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3 years ago

Round 66.42 to the nearest hundredth

Mathematics

1 answer:

Musya8 [376]3 years ago

3 0

Answer:

66.42

Step-by-step explanation:

66.42 is already rounded to the nearest hundredth.

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Assume that the pond contains 90 fish: 64 purple, 25 green, and 1 red. A contestant pays $0.55 to randomly catch one fish and re

Licemer1 [7]

Answer:

The carnival is losing (on average) $0.15 on each play

Step-by-step explanation:

To find out how much the carnival wins or looses in each play one subtract the expected value (EV) from each play from the amount charged by the carnival for each play ($0.55). If the expected value is higher than what the carnival charges, the carnival is losing money.

Expected is the sum of the payouts of each bet multiplied by its likelihood:

$EV= \frac{64*0.40+25*1.15+1*9.00}{90}\\EV= 0.70$

Since the expected value is higher than $0.55, the carnival is losing money, on average, on each play:

$0.55-0.7 = -0.15$

The carnival is losing (on average) $0.15 on each play

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3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

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The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

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2 years ago

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