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[answered] Molly Leonard MA 131, section 601, Fall 2016 Instructor: Br


DUE TONIGHT @ 10:00pm (1 hr and 15 minutes)

I asked this same question earlier, but the "tutor" got ALL the answers wrong, except for Exponential and Logarithmic Functions Question #1. He did not complete?Exponential and Logarithmic Functions Question #3. He got Applications: Exponential, Logarithmic Functions Question #4 WRONG with answer 40588.9. And did not complete any of the two questions in Further Integration - which is the one I really need help on. PLEASE HELP!!!! AND ONLY DO IF YOU ARE CORRECT!!! THANKS!!




Molly Leonard

 

MA 131, section 601, Fall 2016

 

Instructor: Brenda Burns WebAssign Applications: Exponential, Logarithmic Functions (Homework)

 

Current Score : 11 / 14 Due : Tuesday, December 13 2016 11:13 PM EST 1. 3/3 points | Previous Answers The population of a town is growing according to the differential equation The growth constant, k, is equal to 0.13 year -1.

 

The size of the population at the start of the year 2000 was 15 thousand.

 

Since the population is growing exponentially, the population in year t is given by Here, y is measured in thousands and t is measured in years since 2000.

 

a. What is the population of the town at the start of the year 2005?

 

(Enter your answer correct to one decimal place.)

 

Population: 28.7 thousand. b. How many years does it take for the population to double?

 

(Enter your answer correct to two decimal places.)

 

Doubling time : 5.33 years. c. What will be the size of the population after three doubling times have passed? (That is, after

 

three times the correct answer to part b have passed?)

 

(Enter your answer correct to one decimal place.)

 

Population after three doubling times : 120 thousand. 2. 5/5 points | Previous Answers The number of bacteria in a flask grows according to the differential equation In this question, time is measured in hours and the number of bacteria, y, is measured in millions.

 

The number of bacteria at time t = 0 is 5 million.

 

a. Enter a formula for the number of bacteria at time t

 

y = 5e^(0.06t) b. What is the value of the growth constant?

 

Growth constant : 0.06 per hour. c. How long does it take for the number of bacteria to double?

 

(Enter your answer correct to two decimal places.)

 

Doubling time : 11.55 hours. d. How many million bacteria will be present after 4 hours have passed?

 

(Enter your answer correct to one decimal place.)

 

Number present after 4 hours : 6.4 million. 3. 3/3 points | Previous Answers An isotope of a radioactive element has decay constant equal to 0.05 per year.

 

Initially, there are 20 million atoms of the isotope present.

 

Since the isotope decays exponentially, the number of atoms obeys the following equation In this question, time is measured in years and the number of atoms, P, is measured in millions.

 

a. What is the value of the decay constant?

 

Decay constant : 0.05 per year. b. What is the half-life of the isotope?

 

(Enter your answer correct to two decimal places.)

 

Half-life : 13.86 years. c. How many million atoms will be present after 6 years have passed?

 

(Enter your answer correct to one decimal place.)

 

Number present after 6 years : 14.8 million. 4. 0/3 points | Previous Answers An isotope of a radioactive element has half-life equal to 10 thousand years.

 

Imagine a sample that is so old that most of its radioactive atoms have decayed, leaving just 6 percent

 

of the initial quantity of the isotope remaining.

 

How old is the sample?

 

Give your answer in thousands of years, correct to one decimal place.

 

Age : 40597.6 thousand years.

 


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