1X2
b32
Eagles
for the first page
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:
Option (1)
Step-by-step explanation:
By the property of alternate segment theorem,
"Angle formed between the tangent and the chord in a circle measures the half of the measure of the intercepted arc"
m(∠EFG) = 
× m(minor arc FG)
minor arc FG = 2m(∠EFG)
= 2(76°)
= 152°
Therefore, Option (1) will be the correct option.
Answer:
-10
Step-by-step explanation:
-4 + -3 = -7
-7 + -2 = -9
-9 + -1 = -10
