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3 years ago

A lottery game has balls numbered 0 through 13. What is the probability of selecting an even numbered ball or a 7

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

3 0

6/13 of choosing an even number and 1/13 for seven
there's only 6 even numbers and only one 7. out of 13

Rzqust [24]3 years ago

3 0

EDIT: added 0

The possible even numbers are 0,2,4,6,8,10, and 12. With 7 included, this is 8 out of 13 balls, so your probability of selecting one is 8/13 or approximately 62%. Hopefully this helps, please mark as brainliest if it did! :)

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Consider the functions given below. SEE F

Wewaii [24]

Answer:

1. $P(x)$ ÷ $Q(x)$ ---> $\frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ ---> $\frac{2(6x - 1)}{(3x - 1)(-3x + 2)}$

3. $P(x) - Q(x)$ ---> $\frac{-2(12x - 5)}{(3x - 1)(-3x + 2)}$

4. $P(x)*Q(x)$ --> $\frac{12}{(3x - 1)(-3x + 2)}$

Step-by-step explanation:

Given that:

1. $P(x) = \frac{2}{3x - 1}$

$Q(x) = \frac{6}{-3x + 2}$

Thus,

$P(x)$ ÷ $Q(x)$ = $\frac{2}{3x - 1}$ ÷ $\frac{6}{-3x + 2}$

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

$\frac{2}{3x - 1}*\frac{-3x + 2}{6}$

$\frac{2(-3x + 2)}{6(3x - 1)}$

$= \frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ = $\frac{2}{3x - 1} + \frac{6}{-3x + 2}$

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

$\frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}$