Answer each of the following on YOUR OWN PAPER. Show all work. 1. Bacteria grows in a petri dish at a hourly rate of 6.93%. If there are 100 bacterium initially, then how long will it take for it to (a) double (b) triple (c) reach 375 2. The population of Smallville in 1890 is 6250. If the population grows at a rate of 3.25% yearly, then in what year would the population of Smallville be 1 million people? 3. Jakielite is a radioactive substance whose half life is 115 years. (a) What is the annual rate of decay? (b) How much Jakielite will there be left from 100g of it after 400 years? (c) How long will it take for the 100g of Jakielite to decay to 1 g? 4.
Carbon-14 is used to determine the age of objects. It has a half-life of 5700 years. A vase was found in China was tested and it had only 15% of its Carbon-14 remaining. If it was found this year, then in what year was it made?
5. A rumor spreads through a school. The number of people who know the rumor is modeled by the following formula: where P is the number of people knowing the rumor and d is the number of days since the rumor began. (a) (b) (c) (d) 6.
How many people know the rumor at the beginning? How many people know the rumor after 1 week? If there are 1200 students in the school, on which day will half the school know the rumor? How many days will it take before the entire school learns the rumor?
Plutonium-210 (abbreviated P-210) is a radioactive element with a half-life of 140 days. A 10g sample was found in a container (a) How much remains after 12 weeks? (b) How long will it take to decay to 1g? (c) The container had a label on it. On the label it said 5kg of Plutonum-210 (approx 11lbs). In what year what the jar filled with P-210 if it was discovered this year? (d) If another jar contained 10g of P-210 had a label marked 1999, then how much P-210 was in the jar to start?
7. An experiment of the growth rate of fruit flies began. Each day the number of fruit flies were counted and recorded. One day, someone realized that the first page of recordings was missing. The second sheet had day 12 with 200 fruit flies and day 18 with 900 fruit flies. How many fruit flies were there originally? 8. Cars depreciate in value the moment you drive it off the lot. The rate of depreciation is 0. 5% per week. If a one-year old GMC Envoy was appraised at $41,251, then what is the cost of the Envoy when it was bought?
Exponential Growth/Decay - FishTR.COM
Worksheet: Exponential Growth/Decay - B
Answer each of the following on YOUR OWN PAPE...
EXAMPLE: POPULATION GROWTH. âThe population of Silver Run in the year 1890 was 6250. Assume the population increased at a rate of. 2.75% per year. âa) Estimate the population in 1915 and 1940. âb)Approximately when did the population reach 50,0
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Tell whether the population model is an exponential growth function or .... Use the 1900Ã¢â¬â2000 data in Table 3.9 and ex...
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Your Turn: The population of âSmallvilleâ in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. 5. When did the population reach 50,000 ? Your turn: Year 1990 2009 Population USA 248,709,873 307,006,550 6. Assumi
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P(t ) P0 (1 r ) t P0 10 P(t ) 10(1.05) t or 'r' = 0.05 f ( x) 10(1.05) x Your Turn: The population of âSmallvilleâ in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. â¢ What is the population in 1915 ? Modeli
6.28% each year. At that rate, when will the population be. 1 million? 31. Exponential Growth The population of Smallville in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. (a) Estimate the population in 1915 and
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31. Exponential Growth The population of Smallville in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. (a) Estimate the population in 1915 and 1940. (b) Predict when the population reached 50,000. 32. Exponential
Aug 8, 2009 - This means that in the year 1891 the population would have been: 6250 * (1.0275) = 6,421.875 (I'll keep the part of a person for now). It also means that in 1892 the population would be: (6,421.875) * (1.0275) which is equal to (6,250)
Aug 26, 2013 - Exponential Growth The population of Smallville in the year. 1890 was 6250. Assume the population increased at a rate of. 2.75% per year. (a) Estimate the population in 1915 and 1940. 12,315; 24,265. (b) Predict when the population rea
31. exponential growth The population of Smallville in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. (a) Estimate the population in 1915 and 1940. (b) Predict when the population reached 50,000. 32. exponential
This analytical approach involves matrix exponentiation and is referred to here as the Matrix Exponential Time Advancement (META) method. Two algorithms are presented for the META method, one for symmetric and the other for non-symmetric exponent mat
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Answer to Exponential Growth The population of Smallville in the year 1890 was 6250. Assume the population increased at a rate of....
Exponential Growth The population of Smallville in the year 1890 was 6250. Assume the popu to problem 31 ch. 3.2 of Precalculus: Graphical, Numerical, Algebraic 8th Edition.