Answer:
The absolute brightness of the Cepheid star after a period of 45 days is -5.95
Step-by-step explanation:
Since the absolute magnitude or brightness of a Cepheid star is related to its period or length of its pulse by
M = –2.78(log P) – 1.35 where M = absolute magnitude and P = period or length of pulse.
From our question, it is given that P = 45 days.
So, M = –2.78(log P) – 1.35
M = –2.78(log 45) – 1.35
M = –2.78(1.6532) – 1.35
M = -4.60 - 1.35
M = -5.95
So, the absolute magnitude or brightness M of a Cepheid star after a period P of 45 days is -5.95
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.
(2,5)(1,3)
slope(m) = (3 - 5) / (1 - 2) = -2/-1 = 2 (2 units up, one unit to the right)
y - y1 = m(x - x1)....using (1,3)
y - 3 = 2(x - 1)....talia's last step is incorrect because she didn't sub in her slope of 2
There are an infinite number of possibilities, but one of them could be 45/100 or 9/20
