Answer:
Not going to give answer, but let you solve
Step-by-step explanation:
You first have to start by drawing a number line and mark two integers between the fraction lines. Then you have to divide the section between the two integers in an equal number of parts which is equal to the denominator.
Hope I helped
Answer
$540
Step-by-step explanation:
hopes this helps
No, because 1 is not a mixed and there is not another pair of nubers that equal 2 in mixed numbers.
Answer:
Exact Form: z = 0, -2/25
Decimal Form: 0, -0.08
Step-by-step explanation:
I used cymath.com. I highly recommend it. It shows you the steps to the problem you want to solve!
I hope this helps! :)
The last step is to solve for z
Firstly, we'll fix the postions where the
women will be. We have
forms to do that. So, we'll obtain a row like:

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

Since there is no women sitting together, we must write that
. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

The equation (i) can be rewritten as:

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: ![\dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%28n%29%2B%28m-n%2B1%29%5D%21%7D%7B%28n%29%21%28m-n%2B1%29%21%7D%3D%5Cdfrac%7B%28m%2B1%29%21%7D%7Bn%21%28m-n%2B1%29%21%7D)
[I can write the proof if you want]
Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: 
Multiplying all results:
