Can someone pls help, I know it’s a lot but I’ve been stuck on this for a while now

Question

The data in the function can be modeled by an exponential function. Find the function that would fit the data.

Year Population

0 31

1 50

2 80

3 130

4 200

After completing the regression using this information the value of a is (__), the value of b is (__). This is an example of exponential (__). The equation for the situation is

f(x)= (__). Use the equation to predict the population for 8 years.

Answer 1

P = a(b)^rt where a= 31, r = rate of increase, t = time

31 = a(b)^0 = a

a = 31

50= 31b^r

50b = 31b^2r

80=31b^2r subtract from above

50b-80 = 0

b = 8/5 = 1.6

P = 31(1.6)^rt r= about 1

P= 31(1.6)^t

P= 31(1.6)^8 = about 1331 in 8 years

Answer 2

Need to fit the data to f(x) = Ar^x.

First, convert the table to log(y)

Year Population log(Population)

0 31 1.49

1 50 1.7

2 80 1.9

3 130 2.1

4 200 2.3

Doing linear regression to fit the log(data) to a line: y = mx+b

m = 0.202

b = 1.494

r = 10^m = 1.59

A = 10^b = 31.2

f(x) = 31.2(1.59)^x

Checking

year P

0 31.2

1 49,6

2 78.9

3 125.4

4 199.4

f(8) ≈ 31.2(1.59)⁸ ≈ 1274

Answer 3

P(t) = P₀b^t;

t = 0, P(0) = P₀= 31; so P(t) = 31b^t;

t = 1, P(1) = 31b = 50; so b = 50/31;

P(t) = 31(50/31)^t.

Let check how this function works for others years.

t = 2, P(2) = 31(50/31)² ≈ 80

t = 3, P(3) = 31(50/31)³ ≈ 130

t = 4, P(4) = 31(50/31)⁴ ≈ 209

This tell you that model you have to change.

Now if t = 8 P(8) = 31(50/31)⁸ ≈ 1419