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A quantity P is an exponential function of time t, such that P=110 when t=3 and P=80 when t=1. Use the given information about P=P0eH to: (a) Find values for the parameters k and P0. Round your answers to three decimal places. k= P0= (b) Give the initial quantity and the continuous percent rate of growthor decay. Round your answer for the initial quantity to three decimal places. Enter your answer for the growth or decay rate as a positive value. rounded to one decimal place. The initial quantity is and the quantity is at a continuous rate of \% per unit time.
A quantity P is an exponential function of time t, such that P=110 when t=3 and P=80 when t=1. Use the given information about P=P0eH to: (a) Find values for the parameters k and P0. Round your answers to three decimal places. k= P0= (b) Give the initial quantity and the continuous percent rate of growthor decay. Round your answer for the initial quantity to three decimal places. Enter your answer for the growth or decay rate as a positive value. rounded to one decimal place. The initial quantity is and the quantity is at a continuous rate of \% per unit time.
Final answer:To find the values of k and P0, set up two equations using the given information and solve simultaneously. Then, use the values of k and P0 to find the initial quantity and the continuous percent rate of growth or decay.Explanation:To find the values for the parameters k and P0, we can use the given information about P=P0e^H and the given values of P at different values of t. We can set up two equations using these values and solve for k and P0. For part (a), we can substitute the given values and solve simultaneously to find k and P0 to three decimal places. For part (b), once we have the values for k and P0, we can use them to find the initial quantity and the continuous percent rate of growth or decay.Let's start with part (a):For t=3 and P=110, we have 110=P0e^(k*3).For t=1 and P=80, we have 80=P0e^(k*1).We have a system of two equations:110=P0e^(3k)80=P0e^(k)By dividing the two equations, we can eliminate P0:(110/80) = (P0e^(3k))/(P0e^(k))Simplifying, we get:11/8 = e^(2k)Taking the natural logarithm (ln) of both sides, we get:ln(11/8) = 2kSolving for k, we find:k = (ln(11/8))/2Now, substituting the value of k, we can find P0 by rearranging one of the equations:80 = P0e^((ln(11/8)/2)*1)Solving for P0, we get:P0 = 80/e^(ln(11/8)/2)Both k and P0 are rounded to three decimal places.For part (b):Once we have the values for k and P0, we can use them to find the initial quantity and the continuous percent rate of growth or decay.The initial quantity is P0, rounded to three decimal places.The continuous percent rate of growth or decay, denoted by H, can be found using the formula:H = k * 100Since k represents the growth or decay rate as a decimal, multiplying it by 100 gives us the rate as a percentage, rounded to one decimal place.Learn more about Exponential functions here:brainly.com/question/35259468#SPJ11...