A water tank is contaminated with a colony of 20 000 e. Coli bacteria. After 3 days the bacteria has grown to 28 284 e. Coli. What is the doubling time of the bacteria using an equation

6

Step-by-step explanation:

The formula for doubling time is given as:

P(t) = Po(2)^t/k

Where:

P(t) = Population after time t = 28,284

Po = Initial Population = 20,000

t = time = 3 days

k = doubling time = ?

28,284 = 20,000 × (2)^3/k

Divide both sides by 20,000

28,284/20,000 = 20,000/20,000 × (2)^3/k

1.4142 = (2)^3/k

2^1/2 = (2)^3/k

Divide both sides by 2

1/2 = 3/k

Cross Multiply

k × 1 = 2 × 3

k = 6

Therefore, the doubling time(k) of the bacteria is 6

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

answer: 4 (:

step-by-step explanation: