Question #264526

  1. In 8:00 AM, the population of a bacteria is 1000. At 10:30 AM, the number of bacteria triples. At what time will the population become 100 times the initial population of bacteria? What will be the population at 2:00 PM?  Ans. 6:29 PM; 13,967
  2.  If the present population of a certain country is 40 million and in 10 years, the population is 50 million, what will be its population 20 years from now? Ans. 62.5 million

Expert's answer

1) Bacteria at initial state=1000,8:00am

Bacteria at final state=100000

Tripled after 2.5hrs

The exponential form is;

100000=10003(t150)100000=1000*3^{(\frac{t}{150})}

100=3(t150)100=3^{(\frac{t}{150})}

ln 100=t150ln 3ln\ 100=\frac{t}{150}ln\ 3

t150=(ln 100ln 3)\frac{t}{150}=(\frac{ln\ 100}{ln\ 3})

t=(ln 100ln 3)150{t}=(\frac{ln\ 100}{ln\ 3})*150

t=628.77mins629minst=628.77mins \approx629mins

t=10hrs 29minst=10hrs \ 29 mins

Time =8.00 am+10hr 29 mins

=6.29pm


Population at 2.00pm

P=10003(t150)P=1000*3^{(\frac{t}{150})}

t=2.00pm -8.00am=6hrs=360mins

P=10003(360150)P=1000*3^{(\frac{360}{150})}

=41899.83


2)dNdt=kn\frac{dN}{dt}=kn

dNN=kdt\int \frac{dN}{N}=\int kdt

ln N=kt+Cln\ N=kt+C

Given for t=0, N=40million

    ln(40)=C\implies ln(40)=C

And for t=10, N=50million

ln(50)=k*10+ln(40)

k=110(ln 50ln 40)=110ln (1.25)k=\frac{1}{10}(ln\ 50-ln\ 40)=\frac{1}{10}ln\ (1.25)

Hence, ln (N)=110ln (1.25)t+ln (40)ln\ (N)=\frac{1}{10}ln\ (1.25)t+ln\ (40)

For t=20

ln (N)=110ln (1.25)20+ln (40)ln\ (N)=\frac{1}{10}ln\ (1.25)*20+ln\ (40)

=ln (1.25)2+ln (40)=ln (401.252)ln\ (1.25)^2+ln\ (40)=ln\ (40*1.25^2)

N=40*1.252

=62.5million


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS