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grin007 [14]
3 years ago
8

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

Mathematics
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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7.<br>(-3x³)(-2x³y⁴z)(-3z²)÷(4x³z)(-3yz) (-3xyz)​
serg [7]

Answer:

-\frac{x^2y^2}{2}

Step-by-step explanation:

We have an extremely large equation and are asked to divide it, so let's solve it step-by-step :

Remove the parenthesis to make it easier to read :

\frac{-3x^3 *2x^3y^4z *3z^2}{4x^3z*3yz*3xyz}

Multiply the numerators :

\frac{-18x^6y^4z^3}{4*3*3x^3yyzzz}

Multiply the denominators :

\frac{-18x^6y^4z^3}{36x^4y^2z^3}

Apply the negative rule :

-\frac{18x^6y^4z^3}{36x^4y^2z^3}

Cancel the common factor which is 18 :

-\frac{x^6y^4z^3}{2x^4y^2z^3}

Apply the addition exponent rule :

\frac{y^4z^3x^{6-4} }{2y^2z^3}

Subtract :

\frac{x^2y^4z^3 }{2y^2z^3}

Apply the rule for y :

\frac{x^2y^{4-2} z^3 }{2z^3}

Subtract :

\frac{x^2y^2z^3 }{2z^3}

Cancel the common factor of z^3 :

-\frac{x^2y^2}{2}

5 0
3 years ago
NEED HELP ASAP. WILL GIVE BRAINLIEST
Andreas93 [3]

Answer:

A. The magnitudes are 10, and the direction angles are about 18 degrees.

Step-by-step explanation:

Edge 2020

8 0
3 years ago
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The perimeter of this triangle should be 34 units.

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