The number of e. coli bacteria cells in a pond of stagnant water can be represented by the function below, where a represents the number of e. coli bacteria cells per 100 ml of water and t represents the time, in years, that has elapsed.

a(t)=136(1.123)^4t

based on the model, by approximately what percent does the number of e. coli bacteria cells increase each year?

a.

60%

b.

59%

c.

41%

d.

40%

Explanation:

The function that represents the number of E.coli bacteria cells per 100 mL of water as the time t years elapses is:

The base, 1.123, represents the multiplicative constant rate of change of the function, so you just must substitute 1 for t in the power part of the function:

Then, the multiplicative rate of change is 1.590, which means that every year the number of E.coli bacteria cells per 100 mL of water increases by a factor of 1.590, and that is 1.59 - 1 = 0.590 or 59% increase.

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads

where is the diagram? shown below where? ?

step-by-step explanation: