2.3p - 10.1 = 6.5p - 4 - 0.01p
If we multiply this equation by 100 we have
230p - 1010 = 650p - 400 - p
So the right answer is this
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: $0.51
Step-by-step explanation:
12/100 of $4.25 = $0.51
Amount payable plus sales tax for a 6- pack of soda will now be $4.25 + $0.51 = $4.76
I = p*r*t
$3000=p*.06*5=?
Multiply and you would have yor answer(rounded to nearsest cent)!
