A bacteria culture started with 5500 bacteria and increased by 35% in the first
ID: 3427308 • Letter: A
Question
A bacteria culture started with 5500 bacteria and increased by 35% in the first 2 hours. (a) Find an expression for the number of bacteria t hours after it was first observed. Q(t) = 5500e( ln(1.35) 2)t
b) Find the number of bacteria that will be present after 8 hours. (Round your answer to the nearest whole number.)
(c) When will the population reach 35,000? (Round your answer to one decimal place.) hr (d) How long does it take the population to triple in size? (Round your answer to one decimal place.) hr
#2
A bacteria culture started with 5500 bacteria and increased by 35% in the first 2 hours.
(a) Find an expression for the number of bacteria t hours after it was first observed.
(b) Find the number of bacteria that will be present after 8 hours. (Round your answer to the nearest whole number.)
(c) When will the population reach 35,000? (Round your answer to one decimal place.)
hr
(d) How long does it take the population to triple in size? (Round your answer to one decimal place.)
hr
Explanation / Answer
Initial population of bacteria=5500
After 2 hours it increased by 35%
And 35% of 5500= 1925
Therefore after 2 hours,polulation= 7425
We have to use the following formula here
N(t)= N(0) ekt
7425=5500 e2t
7425/5500= e2t
1.35= e2tTaking ln on both sides
t=ln1.35/2
Therefore the required function is N(t) = 5500e(ln(1.35)/2)t
b. t= 8 hrs
N(8)= 5500e(ln(1.35)/2)8 = 5500 e4ln(1.35) = 18268
c. 35000= 5500 e(ln(1.35)/2)t
35000/5500= e(ln(1.35)/2)t
Taking ln on both sides
ln(35000/5500)=t* ln(1.35)/2
2 ln(35000/5500)= t ln 1.35
3.7 = t* ln 1.35
t= 12.3 hrs
d. Initial population =5500
Triple of this population = 3*5500= 16500
16500=5500e(ln(1.35)/2)t
3= e(ln(1.35)/2)t
Taking ln on both sides
ln 3= t* ln(1.35)/2
2 ln 3/ln1.35 = t
t=7.3 hrs
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.