Give the starting value a, the growth factor b, and the growth rate r if Q = abt = a(1 + r)t . Write r as a percent. Q=79(1.002)^t
\(Q = abt = a(1 + r)^t \\Q=79(1.002)^t \\Q=79(1+0.002)^t \\ \\r=0.002\times 100\%=0.2\%\)
Answer: r as a percent will be equal to 0.2%.
Step-by-step explanation:
It is given that Q = \(ab^{t}\) = \(a(1 + r)^{t}\) ........... (1)
Also, Q = \(79(1.002)^{t}\) .......... (2)
Thus, equating both equations (1) and (2), we will get the equation as follows.
\(a(1 + r)^{t}\) = \(79(1.002)^{t}\) ........ (3)
Therefore, from equation (3) we can conclude that the value of a = 79 and (1 + r) = 1.002
Since, 1 + r = 1.002
thus, r = 1.002 - 1
r = 0.002
Now, convert 0.002 into fraction form as follows.
\(\frac{2}{1000}\)
Therefore, r as a percent will be as follows.
\(\frac{2}{1000}\) × 100%
Hence, r as a percent will be equal to 0.2%.