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Marina CMI [18]
2 years ago
12

The frog population on an island is declining at a rate of 3.2% per year. The population was 4500 in the year 2009.

Mathematics
1 answer:
PtichkaEL [24]2 years ago
5 0

Answer:

3636

Step-by-step explanation:

2015 - 2009 = 6

3.2 x 6 = 19.2

19.2/100 x 4500 = 864

4500 - 864 = 3636

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Collin ordered a set of yellow and brown pins. He received 400 pins in all. 340 of the pins were yellow. What percentage of the
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The yellow pins show as 85% of the total number of pins that Collin ordered
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The size of the tank a fish collector needs depends on how many fish he has. Which in the independent variable
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Answer:

Step-by-step explanation:

"<u><em>The size of the tank</em></u> a fish collector needs <u><em>depends</em></u> on <em>how many fish he has</em>."

Dependent variable is the size of the tank.

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2 years ago
3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared
navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

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x=0-87 hope this helps

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