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3 years ago

Organize the following polynomial expressions from least to greatest based on their degree:

Mathematics

2 answers:

bixtya [17]3 years ago

7 0

The correct answer is
III, I, II, IV
8a - 5
6x2
18x3 + 5ab - 6y
4x3y + 3x2 - xy - 4

My name is Ann [436]3 years ago

7 0

Answer:

The correct answer is C)

III, I, II, IV

8a - 5

6x2

18x3 + 5ab - 6y

4x3y + 3x2 - xy - 4

Step-by-step explanation:

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Write an equation of a line that passes through the point (2,0) and is perpendicular to the line y = − 1/2 x + 3.

tatyana61 [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

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A process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of175. A ran

Tom [10]

Answer:

the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

Step-by-step explanation:

since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:

Z= S²*(n-1)/σ²

follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom

since

S² > 3.10 , σ²= 1.75 , n= 20

thus

Z > 33.65

then from χ² distribution tables:

P(Z > 33.65) = 0.02020

therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

8 0

3 years ago

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68.76−5.23 A 16.46 B 18.53 C 63.53 D 73.99

Romashka [77]

The answer is 63.53 subtracts 5.23 from 68.76

6 0

3 years ago

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Slove the inequality 5(x + 1) <25 x < 4 x > -4 x > 4 Pls help meee

viktelen [127]

Algebra

Use the distributive property to make 5x+5.

5x+5<25

Subtract 5 from both sides.

5x<20

Divide both sides by five.

x<4

7 0

3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago