Question 1
1.67 / 1.67 pts
Solve for t:
20=3e
t2
20=3et2
Correct!
t=2ln(
203
)
t=2ln(203)
t=ln(
103
)
t=ln(103)
t=
ln206
t=ln206
t=ln30
t=ln30
t=
12
ln(30)
t=12ln(30)
20203ln(203)t=3e
t2
=e
t2
=t2=2ln(203)
20=3et2203=et2ln(203)=t2t=2ln(203)
Question 2
1.67 / 1.67 pts
A logistic function is a function of the form
A1+Be
−ct
A1+Be−ct, where
A,B,c
A,B,c are constants.
Given the logistic function
s(t)=
151+Be
−4t
,
s(t)=151+Be−4t,
calculate the constant
B
B if
f(0)=2
f(0)=2.

B=−5
B=−5
Correct!
B=
132
B=132
B=3
B=3
B=
32
B=32
B=2
B=2
f(0)=221+BBB=151+Be
−4⋅0
=151+B=152=152−1=132
f(0)=2=151+B
e−4
⋅
02=151+B1+B=152B=152−1B=132
Question 3
0 / 1.66 pts
Atmospheric pressure decays exponentially as a function of altitude and can be
modeled as
P(a)=P
0
e
−
18
a
P(a)=P0e−18a, where $a$ is the altitude above sea
level, in kilometers, and
P
0
P0 is the atmospheric pressure at sea level,
P
0
P0 equals 1 atmosphere (1 atm). At what altitude is atmospheric pressure
P(t)=1/3
P(t)=1/3 atm (1/3 the pressure at sea level)?
You may need to recall that
−ln(
1x
)=ln(x)
−ln(1x)=ln(x)
Correct Answer

a=8ln(3)≈8.79
a=8ln(3)≈8.79
km
a=3ln(8)≈6.24
a=3ln(8)≈6.24
km
You Answered
a=
13
ln(8)≈0.69
a=13ln(8)≈0.69
km
a=
18
ln(3)≈0.14
a=18ln(3)≈0.14
km
a=
13
ln(24)≈1.06
a=13ln(24)≈1.06
km
We need to solve the following equation for
a
a:
13ln(13)−8ln(13)a=8ln3=e
−
18
a
=−18a=a
13=e−18aln(13)=−18a−8ln(13)=a
a=8ln3
This is very close to the altitude at the summit of Mt. Everest. (8.84 km)
Question 4
1.67 / 1.67 pts
Calculate the doubling time for an investment earning 8% interest, compounded
annually.

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- Winter '13
- Mr. Kerry
- Exponential Function, 2Ln, P0e