over time, the number of organisms in a population increases exponentially. the table below shows the…

Explanation:

Step1: Find the growth factor

The general form of an exponential function is (n = ab^{y}). When (y = 1), (n=55), so (n = ab^{1}=55). When (y = 2), (n = 60), so (n=ab^{2}=60). Then (\frac{ab^{2}}{ab}=b=\frac{60}{55}=\frac{12}{11}\approx1.09). And (a=\frac{55}{b}=\frac{55\times11}{12}=\frac{605}{12}\approx50.42). So the function is (n=\frac{605}{12}\times(\frac{12}{11})^{y}).

Step2: Solve for (y) when (n = 600)

Set up the equation (600=\frac{605}{12}\times(\frac{12}{11})^{y}). First, simplify the equation: (600\times\frac{12}{605}=(\frac{12}{11})^{y}). (\frac{7200}{605}=(\frac{12}{11})^{y}), (\frac{1440}{121}=(\frac{12}{11})^{y}). Since (\frac{12}{11}\times\frac{12}{11}\times\cdots\times\frac{12}{11}) ( (y) times). And ((\frac{12}{11})^{y}=\frac{1440}{121}), ((\frac{12}{11})^{y}=\frac{12^{2}\times10}{11^{2}}).

Another way: Use the formula (n = n_{0}(1 + r)^{y}), where (n_{0}) is the initial - value, (r) is the growth rate. (n_{0}\approx50.42), (r=\frac{12}{11}-1=\frac{1}{11}). The equation is (n = 50.42\times(1+\frac{1}{11})^{y}).

Take the natural logarithm of both sides of (600 = 50.42\times(\frac{12}{11})^{y}). (\ln(600)=\ln(50.42)+y\ln(\frac{12}{11})). (y=\frac{\ln(600)-\ln(50.42)}{\ln(\frac{12}{11})}). (\ln(600)\approx6.3969), (\ln(50.42)\approx3.921), (\ln(\frac{12}{11})\approx0.087). (y=\frac{6.3969 - 3.921}{0.087}=\frac{2.4759}{0.087}\approx28.46).

If we use the rule of checking the ratio: (y = 1,n = 55); (y = 2,n = 60); (y = 3,n = 67); (y = 4,n = 75) (y = 5,n=75\times\frac{12}{11}\approx81.82); (y = 6,n\approx81.82\times\frac{12}{11}\approx89.26); (y = 7,n\approx89.26\times\frac{12}{11}\approx97.37); (y = 8,n\approx97.37\times\frac{12}{11}\approx106.23); (y = 9,n\approx106.23\times\frac{12}{11}\approx115.9); (y = 10,n\approx115.9\times\frac{12}{11}\approx126.4); (y = 11,n\approx126.4\times\frac{12}{11}\approx137.9); (y = 12,n\approx137.9\times\frac{12}{11}\approx150.4); (y = 24) (since the growth is exponential, and if we consider the pattern of multiplying by (\frac{12}{11}) each year.

Let's check the ratio method more accurately: We know that (n(y)=n(0)\times(\frac{12}{11})^{y}). Assume (n(0)) is approximately (50) (by back - calculating from (n(1) = 55), (n(0)=\frac{55\times11}{12}\approx50.42)). (n(y)=50.42\times(\frac{12}{11})^{y}). If (y = 24): (n(24)=50.42\times(\frac{12}{11})^{24}) ((\frac{12}{11})^{24}=\left(1+\frac{1}{11}\right)^{24}\approx\frac{12^{24}}{11^{24}}) Using the formula (a^{n}/b^{n}=(a / b)^{n}) (\ln(n(24))=\ln(50.42)+24\ln(\frac{12}{11})) (\ln(50.42)\approx3.921), (24\ln(\frac{12}{11})\approx24\times0.087 = 2.088) (\ln(n(24))\approx3.921 + 2.088=6.009), (n(24)=e^{6.009}\approx407) (approximate value from this method has some error due to approximation of (n(0))).

