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2 years ago

6

In a well-shuffled, standard 52 card deck, what is the probability that a card is a Queen, given that it is a black card

2 answers:

yuradex [85]2 years ago

8 0

Answer:

2/52

Step-by-step explanation:

dlinn [17]2 years ago

7 0

There are two black queens in a standard deck, so the probability is 2/52 or 1/26.

If instead you know you pulled a black card from the deck and you’re wondering what the probably is that it’s a queen, then it’s 2/26, since there are still 2 black queens, but only 26 black cards.

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sammy [17]

Answer:

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Step-by-step explanation:

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3 years ago

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trasher [3.6K]

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3 years ago

A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building

harina [27]

Answer:

17.3

Step-by-step explanation:

To solve this equation, you need to use the pythagorean theorem which is:

$a^{2} +b^{2}= c^{2}$

a= length

b= height

c= hypotenuse

First, substitute in your values for the equation:

$10^2+b^{2} =20^{2}$

Then, use the squares in the equation:

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3 years ago

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IceJOKER [234]

Answer:

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3 years ago

If f(x)=2x-6 and g(x)=x^3 what is (g f)(0)

marin [14]

ANSWER

$(g \circ \: f)(0) = - 216$

EXPLANATION

The functions are:

$f(x) = 2x - 6$

$g(x) = {x}^{3}$

$(g \circ \: f)(x) =g(f(x))$

$(g \circ \: f)(x) =g(2x - 6)$

We substitute f(x) into g(x) to obtain:

$(g \circ \: f)(x) =(2x - 6)^{3}$

We now substitute x=0 to obtain;

$(g \circ \: f)(0) =(2(0) - 6)^{3}$

$(g \circ \: f)(0) =(- 6)^{3}$

This simplifies to:

$(g \circ \: f)(0) = - 216$

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2 years ago