Let A(n+1)=A(n)+(10X16)/2X.1
=10+(160)/2x.1
=10+(16)/2
=10+8
=18 females after first year.
A(n) being amount of present female dogs = 10
Now I need to show A+2 is right (for next year), is that correct? If so thats where I start to struggle.
The question is 10 female dogs average 16 puppies each year with 50% of those puppies being female. Every year the owners remove 60% of the female puppies and all of the male puppies.
What will the growth be of female puppies left and their mothers over 15years?
Only the original mothers and 20% of female puppies can have puppies (each year); the rest will be neutered.
I know this is pretty easy to work out with a calculator if you keep entering the numbers, but is there a equation that will work it out?
Thanks for any help working towards finding it.
Let A(n+1)=A(n)+(10X16)/2X.1
=10+(160)/2x.1
=10+(16)/2
=10+8
=18 females after first year.
A(n) being amount of present female dogs = 10
Now I need to show A+2 is right (for next year), is that correct? If so thats where I start to struggle.