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koban [17]
3 years ago
12

In a standard 52-card deck, half of the cards are red and half are black. The 52 cards are divided evenly into 4 suits: spades,

hearts, diamonds, and clubs. Each suit has three face cards (jack, queen, king), and an ace. Each suit also has 9 cards numbered from 2 to 10. 7.
Dawn draws 1 card, replaces it, and draws another card. Is it more likely that she draws 2 red cards or 2 face cards?​
Mathematics
1 answer:
liubo4ka [24]3 years ago
6 0

Answer:

It is more likely she draws 2 red cards

Step-by-step explanation:

Given that:

Number of red d cards in deck = 26

Number of face cards = 4 * 3 = 12 face cards

If two cards are drawn with replacement ;

Probability = (required outcome / Total possible outcomes)

P(1st card red) = 26 /52) = 1/2

P(2nd card red) = 26 /52 = 1/2

P(1st card red) * P(2nd card red)

1/2 * 1/2 = 1/4 = 0.25

P(1st card face card) = 12/52 = 3/13

P(2nd card, face card) = 12/52 = 3/13

P(1st card face card) * P(2nd card face card)

3/13 * 3 /13 = 9/169 = 0.053

0.25 > 0.053

It is more likely she draws 2 red cards

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Sari has 3 times as many pencil erasers as Sam. Together, they have 28 erasers. How many erasers does Sari have?
BartSMP [9]

Let

x--------> amount of erasers that Sari has

y--------> amount of erasers that Sam has


we know that


x=3y-------> equation 1

x+y=28-----> equation 2


Substitute equation 1 in equation 2

[3y]+y=28\\ 4y=28\\ y=28/4\\ y=7


find the value of x

x=3y\\ x=3*7\\ x=21


therefore


the answer is

Sari has 21 erasers


8 0
3 years ago
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Suppose that 2 ≤ f '(x) ≤ 3 for all values of x. what are the minimum and maximum possible values of f(5) − f(1)?
zzz [600]

Assuming f is differentiable over the interval (1, 5), by the mean value theorem we know that there is some 1 such that

f'(c)=\dfrac{f(5)-f(1)}{5-1}

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8 0
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50 divided by 70 pluse 10
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If 34 percent of 60 million barrels is gasoline and 29 percent of 60 million barrels is desil how many more barrels are there of
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34 - 29 = 5 percent.

60 million times 5% = 60 x 0.05 = 3 million

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