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3 years ago

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A scientist is observing the grizzly bear population in two state parks. The parks each have 150 bears when he begins observing

them. The population in Park A increases by 20% each year. The population in Park B decreases by 20% each year. In each equation, y represents the bear population with respect to a number of years, x. Drag each equation to show whether it can be used to represent the bear population in Park A, Park B, or neither of these.

2 answers:

babunello [35]3 years ago

7 0

Answer:

Step-by-step explanation:

Initial population of bears =150

Park A is given that the population increases by 20% each year

This gives that in year x population would be

P(t) = $150(1+\frac{20}{100} )^x\\=150(1.2)^x$

---

In the II park park B, the population decreases by 20%

Hence have population at time x as

P(x) = $150(1-\frac{20}{100} )^x\\=150(0.8)^x$

Mars2501 [29]3 years ago

5 0

Nothing to drag; nowhere to drag it to.

Park A's population can be modeled by
.. y = 150*1.2^x

Park B's population can be modeled by
.. y = 150*0.8^x

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Check the explanation

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confidence interval is

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There is sufficient evidence to conclude that the two indenters produce different hardness readings.

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Tems11 [23]

Divide through everything by <em>b</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since <em>a/b</em> < <em>c/d</em>, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by <em>b/d</em> to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and <em>a/b</em> < <em>c/d</em> tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

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