3 days after the start of an experiment there were 484 bacteria in a culture. After 5 days there were 1135. Use a system of equations to determine the initial number of bacteria in the culture (c) and the k value for the growth

Step-by-step explanation:

We assume you want your model to be ...

  p = c·e^(kt)

Filling in (t, p) values of (3, 484) and (5, 1135), we have two equations in the two unknowns:

  484 = c·e^(3k)

  1135 = c·e^(5k)

Taking logs makes these linear equations:

  ln(484) = ln(c) +3k

  ln(1135) = ln(c) +5k

Subtracting the first equation from the second, we have ...

  ln(1135) -ln(484) = 2k

  k = ln(1135/484)/2 ≈ 0.42615

Using that value in the first equation, we find ...

  ln(484) = ln(c) +3(ln(1135/484)/2)

  ln(c) = ln(484) -(3/2)ln(1135/484)

  c = e^(ln(484) -(3/2)ln(1135/484)) ≈ 134.8

The initial number in the culture was 135, and the k-value is about 0.42615.

I prefer to start with the model ...

  p = 484·(1135/484)^((t-3)/2)

Then the initial value is that obtained when t=0:

  c = 484·(1135/484)^(-3/2) = 134.778 ≈ 135

The value of k the log of the base for exponent t. It is ...

  ln((1135/484)^(1/2)) = 0.426152

This starting model matches the given numbers exactly. The transformation to c·e^(kt) requires approximations that make it difficult to match the given numbers.

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For this model, the base of the exponent is the ratio of the two given population values. The exponent is horizontally offset by the number of days for the first count, and scaled by the number of days between counts. The multiplier of the exponential term is the first count. The model can be written directly from the given data, with no computation required.

the answer is d. (1.2, -4.7) on edgn

hope this !

answer: 40.95

step-by-step explanation: