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3 years ago

Suppose you draw a card from a well-shuffled deck of 52 cards. What is the probability of drawing a 5 or a queen?

1 answer:

5 0

A 5 or a queen

52 cards per deck
each card repeats 4 times

a 5 or a queen
there are four 5's and four queens
4+4=8

probablity=desiredoutcomes/totalpossible
probablity=8/52
probablity=4/26
probablity=2/13

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Need help ASAP

joja [24]

Part (1) : The solution is $729$

Part (2): The solution is $\frac{1}{16 x^{8}}$

Part (3): The solution is $\frac{2 x^{2}}{3 y z^{7}}$

Explanation:

Part (1): The expression is $3^{2} \cdot3^{4}$

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$ , we get,

$3^{2} \cdot 3^{4}=3^{2+4}$

Adding the exponent, we get,

$3^{2} \cdot3^{4}=3^6=729$

Thus, the simplified value of the expression is $729$

Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

$\frac{1}{2^4x^8}$

Thus, we have,

$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$ , we have,

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

Adding the powers, we get,

$\frac{2x^{2}y^{-1}z^{-7}}{3}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\frac{2 x^{2}}{3 y z^{7}}$

Thus, the value of the expression is $\frac{2 x^{2}}{3 y z^{7}}$

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c. correct

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Suppose you draw a card from a​ well-shuffled deck of 52 cards. What is the probability of drawing a 5 or a queen?

Suppose you draw a card from a well-shuffled deck of 52 cards. What is the probability of drawing a 5 or a queen?