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vichka [17]
3 years ago
15

Measures of variability are used to characterize how the data are _____ around the _____ of the distribution.

Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer: Measures of variability are used to characterize how the data are spread around the mean of the distribution.

Step-by-step explanation:The measures of average such as the median and mean represent the typical value for a dataset.However , In the dataset the actual values usually differ from one another and from the average value itself.

In this case the measure of spread, sometimes also called a measure of dispersion, is used to describe the variability in a sample or population around the mean .

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Please help with 1, 2, and 3
Akimi4 [234]
Don’t know 1 but 2 is J or 3 because it’s y=mc+b and m=slope and parallel lines have the same slope and 3 is B or -4/5 because if you use the slope formula ( y2-y1/x2-x1) you get the slope (-4/5) and parallel lines have the same slope.
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There are 210210210 students in a twelfth grade high school class. 909090 of these students have at least one sister and 1051051
BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

P(X|Y)=\frac{P(A\cap Y)}{P(Y)}

If the events <em>X</em> and <em>Y</em> are independent then,

P(X|Y)=P(X)

(1)

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

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What is the greatest common factor (GCF) of 48 and 32?
Nat2105 [25]

Answer:

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A rectangle has a length of (5x+2) and a width of (3x-1) with an area of 308 square units. Find the perimeter
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See photo. Hope this helps

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