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pantera1 [17]
3 years ago
13

What is the sum of the probability of drawing a green marble and the probability of not drawing a green marble?

Mathematics
2 answers:
otez555 [7]3 years ago
8 0

6/12 to draw a green marble

6/12 to not draw a green marble

Rzqust [24]3 years ago
3 0

There are a total of 6 green marbles.

Along with 3 blue and 3 red.

There is a total of 12 marbles in the bag.

This would mean that you would have 6/12 chance of getting a green marble.

If you simplify this then it would be 1/2 chance of getting a green marble.

3/12 and 3/12 chance of getting a blue or red marble.

Therefore, there is 1/2 a chance of drawing a green marble and 1/2 a chance of not drawing a green marble.

Hope this helps! :3

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In golf, scores are calculated relative to the par of the golf course. Par is 0 on a number line. Given the following informatio
ddd [48]
The answer would be six under par
3 0
3 years ago
What is the result of adding these two equations? \begin{aligned} 2x+3y &= -5 \\\\ 5x-y &= -12 \end{aligned} 2x+3y 5x−y
EleoNora [17]

Answer:

7x+2y = -17

Step-by-step explanation:

We have the following equations:

2x+3y = -5

5x-y = -12

If we want to add these equations, we just need to add x-terms, y-terms and independent terms.

(2x+5x)+(3y-y) = -5-12

7x+2y = -17

I hope it helps you!

7 0
2 years ago
10PTS:
Ymorist [56]
In order to get rid of the fractions, you need to multiply the equation by 12

8 0
3 years ago
Read 2 more answers
. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19
timofeeve [1]
Let's rewrite the binomial as:
(1 - 2x)^{19}

\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}

Using the binomial expansion, we get:
\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}
3 0
3 years ago
Use the discriminant to determine the number of real solutions to the quadratic equation.
gulaghasi [49]

Answer:

There are two real and different roots

Step-by-step explanation:

The discriminant is given by b²-4ac

(-12)²-4(70×1)

144-280

136

Now 136>0

5 0
2 years ago
Read 2 more answers
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