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4 years ago

Factor completely 2x^2+4x-2

Mathematics

1 answer:

Aleks04 [339]4 years ago

8 0

Answer:

2x^2+4x-2 = 2 * (x - 1 - $\sqrt{2}$ ) * (x - 1 + $\sqrt{2}$ )

Step-by-step explanation:

Factor completely 2x^2+4x-2

2x^2+4x-2 = 2* (x^2 + 2x - 1)

x = 1 + root( 4 - 4*(-1))/2

x = 1 + root(2)

and x = 1 - root(2)

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A newspaper published an article about a study in which researchers subjected laboratory gloves to stress. Among 225 vinyl? glov

romanna [79]

Answer:

H₀: ρ₁ ≤ ρ₂

H₁: ρ₁ > ρ₂

Z= $\frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} }$ ≈ N(0;1)

p-value < .00001

Reject the null hypothesis.

The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.

Step-by-step explanation:

Hello!

The objective is to test if the population proportion of vinyl gloves that leak viruses (population 1) is greater then the population proportion of latex gloves that leak viruses (population 2).

Sample 1 (Vinyl)

n₁= 225

Sample proportion 'ρ₁= 0.60

Sample 2 (Latex)

n₂= 225

Sample proportion 'ρ₂= 0.09

Pooled proportion: 'ρ= (x₁+x₂)/(n₁+n₂)= ( 'ρ₁ + 'ρ₂)/2= (0.60 + 0.09)/2= 0.345 ≅0.35

The hypothesis is:

H₀: ρ₁ ≤ ρ₂

H₁: ρ₁ > ρ₂

α: 0.05

The statistic to use is a pooled Z

Z= $\frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} }$ ≈ N(0;1)

The critical region is one-tailed to the right (positive), the critical value is:

$Z_{1-\alpha } = Z_{0.95} = 1.64$

If Z ≥ 1.64, you will reject the null hypothesis.

If Z < 1.64, you will not reject the null hypothesis.

Z= $\frac{(0.6 - 0.09) - (0}{\sqrt{0.35(1-0.35)[\frac{1}{225} + \frac{1}{225} ]} }$ = 11.34

Since the calculated Z- value is greater than the critical value, the decision is to reject the null hypothesis. The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.

P-value approach:

p-value < .00001

If you compare with the level of signification, you reach the same decision.

I hope it helps!

8 0

4 years ago

The image of (6, 9) under a dilation is (4, 6). The scale factor is 0-2 O 2/3 0 -2/3

lisov135 [29]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

To determine the scale factor consider the ratio of the image coordinates to the original coordinates, that is

scale factor = $\frac{4}{6}$ = $\frac{6}{9}$ = $\frac{2}{3}$

7 0

3 years ago

1. 6.5 x 4.1 please help me

Eddi Din [679]

Answer:

26.65 :D

Step-by-step explanation:

(6.5)(4.1)

=26.65

7 0

3 years ago

Read 2 more answers

Slope: 0 and y-intercept:0

Marat540 [252]

Answer:

y=0

Step-by-step explanation:

A slope of 0 is a horizontal line.

4 0

3 years ago

Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

###

9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

###

12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

###

13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

###

14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

3 0

3 years ago