9, 14, 7, 9, 3, 12, 5,4

Mathematics

1 answer:

Rus_ich [418]3 years ago

5 0

Answer:

mean: 65

median: 9

Step-by-step explanation:

You take all the numbers you have and add them up. Make it simple by adding smaller numbers together. Then once you have the total which is 65 you divide it by the total of numbers you have and then round because after dividing you get 8.125 and the safest way to round is to 9.

You might be interested in

In need of geometry help, once again my stupid brain cant remember anything.

melomori [17]

You cannot rely on the drawing alone to prove or disprove congruences. Instead, pull out the info about the sides and angles being congruent so we can make our decision.

The diagram shows that:

Side AB = Side XY (sides with one tick mark)
Side BC = Side YZ (sides with double tickmarks)
Angle C = Angle Z (similar angle markers)

We have two pairs of congruent sides, and we also have a pair of congruent angles. We can't use SAS because the angles are not between the congruent sides. Instead we have SSA which is not a valid congruence theorem (recall that ambiguity is possible for SSA). The triangles may be congruent, or they may not be, we would need more information.

---------------

So to answer the question if they are congruent, I would say "not enough info". If you must go with a yes/no answer, then I would say "no, they are not congruent" simply because we cannot say they are congruent. Again we would need more information.

3 0

3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.