Question

Need help please. Links and empty answers WILL be reported.

Answer 1

Answer:

a) $P=975t-950$

b) 14650

Step-by-step explanation:

First let's find out how much the population increased during those 4 years.

In 1992 it was 1000.

In 1996 it was 4900.

$4900-1000 = 3900$. During 4 years, the population increased by 3900.

Therefore during 1 year, the population increased by $\frac{3900}{4} =975$

Since it's linear, we can write it using the general equation for a straight line, which is $y=mx+c$.

We've already found m, which is the slope of the line (975).

The question asks us to write it in the form: $P=mt+c$. where P is the population and t is the time in years since 1990.

We can sub some values into this formula to work out the constant, c.

Let's put in the values for 1992. Population is 1000 and the time in years after 1990 is 2 (since 1992 - 1990 = 2)..

Therefore:

$1000=975(2)+c$

Re-arrange to get: $c=-950$

Therefore, we have our formula now.

$P=975t-950$

Now for the second part of the question, population of moose in 2006. We need to work out how many years apart 1990 and 2006 are which we can calculate by doing $2006-1990 = 16$.

Now let's put this into our formula.

$P=975(16)-950 = 14650$