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GrogVix [38]
3 years ago
7

Please help it’s for a unit test!

Mathematics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

X=___ or x=8 or one of my other answers in the explanation! I hope this helped!

Step-by-step explanation:

it must be 4 because 4 goes on the side by +4 and its x+4 then the other side of the shape is 8 so that meaning x is 4 then 2x+1 is 2 times X + 1 so 2 times 4... 2,4,6,8 or 4+4=8 so then 2 times + 1 is 8 + 1 so 8+1=9 so 2x+1 might also be 12 because its on the other side of the problem/ shape so you use all that you have so you could maybe plus the other side not the problem side for making it a and c and stuff but also 8+12 so 8+12=20 then x might be 20 but you answer meaning X is 8. X=8 Brainliest? i tried really hard! I am not even near this grade whatever grade this is... Plz! Thanks! Thank you! Have a great day or week or weekend! i answer a lot of questions and help 114 people or something or more! Plz put 5 stars, a heart, and brainliest i would really appreciate that thank you! Enjoy the rest or your day! Bye! <3 :)

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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
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Answer:

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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.

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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)

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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.

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