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2 years ago

15

How do i solve this quesiton 6(x − 2) > 15

1 answer:

trasher [3.6K]2 years ago

7 0

Answer:

Step-by-step explanation:

$\displaystyle\ \!\!6(x-2)>15 \\\\6x-12>15 \\\\6x>27\\\\ \boldsymbol{x>4,5 \ \ or \ \ x\in(4,5\ ; \infty)}$

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Suppose I have an urn with 9 balls: 4 green, 3 yellow and 2 white ones. I draw a ball from the urn repeatedly with replacement.

Mrrafil [7]

Answer:

$E(X_n)=\frac{2(n-1)}{27}$

$E(y)=\frac{14}{9}$

Step-by-step explanation:

From the question we are told that:

Sample size n=9

Number of Green $g=4$

Number of yellow $y=4$

Number of white $w=4$

Probability of Green Followed by yellow P(GY) ball

$P(GY)=\frac{4}{9}*\frac{3}{9}$

$P(GY)=\frac{4}{27}$

Generally the equations for when n is even is mathematically given by

$Probability of success P(S)=\frac{4}{27}$

$Probability of Failure P(F)=\frac{27-4}{27}$

$Probability of Failure P(F)=\frac{23}{27}$

Therefore

$E(X_n)=\frac{n}{2}*P$

$E(X_n)=\frac{n}{2}*\frac{4}{27}$

$E(X_n)=\frac{2n}{27}$

Generally the equations for when n is odd is mathematically given by

$\frac{n-1}{2}$

$E(X_n)=\frac{n-1}{2}*\frac{4}{27}$

$E(X_n)=\frac{2(n-1)}{27}$

b)

Probability of drawing white ball

$P(w)=\frac{2}{9}$

Therefore

$E(w)=\frac{1}{p}$

$E(w)=\frac{1}{\frac{2}{9}}$

$E(w)=\frac{9}{2}$

Therefore

$E(y)=[E(w)-1]\frac{4}{9}$

$E(y)=[\frac{9}{2}-1]\frac{4}{9}$

$E(y)=\frac{14}{9}$

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