13) How do you know when to use the mean or median as the best measure of center?

Mathematics

1 answer:

scoray [572]3 years ago

6 0

Answer:

The mean is the most common measure of center. It is what most people think of when they hear the word "average". However, the mean is affected by extreme values so it may not be the best measure of center to use in a skewed distribution. The median is the value in the center of the data.

Step-by-step explanation:

hope this helps

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Round 0.9874 to the greatest non zero place

Anna71 [15]

0.9874 the greatest non zero place is 9 but the nearest digit is 8 when you round it off to 0.9 the answer becomes 1.0
0.9874~1.0

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3 years ago

The difference of two numbers is 16 and the sum of their squares is 130. What are the numbers.

Korvikt [17]

x - y = 16
x^2 + x^2 = 130

----------------
x^2 - y^2 = 256
+(x^2 + y^2 = 130)
2x^2 = 386
x^2 = 193
x = <span>13.892 (3DP)
----
</span>x - y = 16
13.892 - y = 16
-y = 2.108
y = -2.108
The numbers are: 13.892 and -2.108

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3 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative