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2 years ago

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Mathematics

1 answer:

rjkz [21]2 years ago

7 0

Answer:

UP THE DOWN THAT WHAT I GOT

Step-by-step explanation:

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Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

2 years ago

A chili cook-off has adult tickets (a) and child tickets (c) . A group of Visitors buy three adult tickets and two child tickets

77julia77 [94]

Answer:

Equation 1 would be: 3a+2C

Equation 2 would be: 5a+3C

Step-by-step explanation:

4 0

2 years ago

Can someone help me with this question?

Andrei [34K]

Answer:

I don’t know what 14 and 15 are but I know 16 is 6 ft

Step-by-step explanation:

3 0

3 years ago

masha68 [24]

Answer: Um It’s C?

Step-by-step explanation:

8 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:

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

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:

$n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\
\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}$

4 0

3 years ago

Graph -1,-1 on this graph​

Graph -1,-1 on this graph