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eimsori [14]
3 years ago
6

Solve for p. −4.5 + p6= −18.9

Mathematics
2 answers:
Vesna [10]3 years ago
7 0

Answer:

-2.4

Step-by-step explanation:

Given data

We have the expression before us as

−4.5 + p6= −18.9

Collect like terms

p6= −18.9+4.5

p6= -14.4

Divide both sides by 6

p= -14.4/6

p=-2.4

Hence the value of p= -2.4

qaws [65]3 years ago
4 0

Answer:

That one answer you see below is wrong

Step-by-step explanation:

Thats it

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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
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Step-by-step explanation:

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