Let's use the compound - interest formula (n=n_{1}\times r^{y - 1}) (where (n_{1}=55), (r = \frac{12}{11})) (600=55\times(\frac{12}{11})^{y - 1}) ((\frac{12}{11})^{y - 1}=\frac{600}{55}=\frac{120}{11}) Take the logarithm: ((y - 1)\ln(\frac{12}{11})=\ln(\frac{120}{11})) (\ln(\frac{120}{11})\approx2.47), (\ln(\frac{12}{11})\approx0.087) (y-1=\frac{\ln(\frac{120}{11})}{\ln(\frac{12}{11})}\approx\frac{2.47}{0.087}\approx28.4), (y\approx29.4)

If we use the rule of checking the options: We know that the function is growing exponentially. If (y = 12): Let's calculate the growth step - by - step. From (y = 4) ((n = 75)) (y=5), (n = 75\times\frac{12}{11}\approx81.8) (y = 6), (n\approx81.8\times\frac{12}{11}\approx89.3) (y = 7), (n\approx89.3\times\frac{12}{11}\approx97.4) (y = 8), (n\approx97.4\times\frac{12}{11}\approx106.2) (y = 9), (n\approx106.2\times\frac{12}{11}\approx115.9) (y = 10), (n\approx115.9\times\frac{12}{11}\approx126.4) (y = 11), (n\approx126.4\times\frac{12}{11}\approx137.9) (y = 12), (n\approx137.9\times\frac{12}{11}\approx150.4)

If (y = 24): We can also use the fact that the ratio of consecutive terms is (r=\frac{12}{11}). Let (n(y)) be the number of organisms at year (y). We know that (n(y)) forms a geometric sequence with (a = 55) (when (y = 1)) and (r=\frac{12}{11}). The formula for the (n)th term of a geometric sequence is (n=a\times r^{y - 1}) (600=55\times(\frac{12}{11})^{y - 1}) ((\frac{12}{11})^{y - 1}=\frac{600}{55}=\frac{120}{11}) Take both sides to the power of (\frac{1}{2}) (for a rough check of the magnitude) ((\frac{12}{11})^{\frac{y - 1}{2}}=\sqrt{\frac{120}{11}}\approx3.3) ((\frac{12}{11})^{12}\approx(1.09)^{12}\approx3.06), ((\frac{12}{11})^{13}\approx3.34) (\frac{y - 1}{2}\approx13), (y-1 = 26), (y = 27) (approximate). But among the given options, if we consider the growth rate and the pattern of the geometric sequence (since the growth is exponential and the options are given), and if we assume a less - accurate but option - based approach: We know that the growth is not linear. If we consider the differences: (55,60(+ 5),67(+7),75(+8)) but it's exponential. If we use the rule of (n(y)=n(1)\times(\frac{12}{11})^{y - 1}) When (y = 24) (n(24)=55\times(\frac{12}{11})^{23}) ((\frac{12}{11})^{10}\approx2.59), ((\frac{12}{11})^{20}\approx6.7), ((\frac{12}{11})^{23}\approx(\frac{12}{11})^{20}\times(\frac{12}{11})^{3}) ((\frac{12}{11})^{3}=\frac{1728}{1331}\approx1.3), so ((\frac{12}{11})^{23}\approx6.7\times1.3 = 8.7) (n(24)=55\times8.7 = 478.5\approx480) (approximate due to approximation of powers). But if we consider the fact that the function (n(y)) is (n(y)=n(0)\times(\frac{12}{11})^{y}) (where (n(0)\approx50)) (n(24)=50\times(\frac{12}{11})^{24}) ((\frac{12}{11})^{24}=\left(1+\frac{1}{11}\right)^{24}) Using the binomial approximation ((1 + x)^{n}\approx1+nx+\frac{n(n - 1)}{2}x^{2}) (for (x) small, but this is an exponential function, better to use the fact that among the options, 24 is the most reasonable as 12 is too small ( (n(12)\approx150)), 61 and 82 are too large (the growth rate (\frac{12}{11}) is not that slow to reach 600 at 61 or 82)

Answer:

24