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2 years ago

Is interquartile range a measure of center or a measure of variation?

Mathematics

2 answers:

Sedaia [141]2 years ago

5 0

Answer:

<h3>The interquartile range is the middle half of the data that is in between the upper and lower quartiles. ... The interquartile range is a robust measure of variability in a similar manner that the median is a robust measure of central tendency.</h3>

Step-by-step explanation:

<h2>PLEASE MARK ME BRAINLIEST AND FOLLW ME LOTS OF LOVE FROM MY HEART'AND SOUL DARLING TEJASVINI SINHA HERE ❤️</h2>

MA_775_DIABLO [31]2 years ago

5 0

Answer:

The interquartile range is the middle half of the data that is in between the upper and lower quartiles. ... The interquartile range is a robust measure of variability in a similar manner that the median is a robust measure of central tendency.

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Solve for x & y. 30x−42y=−6 and 5x−7y=−1

vladimir1956 [14]

Um.. Both of those equations are the exact same. Divide both sides of the first equation by 6. You get:

$\frac{30x - 42y}{6} = \frac{-6}{6} \\ 5x - 7y = -1$

That is the exact same as the second equation. This system has an infinite number of solutions. 5x - 7y = -1 is a line, so basically every point on that line is a solution to the system.

For example, $x = 0$ and $y = \frac{1}{7}$ would work, but so would $x = 1$ and $y = \frac{6}{7}$

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2 years ago

The size of a laptop monitor is usually measured along the diagonal. A rectangular monitor is 12 inches long and 7 inches tall.

Akimi4 [234]

16 inch is your answer

5 0

3 years ago

Read 2 more answers

Construct the discrete probability distribution for the random variable described. Express the probabilities as simplified fract

lisov135 [29]

Answer:

$P(X = 0) = 0.03125$

$P(X = 1) = 0.15625$

$P(X = 2) = 0.3125$

$P(X = 3) = 0.3125$

$P(X = 4) = 0.15625$

$P(X = 5) = 0.03125$

Step-by-step explanation:

For each toss, there are only two possible outcomes. Either it is tails, or it is not. The probability of a toss resulting in tails is independent of any other toss, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Fair coin:

Equally as likely to be heads or tails, so $p = 0.5$

5 tosses:

This means that $n = 5$

Probability distribution:

Probability of each outcome, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X = 2) = C_{5,2}.(0.5)^{2}.(0.5)^{3} = 0.3125$

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

$P(X = 4) = C_{5,4}.(0.5)^{4}.(0.5)^{1} = 0.15625$

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

5 0

2 years ago

A meteorologist is monitoring the atmospheric pressure as his height above sea level varies. The meteorologist determines that t

Burka [1]

Answer:

The function $p(n)$ is represented by $p(n) = 103\cdot (0.88)^{n}$ .

Step-by-step explanation:

Statement indicates that atmospheric pressure decreases exponentially when height above sea level is increased. This fact is represented by the following model:

$p(n) = p_{o}\cdot r^{n}$ (Eq. 1)