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iris [78.8K]
2 years ago
12

A wild animal generally stays at least x mi from the edge of a forest. For a rectangular forest preserve that is 2 mi

Mathematics
1 answer:
Butoxors [25]2 years ago
3 0

Answer:

Area = 4x^2 - 14x + 10

Step-by-step explanation:

<em>See Attachment for Complete Question</em>

Given

Width = 5mi --- For the forest

Length = 2mi -- --- For the forest

Required

Determine the area of the habitat

Since the distance between the animal's habitat is x mi on both sides;

The length and width of the habitat is:

Width = 5 - (x + x)

Width = 5 - 2x

Height = 2 - (x +x)

Height = 2 - 2x

The area is then calculated as follows;

Area = Width * Height

Area = (5 - 2x) * (2 - 2x)

Expand

Area = 5(2 - 2x) -2x(2 - 2x)

Open Brackets

Area = 10 - 10x - 4x + 4x^2

Area = 10 - 14x + 4x^2

Reorder

Area = 4x^2 - 14x + 10

Hence; the required polynomial for the habitat area is:

Area = 4x^2 - 14x + 10

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A local library claims it has 2.73x103 books. If there are 42 people in the building, how many books per capita are there?
sveticcg [70]

Answer:

Per\ Capita = 64

Step-by-step explanation:

Given

Books = 2.7  * 10^3

People = 42

Required

Determine the per capita

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Per\ Capita = \frac{Books}{People}

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Per\ Capita = \frac{2.7 * 10^3}{42}

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5 0
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What is the relationship between the unfinished puzzle and the missing piece
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2 years ago
let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo
anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\

a) P(x > 5) = \int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = \int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}

But P(x > a) = 0.05

Therefore:

e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98

a  ≅ 12

6 0
2 years ago
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