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ivolga24 [154]
2 years ago
11

We have all probably played games like monopoly or yahtzee that use standard six sided dice. what does it mean for dice to be fa

ir? a fair six-sided die is defined as a die that will have each of the 6 faces of the die come up one-sixth of the time in the long run. a loaded six-sided die is defined as a die that has one face of the die that comes up more often than one-sixth of the time in the long run. an avid yahtzee player wants to know whether or not his lucky die is loaded so that 6's appear more often than any other number. he throws his lucky die 85 times and noted that he rolled a "six" on 25 of those rolls. what are the hypothesis and conclusion for this experiment?
Mathematics
1 answer:
Nonamiya [84]2 years ago
8 0
1/6 of the time

so to find out

25/85=1/6?
5/17=1/6?
0.29=0.16?
naw

it is loaded
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b) -4

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First, we subtract 4 from 36 to get 32

36-4=32

-8x=32

Since a negative times a negative equals a positive, then the answer has to be negative because 36 is positive.

32 divided by 8 = 4

B) -4

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Answer:

the 4th one is the answer

Step-by-step explanation:

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Evaluate the expression -4x+5y-14 when x=0 and y= 3/5​
emmainna [20.7K]

Answer:

-11

Step-by-step explanation:

Plug in 0 for x and 3/5 for y into the expression

-4(0) + 5(3/5) - 14

Multiply the numbers inside the parentheses

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3 - 14 = -11, which is your answer

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There are 35 men and 25 women on a bus. If 20 married couples leave the bus, the number of men is how many times the number of w
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6 0
3 years ago
Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
1 year ago
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