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Answer :
The correct option is c. 89, 101, 110, 116, 121, 125, 128, 130, 131, 132, 133.
To solve this problem, we will use the given population model:
[tex]\[ P_{n+1} = P_n + r(P_n)(1 - \frac{P_n}{K}) \][/tex]
We will calculate the population for each of the first ten years using the model, rounding to the nearest whole number before each calculation.
Starting with [tex]\( P_0 = 89 \)[/tex], we apply the model iteratively:
[tex]\[ P_1 = P_0 + 1.75(P_0)(1 - \frac{P_0}{135}) \][/tex]
[tex]\[ P_1 = 89 + 1.75(89)(1 - \frac{89}{135}) \][/tex]
[tex]\[ P_1 \approx 89 + 1.75(89)(1 - 0.6593) \][/tex]
[tex]\[ P_1 \approx 89 + 1.75(89)(0.3407) \][/tex]
[tex]\[ P_1 \approx 89 + 1.75(30.2293) \][/tex]
[tex]\[ P_1 \approx 89 + 52.8958 \][/tex]
[tex]\[ P_1 \approx 141.8958 \][/tex]
Since we need to round to the nearest whole number before each calculation, we take [tex]\( P_1 \approx 142 \)[/tex].
However, we notice that the population cannot exceed the carrying capacity [tex]\( K \)[/tex]. Therefore, we must adjust the population to not exceed [tex]\( K \)[/tex]. In this case, since [tex]\( P_1 \)[/tex] exceeds [tex]\( K = 135 \)[/tex], we set [tex]\( P_1 = 135 \)[/tex].
Now, for each subsequent year, we apply the same formula, ensuring that the population does not exceed [tex]\( K \)[/tex]. Here are the calculations for the first ten years:
[tex]\[ P_2 = 135 + 1.75(135)(1 - \frac{135}{135}) \][/tex]
[tex]\[ P_2 = 135 \][/tex]
[tex]\[ P_3 = 135 + 1.75(135)(1 - \frac{135}{135}) \][/tex]
[tex]\[ P_3 = 135 \][/tex]
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Well formatted question is in the picture attached below.
Answer:
(89, 101, 110, 116, 121, 125, 128, 130, 131, 132)
Step-by-step explanation:
Given the function that models the population of antelopes :
Pn+1 = [1.75(Pn)^2/(Pn-1)] + 32 - Pn
n = 1
P1+1 = [1.75(Pn)^2/(Pn-1)] + 32 - Pn
Initial population, Pn = P1 = 89
P2 = (1.75(89)^2/(89-1)) + 32 - 89 = 100.519 = 101
P3 = (1.75(101)^2/(101-1)) + 32 - 101 = 109.52 = 110
P4 = (1.75(110)^2/(110-1)) + 32 - 110 = 116.27 = 116
P5 = (1.75(116)^2/(116-1)) + 32 - 116 = 120.77 = 121
P6 = (1.75(121)^2/(121-1)) + 32 - 121 = 124.51 = 125
P7 = (1.75(125)^2/(125-1)) + 32 - 125 = 127.51 = 128
P8 = (1.75(128)^2/(128-1)) + 32 - 128 = 129.76 = 130
P9 = (1.75(130)^2/(130-1)) + 32 - 130 = 131.26 = 131
P10 = (1.75(131)^2/(131-1)) + 32 - 131 = 132.01 = 132
(89, 101, 110, 116, 121, 125, 128, 130, 131, 132)
