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jonny [76]
2 years ago
14

The data set shows the admission prices at several museums. $20, $20, $16, $12, $15, $25, $11 Find and interpret the range, inte

rquartile range, and mean absolute deviation of the data. The range is . The interquartile range is . The mean absolute deviation is . Question 2 The prices vary by no more than $ . The middle half of the prices vary by no more than $ . The admission prices differ from the mean price by an average of $ . 7 of 8 answered need help?check answer Next
Mathematics
1 answer:
Gemiola [76]2 years ago
8 0
11, 12, 15, 16, 20, 20, 25
Mean: 17

The range is 25-11 = 14 (the numbers in bold)

The interquartile range is 20-12=8 (the numbers underlined)

The mean absolute deviation is: 4; this is found by finding how far each number is from 17 (mean): 6,5,2,1,3,3,8  (28) and dividing by 7.

Part B:

The prices vary by no more that $14 (range).

<span>The middle half of the prices vary by no more than $8 (IQR).
</span>
<span>The admission prices differ from the mean price by an average of $4 (MAD).</span> 

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Ludmilka [50]

Answer:

y = 0.64/x^3

Step-by-step explanation:

If y is inversely proportional to a^3, we can write the equation:

y * a^3 = k

Using a = 2 and y = 10, we have:

10 * 2^3 = k

k = 80

If a is directly proportional to x, we can write the equation:

a = c * x

Using x = 4 and a = 20, we have:

20 = c * 4

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Now, using the value of a = 5*x in the equation y * a^3 = 80, we have:

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8 0
2 years ago
Does anyone know the answer?
laila [671]

Answer:

a. θ = 30°

b. μ = √3 / 15 ≈ 0.115

Step-by-step explanation:

Draw a free body diagram for each scenario (see attached figure).  The body has four forces acting on it:

  • Weight pulling down
  • Normal force perpendicular to the incline
  • Applied force parallel up the incline
  • Friction force parallel to the incline

Remember that friction opposes the direction of motion.  So when the body is sliding up, friction points down the incline.  And when the body is sliding down, friction points up the incline.

Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.

For sliding up, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

P₁ − f − mg sin θ = 0

P₁ − Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₁ − mgμ cos θ − mg sin θ = 0

Now for sliding down, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

And sum of the forces parallel to the incline:

∑F = ma

P₂ + f − mg sin θ = 0

P₂ + Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₂ + mgμ cos θ − mg sin θ = 0

We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.

So we have two equations and two unknowns (μ and θ):

P₁ − mgμ cos θ − mg sin θ = 0

P₂ + mgμ cos θ − mg sin θ = 0

Let's start by adding the equations together:

P₁ + P₂ − 2 mg sin θ = 0

P₁ + P₂ = 2 mg sin θ

sin θ = (P₁ + P₂) / (2 mg)

Plugging in the values:

sin θ = (6 + 4) / (2 × 10)

sin θ = 1/2

θ = 30°

Now we can plug this into either equation and find μ.

P₁ − mgμ cos θ − mg sin θ = 0

6 − (10 cos 30°) μ − 10 sin 30° = 0

6 − 5√3 μ − 5 = 0

1 = 5√3 μ

μ = √3 / 15

μ ≈ 0.115

6 0
3 years ago
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