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2 years ago

What is the difference between standard deviation and standard error

Mathematics

2 answers:

BARSIC [14]2 years ago

7 0

Answer:

Step-by-step explanation:

The standard deviation (SD) can be defined as a entity that measures the amount of dispersion for a set of data from the mean, while the standard error of the mean (SEM) measures the magnitude of deviation of the sample mean of the data from the true population mean.

Natasha_Volkova [10]2 years ago

6 0

Answer:

The standard deviation (SD) is a measure of the variability of a random variable. A standard error (SEM) is a measure of how precise an estimate is.

Step-by-step explanation:

The standard deviation (SD) is a measure of the variability of a random variable. A standard error (SEM) is a measure of how precise an estimate is.

The contrast between these two terms reflects the important distinction between data description and inference.

The SEM is always smaller than the SD.

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D=a+b+c solve for b I need some help

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Hi Walker

d=a+b+c

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a+b+c=d

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b+c=-a+d

Now add -c to both sides

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3 years ago

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3 years ago

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3 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$