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1 year ago

49 Which of the following is a measure

Mathematics

1 answer:

Mariulka [41]1 year ago

4 0

we are ask for measuring of center

there are 3 different measures

mode, when the data value occurs the most frequently

median, is the data value in the middle of a sorted list of data

mean is the average of the numbers

so the answer is A) mean and median

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I just drew a blank can someone please help

PIT_PIT [208]

Answer:

≈ 542,029 copies

Step-by-step explanation:

37,400 copies represents a 6.9% of the total copies sold to date.

Therefore; Copies sold to date will be 100%

Thus;

Total number of copies sold = (37,400/6.9)× 100

= 542,028.99 copies

≈ 542029 copies

3 0

3 years ago

Read 2 more answers

+5 pts Answered The table gives a few (x,y) pairs of a line in the coordinate plane. what is the x-intercept of the line?

bagirrra123 [75]

Answer:

Here maybe this image might help.

5 0

3 years ago

How does the graph of g(x) = 5x – 3 compare to the graph of f(x) = 5*?

Karo-lina-s [1.5K]

Answer:

the graph is shifted down 3 units

Step-by-step explanation:

5x-3

meaning the original graph of 5x going down three units

3 0

3 years ago

the grades recieved by 300students follow a normal distribution. the mean of the grades is 75% and the standard deviation is 8%.

tiny-mole [99]

Answer:

Correct answers ended up being 6.7% for an A. 100- 93.32= 6.68 round one place 6.7 %

For B it's 93.32 - 69.1463 = 24.17 round one place so 24.2%

Step-by-step explanation:

3 0

3 years ago

\int\limits^0_\pi {x*sin^{m} (x)} \, dx

Ket [755]

Let

$I(m) = \displaystyle \int_0^\pi x\sin^m(x)\,\mathrm dx$

Integrate by parts, taking

u = x ==> du = dx

dv = sinᵐ (x) dx ==> v = ∫ sinᵐ (x) dx

so that

$I(m) = \displaystyle uv\bigg|_{x=0}^{x=\pi} - \int_0^\pi v\,\mathrm du = -\int_0^\pi \sin^m(x)\,\mathrm dx$

There is a well-known power reduction formula for this integral. If you want to derive it for yourself, consider the cases where m is even or where m is odd.

If m is even, then m = 2k for some integer k, and we have

$\sin^m(x) = \sin^{2k}(x) = \left(\sin^2(x)\right)^k = \left(\dfrac{1-\cos(2x)}2\right)^k$

Expand the binomial, then use the half-angle identity

$\cos^2(x)=\dfrac{1+\cos(2x)}2$

as needed. The resulting integral can get messy for large m (or k).

If m is odd, then m = 2k + 1 for some integer k, and so

$\sin^m(x) = \sin(x)\sin^{2k}(x) = \sin(x)\left(\sin^2(x)\right)^k = \sin(x)\left(1-\cos^2(x)\right)^k$

and then substitute u = cos(x) and du = -sin(x) dx, so that

$I(2k+1) = \displaystyle -\int_0^\pi\sin(x)\left(1-\cos^2(x)\right)^k = \int_1^{-1}(1-u^2)^k\,\mathrm du = -\int_{-1}^1(1-u^2)^k\,\mathrm du$

Expand the binomial, and so on.

8 0

2 years ago