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3 years ago

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Suppose that 40 deer are introduced in a protected wilderness area. The population of the herd

1 answer:

Scorpion4ik [409]3 years ago

8 0

The horizontal asymptote of the function is the minimum number of deer in the area.

The equation of horizontal asymptote is: $\mathbf{y = 40}$
The horizontal asymptote means that, the number of deer will never be less than 40

The equation is given as:

$\mathbf{f(x) = \frac{40 + 2x}{1 + 0.05x}}$

Expand the numerator

$\mathbf{f(x) = \frac{40(1 + 0.05x)}{1 + 0.05x}}$

Cancel out the common factor

$\mathbf{f(x) = 40}$

Hence, the equation of horizontal asymptote is:

$\mathbf{y = 40}$

The horizontal asymptote means that, the number of deer will never be less than 40

Read more about horizontal asymptotes at:

brainly.com/question/4084552

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yuradex [85]

Triangle FED = Triangle WXY

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3 years ago

The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation

Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

Sample size: n = 20
Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}$

This gives

$P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}$

$P(x = 11) = 0.0011$

Express as percentage

$P(x = 11) = 0.11\%$

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}$

This gives

$P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0$

$P(x = 20) = 0.0388$

Express as percentage

$P(x = 20) = 3.88\%$

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

$\mu = np$

So, we have:

$\mu = 20 \times 85\%$

$\mu = 17$

The standard deviation is calculated as:

$\sigma = np(1 - p)$

So, we have:

$\sigma = 20 \times 85\% \times (1 - 85\%)$

$\sigma = 20 \times 0.85 \times 0.15$

$\sigma = 2.55$

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

Read more about probabilities at:

brainly.com/question/15246027

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3 years ago

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Lady_Fox [76]

the second one I believe

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3 years ago

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max2010maxim [7]

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3 years ago

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Evaluate <br> h(x) = 17+ x/6<br> H(-18)=

yan [13]

The answer is 14

-18/6 is -3. add that to 17 and you get 14.

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3 years ago