Math 110 Lesson 2 Quiz Attempt 1.docx - Question 1 1.67 1.67 pts Solve for t 20=3et220=3et2 Correct t=2ln(203)t=2ln�(203 t=ln(103)t=ln�(103

Math 110 Lesson 2 Quiz Attempt 1.docx - Question 1 1.67...

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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.
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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
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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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