Answer:
Each octave change requires a doubling of the frequency.
Therefore, one octave above 126.63 is 126.63 * 2 which equals
253.26
Step-by-step explanation:
First you would do parenthesis according to the PEMDAS(Parenthesis, Exponents, Multiply, Divide, Add, Subtract) Rule
4 -2(-5a -10) = 30
4 + 10a + 20 = 30
Now you need to add 4 and 20 because they are on the same side
24 + 10a = 30
Now subtract 24 on both sides
24 + 10a = 30
-24. -24
10a = 6
Last divide 10 from both sides
10a/10. = 6/10
a= 6/10
Last simplify (optional but recommended)
a = 3/5 or 0.6
Hope this helped :)
A = 0.6
Let
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}](https://tex.z-dn.net/?f=W%3D%5Cbegin%7Bcases%7D10-1%3D9%26%5Ctext%7Bif%20the%20dice%20sum%20is%20odd%7D%5C%5C5-1%3D4%26%5Ctext%7Bif%20the%20dice%20sum%20is%204%20or%208%7D%5C%5C50-1%3D49%26%5Ctext%7Bif%20the%20dice%20sum%20is%202%20or%2012%7D%5C%5C-1%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
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}](https://tex.z-dn.net/?f=P%28W%3Dw%29%3D%5Cbegin%7Bcases%7D%5Cfrac12%26%5Ctext%7Bfor%20%7Dw%3D9%5C%5C%5Cfrac29%26%5Ctext%7Bfor%20%7Dw%3D4%5Ctext%7B%20or%20%7Dw%3D-1%5C%5C%5Cfrac1%7B18%7D%26%5Ctext%7Bfor%20%7Dw%3D49%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
The expected value of playing the game is then
![E[W]=\displaystyle\sum_ww\,P(W=w)=\frac92+\frac89-\frac29+\frac{49}{18}=\frac{71}9](https://tex.z-dn.net/?f=E%5BW%5D%3D%5Cdisplaystyle%5Csum_ww%5C%2CP%28W%3Dw%29%3D%5Cfrac92%2B%5Cfrac89-%5Cfrac29%2B%5Cfrac%7B49%7D%7B18%7D%3D%5Cfrac%7B71%7D9)
or about $7.89.
Answer:
Step-by-step explanation:
The correct answer is sea otter.