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vova2212 [387]
3 years ago
10

An accounting professor wishing to know how many MBA students would take a summer elective in international accounting did a sur

vey of the class she was teaching. Which kind of sample is this?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
6 0

Answer: Convenience sample.

Step-by-step explanation:

Convenience sample is also known as grab or accident or opportunity samples. It is a example of non probability sample that involves selecting of subjects because of the proximity, convenience and accessibility a researcher as to them. This type of samples are not reliable for data gathering when it involves a very large sample space, let's say a global audience.

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Answer:

42

Step-by-step explanation:

8 0
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If a card is chosen at random from a pack of 52 playing cards, what is the probability of a Queen or a Club?
bonufazy [111]
The answer is b because you have 4 queens in a deck of cards and 13 clubs in a deck so if you add them you would get 17/52 cards
6 0
3 years ago
What is the value of x? sin49°=cosx
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3 0
3 years ago
En la figura, la suma de las medidas p + q + r + s =/In the figure, the sum of the measures p + q + r + s =
sertanlavr [38]

Answer:

I think its 300

Step-by-step explanation

This is from my understanding, so I may be a bit wrong.

1. First observe the amount of triangles that form, you have two main triangles and one isosceles triangle that forms in the middle by how these are joined together.

2. Using this idea we will come up  with the equations of  each triangle that has been formed.

3. We can observe that almost all angles are named with a variable except the one in the middle, being an isosceles we will name the missing variable y.

4. Triangle 1(left side) or T1 equals 180 degrees but what angles form this triangle, these would be p+q+y = 180, T2(right side) = r+s+y = 180

5. After creating these two triangles we notice that we have not taken into account the triangle in the middle, we are given 1 angle of this isosceles. There is a mathematical property which I don't remember the name, but essentially the angle below is the same as the angle above, this being 120 degrees. This will result in T3 = 2y + 120 = 180

6. Solve for y, which gives us y=30, *<em>simple triangle math you know*</em>

7. Finally we fix our equations for p+q+r+s, essentially T1 and T2 will transform into p+q = 180-y and r+s = 180-y, we add them up forming p+q+r+s = 2(180-y)

8. Solve the equation p+q+r+s = 360 - 2y which when adding y becomes p+q+r+s = 360-60 = 300

Notes: when doing these exercises it is important to take into account every triangle, the isosceles triangle is very vital for solving it, as it gives us a way for making an equation that is single handedly dependent on y a constant we can find and use. A good approach for these issues would be writing every piece of information as separate sets and then trying to join them using algebra.

3 0
3 years ago
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

3 0
2 years ago
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