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3 years ago

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Opening a daycare regular regulations mandate 35 ft.² of space per child how much floor space will you need for 15 children will

a space 22’ x 24’ meet the requirement how many square feet will be required for 15 children

1 answer:

NeTakaya3 years ago

6 0

525 ft.^2 for 15 children. No it will not.

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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 $P(0) = 5100$

So

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

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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3 years ago

Sooo..... can anybody help.....

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3-^1 x (4 x 6) x 2-^3

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$\huge{ \boxed{ \mathbf{ \blue{Problem:-}}}}$

$\bold{ {3}^{ - 1 \ } (4 \times 6) \times {2}^{ - 3} }$

$\huge{ \boxed{ \bf{ \blue{Solution:-}}}}$

$\bold{ {3}^{ - 1}(4 \times 6) \times {2}^{ - 3} }$

$= \bold{ \frac{1}{3}^{1} (4 \times 6) \times { \frac{1}{2}^{3} } }$

$= \bold{ \frac{1}{3} \times 24 \times \frac{1}{8} }$

$\bold{= 1}$

<h3><u>〜</u><u>Hope</u><u> it's</u><u> helpful</u></h3>

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