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3 years ago

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There are currently 4000 birds of a particular species in a national park and their number is increasing at a rate of R(t) = 525

e0.05t birds/year. If the proportion of birds that survive t years is given by S(t) = e−0.1t, what do you predict the bird population will be 10 years from now? (Round your answer to the nearest whole number.)

1 answer:

dalvyx [7]3 years ago

6 0

Answer:

It will be 1790.

Step-by-step explanation:

There are 4000 birds of a particular species in a national park.

Now, the number of birds increasing by the relation $R(t) = 525 (e)^{0.05t}$ birds per year.

So, after 10 years the bird population will become, $R(10) = 525(e)^{0.05 \times 10} = 865.58$ plus 4000.

That means the population after 10 years will become (865.58 + 4000) = 4865.58.

Now, the proportion of birds that survive t years is given by $S(t) = e^{- 0.1t}$ .

So, after 10 years the proportion of birds that survive after 10 years will be

$S(10) = e^{- 0.1 \times 10} = 0.368$ .

Therefore, the predicted population of birds after 10 years from now will be = 4865.58 × 0.368 = 1790. (Answer)

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Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),

larisa86 [58]

Answer

$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$ .

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$ .

$f(\pi /2)=c_0+c_1+0c_2=0.5$ .

$f(\pi)=c_0+0sin(t)-c_2=-2.5$ .

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$ .

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$

Where

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$ ,

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$

To determine the values of X, we use the expression

$X=(A^{T}A)^{-1}A^{T}B$

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

Therefore:

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)$

Therefore, the trigonometric function which fits to the given data is:

$f(t)=-0.2+4.1sin(t)+4cos(t)$

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The answer to the questions

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The table below shows the radius y, in inches, created by growing algae in x days: Time (x) (days) 1 4 7 10 Radius (y) (inches)

tatuchka [14]

PART A:
Given the table below showing the radius y, in inches, created by growing algae in x days.

Time (x) (days): 1 4 7 10
Radius (y) (inches): 1 8 11 12

We can find the find the correlation coeficient of the data using the table below:
$\begin{center}
\begin{tabular}
{|c|c|c|c|c|}
x & y & x^2 & y^2 & xy \\ [1ex]
1 & 1 & 1 & 1 & 1\\
4 & 8 & 16 & 64 & 32\\
7 & 11 & 49 & 121 & 77\\
10 & 12 & 100 & 144 & 120\\ [1ex]
\Sigma x=22 & \Sigma y=32 & \Sigma x^2=166 & \Sigma y^2=330 & \Sigma xy=230
\end{tabular}
\end{center}$

Recall that the correlation coefitient is given by the equation:
$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{ \sqrt{(n\Sigma x^2-(\Sigma x)^2)(n\Sigma y^2-(\Sigma y)^2)} } \\ \\ = \frac{4(230)-(22)(32)}{ \sqrt{(4(166)-(22)^2)(4(330)-(32)^2)} } = \frac{920-704}{ \sqrt{(664-484)(1,320-1,024)}} \\ \\ = \frac{216}{ \sqrt{180(296)}} = \frac{216}{ \sqrt{53,280} } = \frac{216}{230.8} =0.94$

[Notice: even without using the formular to find the value of the correlation coeficient, it can be seen that the value of y strictly increases as the valu of x increases. This means that the value of the correlation coeficient is closer to +1 and from the options the value that is closest to +1 is 0.94]

From the value of the correlation coeffeicient, it can be deduced that the radius of the algae has a strong positive relationship with the time.

Recall the for the value of the correlation coeficient closer to +1, the relationship is strong positive, for the value closer to -1, the value is strong negative and for the values closer to zero, either way of zero is a weak positive if it is positive and weak negative if it is negative.

PART B:
Recall that the slope of a straight line passing through two points
$(x_1,y_1)$ and $(x_2,y_2)$
is given by
$m= \frac{y_2-y_1}{x_2-x_1}$

Thus the slope of the graph of radius versus time between 4 and 7 days, [i.e. the line passes through points (4, 8) and (7, 11)] is given by
$m= \frac{11-8}{7-4}= \frac{3}{3} =1$

The value of the slope means that the radius of the algea grows by 1 inch evry day between day 4 and day 7.

PART C:
We can say that the data above represent both correlation and causationg.

Recall that correlation expresses the relationship between two variables while causation expresses that an event is as a result of another event.

From the information above, we have seen that there is a relationship (correlation) between the passing of days and the growth in the radius of the algae.

Also we can conclude that the growth in the radius of algae is a function of the passing of days, i.e. the growth in the radius of algae is as a result of the passing of days.

Therefore, <span>the data in the table represent both correlation and causation.</span>

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Aneli [31]

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3 years ago

Explain the difference between each pair of concepts.

Fynjy0 [20]

Answer:

a. B. Frequency is the number of times a particular distinct value occurs. Relative frequency is the ratio of the frequency of a value to the total number of observations.

b. A. A relative frequency is the same as a percentage expressed as a decimal.

Step-by-step explanation:

Frequency is the number of repetitions of a particular observation.

Relative Frequency is the ratio of the frequency of particular observation to the sum of frequencies of all the observations.

The percentage is the proportion of the whole.

Thus, In question (a)

Option B is the only correct answer.

and in question (b)

Option A is the only correct answer.

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