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Kisachek [45]
2 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]2 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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No links, I just need an explanation on how to do this, like step by step because I don't understand. T-T
Goryan [66]

Answer:

a. Jared and Ali

Step-by-step explanation:

b. Molly make mistake when he/she divide all equation by 4 but left the 8x remain 8x, it should be 2x.

Mark make mistake in changing the sign whitout changing side/ passing trough equal sign. FYI, + can be - and also - can be + if they move to the other side

i.e.

2 + 3 = 5

2 = 5 - 3

7 0
2 years ago
Find the domain of the function
Svetradugi [14.3K]

Answer:

<em>The domain of f is (-∞,4)</em>

Step-by-step explanation:

<u>Domain of a Function</u>

The domain of a function f is the set of all the values that the input variable can take so the function exists.

We are given the function

f(x)=\frac{1}{\sqrt{4-x} }

It's a rational function which denominator cannot be 0. In the denominator, there is a square root whose radicand cannot be negative, that is, 4-x must be positive or zero, but the previous restriction takes out 0 from the domain, thus:

4 - x > 0

Subtracting 4:

- x > -4

Multiplying by -1 and swapping the inequality sign:

x < 4

Thus the domain of f is (-∞,4)

7 0
2 years ago
Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
7 0
3 years ago
Which statement is true about the graphed function?
Semmy [17]

A function is positive if it lies above the x axis.

This function is positive before x = -4, i.e. over the interval (-\infty, -4)

It is negative elsewhere, i.e. over the interval (-4, \infty)

So, the only correct option is the second one.

6 0
3 years ago
Read 2 more answers
Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. A = 56°, a = 12, b = 14
Eva8 [605]
We know that
case 1)
Applying the law of sines
a/Sin A=b/Sin B
A=56°
a=12
b=14
so
a*Sin B=b*Sin A----> Sin B=b*Sin A/a---> Sin B=14*Sin 56°/12
Sin B=0.9672
B=arc sin (0.9672)------> B=75.29°-----> B=75.3°

find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+75.3)----> C=48.7°

find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 48.7°/Sin 56°)
c=10.87-----> c=10.9

the answer Part 1)
the dimensions of the triangle N 1
are
a=12  A=56°
b=14  B=75.3°
c=10.9  C=48.7°

case 2) 
A=56°
a=12
b=14
B=180-75.3----> B=104.7°

find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+104.7)----> C=19.3°

find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 19.3°/Sin 56°)
c=4.78-----> c=4.8

the answer Part 2)
the dimensions of the triangle N 2
are
a=12  A=56°
b=14  B=104.7°
c=4.8    C=19.3°
3 0
3 years ago
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