Answer:
Given- O is the centre of a circle in which an equilateral ΔPQR has been inscribed.
To find out- ∠QOR=?
Solution- The points P, Q & R are on the circumference of the circle since ΔPQR has been inscribed in the circle. Now the chord QR subtends ∠QOR to the centre O and ∠QPR to the circumference at P.
∴ By rule, ∠QOR=2∠QPR.........(i).
Now PQR is an equilateral Δ. Each of its angle is 60
o
. i.e ∠QPR=60
o
.
So, from (i), ∠QOR=2∠QPR=2×60
o
=120
o
.
Ans- Option B.
solution
36°, 72°, 72°
let x be the third angle then each of the other 2 angles is 2x
the sum of the angles in a triangle = 180° thus
x + 2x + 2x = 180
5x = 180 ( divide both sides by 5 )
x = 
= 36
the angles are 36°, 72° and 72°
Answer:
a) margin of error ME = 5.77
b) Margin of error becomes smaller
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
x+/-ME
Where margin of error ME = zr/√n
a)
Given that;
Mean = x
Standard deviation r = 25
Number of samples n = 72
Confidence interval = 95%
z(at 95% confidence) = 1.96
Substituting the values we have;
ME = 1.96(25/√72)
ME = 1.96(2.946278254943)
ME = 5.774705379690
ME = 5.77
b)
For n = 89
ME = 1.96(25/√89)
ME = 1.96(2.649994700015)
ME = 5.193989612031
ME = 5.19
5.19 is smaller than 5.77 in a) above. So,
Margin of error becomes smaller
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