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MariettaO [177]
3 years ago
12

At the start of the recovery period, the waterbuck population contained only 140 individuals. The population had 0.67 births per

year per individual and 0.06 deaths per year per individual. What is the maximum per capita growth rate (r) for this population
Mathematics
1 answer:
Sergio [31]3 years ago
6 0

Answer:

the intrinsic growth rate of the population = 0.67 - 0.06 = 0.61 or 61%

Step-by-step explanation:

since populations increase exponentially, the same will happen here

in 5 years the population should be:

P₅ = P₀e°⁶¹ˣ⁵ = 140e°⁶¹ˣ⁵ = 2,956.15 ≈ 2,956

in 10 years the population will be:

P₁₀ = P₀e°⁶¹ˣ¹⁰ = 140e°⁶¹ˣ¹⁰ = 62,420.09 ≈ 62,420

when you are using exponential growth rates, we have to assume that r will always remain constant

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Construct a quadrilateral abcd in which. ab=4 cm and ad=5cm ,Angle A=60°.Angle B=90°and Angle D=90°​
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Step-by-step explanation:

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2 years ago
A sample of lake water contains 150grams of salt. If 30% of the water sample is salt, find the number of grams the entire sample
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500 grams

Step-by-step explanation:

Let's say that the total number of grams of the sample is x. We know that 30% of x is salt and we also know this quantity is 150 grams. So, we can write this as:

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Hope this helps!

5 0
3 years ago
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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