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icang [17]
2 years ago
13

Suppose that Aces can be either high or low; that is, that {A, 2, 3, 4, 5} is a straight, and so is {10, Jack, Queen, King, Ace}

. Moreover, a hand that is an Ace-high beats a King-high, etc. The number of ways of getting a five card hand that is two pair from a standard deck of cards is (Q8) The probability of being dealt a five card hand that is two pair from a well shuffled standard deck of cards is
Mathematics
1 answer:
zmey [24]2 years ago
8 0

The probability of being dealt a five card hand that is two pair from a well shuffled standard deck of cards is 40.

According to the statement

I have 10 starting cards, from Ace to 10, and 4 suits,

and by this way we get the 40 subsets and because of these subsets there is a probability becomes 40.

So, The probability of being dealt a five card hand that is two pair from a well shuffled standard deck of cards is 40.

Learn more about PROBABILITY here brainly.com/question/251701

#SPJ4

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Determine whether the number is a perfect square P.S there is a lot
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Answer:

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3 years ago
Find a • b.
melamori03 [73]
That is about coordinates and the dot product
we have a has coordinates (4;8)
and b has coordinates (-2;3)
So we have a.b= 4x(-2)+ 8x3= -8+24=16
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6 0
3 years ago
A road is 3/5 of a mile long. A crew needs to repave 3/4 of the road. How long is the section that needs to be repaved? Write yo
storchak [24]

Answer:

9/20 mile

Step-by-step explanation:

Take (calculate) 3/4 of 3/5 mile:

3      3

---- * ---- = 9/20

 4     5

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6 0
3 years ago
A test taker gets 70 on 1st exam, 80 on 2nd exam, 2/3 of 4/5 of his 2nd exam on his 3rd test. If the professor gives 5 points ex
Serga [27]
<h3>Answer:  127</h3>

=================================================

Explanation:

The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.

The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.

Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.

So we have these scores

  • first exam = 70
  • second exam = 80
  • third exam = 43
  • fourth exam = x+5

Adding up these scores and dividing by 4 will get us the average

(sum of scores)/(number of scores) = average

(70+80+43+x+5)/4 = 80

(x+198)/4 = 80

x+198 = 4*80

x+198 = 320

x = 320 - 198

x = 122

So the student got a score of x+5 = 122+5 = 127 on the fourth exam.

7 0
3 years ago
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