Answer:
Claim : Jump distance is less than 16 feet.
A ) 
n = 19
Since n <30 we will use t test
The mean distance of the sample jumps is 13.2 feet. i.e. x = 13.2 feet
The standard deviation of her jump distance is 1.5 ft i.e. s = 1.5
Formula : 
Substitute the values:
B) 

degree of freedom df = n-1 = 19-1 = 18


t critical > t statistics
So, we accept the null hypothesis
C) So, the claim is wrong that ump distance is less than 16 feet.
Hence long jump distance must be greater than or equal to 16
Answer:
Step-by-step explanation:
XY = 15 MI.
YZ = 8 mi.
Area = A
A = 1/2 (XY)(YZ)
A = 1/2 (15)(8)
A = 1/2 (120)
A = 60
Let r be the lesser root and r^2 be the greater.......the sum of the roots = -b/a = -[-6] / 1 = 6
So we have that
r^2 + r = 6 → r^2 + r - 6 = 0
Factor
(r + 3) (r - 2) = 0
So r = -3 or r = 2
Then r^2 = 9 or r^2 =4
And the product of the roots = c/a = k/a = k
So....k = (-3)(9) = -27 or k = (2)(4) = 8
Check
x^2 - 6x - 27 = 0 factors as (x + 3)(x - 9) = 0 and the roots are -3 and 9
x^2 - 6x + 8 = 0 factors as (x -2) (x - 4) =0 and the roots are 2 and 4
Answer:
72
Step-by-step explanation:
you multiply 36 *2 therefore giving you 72centimeters squared