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Answer :
Sure, let's work through this problem step-by-step to find the population of antelope after each of the first ten years. We'll start with the initial population and apply the population model iteratively for each year.
The initial population, [tex]\( p_0 \)[/tex], is given as 89.
The population model is:
[tex]\[ p_{n+1} = \frac{1.75 \left( p_n \right)^2}{p_n - 1} + \left( 32 - p_n \right) \][/tex]
Here's how we calculate the population for each year:
### Year 0:
Initial population, [tex]\( p_0 \)[/tex], is 89.
### Year 1:
[tex]\[ p_1 = \frac{1.75 \left( 89 \right)^2}{89 - 1} + \left( 32 - 89 \right) \][/tex]
[tex]\[ p_1 = \frac{1.75 \times 7921}{88} - 57 \][/tex]
[tex]\[ p_1 \approx 101 \][/tex]
So, the population after 1 year is 101.
### Year 2:
[tex]\[ p_2 = \frac{1.75 \left( 101 \right)^2}{101 - 1} + \left( 32 - 101 \right) \][/tex]
[tex]\[ p_2 = \frac{1.75 \times 10201}{100} - 69 \][/tex]
[tex]\[ p_2 \approx 110 \][/tex]
So, the population after 2 years is 110.
### Year 3:
[tex]\[ p_3 = \frac{1.75 \left( 110 \right)^2}{110 - 1} + \left( 32 - 110 \right) \][/tex]
[tex]\[ p_3 = \frac{1.75 \times 12100}{109} - 78 \][/tex]
[tex]\[ p_3 \approx 116 \][/tex]
So, the population after 3 years is 116.
### Year 4:
[tex]\[ p_4 = \frac{1.75 \left( 116 \right)^2}{116 - 1} + \left( 32 - 116 \right) \][/tex]
[tex]\[ p_4 = \frac{1.75 \times 13456}{115} - 84 \][/tex]
[tex]\[ p_4 \approx 121 \][/tex]
So, the population after 4 years is 121.
### Year 5:
[tex]\[ p_5 = \frac{1.75 \left( 121 \right)^2}{121 - 1} + \left( 32 - 121 \right) \][/tex]
[tex]\[ p_5 = \frac{1.75 \times 14641}{120} - 89 \][/tex]
[tex]\[ p_5 \approx 125 \][/tex]
So, the population after 5 years is 125.
### Year 6:
[tex]\[ p_6 = \frac{1.75 \left( 125 \right)^2}{125 - 1} + \left( 32 - 125 \right) \][/tex]
[tex]\[ p_6 = \frac{1.75 \times 15625}{124} - 93 \][/tex]
[tex]\[ p_6 \approx 128 \][/tex]
So, the population after 6 years is 128.
### Year 7:
[tex]\[ p_7 = \frac{1.75 \left( 128 \right)^2}{128 - 1} + \left( 32 - 128 \right) \][/tex]
[tex]\[ p_7 = \frac{1.75 \times 16384}{127} - 96 \][/tex]
[tex]\[ p_7 \approx 130 \][/tex]
So, the population after 7 years is 130.
### Year 8:
[tex]\[ p_8 = \frac{1.75 \left( 130 \right)^2}{130 - 1} + \left( 32 - 130 \right) \][/tex]
[tex]\[ p_8 = \frac{1.75 \times 16900}{129} - 98 \][/tex]
[tex]\[ p_8 \approx 131 \][/tex]
So, the population after 8 years is 131.
### Year 9:
[tex]\[ p_9 = \frac{1.75 \left( 131 \right)^2}{131 - 1} + \left( 32 - 131 \right) \][/tex]
[tex]\[ p_9 = \frac{1.75 \times 17161}{130} - 99 \][/tex]
[tex]\[ p_9 \approx 132 \][/tex]
So, the population after 9 years is 132.
### Year 10:
[tex]\[ p_{10} = \frac{1.75 \left( 132 \right)^2}{132 - 1} + \left( 32 - 132 \right) \][/tex]
[tex]\[ p_{10} = \frac{1.75 \times 17424}{131} - 100 \][/tex]
[tex]\[ p_{10} \approx 132 \][/tex]
So, the population after 10 years is 132.
Therefore, the populations of antelope after each of the first ten years are:
[tex]\[ 89, 101, 110, 116, 121, 125, 128, 130, 131, 132 \][/tex]
So, option c. [tex]\( 89, 101, 110, 116, 121, 125, 128, 130, 131, 132 \)[/tex] is the correct population sequence.
