Consider the following problem: The population of ants in Chloe's ant farm changes at a rate of r(t)=-15.8*0.9^(t) ants per month (where t is time in months). At time t=0, the ant farm's population is 150 ants. How many ants are in the farm at t=4 ? Which expression can we use to solve the problem? Choose 1 answer: (A) int_(3)^(4)r(t)dt (B) 150+int_(0)^(4)r(t)dt (C) 150+int_(3)^(4)r(t)dt (D) int_(0)^(4)r(t)dt

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Consider the following problem: The population of ants in Chloe's ant farm changes at a rate of  r(t)=-15.8*0.9^(t) ants per month (where  t is time in months). At time  t=0, the ant farm's population is 150 ants. How many ants are in the farm at  t=4 ? Which expression can we use to solve the problem? Choose 1 answer: (A)  int_(3)^(4)r(t)dt (B)  150+int_(0)^(4)r(t)dt (C)  150+int_(3)^(4)r(t)dt (D)  int_(0)^(4)r(t)dt

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Initial Population and Rate Function: To find the number of ants at t=4, we need to account for the initial population and the change in population over time from t=0 to t=4. The change in population is given by the integral of the rate function r(t) over the interval from t=0 to t=4.Calculate Population at t=4: The initial population at t=0 is given as 150 ants. To find the population at t=4, we need to add the initial population to the integral of the rate of change from t=0 to t=4.Calculate Integral of Rate Function: The correct expression to calculate the population at t=4 is the initial population plus the integral of the rate function from t=0 to t=4. This corresponds to option (B): 150 + ∫_(0)^(4)r(t)dt.Antiderivative of Rate Function: Now we will calculate the integral of the rate function r(t) = -15.8*0.9^(t) from t=0 to t=4. To do this, we need to find the antiderivative of r(t) and evaluate it at the bounds t=0 and t=4.Evaluate Antiderivative at t=4 and t=0: The antiderivative of r(t) = -15.8*0.9^(t) can be found using the formula for the integral of an exponential function: ∫a*b^(x)dx = (a/b)*b^(x) + C, where C is the constant of integration. In this case, a = -15.8 and b = 0.9.Calculate Change in Population: The antiderivative of r(t) is (-15.8/ln(0.9))*0.9^(t) + C. We can now evaluate this from t=0 to t=4.Calculate Final Number of Ants at t=4: Evaluating the antiderivative at t=4 gives us (-15.8/ln(0.9))*0.9^(4). Evaluating it at t=0 gives us (-15.8/ln(0.9))*0.9^(0), which simplifies to (-15.8/ln(0.9)).The change in population from t=0 to t=4 is the difference between the antiderivative evaluated at t=4 and t=0: [(-15.8/ln(0.9))*0.9^(4)] - [(-15.8/ln(0.9))].Now we calculate the actual values: [(-15.8/ln(0.9))*0.9^(4)] - [(-15.8/ln(0.9))] = (-15.8/ln(0.9))*(0.9^(4) - 1).Finally, we add the initial population to the change in population to find the total population at t=4: 150 + [(-15.8/ln(0.9))*(0.9^(4) - 1)].Performing the calculation, we get the final number of ants at t=4. Let's calculate the exact value.150 + [(-15.8/ln(0.9))*(0.9^(4) - 1)] = 150 + [(-15.8/-0.10536)*(0.6561 - 1)] = 150 + [150.2096*(-0.3439)] ≈ 150 - 51.68 ≈ 98.32 ants.

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