Answer:
a) 3.128
b) Yes, it is an outerlier
Step-by-step explanation:
The standardized z-score for a particular sample can be determined via the following expression:
z_i = {x_i -\bar x}/{s}
Where;
\bar x = sample means
s = sample standard deviation
Given data:
the mean shipment thickness (\bar x) = 0.2731 mm
With the standardized deviation (s) = 0.000959 mm
The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression
z_i = {x_i -\bar x}/{s}
z_i = {0.2761-0.2731}/{ 0.000959}
z_i = 3.128
b)
From the standardized z-score
If [z_i < 2]; it typically implies that the data is unusual
If [z_i > 2]; it means that the data value is an outerlier
However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.
Answer:
Where is the table
Step-by-step explanation:
I cant answer without it
Y=mx+b
slope=m
yint=b
we are given
slope=-2/3
point=(-3,-1)
y=-2/3x+b
input point -3,-
x=-3
y=-1
find b
-1=-2/3(-3)+b
-1=2+b
minus 3
-3=b
y=-2/3x-3
answer is D
Let
. Then
Meanwhile, since
, you have
.
It follows that
,
, and
, so that the quadratic fit for
that satisfies the given points is
Note that this is just the second-order Taylor polynomial for
about
.
Answer:
1) 49/10 and 490/100
2) 49/1000
Step-by-step explanation:
