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professor190 [17]
3 years ago
13

The values for three sets of data are shown below. Data Data Set Values 1 42, 48, 50, 88, 49 2 63, 29, 35, 28, 30 3 2, 5, 3, 8 W

ithout calculating any statistics, Anna knows that data set 3 would have the least mean absolute deviation among the three sets. Which statement explains how she knows?
Mathematics
2 answers:
nataly862011 [7]3 years ago
8 0
<h2>Answer:</h2>

data set 3 would have the least mean absolute deviation among the three sets since there is less spread of the data and the data values in set-3 lie close to the mean.

<h2>Step-by-step explanation:</h2>

The mean absolute deviation is a  measure of spread of the data.

  • If the data values in the given data set are widely spread then we obtain a higher mean absolute deviation.
  • if the data values of a given data set are close to each other i.e there is a less spread of the data and hence the mean absolute deviation will be low as the data values will lie close to the mean.

We are given three data set as:

set- 1             42, 48, 50, 88, 49

set- 2             63, 29, 35, 28, 30

set- 3                2, 5, 3, 8

Hence, we could observe that the data values in set 1 and set 2 are widely spread.

In set-1 the data value 88 is  much higher value as compared to other data values.

Similarly in set-2 the data value 63 is again a much higher value as compared to other data values.

Whereas in set-3 the data values are all closely related and there is not much spread in the data.

Mazyrski [523]3 years ago
4 0

Answer:

data set 3 would have the least mean absolute deviation among the three sets since there is less spread of the data and the data values in set-3 lie close to the mean.

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Answer:

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Step-by-step explanation:

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A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
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Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

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dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

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The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

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If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

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