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2 years ago

Please Help Me Will Give Brainliest AYUDA

Mathematics

2 answers:

zvonat [6]2 years ago

8 0

Answer:

Option A is right

Step-by-step explanation:

In any data set, we find that the range is the difference between maximum and minimum. The range cannot give full information about the variability of the data set.

Outliers are the ones which are either below Q1-1.5IQR or Q3+1.5IQR

Thus Inter quartile range, the difference between I quartile and III quartile gives a better idea about variability and also outlier.

Hence out of the four options given, we find that option B is the right one.

Option

A. The IQR is the best measure of variability because the distribution has an outlier.

NikAS [45]2 years ago

3 0

Answer: the answer is a

Step-by-step explanation: hope it is correct .

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$