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3 years ago

10

A card is randomly selected from a standard 52-card deck of playing cards. determine the probability that the randomly selected

card:
a. is red

2 answers:

Lilit [14]3 years ago

6 0

Half (or 26) of the cards are red, and the rest are black.

P(randomly selected card is red) = 26/52 = 1/2.

Vesnalui [34]3 years ago

5 0

$|\Omega|=52\\ |A|=26\\\\ P(A)=\dfrac{26}{52}=\dfrac{1}{2}=50\%$

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You make a table to use as a quick reference guide. y = -180 - 9(x)

Lelechka [254]

There all multiples of 9 so just times it by your Y axis

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3 years ago

Which is Bigger?

igomit [66]

A)
$43\%=\frac{43}{100} \\ \\
\frac{24}{60}=\frac{4}{10}=\frac{40}{100} \\ \\
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b)
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3 0

3 years ago

Can i have help please

IgorLugansk [536]

(y + 1)^3 would just be dividing the two

If you want it completely simplified, then it’s the following:

(y + 1)(y + 1)(y + 1)

= y^2 + y + y + 1 (y + 1)

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3 years ago

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are

FrozenT [24]

Answer:

15 possible combinations

Step-by-step explanation:

Given

$Vegetables = 6$

$Selection = 2$

Required

Determine the possible number of combinations

The question emphasizes on "selection" which means "combination".

So; To answer this question, we apply the following combination formula:

$^nC_r = \frac{n!}{(n-r)r!}$

In this case:

$n = 6$

$r = 2$

The formula becomes:

$^6C_2 = \frac{6!}{(6-2)!2!}$

$^6C_2 = \frac{6!}{4!2!}$

$^6C_2 = \frac{6*5*4!}{4!2*1}$

$^6C_2 = \frac{6*5}{2}$

$^6C_2 = \frac{30}{2}$

$^6C_2 = 15$

<em>Hence, there are 15 possible combinations</em>

7 0

3 years ago