We need to find the probability of the event:
having blue eyes or blond hair
This event is the union of the events:
A: having blue eyes.
B: having blond hair.
So, we need to find the probability P(A ∪ B). In order to do so, we can use the following formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
We know that the probability of the intersection of A and B (having both) is
P(A ∩ B) = 24%
Also:
P(A) = 43%
P(B) = 46%
Then, using those values into the above formula, we find:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 43% + 46% - 24%
P(A ∪ B) = (43 + 46 - 24)%
P(A ∪ B) = 65%
Answer:
Step 1
Step-by-step explanation:
You distributed the 2 in 2(x-4) to get 2x-8
She earned 11.5 per hour! And we know this because 247.25 divided by 21.5 = $11.5 per hour ( hope this helps:)
I've Got Your Back:
What do you need help with?
Here's what you wanna do.
Add 6 dollars to 12.75.
You now have $18.75
Now subtract 10
18.75 - 10 = 8.75
Now subtract 3.50
8.75 - 3.50 = 5.25
So, the cotton candy costs $5.25
![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20~%5Chfill%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cboxed%7Ba%5E6%2Bb%5E6%7D%5Cimplies%20a%5E%7B2%5Ccdot%203%7D%2Bb%5E%7B2%5Ccdot%203%7D%5Cimplies%20%28a%5E2%29%5E3%2B%28b%5E2%29%5E3%20%5C%5C%5B2em%5D%20%5Ba%5E2%2Bb%5E2%5D%20%5B%28a%5E2%29%5E2-a%5E2b%5E2%2B%28b%5E2%29%5E2%5D%5Cimplies%20%5Cboxed%7B%28a%5E2%2Bb%5E2%29%28a%5E4-a%5E2b%5E2%2Bb%5E4%29%7D)
about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.
5, 8, 13 are no dice, namely 5² + 8² ≠ 13
25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²
however, 5,12 and 13 are indeed a pythagorean triple
also is 39, 80, 89.
when looking for a pythagorean triple, recall that c² = a² + b².
so the longest leg is the sum of the square of the small ones.
so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.
