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natita [175]
3 years ago
14

The ages of two groups of dance students are shown in the following dot plots: A dot plot shows Age in years on the horizontal a

xis. For Group X, there are 3 dots on 4, 1 dot on 5, 5 dots on 8, 2 dots on 10, 3 dots on 12, and 2 dots on 16. For Group Y, there are 2 dot on 5, 3 dots on 7, 1 dot on 9, 3 dots on 11, 1 dot on 13, 2 dots on 14, 1 dot on 15, 1 dot on 18, 3 dots on 20, 1 dot on 22, and 2 dots on 24. The mean absolute deviation (MAD) for group X is 3.07 and the MAD for group Y is 5.25.
Which of the following observations can be made using these data?
Group X has less variability in the data.
Group X has greater variability in the data.
Group Y has a lower range.
Group Y has a lower mean.
Mathematics
2 answers:
Dimas [21]3 years ago
4 0

Answer:Group X has less variability in the data.


Step-by-step explanation:


lutik1710 [3]3 years ago
3 0
I did the test. The answer is "Group X has less variability in the data."
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The magnitude of the second atom relative to the first = 10⁻¹¹ / 10⁻¹³ = 100

<h3>Equation</h3>

An equation is an expression used to show the relationship between two or more variables or numbers.

The magnitude of the second atom relative to the first = magnitude of second atom / magnitude of first atom

The magnitude of the second atom relative to the first = 10⁻¹¹ / 10⁻¹³ = 100

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8 0
2 years ago
Y varies inversely with x y = 132 when x = 6 What is the value of k, the constant of inverse variation?
malfutka [58]
The answer to this question is i think a or b.
8 0
3 years ago
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Let f(x)=4x^2+6 .
Aleks [24]
1.) g(x)=4x^2+2
2.) g(x)=3/4x^2+7
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3 years ago
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Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ.e^{-\lambda.x}

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - e^{-\lambda.x}

(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - e^{-0.0143.200}

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = \sqrt{\frac{1}{\lambda^{2}} }

σ = \sqrt{\frac{1}{0.0143^{2}} }

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - e^{-0.0143*209.79})

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

1 - e^{-\lambda.m} = 0.5

-e^{-\lambda.m} = - 0.5

ln(e^{-0.0143.m}) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0
4 years ago
A bird at the top of a tree looks down at a field mouse with an angle of depression of 65Á. If the field mouse is 30 meters from
Dafna11 [192]
Correct answer is 33.1 m.

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sin{65^o}= \frac{30}{x} &#10;\\x= \frac{30}{sin{65^o}}&#10;\\x=33.1 &#10;

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3 years ago
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