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2 years ago

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Mathematics

1 answer:

gulaghasi [49]2 years ago

5 0

Step-by-step explanation:

Hope you understand. If you don't ask me anything

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Please solve the following question.

miss Akunina [59]

The expected number of defective sample is 0.25

<h3>The probability distribution</h3>

The given parameters are:

Population, N = 30
Sample, n = 2
Selected, x = 4

Start by calculating the defective proportion using:

$p = \frac{x}{N}$

So, we have:

p = 4/30

p = 0.13

The probability distribution is calculated as:

$P(x) = ^nC_x * p^x *(1 - p)^{n - x}$

So, we have:

$P(0) = ^2C_0 * 0.13^0 *(1 - 0.13)^{2 - 0} = \frac{19}{25}$

$P(1) = ^2C_1 * 0.13^1 *(1 - 0.13)^{2 - 1} =\frac{23}{100}$

$P(2) = ^2C_2 * 0.13^2 *(1 - 0.13)^{2 - 2} = \frac{1}{100}$

So, the probability distribution is:

x 0 1 2

P(x) 19/25 23/100 1/100

<h3>The expected number of defective sample</h3>

This is calculated using:

$E(x) = \sum x * P(x)$

So, we have:

E(x) = 0 * 19/25 + 1 * 23/100 + 2 * 1/100

Evaluate

E(x) = 0.25

Hence, the expected number of defective sample is 0.25

Read more about binomial distribution at:

brainly.com/question/15246027

#SPJ1

8 0

1 year ago

Please solve.this question

bonufazy [111]

Answer:

See explanation

Step-by-step explanation:

Prove equality

$(1-\cot A)^2+(\tan A-1)^2=4\csc 2A(\csc 2A-1)$

Consider left and right parts separately.

$(1-\cot A)^2+(\tan A-1)^2\\ \\=1-2\cot A+\cot^2 A+\tan^2 A-2\tan A+1\\ \\=(1+\cot^2 A)+(1+\tan^2 A)-2(\cot A+\tan A)\\ \\=\dfrac{1}{\sin^2 A}+\dfrac{1}{\cos ^2A}-2\dfrac{\cos^2 A+\sin^2 A}{\cos A\sin A}\\ \\=\dfrac{1}{\sin^2 A}+\dfrac{1}{\cos ^2A}-\dfrac{2}{\cos A\sin A}\\ \\=\dfrac{\cos^2A+\sin^2A-2\cos A\sin A}{\cos ^2A\sin ^2A}\\ \\=\dfrac{1-\sin 2A}{\frac{1}{4}\cdot 4\cdot \cos^2A\sin^2 A}\\ \\=\dfrac{1-\sin 2A}{\frac{1}{4}\cdot \sin^22 A}\\ \\=4\dfrac{1-\sin 2A}{\sin^22 A}$

<u>Right part:</u>

$\csc 2A=\dfrac{1}{\sin 2A}$

Hence

$4\csc 2A(\csc 2A-1)\\ \\=4\cdot \dfrac{1}{\sin 2A}\cdot \left(\dfrac{1}{\sin 2A}-1\right)\\ \\=4\cdot \dfrac{1}{\sin 2A}\cdot \dfrac{1-\sin 2A}{\sin 2A}\\ \\=4\dfrac{1-\sin 2A}{\sin^2 2A}$