N(t)=2000e^(1.1t)
the continuous growth rate is 110%
effective growth rate?__%
Q(t)=65e^(0.09t)
continuous growth rate is 9%
effective growth rate?__%
G(t)=650e^(0.055t)
continuous growth rate is 5.5%
effective growth rate?__%
round to 2 decimal places when necessary for effective growth rate...
is there a formula for effective growth rate
I will do the last one, you do the other two
G(t)=650e^(0.055t) , tells me the continuous rate is 5.5%
with a different base:
G(t) = 650(1+r)t
pick t = 5, (we can choose anything)
in the first:
g(5) = 650 e^(.055(5))
= 650 e^(.275) = 855.7449
in the 2nd:
855.7449 = 650(1+r)^5
1.316530675 = (1+r)^5
take 5th root of both sides
1.05654 = 1+r
r = .05654 -----> 5.654%
Yes, there is a formula for calculating the effective growth rate. The formula is:
Effective Growth Rate = (e^(Continuous Growth Rate) - 1) * 100
Now let's calculate the effective growth rate for each given function:
1) For N(t) = 2000e^(1.1t):
Continuous growth rate = 110%
Effective Growth Rate = (e^(1.1) - 1) * 100
Using a calculator, we find that e^(1.1) ≈ 3.0042
Effective Growth Rate ≈ (3.0042 - 1) * 100 ≈ 200.42%
So, the effective growth rate for N(t) is approximately 200.42%.
2) For Q(t) = 65e^(0.09t):
Continuous growth rate = 9%
Effective Growth Rate = (e^(0.09) - 1) * 100
Using a calculator, we find that e^(0.09) ≈ 1.093
Effective Growth Rate ≈ (1.093 - 1) * 100 ≈ 9.3%
Therefore, the effective growth rate for Q(t) is approximately 9.3%.
3) For G(t) = 650e^(0.055t):
Continuous growth rate = 5.5%
Effective Growth Rate = (e^(0.055) - 1) * 100
Using a calculator, we find that e^(0.055) ≈ 1.056
Effective Growth Rate ≈ (1.056 - 1) * 100 ≈ 5.6%
Therefore, the effective growth rate for G(t) is approximately 5.6%.
Remember to round your answer to 2 decimal places, as requested.