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Irina-Kira [14]

3 years ago

7

a bookstore owner asks every fifth customer on a single day if he or she owns a e-book reader . select yes or no for each questi

on . is this a random sample of the store customers ? yes or no is this a random sample of the general population ? yes or no explain your answers.

1 answer:

Masteriza [31]3 years ago

6 0

Yes, it is random because it isn't convenient for him.

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Function form 8x + 4y =20

balandron [24]

8x + 4y = 20

4y = 20 - 8x

y = 5 - 2x

8 0

3 years ago

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What is the range of the relation?

tia_tia [17]

The correct answer is (-2,1,2,8)

The range are the y- values

The explanation:

The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements "used" by the relation or function constitute the range.

Hope this helped you :D

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3 years ago

Can a trapezoid have four obtuse angles

Pavel [41]

The four angles inside EVERY four-sided figure add up to 360 degrees, no matter what the shape of the figure is.

"Four obtuse angles" means "four angles of more than 90 degrees each".

(four times more-than-90) = more than 360

so NO 4-sided figure can have four obtuse angles inside it.

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4 years ago

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Evaluate 7!.<br> Ο 120<br> Ο 5,040<br> Ο 40,320

timurjin [86]

Answer:

5040 is the answer of it

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3 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

4 years ago