P2m divied 4; use m =9 and p=3

Confidence intervals for the means provide an estimate for where the true mean lies.

stepladder [879]

It is true that the confidence intervals for the mean provide an estimate for where the true mean lies.

In statistics, a confidence interval denotes the likelihood that a population parameter will fall between a set of values for a given proportion of the time. A confidence interval depicts the likelihood that a parameter will fall between two values near the mean. Confidence intervals quantify the degree of uncertainty or certainty in a sampling procedure.

The mean is a basic mathematical average of two or more values. There are two sorts of means that may be calculated: the arithmetic mean and the geometric mean. A mean tells you the average of a bunch of values, which helps you contextualize each data point.

To learn more about confidence interval, visit :

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3 0

1 year ago

If an exterior angle of a regular polygon is 21.2° how many sides does the polygon have?

rosijanka [135]

$\bf \textit{sum of all exterior angles in a polygon}\\\\
n\theta =360\quad 
\begin{cases}
n=\textit{number of sides}\\
\theta =\textit{angle in degrees}\\
----------\\
\theta =21.2^o
\end{cases}\implies n21.2=360
\\\\\\
n=\cfrac{360}{21.2}\implies n\approx 16.981132\implies n\approx 17$

the reason why the value is not an integer, is most likely because the 21.2° is just an approximation.

8 0

3 years ago

Is 3.6 km or 3,600 m is greater or eqaul

SVEN [57.7K]

3.6 km is equal to 3,600 m :)

7 0

4 years ago

Read 2 more answers

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

6 0

3 years ago

What is the point that crosses the y line called?

RUDIKE [14]

Answer

its called x

Step-by-step explanation:

6 0

3 years ago