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3 years ago

Evaluate C(6, 6). A. Undefined B. 0 C. 1

Mathematics

2 answers:

irakobra [83]3 years ago

6 0

Evaluate C (6,6)
Answer: A. Undefined

3241004551 [841]3 years ago

5 0

I believe the answer is A. Undefined

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Write an expression that is equivalent to 3/4 a + 2/3 - 1/2 a - 1/3 + 1/2 a

Maksim231197 [3]

Hi there is this 0pp called Math-way that will help

7 0

3 years ago

A shelf can be packed from end to end with 30 large books or 45 small books.

DerKrebs [107]

Answer:

Step-by-step explanation:

since 30÷45 = 2/3 then

The space taken by a small book is equal to

the space taken by 2/3 a large book.

…………………………………………………

<u><em>Calculating the number of large book corresponding to 23 small books</em></u> :

= 23×(2÷3) + 3

= 18.333333333333 (large book)

Then ,Kevin can pack the shelf with (at most) : 30 - 19 = 11 more large books

4 0

2 years ago

Solve for the missing item in the following. (Do not round intermediate calculations. Round your answer to the nearest cent.)

iogann1982 [59]

9514 1404 393

Answer:

$1904.76

Step-by-step explanation:

The interest formula is ...

I = Prt . . . . interest on principal P at rate r for t years

Solving for P, we find ...

P = I/(rt) = 200/(0.07·1.5) = 200/0.105 ≈ 1904.76

The principal amount was $1904.76.

7 0

3 years ago

A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic

Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

$\mathbf{Bars = 6}$

$\mathbf{Chocolate = 3}$

$\mathbf{Custard= 3}$

The total number of donuts in the box is:

$\mathbf{Total= 6 + 3 + 3}$

$\mathbf{Total= 12}$

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

$\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}$

So, we have:

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}$

Simplify

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}$

Multiply

$\mathbf{Pr = \frac{9}{132}}$

Divide

$\mathbf{Pr = 0.068}$

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068