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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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Probably $87 based on the information I was given.
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The local museum had 20000 visitors in 1990 . The number of visitors has decreased by 3% each year since 1990 . Enter a function
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Answer:

It is m(x) = 20,000(0.97)^x.

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2 years ago
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Darryl deposits $1,500 into a savings account that has a simple interest rate of 2.7%. Lori deposits $1,400 into a savings accou
igor_vitrenko [27]
Darryl:
Answer:
A = $1,905.00

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Equation:
A = P(1 + rt)

Lori:

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2 years ago
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Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not
tangare [24]

Answer:

a.

\bar p_1=0.05\\\bar p_2=0.067

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let p_1  \& p_2 denote processes 1 & 2.

For p_1: T1=10,n1=200

For p_2:T2=20,n2=300

Therefore

\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067

b. To test for hypothesis:-

i.

H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05

ii.For a two sample Proportion test

Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\

iii. for \frac{\alpha}{2}=(-1.96,+1.96) (0.5 alpha IS 0.025),

reject H_o if|Z|>1.96

iv. Do not reject H_o. The noncomforting proportions are not significantly different as calculated below:

z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}

z=-0.78

c.(1-\alpha).100\% for the p1-p2 is given as:

(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt   \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645  \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0
2 years ago
Given: circle O with tangent AB mBC= 2x -16; mCD = x+40; mDE = x; mEB = 60 <br><br>find the x
murzikaleks [220]

Answer:

69

Step-by-step explanation:

All the given arcs cover the entire circle circumference, so their measures add up to a full 360.

(2x - 16) + (x + 40) + x + 60 = 360

4x + 84 = 360

4x = 276

x = 69

6 0
2 years ago
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