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suter [353]
3 years ago
12

A population of 40 deer are introduced into a wildlife sanctuary it is estimated that the sanctuary can sustain up to 500 dear a

bsent constraints the population would grow 50% per year estimate the population after one year​
Mathematics
1 answer:
lions [1.4K]3 years ago
8 0

Answer:

There would be 60 deer after 1 year

Step-by-step explanation:

When a population is growing in a constrained environment with capacity K, and absent constraint grows exponentially with growth rate r, then the population behavior can be represents by the logistic growth model,

P_n = P_{n-1}+ r(1-\frac{P_{n-1}}{K})P{n-1}

Where,

P_{n-1} = previous year population,

r = growth rate per year,

K = Capacity,

Here,

P_0 = 40, r = 50\% = 0.50, K = 2000

Thus, the population of deer after 1 year,

P_1=40+0.50(1-\frac{40}{2000})40

=40+0.50(\frac{2000-40}{2000})40

=40+0.50(\frac{1960}{50})

=40+19.60

=59.6\approx 60

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If u(x) = x^5 - x^4 + x^2 and v (x) = -x^2 which expression is equivalent to ( u/ v) (x)
yawa3891 [41]

Answer:

(u/v)(x)=-x^{3}+x^{2}-1

Step-by-step explanation:

You have the following functions:

u(x)=x^{5}-x^{4}+x^{2}\\v(x)=-x^{2}

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3 years ago
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a coin lands on heads 200 times. the relative frequency of heads is 0.4. how many times was the coin thrown?
Kazeer [188]

Answer:

The number of times coin thrown was <u>500</u>.

Step-by-step explanation:

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A coin lands on heads 200 times. the relative frequency of heads is 0.4.

Now, to find the times coin was thrown.

Let the number of times coin thrown was be x.

Relative frequency = 0.4.

Number of lands on heads = 200.

So, to get the number of times coin was thrown we put formula:

Relative\ frequency=\frac{Number\ of\ lands\ on\ heads}{Number\ of\ times\ coin\ thrown}

0.4=\frac{200}{x}

<em>By cross multiplying we get:</em>

0.4x=200

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3 years ago
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Need help with my homework ​
Volgvan

Answer:

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Step-by-step explanation:

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We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

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\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

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\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

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Operate the parentheses:

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\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

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\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

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