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:
5.23 < 5.235
Step-by-step explanation:
5.23 < 5.235
we know:
5.235 ≈ 5.24
hence 5.235 is greater than 5.23
Answer:
y=2x-4
Step-by-step explanation:
90(squared) + 90 (squared) = C (squared)
(we use 90 because the base path is 90 ft long since the diamond is a square that makes all sides 90 ft long)
8100 + 8100 = c(squared)
16200 = c(squared)
we will get the square root of 16200 to get the diagonal length
C=127.3 ft, and since it is the diagonal distance we will device it to 2, to get the distance from the pitcher ro the 2nd base.
127.3/2 = 67.7 ft.
67.7 ft is the distance from the pitcher<span> to the 2nd base.</span>
The answer is 11:5 ……………….
