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Fantom [35]
3 years ago
10

Trials in an experiment with a polygraph include results that include cases of wrong results and cases of correct results. Use a

significance level to test the claim that such polygraph results are correct less than ​% of the time. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer and Step-by-step explanation:

This is a complete question

Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80​% of the time. Identify the null​hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

The computation is shown below:

The null and alternative hypothesis is

H_0 : p = 0.80

Ha : p < 0.80

\hat p = \frac{x}{ n} \\\\= \frac{74}{97}

= 0.7629

Now Test statistic = z

= \hat p - P0 / [\sqrtP0 \times (1 - P0 ) / n]

= 0.7629 - 0.80 / [\sqrt(0.80 \times 0.20) / 97]

= -0.91

Now

P-value = 0.1804

\alpha = 0.01

P-value > \alpha

So, it is Fail to reject the null hypothesis.

There is ample evidence to demonstrate that less than 80 percent of the time reports that these polygraph findings are accurate.

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37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
4/r=5/7solve for r. Help pls
antiseptic1488 [7]
Hello!

You first cross multiply

4 * 7 = 28

5 * r = 5 * r

r * 5 = 28

Divide both sides by 5

r = 28/5 = 5.6

The answer is 28/5 or 5.6

Hope this helps!
4 0
3 years ago
Read 2 more answers
Mathwiz wya?? pic below i need help asap
n200080 [17]

Answer:

Step-by-step explanation:

First, you gotta work out the hypotenuse of ABC, which is AC.

To do that, you need to figure out the scale factor between the two right-angled triangles. You can do that for this question because this is a similar shapes question.

12.5/5 = 2.5

The scale factor length between the two triangles is 2.5.

You can use 2.5 now to work out AC, so AC would be 13 x 2.5, which gives 32.5.

Now that you've got the hypotenuse and BC of ABC, you can use Pythagoras's theorem to work out the length of AB

Pythagoras's theorem = a^2 + b^2 = c^2

a = BC = 12.5

b = AB = we need to work this out

c = AC (the hypotenuse we just worked out) = 32.5

12.5^2 + b^2 = 32.5^2 Let's both simplify and rearrange this at the same time so that we have our b on one side.

b^{2} = 1056.25 - 156.25

b = \sqrt{(1056.25 - 156.25)}

b = \sqrt{900}

b = AB = 30  We've found b or AB, now we can work out the perimeter of ABC.

Perimeter of ABC = AB + BC + AC

= 30 + 12.5 + 32.5

= 75  Here's the perimeter for ABC.

8 0
2 years ago
Answer all please for me !!!
Andreyy89

Answer:

Step-by-step explanation:

3 0
3 years ago
How do you factor quadratics by grouping
Kaylis [27]
Here's an example Its also call the 'ac' method:-

factor  2x^2 - 13x - 15

Multiply first coefficient by last number  which is 2*-15 = -30

Now we need 2 factors of -30 which will when added give - 13:-

-15 and 2  looks good. So we  can write:-

2x^2 - 13 x - 15

= 2x^2 + 2x - 15x - 15 

factor by grouping first 2 and last 2 terms , we get:-

= 2x(x + 1) - 15(x + 1)   

the x+1 is common so we have:-

(2x - 15)(x + 1)   Answer

Hope this helps.


5 0
3 years ago
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