Web Analytics
Answered

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

Find a quadratic model for the data in the table. Type your answer below. Show your work.


Time (hours)

0

1

2

3

4

5

6

Population (1000s)

5.1

3.03

1.72

1.17

1.38

2.35

4.08

Verified Answer :

syed514

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 = 0.389^2 - 2.459 + 5.1
   = 13.83.

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 \(P=0.38t^2-2.45t+5.1\).

Related Questions