00:01
Let's start the solution in this question we have given p is 18 upon 1 plus 4 e power minus 0 .5p where p is number of fish in thousands and p is measured in yes.
00:29
So part a is put for part a put p equal to 4 we get here p is 18 upon 1 plus 4 e power minus 0 .5 into so here we get 18 upon 1 plus 4 e power minus 0 .5 into 4 is minus and so we get 18 upon 1 plus 4 e power minus 0 .5 into 4 is minus and so we get 18 upon 1 plus 4 e power minus 2 is 1.
01:01
541 3 now divide 18 divide 1 .5 413 is 11 .678 so that means here we have p is 11 .678000 next we proceed part b part b says that fine time so this population reach 9 ,000 fish so that means put p equal to 9 so we get 9 equal to 18 upon 1 plus 4 e power minus 0 .5 into t so here we get by simplifying this we get 1 plus 4 e power minus 0 .5 t equal to 18 upon 9 that is so 1 plus 4 e power minus 0 .5 t t is 2.
02:10
So we get 4 e power minus 0 .5 t is 2 minus 1 is 1 .e power minus 0 .5 t is 1 by 4.
02:20
That is e power minus 0 .25 t is 0 .25 by taking log both sides.
02:28
So here we get minus 0 .5 t and we know that log is 1 equal to log.
02:40
0 .25.
02:42
So we get here t is ln 0 .25 upon minus 0 .5.
02:53
By simplifying this, dividing this, here we get, t is 2 .77 years.
03:09
Now check whether this is correct or not.
03:18
Put the value of t...