1. Remember to download from D2L and print a copy of the Final Group Project. DateSection October 304.4 November 4 4.4 Continued November 6Review for test 3 November 11Test 3 November 135.2 November 185.4 November 205.5 November 25No Class (Fall Break) November 27No Class (Fall Break) December 25.6 December 4Last Day of Class Final Group Project due Dec 9 – 15 Final exam week – specific dates and times will be announced in future

2. The table below shows the average weekly amount of electricity used in five Michigan cities in 2011. Population of city in thousands Amount of electricity in 1000’s of kilowatts 73.5365 85.4395 97.2442 108.0503 120.9584 Additional Practice 1. Make a scatter plot of the data. Based on the graph, would an exponential model be appropriate? 2.Write an equation for an exponential model for the weekly amount of electricity used versus the population. Round coefficients to three decimal places (nearest thousandth). 3. Graph the exponential model. 4. What is the growth rate in weekly electricity use per 1,000 people? 5. Use your model to estimate the weekly amount of electricity used in 2011 in a Michigan city with a population of 90,000 people. 6. The weekly amount of electricity used by the people of another Michigan city in 2011 was 660 thousand kilowatts. Use your graph to estimate the population of the city that year. 7. Solve question 6 algebraically.

3. The table below shows the average weekly amount of electricity used in five Michigan cities in 2011. y = 169.973(1.010 x ) where x = population and y = amount of electricity used. Population of city in thousands Amount of electricity in 1000’s of kilowatts 73.5365 85.4395 97.2442 108.0503 120.9584 Additional Practice 1. Make a scatter plot of the data. Based on the graph, would an exponential model be appropriate? 2.Write an equation for an exponential model for the weekly amount of electricity used versus the population. Round coefficients to three decimal places (nearest thousandth). 3. Graph the exponential model. 4. What is the growth rate in weekly electricity use per 1,000 people? 5. Use your model to estimate the weekly amount of electricity used in 2011 in a Michigan city with a population of 90,000 people. 6. The weekly amount of electricity used by the people of another Michigan city in 2011 was 660 thousand kilowatts. Use your graph to estimate the population of the city that year. 7. Solve question 6 algebraically. 1.0% Approximately 416.2 thousand kw are used.

4. The table below shows the average weekly amount of electricity used in five Michigan cities in 2011. y = 169.973(1.010 x ) where x = population and y = amount of electricity used. Population of city in thousands Amount of electricity in 1000’s of kilowatts 73.5365 85.4395 97.2442 108.0503 120.9584 Additional Practice 1. Make a scatter plot of the data. Based on the graph, would an exponential model be appropriate? 2.Write an equation for an exponential model for the weekly amount of electricity used versus the population. Round coefficients to three decimal places (nearest thousandth). 3. Graph the exponential model. 4. What is the growth rate in weekly electricity use per 1,000 people? 5. Use your model to estimate the weekly amount of electricity used in 2011 in a Michigan city with a population of 90,000 people. 6. The weekly amount of electricity used by the people of another Michigan city in 2011 was 660 thousand kilowatts. Use your graph to estimate the population of the city that year. 7. Solve question 6 algebraically. The population is about 136,337.

5. y = 169.973(1.01 x ) 660 = 169.973(1.01 x ) 169.973 3.8830 = 1.01 x log(3.8830) = log(1.01 x ) log(3.8830) = x log(1.01) Therefore, the population of the Michigan city that generated 660 tons of recycled paper was about 136,338 people. x = log(3.8830) log(1.01) = 136.338 The population is about 136,337.

6. Answers to Common Logarithm Homework Handout Find the value of each of the following. Round your answer to four decimal place accuracy. 1. log 10 450 2.6532 2. log 10 (.75) –.1249 3. log 29.8 1.4742 Write each of the following as an exponential equation and find the value of x to the nearest hundredth. 4.log 10 x = 35. log 10 30 = x6. log x = -2 10 3 = x x = 1000 10 x = 30 x = 1.48 10 -2 = x x =.01

7. Solve each equation for x. Answer to the nearest thousandth. Show work. 7. 8. 9. x = log 59 x = 1.771 log x (log 8.3) = log 2030 x = = 3.599 log 2030 log 8.3 119.0476 = 3.57 x log 119.0476 = log (3.57 x ) log 119.0476 = x (log 3.57) x = = 3.756 log 119.0476 log 3.57 4.2

8. The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit). D = Distance in thousands of miles 05101520 T = Temperature in thousands of degrees 12.511.710.9510.39.7 Use exponential regression to find an exponential model for the data. 10. What is the rate of decrease per thousand miles in the average temperature of the gas emitted by the star? 11. What is the average temperature of the gas emitted by the star when the gas is 12,500 miles from the star? 12. When the average temperature of the gas is 7,000 degrees, how many miles from the star has it traveled? Give a graphic solution. 13. Verify your answer to question 12 by giving an algebraic solution.

