Question

Can anyone help me with this math problem?

Answer 1

$\bold{FULL ANSWERS:}$

In 1993, Moose Population: 3280

In 1999, population became: 4960

P (Population) , t (years)

t = 6 —> 4960 - 3280 = 1680

Average Change —> 1680/6 = 280 moose/year

• In terms of 1990:

t = 3 —> 3280-3 (280)

P(1990) = 2440

P(t) = 2440 + 280t

• In 2003; t = 13

P(13) = 2440 + 280 (13)
P(13) = 2440 + 3640
P(13) = 6080

• Moose population in 2003

= 6080