Find the mean. Round to the nearest tenth if necessary.

Mathematics

2 answers:

myrzilka [38]3 years ago

6 0

Answer:

27 is the answer

Step-by-step explanation:

Took the test and got it right

kondor19780726 [428]3 years ago

5 0

Answer:

Step-by-step explanation:

The actual mean is 27.008333333333, round that as you will.

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The Washington Monument in Washington, D.C., is about 556 ft. tall. A three-dimensional puzzle of the Washington Monument is 24

Umnica [9.8K]

Set up a ratio. But first convert the height of the puzzle to feet so we are comparing the same unit of measure.

STEP 1
convert 24 inches to feet

12 inches= 1 foot

24 inches= 2 feet

STEP 2
set up ratio.

= 2 feet/556 feet or 2:556
simplify both by 2

= 1 ft/278 ft or 1:278

ANSWER: 1/278 or 1:278 (can be written either way)

Hope this helps! :)

8 0

4 years ago

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .