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3 years ago

7

Jennifer has 1.7 liters of lemonade. If she pours 3 tenths of lemonade into a glass how many liters of lemonade are in the glass

?

1 answer:

slavikrds [6]3 years ago

3 0

.3 liters of lemonade

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if Mr hansons motorcycle drives 432 miles on 6 gallons of gas how much gas will he need to drive 738

lozanna [386]

Answer:

10.25 gallons

Step-by-step explanation:

432 miles / 6 gallons=72 miles to gallon

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3 years ago

Jackson says has an odd number of model cars.he has 6 cars on one shel and 8 cars on another shelf.is jackson correct?explain

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I hope this helps! If not I'm sorry.

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3 years ago

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Step-by-step explanation:

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3 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

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

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

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

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3 years ago