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frutty [35]
3 years ago
13

Consider the equation v + 4 + v = 8. What is the resulting equation after the first step in the solution?

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0
<span>The resulting equation after the first step in the solution of the starting equation v + 4 + v = 8 is 2v + 4 = 8. So, we have the equation v + 4 + v = 8. We can group similar terms on the left side: (v + v) + 4 = 8. When we sum these two in the parenthesis, we will get the resulting equation which is: 2v + 4 = 8.</span>
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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.

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
Please please open !
Tomtit [17]

Answer:

1.A  2.A  3.A 4.B  5.A  6.B

6 0
2 years ago
Marcus has hired 3 people to help him with his lawn business. He pays all the same, and he pays them every day. He paid $360 to
deff fn [24]

Hi!

If they each got the same, then they each got 1/3 of the total $360. Therefore, we can divide 360 by 3. We can do this by dividing 36 by 3, 36/3, and that's 12. Then add the 0! 120.

Your sentence could be: "The amount of money each worker got can be represented by 360/3, which is $120."

Hope this helps! :D

3 0
2 years ago
Read 2 more answers
Please answer this 2 multiple choice questions correctly I needed today please show work
vredina [299]

Answer:

21) 12x + 11

hoped that worked

8 0
2 years ago
Find x and y and i just need the answer, no need to show work.
vichka [17]

Answer:

x= 20

y = 60

Step-by-step explanation:

y = 3x

6(3x) = 360

3x = 360/6

3x = 60

x = 20

y = 3*20

y = 60

7 0
1 year ago
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