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djyliett [7]
3 years ago
9

Find the absolute mean deviation for the set {X, 2x, 3x, 4x}

Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0

Given:

Data set = \{x, 2x, 3x, 4x\}

To find:

The absolute mean deviation for the given set.

Solution:

We have,

Data set = \{x, 2x, 3x, 4x\}

Mean of the data set is

Mean=\dfrac{\sum x_i}{n}

Mean=\dfrac{x+2x+3x+4x}{4}

Mean=\dfrac{10x}{4}

Mean=2.5x

Now,

The formula for mean absolute deviation (MAD) is

MAD=\dfrac{\sum |x_i-\overline x|}{n}

MAD=\dfrac{|x-2.5x|+|2x-2.5x|+|3x-2.5x|+|4x-2.5x|}{4}

MAD=\dfrac{|-1.5x|+|-0.5x|+|0.5x|+|1.5x|}{4}

MAD=\dfrac{1.5x+0.5x+0.5x+1.5x}{4}

MAD=\dfrac{4x}{4}

MAD=x

Therefore, the mean absolute deviation for the given set of data is x.

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The braking distance, in feet of a car a Travling at v miles per hour is given.
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The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

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D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

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D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

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At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

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