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:
Answer is 120
Step-by-step explanation:
1. 1
2. 100%
3. 1.00
I belive these are the answers to the chart
Answer: 35
Step-by-step explanation:
Given : At Deb’s Deli, a customer may choose either a sandwich and a salad or a sandwich and a soup for the lunch special.
The number of choices for sandwiches = 5
The number of choices for salad = 4
The number of choices for soup = 3
Now, the number of possible lunch special combinations can be ordered will be :-
Hence, the number of possible lunch special combinations can be ordered will be 35.
Answer:
36°
Hint: 4*unknown + 36 = 180°
