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maxonik [38]
3 years ago
7

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.
Mathematics
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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Step-by-step explanation:

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3 0
4 years ago
Stella wants to buy a remote control helicopter. The one she wants costs $120.
aalyn [17]

.45 * 120 = 54

.072 * 54 = 3.888

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5 0
2 years ago
Read 2 more answers
E
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Answer:

$45 after the discount

Step-by-step explanation:

Set up a proportion.

     \frac{x}{50} = \frac{10}{100}      $50 is the total amount and make the 10% a decimal

x(100) = 50(10)

100 = 500

100 ÷ 500

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4 0
3 years ago
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8 0
3 years ago
Farmer Company purchased equipment on January 1, Year 1 for $82,000. The equipment is estimated to have a 5-year life and a salv
frosja888 [35]

Answer:

option (a) $6,240

Step-by-step explanation:

Given:

Purchasing cost of the equipment = $82,000

Estimated life = 5 years

Salvage value = $4,000

Revised expected life = 8 years

Now,

Depreciation per year = \frac{\textup{82,000-4,000}}{\textup{5}}

therefore,

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= $6,240

Hence,

The correct answer is option (a) $6,240

8 1
3 years ago
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