The answer is W. (Please make me Brainliest.) Hope this helps!
Problem 3
Disjoint events are two (or more) events that cannot happen at the same time. An example would be flipping heads and tails at the same time with the same coin. Another example is selecting an ace of spades and selecting an ace of diamonds, when you only select one card.
In general notation, P(A and B) = 0 if and only if A and B are disjoint.
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.3 + 0.1 - 0
<h3>P(A or B) = 0.4</h3>
=========================================
Problem 5
Same idea, but different numbers
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 1/3 + 1/4 - 0
P(A or B) = 4/12 + 3/12 - 0
<h3>P(A or B) = 7/12</h3>
A.) is you’re best option
Answer:
I believe this is pathgorean theorem, which I did at the beginning of alegbra last year. It is basically finding the missing side of a triangle. (Trust me some of the easiest stuff you do all year) Below are my notes from last year. I hope it helps! Good Luck!!!
Step-by-step explanation:
Given two sides
If you know two other sides of the right triangle, it's the easiest option; all you need to do is apply the Pythagorean theorem:
a² + b² = c²
if leg a is the missing side, then transform the equation to the form when a is on one side, and take a square root:
a = √(c² - b²)
if leg b is unknown, then
b = √(c² - a²)
for hypotenuse c missing, the formula is
c = √(a² + b²)
