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NikAS [45]
1 year ago
15

PLEASE HELP !

Mathematics
1 answer:
Alex_Xolod [135]1 year ago
5 0

The characteristics of similarity of circles is that all circles are similar

The result of the comparison  between the parameters of the original and the replica of the Ferris wheel are as follows;

i) Similar Ferris wheels are similar represent similar circles,

The central angles of both Ferris wheels are equal

ii) The smaller Ferris wheel has a smaller arc length than the original Ferris wheel

iii) The equation of the replica of the Ferris wheel is x² + y² = 900

iv) Please find attached the image of the Ferris wheel

The reason for the above values is presented as follows;

Known;

The given parameters of the original Ferris wheel are;

The Ferris wheel name = Singapore Flyer

Wheel diameter = 150 m

Number of compartments = 28

Wheel circumference ≈ 471 m

Wheel area ≈ 17,662.5 m²

Central angle measure = 12.86°

Central angle measure in radians ≈ 0.224

The distance between two cars (arc length) = 16.8

Sector area of region between two cars = 630 m²

Strategy;

We show that the measure of angles in the similar circles are equal

The required parameter are;

The parameters of smaller replica of the Ferris wheel;

The diameter of the wheel, D = 60 meters

The location of the center = (0, 0)

Number of compartments = 28

Wheel circumference, C = π × D = π × 60 ≈  188.5 m

Wheel area ≈ π × D²/4 = π × (60 m)²/4 ≈ 2827.43 m²

Central angle measure = 360/28 ≈ 12.86°

Central angle measure in radians = 2×π/28 ≈ 0.224

The distance between two cars (arc length) = C/28 = 188.5/28 ≈ 6.73 m

Sector area of region between two cars = 2827.43 m²/28 ≈ 101 m²

i) The original and replica of the Ferris wheel are similar, therefore, the conclusion that can be drawn about the central angle is that they will be equal

ii) The arc length of the smaller Ferris wheel is less than the original Ferris wheel

iii) The general equation of a circle is (x - h)² + (y - k)² = r²

Where;

(h, k) = The center of the circle = (0, 0)

r = The radius of the circle of the Ferris wheel = D/2 = 60 m/2 = 30 m

Therefore, the equation of the smaller replica of the Ferris is presented as follows;

(x - 0)² + (y - 0)² = x² + y² = 30² = 900

The equation of the replica of the Ferris wheel is x² + y² = 900

iv) The drawing of the Ferris wheel created with MS Excel is attached

Learn more about Ferris wheel here;

brainly.com/question/20761245

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Step-by-step explanation:

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Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

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\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

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\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

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\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

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\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

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\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

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