In 1993, the moose population in a park was measured to be 3800. By 1999, the population was measured again to be 5240 If the population continues to change linearly: Find a formula for the moose population, P,in terms of t,the years since 1990. P(t)=square What does your model predict the moose population to be in 2006? square

## Step 1<br />The problem involves a linear function, which can be represented in the form \(y = mx + b\), where \(m\) is the slope of the line and \(b\) is the y-intercept.<br /><br />## Step 2<br />In this case, the slope \(m\) is the rate of change of the moose population per year, and the y-intercept \(b\) is the population at the start year (1990).<br /><br />## Step 3<br />We are given two points (3, 3800) and (9, 5240), which represent the population of moose in 1993 and 1999 respectively. We can use these points to calculate the slope \(m\).<br /><br />### **\(m = \frac{y2 - y1}{x2 - x1} = \frac{5240 - 3800}{9 - 3} = 220\)**<br /><br />## Step 4<br />The y-intercept \(b\) is the population at the start year (1990), which is 1990 - 3 = 1987. So, \(b = 3800 - 3*220 = 3140\).<br /><br />## Step 5<br />Therefore, the formula for the moose population \(P\) in terms of \(t\), the years since 1990, is \(P(t) = 220t + 3140\).<br /><br />## Step 6<br />To predict the moose population in 2006, we substitute \(t = 16\) (since 2006 is 16 years after 1990) into the formula.<br /><br />### **\(P(16) = 220*16 + 3140 = 6760\)**