I think your forgot to add some parts of this question, because you can't do anything with this except simplify it down a little maybe. But, I will provide the simplified version of this.

$\frac{x-4}{x^2-2x}+\frac{4}{x^2-4}$

= $\frac{x-4}{x\left(x-2\right)}+\frac{4}{x^2-4}$

= $\frac{x-4}{x\left(x-2\right)}+\frac{4}{\left(x+2\right)\left(x-2\right)}$

= $\frac{\left(x-4\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x}{\left(x+2\right)\left(x-2\right)x}$

= $\frac{\left(x-4\right)\left(x+2\right)+4x}{x\left(x-2\right)\left(x+2\right)}$

= $\frac{x+4}{x\left(x+2\right)}$

Step-by-step explanation:

Hope this helped!

8 0

3 years ago

A company manufacturers video games with a current defect rate of 0.95%. To make sure as few defective video games are delivered

RSB [31]

There would be about 19 defective products delivered.

First, let's start with the number of defective games.
100000 x 0.0095 = 950

Now, the test will catch 98% of those defects. That means 2% of the defects will get through to the consumers.
0.02 x 950 = 19

3 0

3 years ago

Read 2 more answers

Help on 12 Aswell please

aev [14]

108 divided by each cup of fruit and seltzer will get you how many cups per serving and for C divide how many cups of seltzer with cups of orange juice

6 0

3 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

5 0

4 years ago