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2 years ago

The data set below has a lower quartile of 13 and an upper quartile of 37.

Mathematics

1 answer:

timama [110]2 years ago

6 0

The correct option regarding the outliers of the data-set is given by:

The greatest value, 78, is the only outlier.

<h3>How to use the quartiles of a data-set to identitfy outliers?</h3>

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.

The first quartile is the median of the first half of the data-set.

The third quartile is the median of the second half of the data-set.

The interquartile range is the difference of the third quartile with the first quartile.

Measures that are more than 1.5 IQR from Q1 and Q3 are considered outliers.

The IQR for this problem is:

IQR = 37 - 13 = 24.

Hence the bounds for outliers are:

Less than 13 - 1.5 x 24 = -23.

Greater than 37 + 1.5 x 24 = 73,

The options are:

No outliers.

Only 1 is an outlier.

Only 78 is an outlier.

Both 1 and 78 are outliers.

Hence the correct option is that only 78 is an outlier.

More can be learned about outliers of a data-set at brainly.com/question/17083142

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Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

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enot [183]

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astra-53 [7]

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Step-by-step explanation:

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3 years ago

The measure of 5 of the interior angles of a hexagon are 110, 120, 90, 140, and 125. what is the measure of the smallest exterio

Romashka-Z-Leto [24]

Answer:

A) 40

Step-by-step explanation:

The sum of the interior angles of a regular polygon is:

$(n - 2) \times 180$

For a hexagon, we put n=6

$(6 - 2) \times 180 = 4 \times 180 = 720$

Let x be the 6th interior angle, then

$x + 110 + 120 + 90 + 140 + 125 = 720$

$x + 585 = 720$

$x = 720 - 585 = 135$

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Therefore the least exterior angle will be:

$180 - 140 = 40$

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3 years ago