To find the IQR (Interquartile Range) from the given information: Minimum = 15 min, Q1 = 27 min (25th percentile), Q2 = 31 min (median), Q3 = 32 min (75th percentile), Maximum = 50 min. The IQR calculation is IQR = Q3 - Q1 = 32 - 27 = 5. For identifying outliers, apply the k = 1.5 rule. The expected range is [Q1 - 1.5*IQR, Q3 + 1.5*IQR] = [27 - 1.5*5, 32 + 1.5*5] = [19.5, 39.5]. The actual range is [15, 50]. The minimum value is below 19.5, and the maximum value exceeds 39.5. Hence, outliers are present. Conclusion: Outliers exist.
Answer:
A total of 200 students were present at the basketball game
Step-by-step explanation:
Refer to the full question shown in the attached figure
Let
x be the total number of fourth graders
y represent the total attendees at the game
It is known that
The number of game attendees is four times the number of fourth graders
Thus
The linear expression is
-----> equation A
-----> equation B
Replace equation B into equation A and solve for y
Therefore
200 students were present at the basketball event
The average calorie count for chocolate pie pieces at a dining establishment is 350, with a standard deviation of 20. Due to imprecise pie slicing, calorie distribution follows a Normal distribution. What graph illustrates the percentage of pie pieces exceeding 375 calories? The z score is utilized to assess the number of standard deviations the raw score is from the mean. A positive z score indicates the raw score is higher than the mean, while a negative z score implies it is lower. The z score formula is: Given μ = 350 calories, σ = 20 calories, x > 375. The shaded portion of the graph indicates the proportion of pie pieces containing more than 375 calories. According to the normal distribution table, P(x > 375) = P(z > 1.25) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056 = 10.56%.
Response:9/10/3/5
Detailed explanation: because... believe me
Answer:
$6.48
Step-by-step explanation:
Initially calculate the 10% discount on $1.60, which results in $1.44. Then multiply $1.44 by 4.5 to find the total, yielding $6.48.
