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frutty [35]
3 years ago
8

C=36x+200 R=76x find the break-even point

Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer:

x = 5

Step-by-step explanation:

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4/11 divided by 1/7 <br>​
MA_775_DIABLO [31]
4/11•7/1=28/11
Answer: 28/11 or 2 6/11
5 0
3 years ago
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3. A cone has a heighth and a base with radius r. You want to change the cone so its volume is tripled. What is the new height i
Sphinxa [80]

Answer:

Hence the new height  is 3 times the original height .(h1=3h)

Step-by-step explanation:

Given:

A cone with  has height and base with radius r .

To Find:

What is new height

Solution:

Consider as cone with height h base radius r, and volume v

Here given that only height changes for the cone  i.e. r remains the unchanged or same or constant

The volume for a regular cone is given by ,

V=pie*r^2*h/3

Here V is directly proportional to h i.ee pie ,3 and r being constant

V=K*h

i.e V/h=constant

V1 and h1 are new dimensions for new cone

V/h=V1/h1

Here V1=3V

So V/h=3V/h1

1/h=3/h1

i.e  h1=3h

Hence the height  is 3 times the original height .

4 0
3 years ago
A farmer has 20 boxes of eggs
aliina [53]
I guess that is 1:10
Hope ur help !;(
5 0
3 years ago
A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

5 0
3 years ago
Item 8 Solve for x. Use the quadratic formula. 2x2−5x−9=0 Enter the solutions, in simplified radical form, in the boxes.
Lubov Fominskaja [6]

Answer:

5+√97/4

Also 5-√97/4

Step-by-step explanation:

The Quadratic formula is x=-b+-√b^2-4ac/2a

This means that we should plug the values for A B AND C into the formula

We can work out that

<u><em>A = 2</em></u>

<u><em>B=-5</em></u>

<u><em>C=-9</em></u>

Once we have put these into the formula we get

5+√97/4 (all over 4) aka 3.71

Also 5-√97/4 (all over 4) aka -1.21

7 0
3 years ago
Read 2 more answers
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