Ask question

Ask question

All categories

3 years ago

14

What measure of spread is used when your data distribution is symmetrical?

1 answer:

lbvjy [14]3 years ago

4 0

What measure of spread is used when your data distribution is symmetrical?

Answer: When the data distribution is symmetrical, the appropriate measure of spread is standard deviation. The reason is both sides of symmetrical distribution will have same spread.

Symmetrical distribution: Symmetrical distribution is one when both sides of the distribution are identical. The symmetrical distribution is bell shaped and mean, median and mode are equal in symmetrical distribution.

Standard deviation: Standard deviation is the measure of dispersion. It measures the deviation of a set of numbers from its mean. It is defined as the square root of means of the squared deviations from the arithmetic mean

You might be interested in

Mason opens a savings account by making a $165.85 deposit. Every week, he deposits another $20.50 in the account. The following

monitta

Answer:

370.85

Step-by-step explanation:

20.50×10. +165.85=

4 0

3 years ago

Read 2 more answers

Six friends order lunch. Write an expression that shows the total cost if each person orders on sandwich and one cookie.

vodka [1.7K]

4.75×6=28.5
6 cookies= 6x
The expression is 6x+28.5

8 0

4 years ago

What is 3,200,000,000,000 in scientific notation

sashaice [31]

$3 200 000 000 000= 3.2 \cdot 10^{12}$

5 0

3 years ago

Determine if the set of polynomials {x^2 –2x+1, 2x² + 3x -4,-x2+x+5) is a linearly independent set in P2. Is it a basis for P? W

WITCHER [35]

Answer with Step-by-step explanation:

We are given that a set of polynomials ${x^2-2x+1,2x^2+3x-4,-x^2+x+5}$

We have to find that given set is a linearly independent set in $P_2$

and given set is a basis for $P_2$ or not.

Matrix of given set of polynomials

A= $\left[\begin{array}{ccc}1&-2&1\\2&3&-4\\-1&1&5\end{array}\right]$

Linearly Independent set :If any row or any column is not a linear combination of other rows or columns then the set is linearly independent set.

Any row or column is not a linear combination of other rows or columns.Therefore, given set is a linearly independent set .

We know that

$P_2=x^2$

Element of $P_2$ is of the form

$ax^2+bx+c$

Every element of $p_2$ is a linear combination of given set of polynomials.

Hence, given set is linearly independent in $p_2$ .

If any set is basis for any vector space then it satisfied the following two conditions

1.Given set is linearly independent.

2.Every element of given vector space spanned by the given set.

Given set of polynomials are linearly independent and spanned every element of $P_2$ .

Therefore, given set is as basis for $p_2$ because the set is linearly independent and spanned $P_2$ .

7 0

3 years ago

15 points attached file

sdas [7]

Answer:

x = 3 renders the same output for both functions

Step-by-step explanation:

From the graph, we see that at the point x = 3 both lines intersect. This means that their y value coincide for that value of x.

We check using the functions' expressions to verify that our answer is correct. Then we evaluate both functions separately at x = 3:

$f(3)= -\frac{2}{3} (3) + 1 = -2 + 1 = -1\\g(3) = \frac{1}{3} (3) -2 = 1-2 =-1$

So we conclude that effectively, the input x=3 produces equal output value (-1) for the two given functions.

4 0

4 years ago

Read 2 more answers