Try this solution:
1. for probability of both televisions work:
2. for probability of at least on is defective (it means, the sum = one is defective and two are defective):
- for P(1defective);
- for P(2defective);
totaly for at least one is defective:
The notes with all black are 1 beat (first position for tuba idk about the rest) the line things are pauses so you would pauses for like 1.99 seconds
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Answer:
f(x) = 4(x - 8)
4 (10 -8) = 8
Step-by-step explanation:
<span>[x + 2 - (5x + 5)] ² =
[x + 2 - 5x - 5]² =
[x - 5x + 2 - 5]² =
[-4x - 3]² =
(-4x)² - 2.(-4x).3 + (-3)² =
16x² + 24x + 9
Answer:
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