In 1960, the population was 291000. In 1970, the population was 480000. Assuming exponential growth, what is the annual percent rate? and doubling time?
The answers are 5.13% and 14 years. I do not know how to get this though :( Please help!
Please help! test tomorrow
thanks:)
Let's use the equation
number = a e^kt, where a is the initial value, k is the rate of growth and t is the number of years
so let 1960 correspond to a time of t = 0
then 1970 ----> t = 10
also a = 291000
480000 = 291000 e^10k
1.64948 = e^10k
take ln of both sides
ln 1.64948 = ln e^10k = 10k
k = .50046/10 = .050046 = 5.005 %
(no idea how they got their answer, it is not correct)
check:
291000 e^(10(.050046)) = 479999
pretty close to 480000
their answer:
291000 e^(10(.0513)) = 486056 , too big
for doubling time:
2 = 1(e^.050046t)
ln 2 = .050046t
t = ln 2/.050046 = 13.8 years.
To find the annual percent rate and doubling time, we can use the formula for exponential growth: P(t) = P0 * (1 + r)^t, where P(t) is the population at time t, P0 is the initial population, r is the annual growth rate, and t is the time in years.
Given that the population in 1960 was 291,000, we can plug the values into the formula as follows:
P(0) = 291,000
P(10) = 480,000
Using these values, we can set up the equation:
480,000 = 291,000 * (1 + r)^10
Now, let's solve for r:
(1 + r)^10 = 480,000 / 291,000
Taking the 10th root of both sides:
1 + r = (480,000 / 291,000)^(1/10)
Simplifying the right side:
1 + r = (1.6509)^(1/10)
1 + r ≈ 1.0513 (rounded to four decimal places)
Subtracting 1 from both sides:
r ≈ 0.0513 (rounded to four decimal places)
To find the annual percent rate, we need to convert this to a percentage:
Annual percent rate ≈ 0.0513 * 100 ≈ 5.13%
Now, to find the doubling time, we can use the formula:
t = log(2) / log(1 + r)
Substituting the value of r we found:
t = log(2) / log(1 + 0.0513)
Calculating this expression:
t ≈ 14 years (rounded to the nearest whole number)
Therefore, the annual percent rate is approximately 5.13%, and the doubling time is approximately 14 years.