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:
table b graph a
Step-by-step explanation:
Answer:
Step-by-step explanation:
(2)= 90-53 = 37
(1) = 180-74-37 = 69
(4) = 180-69-90 = 21
(3) = 180-53-21 = 106
Answer:
3
Step-by-step explanation:
Using the rule of logarithms
x = n ⇒ x = 
let 
125 = n , then
125 = 
, that is
5³ =
Since the bases on both sides are equal, both 5 then equate the exponents
n = 3
2•1=2. 2/7= 0.285714. The .285714 is repeating
