200.000 rounded to the nearest tenth i'm 9 and still know this so if you don't you are not smart

Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled t-test and

Vesna [10]

Answer:

okay not clear,

upload for me a word document and i will do this for you at my g mail which is [email protected] for clarity and accuracy please

Download docx

5 0

2 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

2 years ago

Can you please help me out with a question

kozerog [31]

As you can see in the given figure, there are two intersecting chords inside the circle.

Recall that the "Intersecting Chords Theorem" is given by

$AE\cdot EC=BE\cdot DE$

For the given case, we have

AE = 7

BE = 6

EC = 9

Let us substitute these values into the above equation and solve for DE

$\begin{gathered} AE\cdot EC=BE\cdot DE \\ 7\cdot9=6\cdot DE \\ 63=6\cdot DE \\ \frac{63}{6}=DE \\ 10.5=DE \\ DE=10.5 \end{gathered}$

Therefore, the length of DE is 10.5 units.

8 0

7 months ago

The diagonal of a rectangle is 25 inch. the width is 15 in. what is the area of the rectangle?

givi [52]

D= diagonal= 25 in
w= 15 in

A= w√(d^2﹣w^2)

A= (15)√(25^2 - 15^2)
square in parentheses

A= (15)√(625 - 225)
subtract in parentheses

A= (15)√(400)
take square root of 400

A= (15)(20)

A= 300 in^2

ANSWER: The area is 300 inches squared

Hope this helps! :)

6 0

2 years ago

Given a triangle JLM with vertices J(-4,3), L(-2,7), and M(2,7). What would the coordinates be if it was reflected across the li

topjm [15]

Answer: M'(2, - 5), L'(-2, -5), j'(-4, - 1)

Step-by-step explanation:

When we do a reflection over a given line, the distance between all the points (measured perpendicularly to the line) does not change.

The line is y = 1.

Notice that a reflection over a line y = a (for any real value a) only changes the value of the variable y.

Let's reflect the points:

J(-4, 3)

The distance between 3 and 1 is:

D = 3 - 1 = 2.

Then the new value of y must also be at a distance 2 of the line y = 1

1 - 2 = 1

The new point is:

j'(-4, - 1)

L(-2, 7)

The distance between 7 and 1 is:

7 - 1 = 6.

The new value of y will be:

1 - 6 = -5

The new point is:

L'(-2, -5)

M(2,7)

Same as above, the new point will be:

M'(2, - 5)

3 0

2 years ago