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.
Answer:
Step-by-step explanation:
2n+15>3
Answer:
(N-2)x180=sum of interior angles
So (5-2)x180=540
Divided by number of sides=540/4=108... then each angle is 108 divide to triangles ..so they will have 2 angles =54 2 sides =4 the last angle =72 ."of the triangle " area of triangle =1/2xsxsxsin(side in
between)
So 1/2x4x4xsin(72)= then multiply it by 5 as u will have 5 triangles .. so it will be 190.2(probably i tried)
Step-by-step explanation:
10.7
23.7-10=10
that r the maths