Please urgent answer my question.

Mathematics

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer:

X=2

Step-by-step explanation:

nlexa [21]3 years ago

5 0

Answer:

$y + 6 = x( {x}^{2} - 1) \\ y + 6 = {x}^{3} - x \\ y + {x}^{3} - x - 6 \\ y = {x}^{3} x - 6xer$

Step-by-step explanation:

you can download photo math and it will help you

You might be interested in

Which is the solution set of the inequality 15y-9 < 36?

Kipish [7]

15y<36-9 ==>15y<27

y<27/15 ==>y<9/5

answer is B

6 0

3 years ago

Read 2 more answers

Expected Value (50 points)

garri49 [273]

Let $W$ be the random variable representing the winnings you get for playing the game. Then

$W=\begin{cases}10-1=9&\text{if the dice sum is odd}\\5-1=4&\text{if the dice sum is 4 or 8}\\50-1=49&\text{if the dice sum is 2 or 12}\\-1&\text{otherwise}\end{cases}$

First thing to do is determine the probability of each of the above events. You roll two dice, which offers 6 * 6 = 36 possible outcomes. You find the probability of the above events by dividing the number of ways those events can occur by 36.

The sum is odd if one die is even and the other is odd. This can happen 2 * 3 * 3 = 18 ways. (3 ways to roll even with the first die, 3 ways to roll odd for the die, then multiply by 2 to count odd/even rolls)
The sum is 4 if you roll (1, 3), (2, 2), or (3, 1), and the sum is 8 if you roll (2, 6), (3, 5), (4, 4), (5, 3), or (6, 2). 8 ways.
The sum is 2 if you roll (1, 1), and the sum is 12 if you roll (6, 6). 2 ways.
There are 36 total possible rolls, from which you subtract the 18 that yield a sum that is odd and the other 10 listed above, leaving 8 ways to win nothing.

So the probability mass function for this game is

$P(W=w)=\begin{cases}\frac12&\text{for }w=9\\\frac29&\text{for }w=4\text{ or }w=-1\\\frac1{18}&\text{for }w=49\\0&\text{otherwise}\end{cases}$