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3 years ago

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A bag contains an equal number of blue, green, pink, and red balls. If a ball is chosen at random 160 times, with replacement, p

redict the number of times the ball would be the color red.

1 answer:

algol133 years ago

3 0

Answer:

Step-by-step explanation:

Any one color would be 1/4 of the total. So the number of times you should get red is 1/4 of 160 = 40

Notice that this very odd problem does not depend on what the total is. It only depends on the number of times you make the draw. Unusual question.

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3/4 of a number is 27.Whats the number?

vazorg [7]

27*3/4

20.25

+ 27

47.25

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3 years ago

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Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

###

Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

3 years ago

Determine the distance between -8 and 0.

Flauer [41]

This one's simple..
B) -8 units

5 0

4 years ago

The equation of this graph is

san4es73 [151]

Is this what you need or? Y= 1/4x +3

4 0

3 years ago

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1. The county fair charges $1.25 per ticket for the rides. Jermaine bought 25 tickets for the rides and spent a total of $43.75

ryzh [129]

If this were to be graphed, the independent variable would be the price of the ticket for the rides. The dependent variable would be the total cost.

The fair admission is not a variable because it is a constant price for every single person who goes into the fair.

The problem asks to use y to represent the total cost and x to represent the number of ride tickets. In order to fully write out the equation, we have to figure out what the fair admission costs.

43.75 = 1.25(25) + b

*b represents the fair admission

Multiply 1.25 by 25

43.75 = 31.25 + b

Subtract 31.25 to find what b costs.

12.50 = b

The fair admission costs $12.50.

Solution: y = 1.25x + 12.50

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3 years ago

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