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.
The best answer I believe is t - 2
Answer & Explanation:
For this question, you need to figure out the length and the width because the area is directly related to them (A = l*w).
What we do know is that the perimeter is 36 inches. Since the perimeter is equal to P = 2l + 2w, we can say:
P = 36 = 2(3 + 2w) + 2(w), where the length l is 3 + 2w
Solve for w first, then solve for l. Use these values to find the area.
Pls, choose me as brainliest!
25 root 3 divided by 2
Base is 5. Height is 5 root 3
Multiply then divide by 2
