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 is x=700 hope this helps
Answer:
The value of 
Step-by-step explanation:
Being the general expression of the sequence:
We replace the value n=12 to get
Answer:
The last graph.
Step-by-step explanation:
The first thing we look for is the y-intercept at -2. The only graphs where the line passes through the y-intercept at this point are the first and the last.
The slope of the line we want is -2/3. This means the line decreases; the first graph increases, not decreases, so it must be the fourth one..
Checking the fourth graph, from the y-intercept, we go down 2 and over 3 to find the next point; this means the slope is -2/3 and this is the correct line.
The quadratic formula is -b plus minus the square root of b^2-4ac all over 2a.
Here, a=1, b=13, and c=30.
The only option that fills in the values correctly is D
