Ask question

Ask question

All categories

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

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!

You might be interested in

A vehicle travels 2345 m in 315 s toward the west. What is its velocity?

olganol [36]

The velocity is 7.44m/s in a direction to the west

7 0

3 years ago

I need help on this too please I'm literally struggling

Orlov [11]

The probability is 0.2256 or 22.56% chance i think

7 0

3 years ago

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

diamong [38]

2x + 3 + x = 39
3x + 3 = 39
3x = 36
x = 12

One tank holds 12 gallons
The other tank holds 27 gallons

6 0

2 years ago

Read 2 more answers

C.9<br> D.36<br> Geometry math question no Guessing and Please show work

liq [111]

AQ : QE = 2 : 1 ( property of a centroid of a triangle)

Therefore 2/1 = 18 / QE

2QE = 18

QE = 9 (answer)

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