2 years ago

MULTIPLE CHOICE QUESTION PLS HELP ME

Mathematics

1 answer:

marysya [2.9K]2 years ago

4 0

Answer:

The answer is C.

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3 years ago

In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the

natka813 [3]

Answer:

$P(t) = 5100e^{0.0039t}$

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that $P(0) = 5100$

$P(t) = 5100e^{rt}$

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that $P(5) = 5200$ .

We use this to find r.

$P(t) = 5100e^{rt}$

$5200 = 5100e^{5r}$

$e^{5r} = \frac{52}{51}$

$\ln{e^{5r}} = \ln{\frac{52}{51}}$

$5r = \ln{\frac{52}{51}}$

$r = \frac{\ln{\frac{52}{51}}}{5}$

$r = 0.0039$

So the equation for the moose population is:

$P(t) = 5100e^{0.0039t}$

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2 years ago