For these problems you have to use PEMDAS so for problem 15 you would do the parentheses first, (21-3) you get 18 and then exponents, 3^2, you get 9, no multiplication but there is division so take your answers, 18 and 9 and divide them, you get 2, so the answer to 15 is 2.
Answer:
a.2nd quarter with 9 goals
b. 4.8 goals
c. 4 goals
Step-by-step explanation:
a. The mode is defined as the most appearing data point or the data point with the highest frequency..
From our data(for away goals):
- 1st quarter-2
- 2nd quarter-9
- 3rd quarter-7
- 4th quarter-4
Hence, the 2nd quarter has the mode for away goals with 9 goals.
b. Mean is defined as the average of a set of data points.
#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

Hence, the mean number of goals per quarter is 4.8 goals
c. To find the number of more home goals than away goals, we subtract from their summations as:

Hence, there are 4 more home goals than away goals.
The answer is 67.93939393939394
Answer:
91/216
Step-by-step explanation:
The probability of getting a 4 in the first three rolls is 1 minus the probability of not getting a 4 on any of the rolls.
P(at least one 4) = 1 − P(no 4s)
P(at least one 4) = 1 − (5/6)³
P(at least one 4) = 91/216
Alternatively, you can calculate it this way.
The probability of getting a 4 on the first roll is 1/6.
The probability of getting a 4 on the second roll is (5/6) (1/6) = 5/36.
The probability of getting a 4 on the third roll is (5/6) (5/6) (1/6) = 25/216.
The probability of any of the three events is 1/6 + 5/36 + 25/216 = 91/216.