we have
Min=26
Q1=67
Q2=80
Q3=87
Max=100
Mean=76
Mode=100
standard deviation=76
IQR=20
so
In this problem
the mean is equal to the standard deviation
therefore
<h2>The standard deviation is incorrectly</h2>
Answer:
C, 6
Step-by-step explanation:
31-13 is 18, 18/3 is 6.
Answer:
Slope = 46.000/2.000 = 23.000
x-intercept = 6/23 = 0.26087
y-intercept = -6/1 = -6.00000
Answer:
Data set 3 has the highest IQR, so the correct option is {13,17,12,21, 18,20}
Step-by-step explanation:
The question is:
Which data set has the greatest spread for the middle 50% of its data
We have given the data sets:
{18,13,22, 17, 21, 24}
{17,19,22,26,17,14}
{13,17,12,21, 18,20}
{18,21,16,22,24,15}
To find the greatest spread for the middle we need to find the IQR for all the data sets and check which one is highest.
So,
For the first set: 22-17 = 5
For the second: 22-17 = 5
For the third set = 20-13 = 7
For the fourth set = 22-16 = 6
Since data set 3 has the highest IQR, so the correct option is {13,17,12,21, 18,20} ....
Step-by-step explanation:
the area of a rectangle is
length × width
in our case
length × width = 70 m²
length = 3×width - 11
we can use this second equation in the first and get
(3×width - 11) × width = 70
3×width² - 11×width - 70 = 0
the general solution to a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 3
b = -11
c = -70
x = (11 ± sqrt((-11)² - 4×3×-70))/(2×3) =
= (11 ± sqrt(121 + 840))/6 =
= (11 ± sqrt(961))/6
x1 = (11 + 31)/6 = 42/6 = 7
x2 = (11 - 31)/6 = -20/6 = -10/3
the negative solution is not applicable for a length in an object.
so, only x1 = 7 m is our solution.
the width of the rectangle is 7 m.
the length = 3×width - 11 = 3×7 - 11 = 21-11 = 10 m
