Find the median, first quartile, third quartile, interquartile range, and any outliers for each set of data. 111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112 A. Median: 102, Q1: 87, Q3: 113, IR: 28, outliers: none B. Median: 102, Q1: 87, Q3: 115, IR: 28, outliers: 68 C. Median: 102, Q1: 87, Q2: 115, IR: 28, outliers: none D. Median: 102, Q1: 87, Q3: 115, IR: 27, outliers: none

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

Step-by-step explanation:

Given the data:

111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112

Ordered data : 68, 74, 84, 87, 88, 88, 93, 102, 105, 111, 112, 115, 119, 151, 152

Sample size, n = 15

The first quartile, Q1 = 1/4(n+1)th term

Q1 = 1/4(15+1)th term

Q1 = 1/4(16) = 4th term

Q1 = 87

The Median , Q2 = 1/2(n+1)th term

Q2 = 1/2(15+1)th term

Q2 = 1/2(16) = 8th term

Q2 = 102

The third quartile, Q3 = 3/4(n+1)th term

Q3 = 1/4(15+1)th term

Q3 = 3/4(16) = 12th term

Q3 = 115

The interquartile range, IQR = Q3 - Q1

IQR = (115 - 87) = 28

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.