P(t) = P₀ e^(kt)
<span>Where P₀ is the initial population, </span>
<span>P(t) is the population after "t" time. </span>
<span>t is your rate (can be hours, days, years, etc. in this case, hours) </span>
<span>k is the growth constant for this particular problem. </span>
<span>So using the information given, solve for k: </span>
<span>P₀ = 2000 </span>
<span>P(4) = 2600 </span>
<span>P(t) = P₀ e^(kt) </span>
<span>2600 = 2000e^(k * 4) </span>
<span>1.3 = e^(4k) </span>
<span>Natural log of both sides: </span>
<span>ln(1.3) = 4k </span>
<span>k = ln(1.3) / 4 </span>
<span>Now that we have a value for "k", use that, the same P₀, then solve for P(17): </span>
<span>P(t) = P₀ e^(kt) </span>
<span>P(17) = 2000 e^(17ln(1.3) / 4) </span>
<span>Using a calculator to get ln(1.3) then to simplify from there, we get: </span>
<span>P(17) ≈ 2000 e^(17 * 0.262364 / 4) </span>
<span>P(17) ≈ 2000 e^(4.460188 / 4) </span>
<span>P(17) ≈ 2000 e^(1.115047) </span>
<span>P(17) ≈ 2000 * 3.0497 </span>
<span>P(17) ≈ 6099.4 </span>
<span>Rounded to the nearest unit: </span>
<span>P(17) ≈ 6099 bacteria hope i could help =)))</span>
Answer:c is the correct answer
Step-by-step explanation:
A repair company's charge for repairing a certain type of copy machine fits the model y = 47.38 + 0.617x
where y is the number of dollars charged and
x is the number of minutes the repair person is on the job.
Therefore, to determine the number of minutes that it would take for the cost of repair to reach $130, we would substitute y =130 into the given model. It becomes
130 = 47.38 + 0.617x
0.617x = 130 - 47.38 = 82.62
x = 82.62/0.617 = 134 minutes
Answer:
no solution
Step-by-step explanation:
