Answer:
d is your Answer .............
The measures of spread include the range, quartiles and the interquartile range, variance and standard deviation. Let's consider each one by one.
<u>Interquartile Range: </u>
Given the Data -> First Quartile = 2, Third Quartile = 5
Interquartile Range = 5 - 2 = 3
<u>Range:</u> 8 - 1 = 7
<u>Variance: </u>
We start by determining the mean,

n = number of numbers in the set
Solving for the sum of squares is a long process, so I will skip over that portion and go right into solving for the variance.

5.3
<u>Standard Deviation</u>
We take the square root of the variance,

2.3
If you are not familiar with variance and standard deviation, just leave it.
<span>Dawn was at 6 am.
Variables
a = distance from a to passing point
b = distance from b to passing point
c = speed of hiker 1
d = speed of hiker 2
x = number of hours prior to noon when dawn is
The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as
a = cx = 9d
cx = 9d
x = 9d/c
The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as
b = dx = 4c
dx = 4c
x = 4c/d
We now have two expressions for x, set them equal to each other.
9d/c = 4c/d
Multiply both sides by d
9d^2/c = 4c
Divide both sides by c
9d^2/c^2 = 4
Interesting... Both sides are exact squares. Take the square root of both sides
3d/c = 2
d/c = 2/3
We now know the ratio of the speeds of the two hikers. Let's see what X is now.
x = 9d/c = 9*2/3 = 18/3 = 6
x = 4c/d = 4*3/2 = 12/2 = 6
Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am.
We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
Answer:
y=-2/19x
Step-by-step explanation:
slope=(y1-y2)/(x1-x2)=-2/19
y-intercept=0
y=-2/19x+0
Step-by-step explanation:
An equivalent fraction can be found by either multiplying the numerator and denominator by the same number or dividing the numerator and the denominator by the same number.