Answer :
Let the equation be : at^2 + bt + c = P,
where t = time (hrs),
P = population (1000's).
When t = 1, P = 3.03.
When t = 2, P = 1.72.
When t = 3, P = 1.17.
Substitute these into the equation to obtain these 3 simultaneous equations : a + b + c = 3.03
4a + 2b + c = 1.72
9a + 3b = c = 1.17
Solving gives :
a = 0.38,
b = -2.45,
c = 5.1.
The equation is therefore,
P = 0.38t^2 - 2.45t + 5.1
Testing with t = 0 to 6 gives the population values as provided, so it seems to be a valid model.
At t = 9 hrs,
P = 0.38*9^2 - 2.45*9 + 5.1
= 13.83.
where t = time (hrs),
P = population (1000's).
When t = 1, P = 3.03.
When t = 2, P = 1.72.
When t = 3, P = 1.17.
Substitute these into the equation to obtain these 3 simultaneous equations : a + b + c = 3.03
4a + 2b + c = 1.72
9a + 3b = c = 1.17
Solving gives :
a = 0.38,
b = -2.45,
c = 5.1.
The equation is therefore,
P = 0.38t^2 - 2.45t + 5.1
Testing with t = 0 to 6 gives the population values as provided, so it seems to be a valid model.
At t = 9 hrs,
P = 0.38*9^2 - 2.45*9 + 5.1
= 13.83.
Answer:
The required equation is [tex]P=0.38t^2-2.45t+5.1[/tex].
Explanation:
Consider the provided data.
We need to find a quadratic model.
Quadratic polynomial can be written as:
[tex]at^{2}+bt+c=P[/tex]
Here, t represents time and P represents population.
Consider the given data,
At t = 0 the population P = 5.1.
Substitute t = 0 and P = 5.1 in above quadratic polynomial.
[tex]a(0)^{2}+b(0)+c=5.1[/tex]
[tex]c=5.1[/tex]
From the given data, at t = 1 the population P = 3.03.
Substitute t = 1, c = 5.1, and P = 3.03 in quadratic polynomial.
[tex]a(1)^{2}+b(1)+5.1=3.03[/tex]
[tex]a+b+5.1=3.03[/tex]
[tex]a+b=-2.07[/tex]
[tex]a=-2.07-b[/tex]
From the given data, at t = 2 the population P = 1.72.
Substitute t = 2, c = 5.1, and P = 1.72 in quadratic polynomial.
[tex]a(2)^{2}+b(2)+5.1=1.72[/tex]
[tex]4a+2b+5.1=1.72[/tex]
[tex]4a+2b=-3.38[/tex]
Now, substitute the value of a in above equation.
[tex]4(-2.07-b)+2b=-3.38[/tex]
[tex]-8.28-4b+2b=-3.38[/tex]
[tex]-2b=-3.38+8.28[/tex]
[tex]-2b=4.9[/tex]
[tex]b=-2.45[/tex]
Substitute [tex]b=-2.45[/tex] in [tex]a=-2.07-b[/tex].
[tex]a=-2.07-(-2.45)[/tex]
[tex]a=-2.07+2.45[/tex]
[tex]a=0.38[/tex]
Thus, the value of a = 0.38, b = -2.45, and c = 5.1.
Therefore, the required equation is [tex]P=0.38t^2-2.45t+5.1[/tex].