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2 years ago

9

I need these pls!!!!!

1 answer:

Allisa [31]2 years ago

4 0

Answer:

1) $38.40 2) $255. 3) $5.80 4) $3.12

Step-by-step explanation:

So you have to multiply the years by the amount of money you have. Then you multiply that by the percent and divide by 100. (For the last one: because its a month you have to make it a fraction so it was 8/12)

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3 years ago

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2 years ago

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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