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Kryger [21]
3 years ago
8

if a player has the cards 1, -7, and 4, find two ways to get five. explain and give the equation (absolute value can be used)

Mathematics
1 answer:
dimulka [17.4K]3 years ago
3 0
-7+4+1 Hope u appreciate my answer


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On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1
muminat

Answer:

Probability = \frac{1}{24}

Step-by-step explanation:

Given

Oranges = 3

Apples = 5

Banana = 2

Total = 10

Required

Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

Probability = P(Orange) * P(Apple) * P(Banana)

Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}

<em>The difference in the numerator is as a result of picking the fruit without replacement</em>

Probability = \frac{30}{720}

Probability = \frac{1}{24}

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2 years ago
Indica el valor del 3 en cada uno de los siguientes números 35.097 _ 753.978 _ 7.543 _ 312.501
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3 years ago
Hot Shot Electronics is designing a packing box for its new line of Acoustical Odyssey speakers. The box is a rectangular prism
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V= W × H × L
81000 = 30 × H × 45
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3 years ago
Select the equation that represents the problem let x represent the unknown number
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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
2 years ago
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