Math help?????

Question

After the Industrial Revolution, people began moving out of rural areas and into cities. In 1890, only 64.9% of people in he U.S lived in rural area as opposed to 71.8% in 1880.

a) Write an exponential function to model the rural population as a percent of the tatal U.S population.

b) Estimate the percent of people living in rural areas in 1870.

Please explain in detail.

Answer 1

a) 71.8* (64.9/71.8)^ (t/10)
b) put in t=-10 and get 71.8* (64.9/71.8)^ -1 = 79.4 % approx.

Self explanatory.

When t=0 {in 1880} you get 71.8* (64.9/71.8)^ 0 =71.8 * 1 = 71.8 Check

When t=10 {in 1890} you get 71.8* (64.9/71.8)^ 1 = 71.8 * 64.9/ 71.8 = 64.9 Check

So it is an exponential function that satifies the requirements. Then t=-10 corresponds to 1870

Answer 2

Exponential function (in general):

f(x) = C a ^ x

To find a:

EDIT (I had the wrong formula previously):

a = (y2/y1) ^ (x2 - x1)

In this case:

EDIT (I had the wrong formula previously):

a= (64.9 / 71.8) ^ (1890 - 1880)

You do the math on that. After finding a (and plugging into the general formula), you must find C. To do this, just plug either (90, 64.9) or (80, 71.8) into the formula and solve for C (C should be the only unknown once you plug in either point, also, I'm treating x as time since 1800).

This will answer part a.

For part b, once you have your equation, with a and C plugged in, simply plug in 70 for x and solve for f(x) to find the estimated percentage.

Hope this helps.

Answer 3

.718=.649e^kt
e^kt = 1.10632
ln e^kt = ln 1.10632
kt = ln1.10632
For t= 10 years we get, k = (ln1.010632)/10 = .0101
So P = .649e^.0101t
for 1870, t= -10
p= .649e^ -.101 =.587 = 58.7%