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Nina [5.8K]
3 years ago
9

Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select

ed. Find the probability of selecting a six and then selecting a nine.
The probability of selecting a six and then selecting a nine is
(Round to three decimal places as needed.)
Mathematics
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

0.006

Step-by-step explanation:

The probability that the first card is a six is 4/52.

There's one less card now, so the probability that the second card is nine is 4/51.

The total probability is:

P = (4/52) (4/51)

P ≈ 0.006

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What two numbers when added equal 25 and when multiplied = -200
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The numbers are irrational, so I can't write them exactly with digits.
When rounded to the nearest ten-thousandth, they are

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3 years ago
3 (182 + 2) – 172 = -17
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Answer:

380≠−17

Step-by-step explanation:

3(182+2)−172 ?= −17

380≠−17

5 0
3 years ago
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What is the solution to the equation<br> 3/5[x+4/3]
aalyn [17]
The answer is x = 1/3

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T+2/3t-1/3 I really need this question to be answered
Murrr4er [49]

<u> Equation:</u>

x(x      -     5)     +     3(x     +     5)

<u>Steps:</u>

x(x      -     5)     +     3(x     +     5)

<u>Expand:</u>

x(x     -     5 ):           x^2      -    5x

x^2    -    5x     +   3(x      +     5)

<u>Expand:</u>

3(x     +    5):        3x   +    15

x^2     -     5x     +     3x       +     15

<u>Add Similar Elements:   </u>  

-5x      +    3x    

=         -2x

Answer        x^2    -     2x      +    15    Doesn't Factor






Hope that helps!!!                                  : )


         


5 0
3 years ago
Prove that the sum of the squares of any two odd numbers is always even
ValentinkaMS [17]

Answer:

proof below

Step-by-step explanation:

Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)

Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;

(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2

= 2(2a^2 + 2a + 2b^2 + 2b + 1)

Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.

6 0
3 years ago
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