Answer:
![m\angle ECB+m\angle EFB=180](https://tex.z-dn.net/?f=m%5Cangle%20ECB%2Bm%5Cangle%20EFB%3D180)
![m\angle CDB\cong m\angle EDF](https://tex.z-dn.net/?f=m%5Cangle%20CDB%5Ccong%20m%5Cangle%20EDF)
![m\angle CDB+m\angle DCB+m\angle CBD=180 \degree](https://tex.z-dn.net/?f=m%5Cangle%20CDB%2Bm%5Cangle%20DCB%2Bm%5Cangle%20CBD%3D180%20%5Cdegree)
Step-by-step explanation:
From the diagram, quadrilateral BCEF is a cyclic quadrilateral.
Opposite angles if a cyclic quadrilateral sum up to 180°
![m\angle ECB+m\angle EFB=180](https://tex.z-dn.net/?f=m%5Cangle%20ECB%2Bm%5Cangle%20EFB%3D180)
The diagonals intersect at D to form two pairs of vertical angles, and vertical angles are congruent.
![m\angle CDB\cong m\angle EDF](https://tex.z-dn.net/?f=m%5Cangle%20CDB%5Ccong%20m%5Cangle%20EDF)
Also sum of angles in triangle CBD is 180°.
![m\angle CDB+m\angle DCB+m\angle CBD=180 \degree](https://tex.z-dn.net/?f=m%5Cangle%20CDB%2Bm%5Cangle%20DCB%2Bm%5Cangle%20CBD%3D180%20%5Cdegree)
Answer:
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Step-by-step explanation:
Given;
Standard deviation r= 25mg
Width of confidence interval w= 10mg
Confidence interval of 95%
Margin of error E = w/2 = 10mg/2 = 5mg
Z at 95% = 1.96
Margin of error E = Z(r/√n)
n = (Z×r/E)^2
n = (1.96 × 25/5)^2
n = (9.8)^2
n = 96.04
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Answer:
96^2in
Step-by-step explanation:
4x4=16
16x6=96