Barbara’s equation did not consider the number of bottles of iced tea.
Answer:
Step-by-step explanation:
we know that
The probability of an event is the ratio of the size of the event space to the size of the sample space.
The size of the sample space is the total number of possible outcomes
The event space is the number of outcomes in the event you are interested in.
Let
x---------> size of the event space
y-------> size of the sample space
In this problem we have
(because is only one number to think)
(there are numbers between and )
substitute
b. The median of the chosen number is 91, is there an limit to how large the aerge of the chosen numbers can be? If so, what is the largest the average can be?
c. The average of the chosen number is 91, what is the smallest the median of the 9 chosen numbers could be?
d. The average of the chosen numbers is 91. What is the largest the median of the chosen numbers could be?
Answer:
a) 1
b) There is no limit to which the largest number can be because we are only given information about the median.
c) 1
d) 90
The smallest average is 49 and the largest average is 91. The smallest median is 91 and the largest median is also 91.
a. Since the median is 91, at least 5 friends must choose numbers greater than or equal to 91, and at most 4 friends can choose numbers less than 91. To minimize the average, let's assume the four friends choose the smallest possible numbers less than 91 (1, 2, 3, and 4). The remaining five friends can then choose 91, 91, 91, 91, and 91. The average of the nine chosen numbers is (1 + 2 + 3 + 4 + 91 + 91 + 91 + 91 + 91)/9 = 49.
b. There is no limit to how large the average of the chosen numbers can be. The nine friends can all choose the same number, such as 91, which would make the average 91.
c. Since the average is 91, let's assume the eight friends choose the smallest possible numbers less than 91 (1, 2, 3, ..., 8). The remaining friend can then choose a number greater than or equal to 91. To minimize the median, the friend can choose the smallest possible number greater than or equal to 91, which is 91. So, the smallest median would be 91.
d. Since the average is 91, let's assume the eight friends choose the largest possible numbers less than 91 (84, 85, ..., 91). The remaining friend can then choose a number greater than or equal to 91. To maximize the median, the friend can choose the largest possible number greater than or equal to 91, which is 91. So, the largest median would also be 91.
#SPJ2
y=2/3x-3
y=-3/2x+2
Answer:
Step-by-step explanation:
3y = 2x - 9
2y = -3x + 4
3x + 2y = 4
-2x + 3y = -9
6x + 6y = 8
-6x + 9y = -27
15y = -15
y = -1
3(-1) = 2x - 9
-3 = 2x - 9
6 = 2x
3 = x
(3, -1)
Answer: 20 pounds
Step-by-step explanation: