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2 years ago

10 points

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

8 0

Step-by-step explanation:

The mode is 7 because it is 5 times

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2 years ago

two bird populations were studied and formulas were developed for each populations growth. the populations p1 and p2 are given b

STatiana [176]

Answer:

$20 e^{k*10} = 60 e^{0.2*10}$

We can divide both sides by 20 and we got:

$e^{10k}= 3 e^{2}$

Now we can appply natural log on both sides and we got:

$10 k = ln(3e^2)$

Now we can divide both sides of the equation by 10 and we got:

$k = \frac{ln(3e^2)}{10}= 0.30986$

So then the aproximate value of k is 0.30986 and rounded would be 0.31

Step-by-step explanation:

We have the following two expression:

$p_1 = 20 e^{kt}$

$p_2 = 60 e^{0.2t}$

We know that the two populations were equal 10 years after the start of the study, so then we can create the following equation:

$20 e^{k*10} = 60 e^{0.2*10}$

We can divide both sides by 20 and we got:

$e^{10k}= 3 e^{2}$

Now we can appply natural log on both sides and we got:

$10 k = ln(3e^2)$

Now we can divide both sides of the equation by 10 and we got:

$k = \frac{ln(3e^2)}{10}= 0.30986$

So then the aproximate value of k is 0.30986 and rounded would be 0.31

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3 years ago