Step 1ask a research questionstep 2design a study and collect datastep 3explore the datastep 4draw inferences beyond the datastep 5formulate conclusionsstep 6<span>look back and ahead</span>
Answer:
The answer to your question is the first graph.
Step-by-step explanation:
It seems that the first graph is correct because the height of the bars is equal to the grade get by the students.
The second graph is incorrect because the height of the bars is different from the grade the students got. Only the bar of Levi is correct.
The third graph is incorrect, just observe the height of John's bar to determine that the graph is wrong.
Also, the last graph is incorrect, in this graph, all the bars are different from the grade the students got.
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 should be A hope it helps
Answer:
21
69
Step-by-step explanation:
