which is about right there. In 1990 there were 1.2 million, which is gonna be higher about right here. In 2000 there were 1.8. So just about here. And in 2008 there were 2.2 will draw that right there. So when we connect thes, we see something interesting. Pretend that that is a straight line. It went through all the points. We see that the slope is neither increasing nor decreasing. We know that if we had an exponential function, the slope would look like that. And if we had a lock rhythmic function, it would look something like this. But since we have neither of these, we know that this relationship is linear. So are our relationship. Here is linear and that is your final answer.

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