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4 years ago

How do you find the range on a box and whisker plot

Mathematics

1 answer:

faust18 [17]4 years ago

5 0

<span>the difference between the largest and smallest data values in a data set.</span>

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Find the value of |-4| - 5 · 2(-1 + 3).

vredina [299]

Use PEMDAS

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

and |a| = a for a ≥ 0 or |a| = -a for a < 0.

|-4| - 5 · 2(-1 + 3) =-(-4) - 5 · 2(2) = 4 - 10(2) = 4 - 20 = -16

6 0

3 years ago

Read 2 more answers

In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0 respectively. If the coefficient sk

RUDIKE [14]

Answer:

a) The third quartile Q₃ = 56.45

b) Variance $\mathbf{ \sigma^2 =2633.31}$

Step-by-step explanation:

Given that :

$Q_1$ = 30.8

Median $Q_2$ = 48.5

Mean = 42

a) The mean is less than median; thus the expression showing the coefficient of skewness is given by the formula :

$SK = \dfrac{Q_3+Q_1-2Q_2}{Q_3-Q_1}$

$-0.38 = \dfrac{Q_3+30.8-2(48.5)}{Q_3-30.8}$

$-0.38Q_3 + 11.704 = Q_3 +30.8 - 97$

$1.38Q_3 = 77.904$

Divide both sides by 1.38

$Q_3 = 56.45$

b) The objective here is to determine the approximate value of the variance;

Using the relation

$SK_p = \dfrac{Mean- (3*Median-2 *Mean) }{\sigma}$

$-0.38= \dfrac{42- (3 *48.5-2*42) }{\sigma}$

$-0.38= \dfrac{(-19.5) }{\sigma}$

$-0.38* \sigma = {(-19.5) }{}$

$\sigma =\dfrac {(-19.5) }{-0.38 }$

$\sigma =51.32$

Variance = $\sigma^2 =51.32^2$

$\mathbf{ \sigma^2 =2633.31}$

7 0

3 years ago

The quotient of

vova2212 [387]

$\displaystyle{ \frac{y^2-16}{4y-16} \div \frac{y+4}{16y}= \frac{y^2-16}{4y-16}\cdot\frac{16y}{y+4}\\\\$

$\displaystyle{ = \frac{y^2-4^2}{4(y-4)} \cdot\frac{16y}{y+4}= \frac{(y-4)(y+4)}{4(y-4)}\cdot\frac{16y}{y+4} = \frac{16y}{4}=4y$

Answer: 4y

Remark: $y^2-4^2=(y-4)(y+4)$ by the difference of squares identity, or formula.