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4 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]4 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]4 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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4 years ago

11. A paint color requires the ratio of green paint to yellow paint to be 4 : 9

neonofarm [45]

Answer:

Step-by-step explanation:

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3 years ago

1/5 ( y + 10 ) ≤ - 25

uranmaximum [27]

Answer:

y≤−135

Step-by-step explanation:

1/5 ( y + 10 ) ≤ - 25

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1/5y+2−2≤−25−2

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3 years ago

: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =

Serhud [2]

Answer:

$y = \tan(x + \frac{x^2}{2})$

Step-by-step explanation:

Poorly formatted question; The complete question requires that we prove that $y=\tan(x+\frac{x\²}{2})$

When

$\frac{dy}{dx} =1+xy\²+x+y\²$ and $y(0)=0$

We have:

$\frac{dy}{dx} =1+xy\²+x+y\²$

Rewrite as:

$\frac{dy}{dx} =1+x+xy\²+y\²$

Factorize

$\frac{dy}{dx} = (1+x)+y\²(x+1)$

Rewrite as:

$\frac{dy}{dx} = (1+x)+y\²(1+x)$

Factor out 1 + x

$\frac{dy}{dx} = (1+y\²)(1+x)$

Multiply both sides by $\frac{dx}{1 + y^2}$

$\frac{dy}{1+y\²} = (1+x)dx$

Integrate both sides

$\int \frac{dy}{1+y\²} = \int (1+x)dx$

Rewrite as:

$\int \frac{1}{1+y\²} dy = \int (1+x)dx$

Integrate the left-hand side

$\int \frac{1}{1+y\²} dy = \tan^{-1}y$

Integrate the right-hand side

$\tan^{-1}y = x + \frac{x^2}{2} + c$

$y(0)=0$ implies that: $(x,y) = (0,0)$

So:

$\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}(0) = 0 + \frac{0^2}{2} + c$

This gives:

$0 = 0 +0 + c$

$0 =c$

$c = 0$

The equation $\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}y = x + \frac{x^2}{2} + 0$

$\tan^{-1}y = x + \frac{x^2}{2}$

Take tan of both sides

$y = \tan(x + \frac{x^2}{2})$ --- Proved

8 0

3 years ago