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Kisachek [45]
3 years ago
11

It has been hypothesized that silicone breast implants cause illness. In one study it was found that women with implants were mo

re likely to smoke, to be heavy drinkers, and to use hair dye than were women in a comparison group who did not have implants. Use the language of statistics explain why this study casts doubt on the claim that implants cause illness
Mathematics
1 answer:
ZanzabumX [31]3 years ago
3 0

Answer:

This study  casts doubts as the hypothesis does not support the women who are smokers, drinkers or use dyes besides the difference of breasts implants.

Step-by-step explanation:

This study  casts doubts as the hypothesis does not support the women who are smokers, drinkers or use dyes.

If the null and alternate hypothesis are formulated as

H0 : silicone breasts implants cause illness in smokers, drinkers and hair dyed people

against the claim

Ha: silicone breasts implants do not cause illness in smokers, drinkers and hair dyed people

then there would be no doubt .

But as the null and alternate hypothesis are

H0: silicone breast implants cause illness in women

against the claim

Ha: silicone breast implants do not cause illness in women

This suggests that both categories of women have same characteristics.

But actually both categories do not have same characteristics. The breast implant women also , smoke, drin and dye hair.

So when the characteristics are changed the illness maybe caused by some other characteristics.

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Which one is the correct answer, and why?
alexdok [17]

Answer:

D  f(n) = 7n-12

Step-by-step explanation:

The first term is -5

The second term is -5+7  =2

The common difference is 7

an = a1 +d(n-1)    where d is the common difference

     = -5 +7(n-1)

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If line segment RU is considered the base of parallelogram RSTU, what is the corresponding height of the parallelogram?
klasskru [66]
Given that line segment RU with vertices R(1, 1) and U(4, 5) is considered the base of parallelogram RSTU.

Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.

The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.

The equation of the line segment RU is given by
\frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3}  \\  \\ 3(y-1)=4(x-1) \\  \\ 3y-3=4x-4 \\  \\ 3y=4x-1 \\  \\ y= \frac{4}{3} x- \frac{1}{3}

Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
\frac{4}{3} m=-1 \\  \\ m=- \frac{3}{4}

Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
y-1=- \frac{3}{4} (x-1) \\  \\ 4(y-1)=-3(x-1) \\  \\ 4y-4=-3x+3 \\  \\ 4y=-3x+7 \\  \\ y=- \frac{3}{4} x+ \frac{7}{4}

The equation of the line segment ST is given by
\frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3}  \\  \\ 3y=4(x-7)=4x-28 \\  \\ y= \frac{4}{3} x- \frac{28}{3}

The line perpendicular to line segment RU intersected line segment ST at the point given by
- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\  \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\  \\  \frac{25}{12} x= \frac{133}{12}  \\  \\ x= \frac{133}{25}  \\  \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25}

Thus, the corresponding height of the parallelogram is the line with endpoints
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)

Recall that the length of a line passing through points
(x_1,y_1) \ and \ (x_2,y_2)
is given by
l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Thus, the length of the line passing through points
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)
is given by
l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2}  \\  \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} }  \\  \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4

Therefore, <span>the corresponding height of the given parallelogram is 5.4 units</span>
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Answer:

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2 years ago
A complex number, represented by z = x + iy, may also be visualized as a 2 by 2 matrix
Marat540 [252]

Answer:

Step-by-step explanation:

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z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}

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z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})

On the other hand, if we sum the matrix visualizations of z \quad \text{and} \quad \tilde{z} we get

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which is the matrix visualization of z + \tilde{z}.

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B)  Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number z=x+iy.

We are looking for the complex number z^{-1}=(x+iy)^{-1} which in terms of matrices is equivalent to find the matrix

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Hence,

z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)

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