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worty [1.4K]
3 years ago
13

I need help pls it’s pretty hard

Mathematics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

x   8  -5

9  72   -45

-6 -48  30

Step-by-step explanation:

We have to complete the multiplication grid, where each cell is the result of multiplying the column lable by the row label.

For the blank in the first column, we have to multiply:

8×(-6)=-48

We now have to find the column label for the second column.

If we call this label x, we know that -6 times x is equal to 30.

Then, we can find x as:

(-6)×x=30

x=\frac{30}{-6}

x=-5

Then, we can complete the first row by multiplying -5 times 9:

-5×9=-45

Then, the table becomes:

x   8  -5

9  72   -45

-6 -48  30

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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
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(a) The value of P (X ≤ 2) is 0.8729.

(b) The value of P (X ≥ 5) is 0.0072.

(c) The value of P (1 ≤ X ≤ 4) is 0.7154.

(d) The probability that none of the 25 boards is defective is 0.2774.

(e) The expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defective boards.

The probability that a circuit board is defective is, <em>p</em> = 0.05.

The sample of boards selected is of size, <em>n</em> = 25.

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The probability mass function of <em>X</em> is:

P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...

(a)

Compute the value of P (X ≤ 2) as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729

Thus, the value of P (X ≤ 2) is 0.8729.

(b)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = 1 - P (X < 5)

              =1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072

Thus, the value of P (X ≥ 5) is 0.0072.

(c)

Compute the value of P (1 ≤ X ≤ 4) as follows:

P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

                   =\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154

Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.

(d)

Compute the value of P (X = 0) as follows:

P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774

Thus, the probability that none of the 25 boards is defective is 0.2774.

(e)

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

E(X)=np=25\times 0.05=1.25

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09

Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

8 0
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22:33 in its simplest form
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divide both numbers by gcf
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33/11=3

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