37. For each location.write a linear model to represent the percent at time t,where t represents the number of years since 1940

38.graph the linear equations you wrote

39.from your graphs ,estimate when the percent of people living near the coast equaled the percent living inland.

Yes, no problem answering those questions if you provide the picture with the needed information. Post that and I'll be happy to answer. Thanks.

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Ok, thanks. I'll get right on this.

37.

Let P1 = percentage of population living within 50 miles of the coastal shoreline since 1940

Let P2 = percentage of population living further inland.

So, then for P1, at t = 0 we begin with 46%

that's the y-intercept of our graph

the slope is (53 - 46) / (1997 - 1940) = 7 / 57

Then the equation is

P1 = 7/57 t + 46

For the population living further inland:

we begin with 54% of the population, so the y-intercept is 54

the slope = (47 - 54) / (1997 - 1940) = -7 / 57

So, the equation for that population is

P2 = -7/57 t + 54

That answers question 37. Now I'll answer question 38.....

38. Here's the graph of the two lines. The x axis represents time t, the y axis is the percentage of the population

studypool graph.png

Now for question 39...

39. From the graph it looks like the line intersect at about 33

So then, about 33 years after 1940 the populations were equal.

So that brings us to 1940 + 33 = 1977 (the year the populations were equal)

For 39, I meant the LINES (plural), not the line. Typo. ...the lines intersect....

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