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2 years ago

In the above diagram, you see a standard deck of cards. Answer each of the following questions using this deck of cards.

Mathematics

1 answer:

Novay_Z [31]2 years ago

7 0

Answer:

See below.

Step-by-step explanation:

52 as there is a total of 52 in the pack.

Probability (drawing a red card) =26/52 = 1/2.

4 aces.

Probability(drawing an ace) = 4/52 = 1/13.

Probability(drawing a numbered card) = 40/52 = 10/13.

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What is the minimum number of feet you should be to avoid being electrocuted by overhead power lines up to 50kV? A. 10 feet B. 1

Andrei [34K]

Answer:

I believe the answer is A 10 feet

Step-by-step explanation:

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3 years ago

Evaluate 1x + y, for x = 8 and y=-15.<br> 7 <br> -7<br> 23<br> -23

statuscvo [17]

Answer:

B) -7

Step-by-step explanation:

x+y

8+(-15)=8-15=-7

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3 years ago

How many times larger is the broadcast area of channel 19 than the broadcast area of channel 36?

GaryK [48]

Answer:

Step-by-step explanation:

You will have to subtract 36 and 19 and than you will get 17

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3 years ago

a corn vendor at a farmers market was selling a bag of 8 ears of corn for $2.56 anithert vendor was selling a bad of 12 for 4.32

maria [59]

Answer:

8 for 2.56

Step-by-step explanation:

THE ANSWER IS IN CENTS.

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3 0

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

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

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

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

5 0

3 years ago