In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without repl
acement, what is the probability that both cards are not blue?
1 answer:
The exact answer in fraction from is 2/15
The approximate answer in decimal form is 0.1333 (rounded to four decimal places)
The decimal value 0.1333 is equivalent to 13.33%
These are three ways to say the same basic answer.
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Explanation:
We have 4 red cards and 6 blue cards giving 4+6 = 10 total.
The probability of picking not blue (aka picking red) is 4/10 because there are 4 red out of 10 total.
If we don't replace the card we picked, then we have 3 red left over out of 9 total. The probability of picking red again is 3/9
Multiply those fractions
(4/10)*(3/9) = (4*3)/(10*9) = 12/90
Now reduce
12/90 = (6*2)/(6*15) = 2/15
note how the GCF 6 cancels
Using a calculator,
2/15 = 0.1333 approximately
The '3's after the decimal point go on forever.
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<h3>given:</h3>

<h3>to find:</h3>
the radius of the given ball (sphere).
<h3>solution:</h3>
![r = \sqrt[3]{ \frac{3v}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3v%7D%7B4%5Cpi%7D%20%7D%20)
![r = \sqrt[3]{ \frac{3 \times 905}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20905%7D%7B4%5Cpi%7D%20%7D%20)

<u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>radius</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ball</u><u> </u><u>is</u><u> </u><u>6</u><u> </u><u>cm</u><u>.</u>
note: refer to the picture I added on how you can change r as the subject of the formula.
Search it up and you shall find an answer believe me it is there
Answer:
35/4 or 8,3/4
Step-by-step explanation:
Flip the second fraction and multiply.
15x plus 7
This is because you just simplify the like terms
Answer:
$29.7
Step-by-step explanation:
10 percent of 27 is 2.7 then you add 27 and 2.7 and you get 29.7