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3 years ago

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What is the probability that 2 cards selected from a standard deck of 52 cards without replacement are both non-face cards?

1 answer:

Orlov [11]3 years ago

5 0

Answer:

B. 0.588

Step-by-step explanation:

There are 40 non-face cards in the deck, so the probability of drawing the first one is 40/52. After doing that, the probability of drawing the second one is 39/51, since the number of non-face cards is 1 fewer, as is the size of the deck.

The joint probability is then ...

(40/52)·(39/51) ≈ 0.588

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I need help on this question

Alenkinab [10]

Answer:

B. 63 degrees

Step-by-step explanation:

Its the same angle as the one above it.

I hope this helps :)

Have a nice day!

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2 years ago

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Help me asap (please please)

CaHeK987 [17]

Area of a circle = pi*r^2.

Here, A = 380.13 mi. Unfortunately, "mi" is NOT a unit of area.

So I will assume that the problem should read A = 380.13 sq. miles.
380.12 mi^2
Thus, 380.13 mi^2 = 3.14*r^2, and r^2 = ------------------- 121.057 mi^2
3.14

The radius is the sqrt of 121.057 mi^2: 11.00 miles.

The diam. is twice that, or d = 22.00 miles.

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3 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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3 years ago

Evaluate 7!.<br> Ο 120<br> Ο 5,040<br> Ο 40,320

timurjin [86]

Answer:

5040 is the answer of it

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3 years ago

I WILL GIVE BRAINLY! The surface of a table to be built will be in the shape shown below. The distance from the center of the sh

ki77a [65]

Answer:

draw lines center to vertex
43.5 square inches
261 square inches

Step-by-step explanation:

<u>Part A</u>:

The area can be decomposed into triangles by drawing a line from the center to each vertex of the hexagon.

__

<u>Part B</u>:

The area of each triangle is given by the triangle area formula:

A = (1/2)bh

A = (1/2)(10 in)(8.7 in) = 43.5 in²

__

<u>Part C</u>:

The area of the table top is 6 times the area of one triangle:

surface area = 6(43.5 in²) = 261 in²

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3 years ago

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