Simply the expression below

Mathematics

2 answers:

Inessa [10]3 years ago

4 0

Answer:

Step-by-step explanation:

Ira Lisetskai [31]3 years ago

4 0

Answer:

D. (w-3)/(w-7)

Step-by-step explanation:

(w+3)(w-3)/(w-7)(w+3)

Hence, removing (w+3),

(w-3)/(w-7)

<em>Feel free to mark this as brainliest :D</em>

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Which of the following is not an example of like terms?

Viktor [21]

Answer:

A. 2 and-6

Step-by-step explanation:

Like terms are terms that contain the same variables raised to the same powers. In other words, they have the same variables with the same exponents.

So in option A, we have the terms 2 and -6. These terms are not like terms because they do not have the same variables.

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2 years ago

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In a box there are two prizes that are worth $4, a single prize worth $20, and a single prize worth $200. A player will reach in

Ronch [10]

Answer:

Step-by-step explanation:

The player has a 1/4 chance of drawing any of the 4 prizes. This means that the probability of drawing a prize of $4 is 1/2 because there are 2 prizes worth of $4. The probability of drawing a prize of $20 is 1/4 and the probability of drawing a prize of $200 is also 1/4. To find the fair price of the game, we have to calculate the expected value that the player will gain. This can be obtained by multiplying any possible value of a price for the probability of drawing a prize of that value and adding all the obtained values togueter. Thus, the fair price of the game is

$4 * \frac{1}{2} + 20 * \frac{1}{4} + 200 * \frac{1}{4} = 57$

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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

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