A flower shop has 744 carnations to tie into bunches of 12 each and 1,896 roses to tie into bunches of 24 each.

Mathematics

1 answer:

Mama L [17]3 years ago

8 0

744 Carnations. 12 of these go into a bunch.

1896 roses, 24 go into a bunch.

744/12=62

1896/24=79

62 bunches of carnations and 79 bunches of roses.

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The sum of three consecutive terms in an arithmetic sequence is 27, and their product is 585. find the three terms.

sammy [17]

$\text{the three consecutive terms in an arithmetic sequence}\\a;\ a+d;\ a+2d\\\\\text{system of equals}\\\\ \left\{\begin{array}{ccc}a+a+d+a+2d=27\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}3a+3d=27&|:3\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}a+d=9&\to d=9-a\\a(a+d)(a+2d)=585\end{array}\right\\$
$\text{substitute to the second equation}\\\\a(a+9-a)(a+2(9-a))=585\\\\a(9)(a+18-2a)=585\\9a(18-a)=585\ \ \ |:9\\a(18-a)=65\\18a-a^2=65\ \ \ |\text{change the signs}\\a^2-18a=-65\\a^2-2a\cdot9=-65\ \ \ |+9^2\\\underbrace{a^2-2a\cdot9+9^2}_{(a-b)^2=a^2-2ab+b^2}=-65+9^2\\(a-9)^2=16\to a-9\pm\sqrt{16}\\a-9=-4\ \vee\ a-9=4\ \ \ |+9\\a=5\ \vee\ a=13\\\\d=9-5=4\ \vee\ d=9-13=-4$
$Answer:\ a=5;\ a+d=9; a+2d=13\ vee\ a=13;\ a+d=9;\ a+2d=5$

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Ivahew [28]

Answer:

Step-by-step explanation:

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3 years ago

Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

erastova [34]

Answer:

The probability is $\frac{56!}{64!}$

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, $64 \choose 8 .$

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function $f : A \rightarrow A ,$ with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}$