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3 years ago

6

a bag has 7 red marbles 8 blue marbles and 3 black marbles what is the probability of picking a black marble replacing it and th

en picking a red marble

1 answer:

Viktor [21]3 years ago

6 0

1/18 is the probablity of picking a black marble

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2/3 × n= -22 what is n?

tekilochka [14]

n would equal <em>-33</em>.

<em>n=-33</em>

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3 years ago

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Meagan is taking piano gymnastics lessons.She has a piano lesson every 7 days and gymnastics lessons.She every 3 days.How obtene

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3 years ago

In slope-intercept form, what is the equation of the line passing through the points

Murljashka [212]

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03) y≈-3x±11

Step-by-step explanation:

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2 years ago

Pls answer precisely

Nata [24]

Answer:

The third one

Step-by-step explanation:

Its asking if its positive in the interval [-3, -2]

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2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago