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2 years ago

Helppppp plzzzzzzz nowww

Mathematics

1 answer:

Harlamova29_29 [7]2 years ago

8 0

Answer:

C the 90 degrees

Step-by-step explanation:

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15,23,31,39, find the 60th term

ra1l [238]

Answer:

487

Step-by-step explanation:

We can use this formula to find the 60th term:

8n+7

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2 years ago

Find the slope of a line through the points (7,1) and (4,0)

Zanzabum

Answer:

3,9

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Select the correct answer. which data set is the farthest from a normal distribution? a. 2, 3, 3, 4, 4, 4, 5, 5, 6 b. 3, 4, 5, 6

tigry1 [53]

The answer choice which is the farthest from a normal distribution is; Choice E; 2, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10.

<h3>Which data set is farthest from a normal distribution?</h3>

A normal distribution, is a data set which when graphed must follow a bell-shaped symmetrical curve centered around the mean. Additionally, such distribution must adhere to the empirical rule that indicates the percentage of the data set that falls within (plus or minus) 1, 2 and 3 standard deviations of the mean.

On this note, upon evaluation of the data sets, it follows that answer choice E represents the data set that's most farthest from a normal distribution.

Read more on normal distribution;

brainly.com/question/26678388

#SPJ4

4 0

2 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

You are hanging 3 display shelves with the same width on a wall so that there is 18 inches of space above each shelf for placing

Kazeer [188]

Answer:

You need to have a minimum of 16 inches of clearance.

Step-by-step explanation:

4 0

3 years ago