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WARRIOR [948]
3 years ago
5

What are the next three terms in the sequence?

Mathematics
2 answers:
Ne4ueva [31]3 years ago
7 0
-1 + 10 = 9 +10 = 19 + 10 = 29 so the next would be 39 because you add ten to it and there is only one answer with 39 so it's C
eduard3 years ago
6 0
In this sequence it looks as each number is increasing by 10

-1 
+10
9
+10
19
+10
29

And so on.

So following this pattern it looks as if the answer is 
C : 39, 49, 59

Hope this helps!
Brainliest is always appreciated if you feel its deserved.
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irinina [24]

Answer:

I believe the answer is 250. I hope this helps Bye!

6 0
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
Find the measure of CB.
mamaluj [8]

Answer:

I'm Pretty Sure Its 40.

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Help me with this question
steposvetlana [31]

Answer:

we cant see the qeastion

Step-by-step explanation:

4 0
3 years ago
PLEASE HELP I AM DESPERATE I WILL GIVE THANKS, 5 STARS, AND THE BRAINLIEST!!!
Amanda [17]

g(-2) + g(2) = 22

Step-by-step explanation:

For x = -2, g(x) = x^2 - 3x

or

g(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10

For x = 2, g(x) = 12, therefore

g(-2) + g(2) = 10 + 12 = 22

3 0
3 years ago
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