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3 years ago

Sofia asked 50 students at her school whether they prefer music class or art class as an

Mathematics

1 answer:

Archy [21]3 years ago

3 0

Answer:

1 and 4

or say it as

More seventh graders prefer art class to music class as an elective

There is no association between a student's grade level and the electives they prefer

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Brainliest! Please help me

lapo4ka [179]

Hey there!

1/2 x 4/4 = 4/8

So , ? = 4

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2 years ago

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What of these types of income is taxed?

geniusboy [140]

In most cases, income tax brackets are progressive, meaning that the greater the income, the higher the rate of taxation. Federal rates for the 2013 tax year range from 10 to 39.6 percent. State and city rates are generally much lower.

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4 years ago

Use the trigonometric subtraction formula for sine to verify this identity: sin((π / 2) – x) = cos x

KIM [24]

Answer:

Step-by-step explanation:

$sin (\frac{\pi}{2} - x) = cos x \\\\ sin (a - b) = sin a.cos b - sin b.cos a \\\\ sin (\frac{\pi}{2} - x) = sin \frac{\pi}{2}.cos x - sin x.cos \frac{\pi}{2} \\\\ sin \frac{\pi}{2} = 1; cos \frac{\pi}{2} = 0 \\\\ sin (\frac{\pi}{2} - x) = 1.cos x - sin x.0 \\\\ sin (\frac{\pi}{2} - x) = cos x$

I hope I helped you.

4 0

3 years ago

Find an equation of the line that passes through (3,4) and parallel to 4x-2y=5

Alex73 [517]

4x - 2y = 5
4x - 4x - 2y = -4x + 5
-2y = -4x + 5
-2 -2
y = 2x - 2¹/₂

y - y₁ = m(x - x₁)
y - 4 = 2(x - 3)
y - 4 = 2(x) - 2(3)
y - 4 = 2x - 6
+ 4 + 4
y = 2x - 2

4 0

3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago