1] Given that the value of x has been modeled by f(x)=12500(0.87)^x, then: the rate of change between years 1 and 5 will be: rate of change is given by: [f(b)-f(a)]/(b-a) thus: f(1)=12500(0.87)^1=10875 f(5)=12500(0.87)^5=6230.3 rate of change will be: (6230.3-10875)/(5-1) =-1161.2 rate of change in years 11 to 15 will be: f(11)=12500(0.87)^11=2701.6 f(15)=12500(0.87)^15=1,547.74 thus the rate of change will be: (1547.74-2701.6)/(15-11) =-288 dividing the two rates of change we get: -288/-1161.2 -=1/4 comparing the two rate of change we conclude that: The average rate of change between years 11 and 15 is about 1/4 the rate between years 1 and 5. The answer is D] 2] Given that the population of beavers decreases exponentially at the rate of 7.5% per year, the monthly rate will be: monthly rate=(n/12) where n is the number of months =7.5/12 =0.625 This is approximately equal to 0.65%. The correct answer is A. 0.65%
