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:
The number of deliveries that are predicted to be made to homes during a week with 50 deliveries to business is 87 deliveries
Step-by-step explanation:
The data categorization are;
The number of home deliveries = x
The number of delivery to businesses = y
The line of best fit is y = 0.555·x + 1.629
The number of deliveries that would be made to homes when 50 deliveries are made to businesses is found as follows;
We substitute y = 50 in the line of best fit to get;
50 = 0.555·x + 1.629 =
50 - 1.629 = 0.555·x
0.555·x = 48.371
x = 48.371/0.555= 87.155
Therefore, since we are dealing with deliveries, we approximate to the nearest whole number delivery which is 87 deliveries.
6. (5y+2)+14x=5y+(2+14x) (1point) A.associative property of addition B.associative property of multiplication C.commutative property of addition D.commutative property of multiplication 7. (m*n)*p=m*(n*p) (1 point) A. commutative property of addition B.commutative property of multiplication C.associative property of addition D. associative property of multiplication
Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.
Since there are two dice, the total possible outcome is 36.
Hence, the probability of getting one number each is 1/36
Hence, the principal used a fair method because each result is an equally likely possible outcome.
Option B is correct.
We can tell from the data that there is a midpoint between the lowest and highest point of the clock, which is at a height of 9.5 feet.
Moreover, the lowest point occurs at 6 o clock, and the highest occurs at 12 o clock.
The amplitude of variation from the mid-point is 0.5 feet given by (10 - 9) / 2.
Finally, the time period for the equation is 12 hours. Thus, the answer is:
h = 0.5cos(πt/6) + 9.5, option B
