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SVETLANKA909090 [29]
3 years ago
7

A construction crew can dig a pit at a rate of

Mathematics
1 answer:
makvit [3.9K]3 years ago
3 0
The answer would be F
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A pack of water bottles contains 24 bottles. A man decided to buy 3 bottles of water. How much bottles of water did he buy?
lukranit [14]

Answer: 1/8 pack

Step-by-step explanation:

Pack of bottles = 24 bottles

Bought = 3 bottles

24 bottles ___ 1 pack

3 bottles ___ x= 3/24 pack

(3 bottles * 1 pack) / ( 24 bottles) = 3/24 packs = 1/8 pack

3 0
3 years ago
Mindy and Troy combined ate 999 pieces of the wedding cake. Mindy ate 333 pieces of cake and Troy had 1/4 of the total cake. ​ W
lubasha [3.4K]

Answer:

Step-by-step explanation:

333+p/4=999

p/4=666

p=2664

6 0
3 years ago
Read 2 more answers
Which of the following multiplication equations doesn't have the same solution as the others?
n200080 [17]
The answer is option 2: 4b = 7. The other options all equal 4.
3 0
3 years ago
Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and
ollegr [7]

Firstly, we'll fix the postions where the n women will be. We have n! forms to do that. So, we'll obtain a row like:

\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}

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

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)

Since there is no women sitting together, we must write that x_2,x_3,...,x_{n-1},x_n\ge1. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}

The equation (i) can be rewritten as:

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)

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)!}

[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: m!

Multiplying all results:

n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

4 0
3 years ago
What is the answer to...<br> 70 - 6 × (4 - 1)2
elixir [45]

Answer:

34

Step-by-step explanation:

5 0
3 years ago
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