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2 years ago

What is the sum of (x - 8) - (2x + 2)

Mathematics

2 answers:

Dahasolnce [82]2 years ago

7 0

Answer:

the answer is 4

vitfil [10]2 years ago

7 0

Answer:

$-z-10$

Step-by-step explanation:

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Find the midpoint of the segment with the following endpoints. (7,-6) and (4, -10)

Firlakuza [10]

Answer:

(5.5,-8)

Step-by-step explanation:

(7 + 4)/2 = 5.5

(-6 + -10)/2 = -8

8 0

3 years ago

Find the roots of f(x)=2x^2+7x+3 by using the quadratic formula

Oksana_A [137]

Answer:

f(x) = 2 ( x + 7/4)2 - 25/8

Step-by-step explanation:

4 0

2 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

3 years ago

If a drink costs 2.32, how many drinks can you buy with 13 dollars?

Llana [10]

Answer:

5 drinks

Step-by-step explanation:

If the drinks cost $2.32 and you have $13 you can divide 13 by 2.32

13/2.32+ 5.6034...

You cannot buy a partal drink so you look at only the whole number which is 5.

6 0

3 years ago

Read 2 more answers

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