Answer:
y = a[1 + 0.26]^t
Rate = 0.26
This is a growth rate
Explanation:
If we have the following equation modelling a growth or decay.
[tex]y=a[1+r]^t[/tex]then if r < 0, then the above equation models decay. If a > 0, then the above equation models growth.
We can rewrite our equation y = a(4)¹/6 as
[tex]a(4^{1/6})^t[/tex]Now
[tex]4^{1/6}=(2^2)^{1/6}=2^{2/6}=2^{1/3}=\sqrt[3]{2}\approx1.26[/tex]Therefore,
[tex]y=a(\sqrt[6]{4})^t\approx a[1.26]^t[/tex]which can also be written as
[tex]\boxed{y=a\left[1+0.26\right]^t.}[/tex]which is our answer!
As can be seen from the above equation, the rate is r = 0.26.
Now r = 0.26 > 0, which means that the equation above models a growth.
Hence, to summerise
y = a[1 + 0.26]^t
Rate = 0.26
This is a growth rate
