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3 years ago

10

Ms. Huggins collected data from each of her four math classes to see if having an after-school job was independent or dependent

of gender. Based on her findings, are the two independent or dependent? Explain your answer.

1 answer:

nikitadnepr [17]3 years ago

8 0

Answer: B) independent because P(Boy)= P(Boy|Job)

Step-by-step explanation:

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What's five divided by two

bogdanovich [222]

The answer to this problem would be 2.5.

4 0

3 years ago

Read 2 more answers

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by <em>b</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since <em>a/b</em> < <em>c/d</em>, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by <em>b/d</em> to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and <em>a/b</em> < <em>c/d</em> tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

5 0

3 years ago

1 For each line

Elena-2011 [213]

Step-by-step explanation:

I'll do line A for you and you can use the formulas to solve lines B and C yourself, since its good for you to practice doing these questions yourself

a)

The gradient, m, is calculated using m = (y2-y1)/(x2-x1) where x1,x2,y1 and y2 can be any ordered pairs on the line. I'm going to use (4,0) and (7,3) as the 2 points.

m = (3-0)/(7-4) = 3/3 = 1

b)

The y-intercept is where the line intersects with the x-axis. In this case (0,-4)

c)

The equation of a linear line is y=mx+b (or c depending on which country you are from)

y = 1x-4

y=x-4

Now try the other 2 lines yourself!

If this answer has helped you, considered making this the brainliest answer!

5 0

3 years ago

I’m about to take the test please hurry! I will give brianliest.

scZoUnD [109]

Answer:

i think its 358

Step-by-step explanation:

5 0

3 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

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4 years ago