Question
1.A population of bacteria follows the continuous exponential growth modelP(t)=P0 e kt, where t is in days. The relative(daily) growth rate is 4.4%. The current
1.A population of bacteria follows the continuous exponential growth modelP(t)=P0ekt, where t is in days. The relative(daily) growth rate is 4.4%. The current population is 976.
A) Find the growth model.(the function that represents the population after tdays).
P(t) =
nothing
B) Find the population exactly 2 weeks from now.Round to the nearest bacterium.
The population in 2 weeks will be
nothing
.
C) Find the rate of change in the population exactly 2 weeks from now.
Round to the nearest unit.
The population will be increasing by about
nothing
bacteria per day exactly 2 weeks from now.
D) When will the popualtion reach 6928? ROUND TO 2 DECIMAL PLACES.
The population will reach 6928 about
nothing
days from now.
2.At the beginning of the year1995, the population of Townsville was 3330. By the beginning of
the year 2012, the population had reached 4569. Assume that the population is growingexponentially, answer the following.
A) Estimate the population at the beginning of the year 2019.
ROUND TO THE NEAREST PERSON.
The population at the beginning of 2019 will be about
nothing
.
B) How long(from the beginning of1995) will it take for the population to reach9000?
ROUND TO 2 DECIMAL PLACES.
The population will reach 9000 about
nothing
years after the beginning of 1995.
C) In what yearwill/did the population reach9000?
The population will(or did) hit 9000 in the year
nothing
3.A virus is spreading across an animal shelter. The percentage of animals infected after t days
is given byV(t)=1001+99e0.182t.
A) What percentage of animals will be infected after 10 days?
ROUND YOUR ANSWER TO 2 DECIMAL PLACES. (i.e. 12.34%)
About
nothing
% of the animals will be infected after 10 days.
B) How long will it take until exactly90% of the animals areinfected?
ROUND YOUR ANSWER TO 2 DECIMAL PLACES.
90% of the animals will be infected after about
nothing
days.
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