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BabaBlast [244]

2 years ago

13

A cohort study was conducted to study the association of coffee drinking and anxiety in a population-based sample of adults. Amo

ng 10,000 coffee drinkers, 500 developed anxiety. Among the 20,000 non-coffee drinkers, 200 cases of anxiety were observed. What is the relative risk of anxiety associated with coffee use?

1 answer:

azamat2 years ago

6 0

Answer: Relative risk of anxiety associated with coffee use is 5 times the anxiety associated with non coffee use.

Step-by-step explanation:

Since we have given that

Number of coffee drinkers = 10000

Number of whom developed anxiety = 500

Number of non coffee drinkers = 20000

Number of whom developed anxiety = 200

So, probability of getting anxiety from coffee drinkers = $\dfrac{500}{10000}=0.05$

Probability of getting anxiety from non coffee drinkers = $\dfrac{200}{20000}=0.01$

so, Relative risk of anxiety associated with coffee use is given by

$\dfrac{0.05}{0.01}=5$

Hence, Relative risk of anxiety associated with coffee use is 5 times the anxiety associated with non coffee use.

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A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3

fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

$\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7$

8 0

1 year ago

Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

7 0

3 years ago

Help me please how would you do this this is confusing

Mice21 [21]

Answer:

10/15-3/15 = 7/15

7 0

2 years ago

I need help it’s not c I tried but it’s not that plz someone help me with this I need help not sure what to do

Kobotan [32]

Answer:

a is the answer because it is

8 0

3 years ago

What is the equation of a line that passes through the point (0, 5) and has a slope of 1/3

Mariana [72]

Y - y1 = m (x - x1)

y1 being the y coordinate
x1 being the x coordinate
m being the gradient (or slope)

y - 5 = 1/3 (x - 0)
y - 5 = 1/3x
y = 1/3x + 5

6 0

2 years ago