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hammer [34]
3 years ago
12

xplain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that the

standard deviation has over the interquartile​ range? Choose the correct answer below. A. The interquartile range is preferred when the data are bell shaped. An advantage of the standard deviation is that it is resistant to extreme values. B. The interquartile range is preferred when the data are bell shaped. An advantage of the standard deviation is that it increases as the dispersion of the data increases. C. The interquartile range is preferred when the data are not skewed or no have outliers. An advantage of the standard deviation is that it uses all the observations in its computation. D. The interquartile range is preferred when the data are skewed or have outliers. An advantage of the standard deviation is that it uses all the observations in its computation. E. The interquartile range is preferred when the distribution is symmetric. An advantage of the standard deviation is that it increases as the dispersion of the data increases. F. The interquartile range is preferred when the distribution is symmetric. An advantage of the standard deviation is that it is resistant to extreme values.
Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0

Answer: The interquartile range is preferred when the data are skewed or have outliers. An advantage of the standard deviation is that it uses all the observations in its computation.

Step-by-step explanation:

The interquartile range are resistant to outliers and can be used for data having glaring outliers.

Outliers are values that are too far away from the central value.

Interquartile range is a better option than standard deviation when there are a few extreme values and the distribution is skwed.

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Find the following GCD and LCM: (3.a) GCD(343,550), LCM(343, 550). (3.b) GCD(89, 110), LCM(89, 110). (3.c) GCD(870, 222), LCM(87
Ierofanga [76]

Answer with Step-by-step explanation:

(3.a) GCD(343,550), LCM(343, 550).

343=7×7×7

550=5×5×2×11

GCD(343,550)=1

LCM(343,550)=7×7×7×5×5×2×11=188650

(3.b) GCD(89, 110), LCM(89, 110).

89=1×89

110=5×2×11

GCD(89, 110)=1

LCM(89, 110)=89×5×2×11=9790

(3.c) GCD(870, 222), LCM(870, 222).

870=2×3×5×29

222=2×3×37

GCD(870, 222)=2×3=6

LCM(870, 222)=2×3×5×29×37=32190

5 0
3 years ago
In both of the problems where would you put parentheses to make it statement true?
Darina [25.2K]
In the first one it would be (21-8)-6=7.
and in the second one it would be 21-(8-6)=19
5 0
3 years ago
Yolanda makes wooden boxes for a craft fair. she makes 100 boxes like the one shown and she wants to paint all the outside faces
Masteriza [31]
Honestly... i don't even know

5 0
3 years ago
Read 2 more answers
PLEASEE HELP!
melamori03 [73]

Answer:

Two non zero vectors, a and b are parallel when they are scalar multiples of each other such that a = c·b where c is a scalar quantity.

Therefore, in order to find a vector that is parallel to the vector, b = (-2, -1), we multiply the vector, b by a scaler quantity

Step-by-step explanation:

Given that the vector b = (-2, -1) can be written as follows;

b = -2·i - j, we have;

\left | b \right | = √((-2)² + (-1)²) = √5

Therefore, we have;

The coordinates of the endpoint of the vector are (-2, 0) and (0, -1)

Therefore, the slope of the vector = (-1 - 0)/(0 - (-2)) = -1/2

The slope of parallel vectors are equal, which gives the slope of the parallel vector = -1/2 = (λ × (-1 - 0))/(λ ×(0 - (-2))

Therefore, a parallel vector is obtained from a vector by multiplying with a scaler product.

7 0
2 years ago
Work out the circumference of this circle.
tigry1 [53]
C=pid
Since d=9m and pi=3.142, C=3.142*9m=28.278
Rounded to 1 decimal point =28.3
3 0
3 years ago
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