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.
The answer would be A.
if he lost 8 seconds off of 3 pit stops that means he lost 24 seconds. if he can’t 10 seconds for the 2 he made, he gained 20. so therefore, he is only 4 seconds behind the leader. I hope this helps :)
<span>(x) - (0.03)(200) = 2.00
Let x be the </span><span>company charge to make the profit
solving for x gives 8$</span>
23,039.
y=13,000(1.045)^t
t=13
You get 1.045 because you add 1+ the percentage in decimal form.
