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USPshnik [31]
3 years ago
7

To select a sample of five children from a class of 30, a researcher puts each child’s name on a slip of paper, then shuffles th

e papers in a hat and randomly picks five names. What kind of sampling is being used?
Mathematics
1 answer:
Ainat [17]3 years ago
7 0

Answer:

We conclude that  a researcher  used simple random sample.

Step-by-step explanation:

We know that select a sample of five children from a class of 30, a researcher puts each child’s name on a slip of paper, then shuffles the papers in a hat and randomly picks five names.  

We conclude that  a researcher  used simple random sample.

Because every children  has an equal chance of being included in the sample.

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One number is nine than another number. if the sum of the numbers is 65, find both numbers.​
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If your saying one number is 9 more than the other then the two numbers are 28 and 37
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Please, I really need help-
posledela
<h3>Answer: 48</h3>

=====================================================

Explanation:

Triangle ABC is isosceles because AC = BC.

The angles opposite these congruent sides are angle ABC and angle BAC. These are the base angles.

For any isosceles triangle, the base angles are congruent.

Angle BAC = 69 degrees is given. So angle ABC = 69 as well.

-------------------

The missing angle must add to the two other angles so that all three angles in a triangle add to 180

(angle ABC) + (angle BAC) + (angle BCA) = 180

69 + 69 + (angle BCA) = 180

138 + (angle BCA) = 180

angle BCA = 180-138

angle BCA = 42

-------------------

Angle DCE is congruent to angle BCA because they are vertical angles.

Triangle DCE is a right triangle. The missing angle is 90-(angle BCA) = 90-42 = 48

angle EDC = 48 degrees

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4 years ago
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Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4
anyanavicka [17]

Answer:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Step-by-step explanation:

Given

f(x) = \sin x\\

c = \frac{3\pi}{4}

Required

Find the Taylor series

The Taylor series of a function is defines as:

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

We have:

c = \frac{3\pi}{4}

f(x) = \sin x\\

f(c) = \sin(c)

f(c) = \sin(\frac{3\pi}{4})

This gives:

f(c) = \frac{1}{\sqrt 2}

We have:

f(c) = \sin(\frac{3\pi}{4})

Differentiate

f'(c) = \cos(\frac{3\pi}{4})

This gives:

f'(c) = -\frac{1}{\sqrt 2}

We have:

f'(c) = \cos(\frac{3\pi}{4})

Differentiate

f"(c) = -\sin(\frac{3\pi}{4})

This gives:

f"(c) = -\frac{1}{\sqrt 2}

We have:

f"(c) = -\sin(\frac{3\pi}{4})

Differentiate

f"'(c) = -\cos(\frac{3\pi}{4})

This gives:

f"'(c) = - * -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

So, we have:

f(c) = \frac{1}{\sqrt 2}

f'(c) = -\frac{1}{\sqrt 2}

f"(c) = -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

becomes

f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Rewrite as:

f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Generally, the expression becomes

f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Hence:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

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The tools essential in performing geometric constructions is B
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