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Fantom [35]
3 years ago
15

A spinner has five equal parts labeled from 1 to 5. The spinner is spun twice. What is the probability of getting a 4 twice in a

row?
Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
8 0

|\Omega|=5^2=25\\ |A|=1\\\\ P(A)=\dfrac{1}{25}=4\%

ella [17]3 years ago
7 0

Answer:

\frac{1}{25}  

Step-by-step explanation:

Given : A spinner has five equal parts labeled from 1 to 5.

           The spinner is spun twice.

To Find: What is the probability of getting a 4 twice in a row?

Solution:

Numbers on spinner = {1,2,3,4,5}=5

Favorable outcome = {4}=1

So, probability of getting 4 on first spin =\frac{1}{5}

Now probability of getting 4 on second spin =\frac{1}{5}

So,  the probability of getting a 4 twice in a row = \frac{1}{5} \times \frac{1}{25}

                                                                               = \frac{1}{25}  

Hence the probability of getting a 4 twice in a row is  \frac{1}{25}  

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So, we are looking for x and y such that x^2 < 142 and y^2 > 142.

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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

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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,

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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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