Question

12In an experiment, a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again.

Answer 1

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.

Answer 2

Answer:

The required equation is $P=0.38t^2-2.45t+5.1$.

Explanation:

Consider the provided data.

We need to find a quadratic model.

Quadratic polynomial can be written as:

$at^{2}+bt+c=P$

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.

$a(0)^{2}+b(0)+c=5.1$

$c=5.1$

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.

$a(1)^{2}+b(1)+5.1=3.03$

$a+b+5.1=3.03$

$a+b=-2.07$

$a=-2.07-b$

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.

$a(2)^{2}+b(2)+5.1=1.72$

$4a+2b+5.1=1.72$

$4a+2b=-3.38$

Now, substitute the value of a in above equation.

$4(-2.07-b)+2b=-3.38$

$-8.28-4b+2b=-3.38$

$-2b=-3.38+8.28$

$-2b=4.9$

$b=-2.45$

Substitute $b=-2.45$ in $a=-2.07-b$.

$a=-2.07-(-2.45)$

$a=-2.07+2.45$

$a=0.38$

Thus, the value of a = 0.38, b = -2.45, and c = 5.1.

Therefore, the required equation is $P=0.38t^2-2.45t+5.1$.