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3 years ago

From a deck of 52 cards, one card is drawn at random. Match the following subsets with their correct probabilities.

Mathematics

2 answers:

zhenek [66]3 years ago

8 0

1.3/13(if you mean jacks queens and kings)
2.1/52
3.1/2
4.1/13
5.1/4

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

P(face card)=3/13

P(seven of hearts)=1/52

P(no black)=1/2

P(king)=1/13

P(diamond)=1/4

Step-by-step explanation:

We know that there are a total 52 cards out of which:

There are 12 face cards ( 4 kings,4 queen and 4 jack)

There are 4 pack:

13- spades 13- club 13-heart 13-diamond.

Out of which there are 26 black cards( 13 spade and 13 club)

There are 26 red cards( 13 heart and 13 diamond)

Now , we are asked to find the probability of each of the following,

P(face card)

Since there are total 12 face cards out of 52 playing cards.

Hence,

P(face card)=12/52=3/13

P( seven of hearts)

As there is just 1 seven of heart out pf 52 cards.

Hence, P(seven of hearts)=1/52

P(no black)

This means we are asked to find the probability of red card.

As there are 26 red card.

Hence P(no black)=26/52=1/2

P(king)

As there are 4 kings out of 52 cards.

Hence, P(king)=4/52=1/13

P(diamond)

As there are total 13 cards of diamond.

Hence,

P(diamond)=13/52=1/4

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A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

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