Answer:
a) 13032
b)6.4 days
Explanation:
The formula for the population growth is given by
![A=Pe^(rt)](https://img.qammunity.org/2020/formulas/mathematics/high-school/2voktn38sksrm6c2zcfbnfou5ed95ywtlr.png)
Here, A = final population, P = initial population, r = growth rate and t = time
From the given directions, we have
A = 1900, P = 1000, t = 1, r = ?
Substituting these values in the above formula to find r
![1900=1000e^(r\cdot1)\\\\1900=1000e^r](https://img.qammunity.org/2020/formulas/mathematics/high-school/5jqopwhsltl33i1g0sey55v64d8ozjtwr2.png)
Divide both sides by 1000
![(1900)/(1000)=e^r\\\\e^r=(19)/(10)](https://img.qammunity.org/2020/formulas/mathematics/high-school/wrvl2e8ld6bdu756yo4bl045gc7hvbj54z.png)
Take natural log both sides
![\ln(e^r)=\ln((19)/(10))\\\\r=0.64185](https://img.qammunity.org/2020/formulas/mathematics/high-school/dz6goa5lt7ijy5i668z4r234tfvpdcr9q4.png)
Therefore, the population model is given by
![A=1000e^(0.64185t)](https://img.qammunity.org/2020/formulas/mathematics/high-school/q3atkcaubmxzd0lg5khlbl3m3o08jtgnlg.png)
(a) The size of the colony after 4 days is given by
![A=1000e^(0.64185\cdot4)\\\\A=13032](https://img.qammunity.org/2020/formulas/mathematics/high-school/3pz0vhy4g2vbkhzqfu2ejl2ifb2hx9b1sy.png)
(B) The time for the number of mosquitoes to be 60,000 is
![60000=1000e^(0.64185t)](https://img.qammunity.org/2020/formulas/mathematics/high-school/948jeb3lhxahcte9nyd7p6c73ck6b8t3vh.png)
Divide both sides by 1000
![(60000)/(1000)=e^(0.64185t)\\\\e^(0.64185t)=60](https://img.qammunity.org/2020/formulas/mathematics/high-school/bozalzr6udq4xu5l3fa5ngeuogvpb2szrw.png)
Take natural log both sides
![\ln(e^(0.64185t))=\ln60\\\\0.64185t=\ln60\\\\t=(\ln60)/(0.64185)\\\\t=6.4](https://img.qammunity.org/2020/formulas/mathematics/high-school/cl2kbfaum2063jnpcm7pzwm13b4hoygehi.png)
Hence, it will take 6.4 days to the population of mosquitoes to be 60000.