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Ksenya-84 [330]
3 years ago
12

Lines a and b are perpendicular. The equation of line a is y=1/3x+3. What is the equation of line b

Mathematics
1 answer:
kykrilka [37]3 years ago
6 0
Perpendicular lines have slopes that are negative reciprocals to each other, that is, if one slope is a/b, the other is -b/a
in this case, the given slope is 1/3, so the other line's slope is -3/1, which is -3
from the graph we see that the y intercept of line b is -4, so the equation of line b is: y=-3x-4
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Step-by-step explanation:

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2 years ago
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.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

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

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first 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}  ^4

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

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

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

= (\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

8 0
3 years ago
A rectangles length is 1.5 times its width. The perimeter of the rectangle is four more than 3 times the length. What are the di
vovikov84 [41]

Answer:

  8 by 12 units

Step-by-step explanation:

Let w represent the width of the rectangle. Then the length is 1.5w. The perimeter is 4 more than 3 times this, so is (3(1.5w) +4) = 4.5w+4

The perimeter is given by the formula ...

  P = 2(length + width)

Filling in the given values for the variables, we have ...

  4.5w +4 = 2(w +1.5w)

  4 = 0.5w . . . . . . subtract 4.5w and collect terms

  8 = w . . . . . . . . . multiply by 2

  length = 1.5×8 = 12

The rectangle is 8 units wide and 12 units long. The perimeter is 40 units.

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2 years ago
I need help with this
adoni [48]
Answer is D
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2 years ago
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