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Komok [63]
4 years ago
14

A professor let's his students pick 3 out of 8 assignments to complete. How many combinations of the 3 assignments are possible

Mathematics
2 answers:
sertanlavr [38]4 years ago
5 0
It is indeed justifiable that the operation involved in this task is combination because the arrangement of the assignments are not important. The combination may be solved directly from a scientific calculator using the function,
                                   nCr
where in this item the value of n is 8 and that of r is 3. Substituting and solving the combination,
                                8C3 = 56
Therefore, there are exactly 56 combinations possible. 
PilotLPTM [1.2K]4 years ago
3 0

Answer:

your answer will be 56 hopefully this helps!

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I need help on this too please I'm literally struggling
Orlov [11]
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Two tanks hold a total of 39 gallons of a toxic solvent one tank holds 3 gallons more than twice the amount in the other how man
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C.9<br> D.36<br> Geometry math question no Guessing and Please show work
liq [111]

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8 0
4 years ago
Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p
posledela

Answer:

<em>P=0.0000037</em>

<em>P=0.00037%</em>

Step-by-step explanation:

<u>Probability</u>

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

\displaystyle \frac{4}{52}=\frac{1}{13}

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

\displaystyle \frac{3}{51}=\frac{1}{17}

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

\displaystyle \frac{2}{50}=\frac{1}{25}

Finally, the probability of getting the remaining ace out of the 49 cards is:

\displaystyle \frac{1}{49}

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}

\displaystyle P= \frac{1}{270,725}

P=0.0000037

P=0.00037%

3 0
3 years ago
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