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# The table gives the population of the world $P(t),$ in millions, where $t$ is measured in years and $t = 0$ corresponds to the year 1900.(a) Estimate the rate of population growth in 1920 and in 1980 by averaging the slopes of two secant lines.(b) Use a graphing device to find a cubic function (a third- degree polynomial) that models the data.(c) Use your model in part (b) to find a model for the rate of population growth.(d) Use part (c) to estimate the rates of growth in 1920 and 1980. Compare with your estimates in part (a).(e) In Section 1.1 we modeled $P(t)$ with the exponential function$f(t) = (1.43653 \times 10^9) \cdot (1.01395)^t$Use this model to find a model for the rate of population growth.(f) Use your model in part (e) to estimate the rate of growth in 1920 and 1980. Compare with your estimates in parts (a) and (d).(g) Estimate the rate of growth in 1985.

## a) 78.5 million/yearb) $a \approx-0.0002849003, b \approx 0.52243312243$, $c \approx-6.395641396,$ and $d \approx 1720.586081$c) $P^{\prime}(t)=3 a t^{2}+2 b t+c($ in millions of people per year)d) 14.16 mil/yr in 1920, 71.72 mil/yr in 1980, both values are smaller than the estimates from part (a)e) $f^{\prime}(t)=p q^{t} \ln q$f) $f^{\prime}(20) \approx 26.25$ million/year, $f^{\prime}(80) \approx 60.28$ million/yearg) $P^{\prime}(85) \approx 76.24$ million/year and $f^{\prime}(85) \approx 64.61$ million/year

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So for the given problem, we're going to be looking at rates of change, especially in this case, we want to be estimating the rate of population growth in 1920 in 1980 by averaging the slopes of the two second lines. So in this case looking at the change in time, there's 60 year change. Um and then subtracting the population over that 60 year time period, we see there's going to be a growth of 78.5 million per year. And then we want to use the graphing device to model the data so that's going to be are a value. So this is going to be a X. Cubed plus three X squared plus C. X plus D. And we get that A is negative 0.2849 B. Is 0.52 to four. Uh See is a negative 6.39564 And then d will be 17 20 plus 0.586 So that right there gives us we'll find this using possibly Excel or some other computing tool that we can model the data.

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