Have I concluded a good formula

The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was *K*, is the following formula connecting the current number, *N*, with time *t* in years since the initial count a correct formula to use?

N = K x 11 x t

Re: Have I concluded a good formula

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**David Green** The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was *K*, is the following formula connecting the current number, *N*, with time *t* in years since the initial count a correct formula to use?

N = K x 11 x t

1. Your formula is not correct.

2. In the beginning, that means t = 0, you have K animals.

After one year, t = 1, you have $\displaystyle N(1)= K + K \cdot \underbrace{0.11}_{11\%} = K(1+0.11)$

Next year, t = 2, you have $\displaystyle N(2)= K(1+0.11) + K (1+0.11)\cdot 0.11 = K(1+0.11)(1+0.11)=K(1+0.11)^2$

After x years, t = x, you have $\displaystyle N(x)=K (1+0.11)^x$

Re: Have I concluded a good formula

Quote:

Originally Posted by

**David Green** The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was *K*, is the following formula connecting the current number, *N*, with time *t* in years since the initial count a correct formula to use?

N = K x 11 x t

At the start $\displaystyle N_0=K$.

Then after one year $\displaystyle N_1=N_0+0.11N_0=1.11K$

After two years $\displaystyle N_2=N_1+0.11N_1+=(1.11)^2K$

So after t years $\displaystyle N_t=~?$

Re: Have I concluded a good formula

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Originally Posted by

**earboth** 1. Your formula is not correct.

2. In the beginning, that means t = 0, you have K animals.

After one year, t = 1, you have $\displaystyle N(1)= K + K \cdot \underbrace{0.11}_{11\%} = K(1+0.11)$

Next year, t = 2, you have $\displaystyle N(2)= K(1+0.11) + K (1+0.11)\cdot 0.11 = K(1+0.11)(1+0.11)=K(1+0.11)^2$

After x years, t = x, you have $\displaystyle N(x)=K (1+0.11)^x$

Thanks for your help, I read it as; N(x) = K x 1.11^x