Answer:
I believe it would be C
Step-by-step explanation:
In a large population, 61% of the people are vaccinated, meaning there are 39% who are not. The problem asks for the probability that out of the 4 randomly selected people, at least one of them has been vaccinated. Therefore, we need to add all the possibilities that there could be one, two, three or four randomly selected persons who were vaccinated.
For only one person, we use P(1), same reasoning should hold for other subscripts.
P(1) = (61/100)(39/100)(39/100)(39/100) = 0.03618459
P(2) = (61/100)(61/100)(39/100)(39/100) = 0.05659641
P(3) = (61/100)(61/100)(61/100)(39/100) = 0.08852259
P(4) = (61/100)(61/100)(61/100)(61/100) = 0.13845841
Adding these probabilities, we have 0.319761. Therefore the probability of at least one person has been vaccinated out of 4 persons randomly selected is 0.32 or 32%, rounded off to the nearest hundredths.
X would equal -9
Because -9 times -2 equals 18
Answer:
14 + 1m + 7n
Explanation:
I combined like terms 5m - 4m and 22 + -8. After I simplified the equation from there.
<h3>
<u>·</u><u>Hope</u><u> </u><u>I</u><u> </u><u>helped</u><u>,</u><u> </u><u>-</u><u>SunnySideHelper·</u></h3>
Answer:
1) -17x-9
Step-by-step explanation:
1)
(-2x^3+x-5)+(x^3-3x-4)
Combine like terms using P.E.M.D.A.S:
(-2x^3+x-5)+(x^3-3x-4)
(-9x-5)+(-8x-4)
Combine like terms:
(-2x^3+x-5)+(x^3-3x-4)
(-9x-5)+(-8x-4)
-17x-9
I hope this helps a bit!☺
