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Law Incorporation [45]

3 years ago

8

Help please, will award brainliest

1 answer:

Blizzard [7]3 years ago

4 0

2x²-2x-12=0
x²-x-6=0
(x-3)(x+2)=0
x-3 =0, x+2=0
x(1)=3, and x(2)=-2
The solutions make the equation true, and the solutions give x-intercepts of the graph of the equation.
(-2, 0) and (3,0) are x -intercepts for the graph of y = 2x²-2x-12

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4 years ago

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of <em>X</em> as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

4 0

3 years ago