Given that mean of quiz scores = 6.4 and standard deviation = 0.7
And we need to use Chebyshev's theorem to find the range in which 88.9% of data will reside.
Chebyshev's theorem states that "Specifically, no more than 
of the distribution's values can be more than k standard deviations away from the mean".
That is 
k = 3
So, we want the range of values within 3 standard deviations of mean.
Hence range is [mean -3*standard deviation, mean +3*standard deviation]
= [6.4 - 3*0.7 , 6.4+3*0.7]
= [6.4 - 2.1 , 6.4+2.1] = [4.3,8.5]
Answer:
5 cm
Step-by-step explanation:
Multiply every number by 2. 3x2 = 9, 2x2 = 4, and 5x2 = 10
Answer:
V= 8/3
Step-by-step explanation:
Answer:
Relative error = 0.0254 %
Step-by-step explanation:
Given details
Height of equipment from base is 104 ft
Elevation angle is 27.5 degree
From figure we have
.........1
x = 54.13 ft
differentiating 1st eq wrt x
![\frac{d}{dx} tan\theta = \frac{d}{dx} [\frac{x}{104}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20tan%5Ctheta%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B%5Cfrac%7Bx%7D%7B104%7D%5D)
Taken 
for minute change
from 
Reltaive error
Relative error = 0.0254 %
Answer:
y=2*0.68^x = exponential decay
Step-by-step explanation:
