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Romashka [77]
2 years ago
11

A biologist is researching the population density of antelopes near a watering hole. The biologist counts 32 antelopes within a

radius of 34 km of the watering hole. What is the population density of antelopes? Enter your answer in the box. Use 3.14 for pi and round only your final answer to the nearest whole number.
Mathematics
1 answer:
Setler [38]2 years ago
7 0

Answer:

=0.3 per km²

Step-by-step explanation:

Population density is the population per uni area.

Area=πr² Where A is area and r is the radius of the circle.

A=3.14×34

=106.76km²

Population density = Population/Area

=32 antelopes/106.76 km²

=0.3 per km²

The population of antelopes within the given radius is 0.3 per km²

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lesantik [10]
1/4 is the anwers because you need to reduce the fraction
3 0
2 years ago
Read 2 more answers
An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c
Ad libitum [116K]

Answer:

The correct options is A.

Step-by-step explanation:

The each diagonal of an irregular parallelogram divides the parallelogram in two equal and congruent parts. The diagonal bisects each other.

It means, if a irregular parallelogram rotates 180° about the midpoint of its diagonal, then the image of the parallelogram coincide with its preimage during the rotation.

If a irregular parallelogram rotates 360° about the midpoint of its diagonal, then the image of the parallelogram coincide 2 times with its preimage during the rotation.

Therefore the correct option is A.

6 0
3 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
An automotive part manufacturer regularly makes shipments of 400 parts to merks auto repair. On average, 14 of the 400 parts are
Radda [10]

Answer:

no matter how many parts are in the shipment, 3.5% of the parts will be defective.

hope this helped, if so hit that thanks button ;D

Step-by-step explanation:


4 0
2 years ago
Help ..........................................
Aleksandr [31]
The correct answer is 2.5×10^2
Hope this helped :)
6 0
2 years ago
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