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natima [27]
3 years ago
5

15 less than twice a number in algebraic expression

Mathematics
2 answers:
hoa [83]3 years ago
4 0
I think it’s 2x-15, sorry if I’m wrong
gladu [14]3 years ago
4 0

The answer is 2x - 15


Hope this helps!

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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
4.89; 4.089; 4.870; 4.881 - Put these in order from least to greatest. Your answer​
motikmotik
Answer: 4.089; 4.870; 4.881; 4.89
3 0
2 years ago
Read 2 more answers
21 is what percent of 40
-Dominant- [34]
To solve for this, set up a proportion.
21/40 = x/100.
Cross multiply the fraction's numerators with the denominators; 
21 x 100, 40 and x.
2100/40x.
Divide both numbers by 40 to get x.
2100/40 = 52.5
21 is 52.5% of 40.
I hope this helps!
5 0
3 years ago
What’s the domain and range for these two?
user100 [1]

Step-by-step explanation:

1) the domain is from -3 to 5 D [-3,5]

the range is from -5 to -2. R [-5,-2]

2)the domain is from -3 to 6 but not including 6 D [-3,6)

the range is from -5 but not including -5 to 2. R (-5,2]

4 0
3 years ago
Which statement is correct? (If you can answer 1 and 3 it would help a lot too)
Lera25 [3.4K]

Answer:

the answer is 3 because 1x3=3

5 0
3 years ago
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