In the year 2005 , a total of 750 fish were introduced into a manmade lake . the fish population was expected to grow at a rate of 8.3% every year .

what will the fish population be in 2050 ?
round to the nearest whole number . hurry

The fish population will be in 2050 = 27,123

Step-by-step explanation:

Given,

In 2005, the total number of fish were introduced (fish population), PV = 750

The growth rate, i = 8.3% = 0.083

The difference between 2050 and 2005 is (2050 - 2005), n = 45 years

We need to solve this problem with the help of future value of compounding. We know, FV = PV (1+i)^n, where, PV = present value, i = interest rate, n = period.

Therefore, the fish population will be in 2050 = 750 x (1 + 0.083)^45

The fish population will be in 2050 = 27,123.26

2050 fish population = 27,123 (Nearest to whole number).

27,123.26 which you round down to 27,123

Step-by-step explanation:

I took the test and got it right.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right