A lake initially contains 1000 fish. Suppose that in the absence of predators or other causes ofremoval, the fish population increases by 10% each month. However, factoring in all causes, 80 fishare lost each month. Give a recurrence relation for the population of fish afternmonths. How many fish are there after5 months? If your fish model predicts a non-integer number of fish, round down to the next lowerinteger

A) P_n = 1.06(P_(n-1)) - 80

B) 887 fishes

Step-by-step explanation:

A) We are told that the lake initially contains 1000 fishes.

Thus, P_o = 1000

Now, the number of fishes increases by 6% each month

Thus, after n months, we have;

P_n = P_(n-1) + 0.06P_(n-1)

P_n = 1.06P_(n-1)

Where P_(n-1) is the population of fish in the previous month.

We are told that 80 fishes are lost each month.

Thus;

P_n = 1.06(P_(n-1)) - 80

B) We want to find out how many fishes we have after 5 months.

Thus;

P_5 = 1.06(P_(5-1)) - 80

P_5 = 1.06(P_4) - 80

We don't know P_4,thus;

P_o = 1000

P_1 = 1.06(1000) - 80 = 980

P_2 = 1.06(980) - 80 = 958.8

P_3 = 1.06(958.8) - 80 = 936.328

P_4 = 1.06(936.328) - 80 = 912.50768

Thus,

P_5 = 1.06(912.50768) - 80 = 887.2581408 ≈ 887

clock

step-by-step explanation:

the correct answer would be d because the slope is -2/2=1 and the y intercept is at 2 so that is the correction equation