The "sample" with "mean absolute deviation" indicate about a sample mean absolute deviation is being used as an estimator of the mean absolute deviation of a population
- Mean of the sample MAD=3.3
- Population MAD =6.4
<h3>What does this indicate about a sample mean absolute deviation used as an estimator of the mean absolute deviation of a population?</h3>
Generally, The MAD measures the average dispersion around the mean of a given data collection.

In conclusion, for the corresponding same to mean
the sample mean absolute deviation
7,7 ↔ 0
7,21 ↔ 7
7,22 ↔ 7.5
21,7 ↔ 7
21,21 ↔ 0
21,22 ↔ 0.5
22,7 ↔ 7.5
Therefore
- Mean of the sample MAD=3.3
- Population MAD =6.4
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<em>f(d)=86,400·d</em>
if you set 1 day (d=1) you get f(1)=86,400 sec
if you set 1 day (d=2) you get f(2)=172,800 sec
...etc.
Y = -4x - 2
P (-16, -11)
a = -4, b = 0, c = -2
m = -a/b
m = -(-4)/(-2)
m = 4/(-2)
m = -2
y-yo = m(x-xo)
y-(-11) = -2[x-(-16)]
y+11 = -2(x+16)
y+11 = -2x-32
2x+y+11+32 = 0
2x+y+43 = 0
A outra questão é semelhante.
D = rt, Distances Equal
72t = 90(t-1)
90 = 18t
90/18 = t = 5hr
second car catch up to the first in 4hrs
Y = 12x - 5x - 2
first simplify the equation by subtracting like terms (in this case):
12x - 5x = 7x
y = 7x - 2
Since you are finding the x, you must isolate the x. Do the opposite of PEMDAS.(Note: because there is a equal sign, what you do to one side, you do to the other)
y = 7x - 2
y (+2) = 7x - 2 (+2)
y + 2 = 7x
(y + 2)/7 = 7x/7
x = (y + 2)/7
x = (y + 2)/7 is your answer
hope this helps