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nordsb [41]
2 years ago
5

Imagine the center of the Singapore Flyer ferris wheel is located at (0, 0) on a coordinate grid, and the radius lies on the x-a

xis. Write an equation of a circle for your ferris wheel, and sketch an image of what your ferris wheel would look like on the grid. Now graph a circle that is similar to the Ferris Wheel. Discuss the transformations needed to show that both circles on your graph are similar.
Mathematics
1 answer:
Nataly_w [17]2 years ago
7 0

The equation of the Ferris wheel is x² + y² = 25, and the transformation that shows that the circles are similar is the dilation transformation

<h3>How to determine the equation of the Ferris wheel?</h3>

From the question, we have the following parameters:

Center (a,b) = (0,0)

Radius, r = 5

The equation of a circle is represented as:

(x - a)² + (y - b)² = r²

So, we have:

(x - 0)² + (y - 0)² = 5²

Evaluate

x² + y² = 25

Hence, the equation of the Ferris wheel is x² + y² = 25

See attachment for the image of the Ferris wheel

Transformation to a similar circle

Assume that the similar circle has the following parameters:

Center (a,b) = (0,0)

Radius, r = 10

The equation of the circle would be:

(x - 0)² + (y - 0)² = 10²

Evaluate

x² + y² = 100

Since both circles have the same center, the transformation that shows that they are similar is the dilation transformation

Read more about circle equations at:

brainly.com/question/1559324

#SPJ1

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Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{340119}{10} = 34011.9

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

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Confidence interval:

\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621

34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)

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Answer:

some

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no

Step-by-step explanation:

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