9. The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit). D = Distance in thousands of miles 05101520 T = Temperature in thousands of degrees 12.511.710.9510.39.7 Use exponential regression to find an exponential model for the data. The exponential regression model is T = 12.472(.987 D )

10. The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit). D = Distance in thousands of miles 05101520 T = Temperature in thousands of degrees 12.511.710.9510.39.7 Use exponential regression to find an exponential model for the data. 10. What is the rate of decrease per thousand miles in the average temperature of the gas emitted by the star? T = 12.472(.987 D ) 1 –.987 =.013 The rate of decrease is 1.3%

11. The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit). D = Distance in thousands of miles 05101520 T = Temperature in thousands of degrees 12.511.710.9510.39.7 11. What is the average temperature of the gas emitted by the star when the gas is 12,500 miles from the star? The average temperature of gas 12,500 miles from the star is 10,590 degrees.

12. The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit). D = Distance in thousands of miles 05101520 T = Temperature in thousands of degrees 12.511.710.9510.39.7 12. When the average temperature of the gas is 7,000 degrees, how many miles from the star has it traveled? Give a graphic solution. The temperature is 7,000 degrees at a distance of 44,140 miles from the star.

13. 13. Verify your answer to question 12 by giving an algebraic solution. T = 12.472(.987 D ) 7 = 12.472(.987 D ).5613 = (.987 D ) log(.5613) = log(.987 D ) log(.5613) = Dlog(.987) D = = 44.13 log(.5613) log(.987) The temperature is 7,000 degrees at a distance of 44,130 miles from the star.

14. 131.77 mg remain after 2 hours 1-.989=.011 or 1.1% y = 496.888(.989 x ) where x = time in minutes and y = # of mg remaining in blood Time (in minutes) 102030405060 Amount of active ingredient left in bloodstream (in mgs) 445402360320290260 14. Write an equation for an exponential model for the amount of active ingredient left in the bloodstream. Round all coefficients to three decimal places (nearest thousandth). 15. What is the rate of decrease of the active ingredient in the bloodstream? 16. How much of the 500 mg of the active ingredient is left in the bloodstream after 2 hrs? 17. After how many minutes will exactly 100 mg of the active ingredient be left in the bloodstream? Round your answer to the nearest hundredth of a minute. 18. Solve question 17 algebraically. 144.94 min Squibb Pharmaceuticals is testing a new pain reliever that comes in pill form, each pill containing 500 mg of the active ingredient. In order to determine how long it takes for the active ingredient to leave the blood stream and be absorbed into the body, the company conducts clinical trials. In the trials, blood samples are taken from subjects who have taken one of the pills. The table below shows the average amount of the active ingredient that remains in the bloodstream after each 10 minute interval. 100

15. 100 = 496.888(.989 x ).2013 =.989 x log(.2013) = log(.989 x ) Time (in minutes) 102030405060 Amount of active ingredient left in bloodstream (in mgs) 445402360320290260 x = 144.94 496.888 x = = 144.92 log(.2013) = x (log.989)

16. Natural Logarithms use a different base, the number e. Logarithms using base e are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus, Last time, we examined common logarithms (logarithms using base 10). Use your calculator to find the value of e 3.69 to the nearest whole number. e 3.69  40 Express this equation using logarithms ln 40  3.69 ln xmeans log e x. log Use the ln button on the calculator to verify this last answer.

17. 100 = 496.888(.989 x ).2013 =.989 x log(.2013) = log(.989 x ) Time (in minutes) 102030405060 Amount of active ingredient left in bloodstream (in mgs) 445402360320290260 x = 144.94 496.888 x = = 144.92 log(.2013) = x (log.989)

18. 100 = 496.888(.989 x ).2013 =.989 x log(.2013) = log(.989 x ) Time (in minutes) 102030405060 Amount of active ingredient left in bloodstream (in mgs) 445402360320290260 x = 144.94 496.888 x = = 144.92 log(.2013) = x (log.989) ln ln ln ln ln ln

19. YearCensus Population 18501,610 18604,385 18705,728 188011,183 189050,395 1900102,479 1910319,198 1920576,673 19301,238,048 Source: http://en.wikipedia.org/wiki/Los_Angeles Shown is a table of the population of the City of Los Angeles as reported in the census from 1850 to 1930. 1.Make a scatter plot of the population using years since 1850. 2.Would an exponential model be appropriate for the data? 3. Find an exponential model for the data.

20. r is called the correlation coefficient. r 2 is called the coefficient of determination. Both r and r 2 are measures of how good a fit a model is to a set of data. r will always be between -1 and 1, and r 2 will always be between 0 and 1. The closer they are to -1 or 1, the better the fit.

21. Year20022003200420052006 Surface Area in sq. miles 26,57929,10431,86834,89638,211

22. Homework: Download, print, and complete the Practice Test for Chapter 4.