Population In Exercises 51-54, the population (in millions)of a country in 2011 and the expected continuous annual rateof change k of the population are given. (Source: U.S. CensusBureau, International Data Base)(a) Find the exponential growth modelP = Cekt%3Dfor the population by letting t = 0 correspond to 2010.(b) Use the model to predict the population of the country in2020.(c) Discuss the relationship between the sign of k and thechange in population for the country.Country2011 Populationk51. Latvia2.2-0.00652. Egypt82.10.02053. Uganda34.60.03654. Hungary10.0-0.002 55. Modeling Data One hundred bacteria are started in aculture and the number N of bacteria is counted each hour for5 hours. The results are shown in the table, where t is the timein hours.1234100126151198243297(a) Use the regression capabilities of a graphing utility to findan exponential model for the data.(b) Use the model to estimate the time required for thepopulation to quadruple in size.56. Bacteria Growth The number of bacteria in a culture isincreasing according to the law of exponential growth. Thereare 125 bacteria in the culture after 2 hours and 350 bacteriaafter 4 hours.(a) Find the initial population.(b) Write an exponential growth model for the bacteriapopulation. Let i represent time in hours.(c) Use the model to determine the number of bacteria after8 hours.(d) After how many hours will the bacteria count be 25,000?

51. The 2011 population of Latvia is 2.2. The value of k is -0.006.

Population In Exercises 51-54, the population (in millions) of a country in 2011 and the expected continuous annual rate of change k of the population are given. (Source: U.S. Census Bureau, International Data Base) (a) Find the exponential growth model P = Cekt for the population by letting t = 0 correspond to 2010. (b) Use the model to predict the population of the country in 2020. (c) Discuss the relationship between the sign of k and the change in population for the country. Country 2011 Population k 51. Latvia 2.2 -0.006