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slamgirl [31]
3 years ago
9

What is the slope of the line that passes through the points (-3,1) and (7,-14)

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

-  \frac{3}{2}

Step-by-step explanation:

\boxed{gradient =  \frac{y1 - y2}{x1 - x2} }

Using the formula above,

slope of line

=  \frac{1 - ( - 14)}{ - 3 - 7}  \\  =  \frac{1 + 14}{ - 10}  \\  =  \frac{15}{ - 10}  \\  =  -  \frac{3}{2}

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A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

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The green die has faces numbered 1, 2, 3, 4, 4, and 4.

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= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

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= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

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