The initial population, [tex]\( p_0 \)[/tex], is given as 89.
The population model is:
[tex]\[ p_{n+1} = \frac{1.75 \left( p_n \right)^2}{p_n - 1} + \left( 32 - p_n \right) \][/tex]
Here's how we calculate the population for each year:
### Year 0:
Initial population, [tex]\( p_0 \)[/tex], is 89.
### Year 1:
[tex]\[ p_1 = \frac{1.75 \left( 89 \right)^2}{89 - 1} + \left( 32 - 89 \right) \][/tex]
[tex]\[ p_1 = \frac{1.75 \times 7921}{88} - 57 \][/tex]
[tex]\[ p_1 \approx 101 \][/tex]
So, the population after 1 year is 101.
### Year 2:
[tex]\[ p_2 = \frac{1.75 \left( 101 \right)^2}{101 - 1} + \left( 32 - 101 \right) \][/tex]
[tex]\[ p_2 = \frac{1.75 \times 10201}{100} - 69 \][/tex]
[tex]\[ p_2 \approx 110 \][/tex]
So, the population after 2 years is 110.
### Year 3:
[tex]\[ p_3 = \frac{1.75 \left( 110 \right)^2}{110 - 1} + \left( 32 - 110 \right) \][/tex]
[tex]\[ p_3 = \frac{1.75 \times 12100}{109} - 78 \][/tex]
[tex]\[ p_3 \approx 116 \][/tex]
So, the population after 3 years is 116.
### Year 4:
[tex]\[ p_4 = \frac{1.75 \left( 116 \right)^2}{116 - 1} + \left( 32 - 116 \right) \][/tex]
[tex]\[ p_4 = \frac{1.75 \times 13456}{115} - 84 \][/tex]
[tex]\[ p_4 \approx 121 \][/tex]
So, the population after 4 years is 121.
### Year 5:
[tex]\[ p_5 = \frac{1.75 \left( 121 \right)^2}{121 - 1} + \left( 32 - 121 \right) \][/tex]
[tex]\[ p_5 = \frac{1.75 \times 14641}{120} - 89 \][/tex]
[tex]\[ p_5 \approx 125 \][/tex]
So, the population after 5 years is 125.
### Year 6:
[tex]\[ p_6 = \frac{1.75 \left( 125 \right)^2}{125 - 1} + \left( 32 - 125 \right) \][/tex]
[tex]\[ p_6 = \frac{1.75 \times 15625}{124} - 93 \][/tex]
[tex]\[ p_6 \approx 128 \][/tex]
So, the population after 6 years is 128.
### Year 7:
[tex]\[ p_7 = \frac{1.75 \left( 128 \right)^2}{128 - 1} + \left( 32 - 128 \right) \][/tex]
[tex]\[ p_7 = \frac{1.75 \times 16384}{127} - 96 \][/tex]
[tex]\[ p_7 \approx 130 \][/tex]
So, the population after 7 years is 130.
### Year 8:
[tex]\[ p_8 = \frac{1.75 \left( 130 \right)^2}{130 - 1} + \left( 32 - 130 \right) \][/tex]
[tex]\[ p_8 = \frac{1.75 \times 16900}{129} - 98 \][/tex]
[tex]\[ p_8 \approx 131 \][/tex]
So, the population after 8 years is 131.
### Year 9:
[tex]\[ p_9 = \frac{1.75 \left( 131 \right)^2}{131 - 1} + \left( 32 - 131 \right) \][/tex]
[tex]\[ p_9 = \frac{1.75 \times 17161}{130} - 99 \][/tex]
[tex]\[ p_9 \approx 132 \][/tex]
So, the population after 9 years is 132.
### Year 10:
[tex]\[ p_{10} = \frac{1.75 \left( 132 \right)^2}{132 - 1} + \left( 32 - 132 \right) \][/tex]
[tex]\[ p_{10} = \frac{1.75 \times 17424}{131} - 100 \][/tex]
[tex]\[ p_{10} \approx 132 \][/tex]
So, the population after 10 years is 132.
Therefore, the populations of antelope after each of the first ten years are:
[tex]\[ 89, 101, 110, 116, 121, 125, 128, 130, 131, 132 \][/tex]
So, option c. [tex]\( 89, 101, 110, 116, 121, 125, 128, 130, 131, 132 \)[/tex] is the correct population sequence.
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