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3 years ago

Rubin has 6 rolls of pennies containing 50 coins each, 5 rolls of nickels

Mathematics

2 answers:

MArishka [77]3 years ago

7 0

Answer:

C. $63.00

Step-by-step explanation:

$(6*50*0.01)+(5*40*0.05)+(4*50*0.10)+(3*40*0.25)=\\(300*0.01)+(200*0.05)+(200*0.10)+(120*0.25)=\\3+10+20+30=\\13+20+30=\\33+30=\\63$

He has C. $63.00.

creativ13 [48]3 years ago

4 0

Answer:

pennies = 1cent

50 cents per roll

that's three dollars

nickels = 5cents

200 cents per roll or 2dollars

that's ten dollars

dimes = 10cents

500 cent or five dollars

20 dollars

quarter = 25cents

four quarters equal a dollar

4/40= $10x3=$30

30+20+10+3=$63

Step-by-step explanation:

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Answer:

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4 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

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2 years ago

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pshichka [43]

Yes, your answer is correct.

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4 years ago